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c for some A and c depending on the chart. By composing this restriction WIth any fixed linear isomorphism olR~>c ~ IRn -l, we obtain a bijection, say X8M, of Un oM onto an open subs-;;t of IRn - 1 which provides a chart (Ua: noM, X8M) for oM. The family of charts obtained in this way is an atlas for oM. The overlaps are smooth and so we have the following: Proposition 1.124. If M is an n-manifold with boundary, then oM is an (n - 1) -manifold. Exercise 1.125. Show that the overlap maps for the atlas just constructed for oM are smooth. Exercise 1.126. The closed unit ball B(p, 1) in IR n is a smooth manifold with boundary oB(p, 1) = sn-l. Also, the closed hemisphere Sf- = {x E sn : x n+ 1 ~ O} is a smooth manifold with boundary. Exercise 1.127. Is the Cartesian product of two smooth manifolds with boundary necessarily a smooth manifold with boundary? Exercise 1.128. Show that the concept of smooth partition of unity makes sense for manifolds with boundary. Show that such exist.
51
Problems
Problems (1) Prove Proposition 1.32. The online supplement [Lee, Jeff] outlines the proof.
(2) Prove Lemma 1.11. (3) Check that the manifolds given as examples are indeed paracompact and Hausdorff. (4) Let M I , M2 and M3 be smooth manifolds. (a) Show that (MI XM2) x M3 is diffeomorphic to MI X(M2 x M 3 ) in a natural way. (b) Show that f : M ---+ MI XM2 is Coo if and only if the composite maps pri 0 f : M ---+ MI and pr2 0 f : M ---+ M2 are both Coo.
(5) Show that a CT manifold M is connected as a topological space if and only it is C T path connected in the sense that for any two points PI, P2 E M there is a CT map c: [0,1] ---+ M such that c(o) = PI and c(l) = P2.
(6) A k-frame in ffi.n is a linearly independent ordered set of vectors Show that the set of all k-frames in ffi.n can be given the structure of a smooth manifold. This kind of manifold is called a Stiefel manifold. (VI, ... , Vk)'
(7) For a product manifold M x N, we have the two projection maps pri : M x N ---+ M and pr2 : M x N ---+ N defined by (x, y) f------7 x and (x, y) f------7 y respectively. Show that if we have smooth maps h : P ---+ M and 12 : P ---+ N, then the map (J,g) : P ---+ M x N given by (J,g)(p) = (J (p), 9 (p)) is the unique smooth map such that pr I 0 (J, g) = f and pr 2 0 (J, g) = g. (8) Prove (i) and (ii) of Lemma 1.35.
(9) Show that the atlas obtained for a regular submanifold induces the relative topology inherited from the ambient manifold.
(10) The topology induced by a smooth structure is not necessarily Hausdorff: Let S be the subset of ffi.2 given by the union (ffi. x 0) U {(O, I)}. Let U be ffi. x and let V be the set obtained from U by replacing the point (0,0) by (0,1). Define a chart map x on U by x(x, 0) = x and a chart y on V by
°
y(x, 0)
= {
~
if x if x
# 0, = 0.
Show that these two charts provide a Coo atlas on S, but that the topology induced by the atlas is not Hausdorff.
52
1. Differentiable Manifolds
(11) As we have defined them, manifolds are not required to be second countable and so may have an uncountable number of connected components. Consider the set jR2 without its usual topology. For each a E jR, define a bijection cPa : jR X {a} --+ jR by cPa (x, a) = x. Show that the family of sets of the form U x {a} for U open in jR and a E jR provide a basis for a paracompact topology on jR2. Show that the maps cPa are charts and together provide an atlas for jR2 with this unusual topology. Show that the resulting smooth manifold has an uncountable number of connected components (and so is not second countable). (12) Show that every connected manifold has a countable atlas consisting of charts whose domains have compact closure and are simply connected. Hint: We are assuming that our manifolds are paracompact, so each connected component is second countable. (13) Show that every second countable manifold has a countable fundamental group (a solution can be found in [Lee, John] on page 10). (14) If C x C is identified with jR4 in the obvious way, then S3 is exactly the subset ofCxC given by {(ZI,Z2): IZ112+lz212 = I}. Letp,q be coprime integers and p > q ~ O. Let w be a primitive p-th root of unity so that Zp = {I, w, . .. , wp- l }. For (ZI' Z2) E S3, let w·(zI, Z2) := (WZl' wQz2) and extend this to an action of Zp on S3 so that wk. (ZI' Z2) = (WkZl' wQkZ2). Show that this action is free and proper. The quotient space Zp \S3 if called a lens space and is denoted by L(p; q). (15) Let SI be realized as the set of complex numbers of modulus one. Defin{ a map 0: SI xS l --+ SI xS l by O(z, w) = (-z, w) and note that 000 = id Let G be the group {id,O}. Show that M := (SI X SI) IG is a smootl 2-manifold. (16) Show that if S is a regular k-dimensional submanifold of an n-manifol( M, then we may cover S by special single-slice charts from the atlas c M which are of the form x : U --+ VI X V2 C jRk X jRn-k = jRn with
x(unS) = VI x {O} for some open sets VI C jRk, V2 C jRn-k. Show that we may arrange f( VI and V2 to both be Euclidean balls or cubes. (This problem should t easy. Experienced readers will likely see it as merely an observation.) (17) Show that
1rl (M
x N, (p, q)) is isomorphic to
1rl (M,
p) x
1rl (N,
q).
(18) Suppose that M = U U V, where U and V are simply connected ar open. Show that if UnV is path connected, then M is simply connecte (19) Prove the three properties about maps involving the model half-spac ajR~~o listed in Section 1.8.
Problems
53
Px(-l,l)
(]I)
Figure 1.9, Smoothly connecting manifolds
(20) Let !vi and N be smooth n-manifolds with boundaries aM and aN. Let P be a smooth manifold diffeomorphic to both aAf and aN via maps 0: and /3. Suppose that there are open neighborhoods U and V of aM and aN respectively and diffeomorphisms
c with p in its domain, Tx- I (Tx(p)IR~=c) is a subspace of TpM. This is the subspace of vectors tangent to the boundary and is identified with Tp8M, the tangent space to 8M (also a manifold). Exercise 2.26. Show that this subspace does not depend on the choice of chart.
If one traces back through the definitions, it becomes clear that because of the way charts and differentiability are defined for manifolds with boundary, any smooth function defined on a neighborhood of a boundary point can be thought of as being the restriction of a smooth function defined
2. The Tangent Structure
72
Figure 2.2. Tangents at a boundary point
slightly "outside" M. More precisely, the representative function always has a smooth extension from a (relatively open) neighborhood in 1R~>c to a neighborhood in IRn. The derivatives of the extended function at points of alR~>c = 1R~=c are independent of the extension. These considerations can be -used to show that tangent vectors at boundary points of a smooth n-manifold with boundary can still be considered as derivations of germs of smooth functions. A closed interval [a, b] is a one-dimensional manifold with boundary, and with only minor modifications in our definitions we can also consider equivalence classes of curves to define the full tangent space at a boundary point. It also follows that if c is a smooth curve with domain [a, b], then we can make sense of the velocities c(a) and c(b), and this is true even if c(a) or c(b) is a boundary point. The major portion of the theory of manifolds extends in a natural way to manifolds with boundary.
2.4. Tangents of Products Suppose that f : MI x M2 ~ N is a smooth map. For fixed p E MI and fixed q E M2, consider the "insertion" maps &p : y H (p, y) and &q : x H (x, q). Then fo&q and fo&p are the maps sometimes denoted by f(-,q) and f(p, .).
Definition 2.27. Let f : MI x M2 tangent maps ad and a2f by
~
N be as above. Define the partial ~
(ad) (p, q)
:=
Tp (f
0
&q) : TpMI
(a2f) (p, q)
:=
Tq (f
0
&p) : TqM2 ~ Tf(p.q)N.
Tf(p,q)N,
Next we introduce another natural identification. It is obvious that a curve c: I ~ MI X M2 is equivalent to a pair of curves CI :
I
~
MI,
C2:
I
~
M2.
2.4. Tangents of Products
73
The infinitesimal version of this fact gives rise to a natural identification on the tangent level. If c(t) = (CI(t),C2(t)) and c(O) = (p,q), then the map T(P,q)prl x T(p,q)pr2 : T(p,q)(MI x M2) --t TpMI x TqM2 is given by [c] f---t ([CI], [C2]), which is quite natural. This map is an isomorphism. Indeed, consider the insertion maps ~p : q f---t (p, q) and ~q : p t--+ (p, q),
We have linear monomorphisms T~q(p) Ttp(q) : TqM2 --t T(p,q)(M I x M 2),
TpMI --t T(p,q)(MI x M 2) and
Then, we have the map
Ti q + Tip: TpMI
X
TqM2 --t T(p,q)(M I x M2),
which sends (v,w) E TpMI X TqM 2 to Tiq(V) + Tip(W). It can be checked that this map is the inverse of [c] f---t ([CI], [C2])' Thus we may identify T(p,q)(M I x M 2) with TpMI x TqM 2. Let us say a bit about the naturalness of this identification. In the smooth category, there is a direct product operation. The essential point is that for any two manifolds MI and M 2 , the manifold MI x M2 together with the two projection maps serves as the direct product in the technical sense that for any smooth maps f : N ~ MI and g: N ~ M2 we always have the unique map f x 9 : N ~ MI X M2 which makes the following diagram commute: N
~gJ~
MI ~ MI
X
M2 ---- M2
For a point x E N, write p = f(x) and q = g(x). On the tangent level we have
Tx(fxg)
Txg
which is a diagram in the vector space category. In the category of vector spaces, the product of TpMI and TpM2 is TpMI X TpM2 together with the
74
2. The Tangent Structure
projections onto the two factors. Corresponding to the maps Tpf and Tqg we have the map Tpf x Tqg. But
(T(P,q)prl x T(P,q)pr2)
0
Tx (f x g)
= (T(P,q)prl 0 Tx (f x g)) x (T(P,q)pr2 0 Tx (f x g)) =
Tx (prl
(f x g)) x Tx (pr2
=
Txf x Txg.
0
0
(f x g))
Thus under the identification introduced above the map Tx (f x g) corresponds to Txf x Txg. Now if v E TpMl and w E TqM2' then (v, w) represents an element of T(p,q) (Ml x M 2), and so it should act as a derivation. In fact, we can discover how this works by writing (v,w) = (v,O) + (O,w). If Cl is a curve that represents v and C2 is a curve that represents w, then (v,O) and (O,w) are represented by t f-7 (Cl(t),q) and t f-7 (p,C2(t)) respectively. Then for any smooth function f on Ml x M2 we have
(v,w)f = (v,O)f =
v[f 0
+ (O,w)f =
~ql
+ w[f 0
:tl
o f(Cl(t),q)
+
:tl
o f(p,C2(t))
~pl.
Lemma 2.28 (Partials lemma). For a map f : Ml x M2 -+ N, we have
T(p,q)f· (v, w) = (8d)(p,q) . v + (82f)(p,q) . w, where we have used the aforementioned identification T(p,q)(Ml x M 2) TpMl x Tq M 2 .
=
Proving this last lemma is much easier and more instructive than reading the proof so we leave it to the reader in good conscience.
2.5. Critical Points and Values Definition 2.29. Let f : M -+ N be a Cr-map and p E M. We say that p is a regular point for the map f if Tpf is a surjection. Otherwise, p is called a critical point or singular point. A point q in N is called a regular value of f if every point in the inverse image f-l{q} is a regular point for f. This includes the case where f-l{q} is empty. A point of N that is not a regular value is called a critical value. Most values of a smooth map are regular values. In order to make this precise, we will introduce the notion of measure zero on a second countable smooth manifold. It is actually no problem to define a Lebesgue measure on such a manifold, but for now the notion of measure zero is all we need. Definition 2.30. A subset A of ~n is said to be of measure zero if for any E > there is a sequence of cubes {Wi} such that A c UWi and L vol(Wi ) < E. Here, vol(Wi) denotes the volume of the cube.
°
2.5. Critical Points and Values
75
In the definition above, if the Wi are taken to be balls, then we arrive at the very same notion of measure zero. It is easy to show that a countable union of sets of measure zero is still of measure zero. Our definition is consistent with the usual definition of Lebesgue measure zero as defined in standard measure theory courses. Lemma 2.31. Let U c ]Rn be open and f : U has measure zero, then f(A) has measure zero.
-7
]Rn a C 1 map. If A
cU
Proof. Since A is certainly contained in the countable union of compact balls (all of which are translates of a ball at the origin), we may as well assume that U = B(O, r) and that A is contained in a slightly smaller ball B(O, r-8) c B(O, r). By the mean value theorem (see Appendix C), there is a constant C depending only on f and its domain such that for x, y E B(O, r) we have Ilf(y) - f(x)11 :S c Ilx - YII. Let E > be given. Since A has measure zero, there is a sequence of balls B(Xi' Ei) such that A c U B(Xi' Ei) and
°
'"' vol(B(xi' Ei)) < _E_. n n ~
Thus f(B(Xi, Ei)) also have vol
c
2 c B(j(Xi),2cEi), and while f(A) C
UB(j(Xi), 2CEi),
we
(U B(j(Xi), 2CEi)) :S L vol(B(j(xi), 2CEi)) :S
L vol(Bl) (2CEi)n :S 2ncn L vol(B(xi' Ei)) :S E,
where Bl = B(O, 1) is the ball of radius one centered at the origin. Since E was arbitrary, it follows that A has measure zero. D The previous lemma allows us to make the following definition: Definition 2.32. Let M be an n-manifold that is second countable. A subset A C M is said to be of measure zero if for every admissible chart (U, x) the set x( A n U) has measure zero in ]Rn. In order for this to be a reasonable definition, the manifold must be second countable so that every atlas has a countable subatlas. This way we may be assured that every set that we have defined to be measure zero is the countable union of sets that are measure zero as viewed in some chart. It is not hard to see that in this more general setting it is still true that a countable union of sets of measure zero has measure zero. Also, we still have that the image of a set of measure zero under a smooth map, has measure zero. Proposition 2.33. Let M be second countable as above and A = {(Uo:, xo:)} a fixed atlas for M. If xo:(A n Uo:) has measure zero for all a, then A has measure zero.
2. The Tangent Structure
76
Proof. The atlas A has a countable subatlas, so we may as well assume from the start that A is countable. We need to show that given any admissible chart (U, x) the set x(A n U) has measure zero. We have
x(A n U) =
Ux(A nUn Un) (a countable union).
Since xn(An U nUn) C xn(An Un), we see that xn(An U n Un) has measure zero for all cx. But x(A nUn Un) = X 0 x~l 0 xn(A nUn Un), and so by the lemma above, x(A nUn Un) also has measure zero. Thus x(A n U) has 0 measure zero since it is a countable union of sets of measure zero. We now state the famous and useful theorem of Arthur Sardo Theorem 2.34 (Sard). Let N be an n-manifold and M an m-manifold, both assumed second countable. For a smooth map f : N ---+ M, the set of critical values has measure zero. The somewhat technical proof may be found in the online supplement [Lee, Jeff] or in [Bro-Jan]. Corollary 2.35. If M and N are second countable manifolds, then the set of regular values of a smooth map f : M ---+ N is dense in N. 2.5.1. Morse lemma. If we consider a smooth function f : M ---+ JR, and assume that M is a compact manifold (without boundary), then f must achieve both a maximum at one or more points of M and a minimum at one or more points of M. Let Pe be one of these points. The usual argument shows that dfl pe = 0. (Recall that under the usual identification of JR with any of its tangent spaces we have dfl pe = TpJ.) Now let p be some point for which dflp = 0, i.e. p is a critical point for J. Does f achieve either a maximum or a minimum at p? How does the function behave in a neighborhood of p? As the reader may well be aware, these questions are easier to answer in case the second derivative of f at p is nondegenerate. But what is the second derivative in this case? Definition 2.36. The Hessian matrix of f at one of its critical points p and with respect to coordinates x = (x\ ... , x n ), is the matrix of second partials:
where Xo = x(p). The critical point p is called nondegenerate if H is nonsingular.
2.5. Critical Points and Values
77
Any such matrix H is symmetric, and by Sylvester's law of inertia, it is congruent to a diagonal matrix whose diagonal entries are either 0 or 1 or -1. The number of -1 's occurring in this diagonal matrix is called the index of the critical point. According to Problem 13 we may define the Hessian H f,p : TpM x TpM -t JR, which is a symmetric bilinear form at each critical point p of f, by letting Hf,p(v,w) = Xp(Yf) = Yp(Xf) for any vector fields X and Y which respectively take the values v and w at p. Thus, we may give a coordinate free definition of a nondegenerate point for f. Namely, p is a nondegenerate point for f if and only if Hf,p is a nondegenerate bilinear form. The form H f,p is nondegenerate if for each fixed nonzero v E TpM the map Hf,p(v,·) : TpM -t JR is a nonzero element of the dual space M.
T;
Exercise 2.37. Show that the nondegeneracy is well-defined by either of the two definitions given above and that the definitions agree. Exercise 2.38. Show that nondegenerate critical points are isolated. Show by example that this need not be true for general critical points. The structure of a function near one of its nondegenerate critical points is given by the following famous theorem of M. Morse: Theorem 2.39 (Morse lemma). Let f : M -t JR be a smooth function and let Xo be a nondegenerate critical point for f of index 1/. Then there is a local coordinate system (U, x) containing Xo such that the local representative fu := f 0 x-I for f has the form
fu(x I , . .. ,xn)
= f(xo) + L hijxix j i,j
and it may be arranged that the matrix h = (h ij ) is a diagonal matrix of the form diag( -1, ... , -1, 1, ... ,1) for some number (perhaps zero) of ones and minus ones. The number of minus ones is exactly the index 1/. Proof. This is clearly a local problem and so it suffices to assume that f : U -t JR for some open U C JRn and also that f(O) = O. Our task is to show that there exists a diffeomorphism ¢ : JRn -t JRn such that f 0 ¢( x) = xt hx for a matrix of the form described. The first step is to observe that if 9 : U c JRn -t JR is any function defined on a convex open set U and g(O) = 0, then
2. The Tangent Structure
78
Thus 9 is of the form 9 = 2:~=1 Uigi for certain smooth functions gi, 1 :::; i :::; n with the property that Oig(O) = gi(O). Now we apply this procedure first to f to get f = 2:~1 Udi where od(O) = fi(O) = and then apply the procedure to each fi and substitute back. The result is that
°
n
f(Ul, ... ,Un ) =
(2.3)
L uiujhij(ul"",un) i,j=l
for some functions hij with the property that hij is nonsingular at, and therefore near 0. Next we symmetrize the matrix h = (h ij ) by replacing hij with ~(hij + h ji ) if necessary. This leaves (2.3) untouched. The index of the matrix (hij(O)) is v, and this remains true in a neighborhood of 0. The trick is to find a matrix C(x) for each x in the neighborhood that effects the diagonalization guaranteed by Sylvester's theorem: D = C(x)h(x)C(X)-l. The remaining details, including the fact that the matrix C(x) may be D chosen to depend smoothly on x, are left to the reader.
2.6. Rank and Level Set Definition 2.40. The rank of a smooth map rank of Tpf.
f at
p is defined to be the
If f : M -t N is a smooth map that has the same rank at each point, then we say it has constant rank. Similarly, if f has the same rank for each p in a open subset U, then we say that f has constant rank on U.
Theorem 2.41 (Level submanifold theorem). Let f : M -t N be a smooth map and consider the level set f-l(qO) for qo E N. If f has constant rank k on an open neighborhood of each p E f-l(qo), then f-l(qO) is a closed regular submanifold of codimension k. Proof. Clearly f-l(qO) is a closed subset of M. Let Po E f-l(qo) and consider a chart (U, cp) centered at Po and a chart (V, 'lj;) centered at qo with f(U) c V. We may choose U small enough that f has rank k on U. By Theorem C.5, we may compose with diffeomorphisms to replace (U, cp) by a new chart (U' , x) also centered at Po and replace (V, 'lj;) by a chart (V', y) centered at qo such that J := yo f 0 x-I is given by (a l , ... , an) H (a l , ... ,ak,O, ... ,0), where n = dim(M). We show that U'
If p E U' or
n f-l(qo) = {p E U ' : Xl(p) = ... = xk(p) = a}.
n f-l(qo),
then yo f(p) =
°
and yo f 0 x-1(X1(p), ... , xn(p)) =
°
Xl(p) = ... = xk(p) = 0. On the other hand, suppose that p E U' and xl (p) = ... = xk (p) = 0. Then we can reverse the logic to obtain that yo f(p) = and hence f(p) = qo.
°
2.6. Rank and Level Set
79
Since Po was arbitrary, we have verified the existence of a cover of f-l(qo) by single-slice charts (see Section 1.7). D Proposition 2.42. Let M and N be smooth manifolds of dimension m and n respectively with n > m. Consider any smooth map f : M ---+ N. Then if q E N is a regular value, the inverse image set f- 1 (q) is a regular submanifold. Proof. It is clear that since f must have maximal rank in a neighborhood of f-l(q), it also has constant rank there. We may now apply Theorem 2.41. D Example 2.43 (The unit sphere). The set 5 n - 1 = {x E JRn : L (x i )2 = I} is a codimension 1 sub manifold of JRn. For this we apply the above proposition with the map f : JRn ---+ JR given by x J---7 L (xi) 2 and with the choice q = 1 E lR.. Example 2.44. The set of all square matrices Mnxn is a manifold by virtue of the obvious isomorphism Mnxn ~ JRn 2 • The set sym(n, JR) of all symmetric matrices is a smooth n(n + 1)/2-dimensional manifold by virtue of the obvious 1-1 correspondence sym(n, JR) ~ JRn(n+1)j2 given by using n(n+ 1)/2 entries in the upper triangle of the matrix as coordinates. It can be shown that the map f : Mnxn ---+ sym(n, JR) given by A J---7 At A has full rank on O(n, JR) = f-l(I) and so we can apply Proposition 2.42. Thus the set O(n, JR) of all n x n orthogonal matrices is a submanifold of Mnxn. We leave the details to the reader, but note that we shall prove a more general theorem later (Theorem 5.107).
The following proposition shows an example of the simultaneous use of Sard's theorem and Proposition 2.42. Proposition 2.45. Let 5 be a connected submanifold of JRn and let L be a codimension one linear subspace of JR n . Then there exist x E JRn such that (x + L) n 5 is a submanifold of 5. Proof. Start with a line l through the origin that is normal to L. Let pr: JRn ---+ 5 be orthogonal projection onto l . The restriction 7r := prls ---+ l is easily seen to be smooth. If 7r(5) were just a single point x, then 7r- 1 (x) = (x + L) n 5 would be all of 5, so let us assume that 7r (5) contains more than one point. Now, 7r(5) is a connected subset of l ~ JR, so it must contain an open interval. This implies that 7r(5) has positive measure. Thus by Sard's theorem there must be a point x E 7r(5) c l that is a regular value of 7r. Then Theorem 2.42 implies that 7r- 1 (x) is a submanifold of 5. But this is D the conclusion since 7r- 1 (x) = (x + L) n S.
We can generalize Theorem 2.42 using the concept of transversality.
2. The Tangent Structure
80
Definition 2.46. Let f : M --t N be a smooth map and SeN a submanifold of N. We say that f is transverse to S if for every p E f- 1 (S) we have Tf(p)N = Tf(p)S + Tpf(TpM). If f is transverse to S, we write f rh S. Theorem 2.47. Let f : M --t N be a smooth map and SeN a submanifold of N of codimension k and suppose that f rh Sand f- 1(S) i= 0. Then f- 1(S) is a submanifold of M with codimension k. Furthermore we have Tp(f-l(S)) = Tf-l(Tf (p)S) for all p E f-l(S). Proof. Let q = f(p) E S and choose a single-slice chart (V, x) centered at q E V so that x(S n V) = x(V) n (II~n-k x 0). Let U := f-l(V) so that p E U. If 7f : IR n - k x IRk --t IRk is the second factor projection, then the transversality condition on U implies that 0 is a regular value of 7f 0 x 0 flu. Thus (7f 0 x 0 flu )-1 (0) = f- 1 (S) n U is a submanifold of U of codimension k. Since this is true for all p E f- 1 (S), the result follows. D We can also define when a pair of maps are transverse to each other: Definition 2.48. If h : Ml --t Nand 12 : M2 say that hand 12 are transverse at q E N if
Tf(p)N = TP1h(TPIM)
+ Tp212(Tp2M)
--t
N are smooth maps, we
whenever h(pd = 12(P2) = q.
(N ote that h is transverse to 12 at any point not in the image of one of the maps h and h.) If hand 12 are transverse for all q E N, then we say that hand 12 are transverse and we write h rh 12. One can check that if f : M of N, then f and the inclusion according to Definition 2.47.
--t L :
N is a smooth map and S is a submanifold S y N are transverse if and only if f rh S
If h : Ml --t Nand 12 : M2 consider the set
--t
N are smooth maps, then we can
(h x 12)-1 (~) := ((pl,P2) E Ml x M2 : h(Pl) = 12(P2)}, which is the inverse image of the diagonal
~ :=
{(ql, q2)
E
N x N : ql = q2}.
Corollary 2.49 (Transverse pullbacks). If h : Ml --t Nand 12 : M2 --t N are transverse smooth maps, then (h x 12)-1 (~) is a submanifold of lvIt x M 2. If gl : P --t Ml and g2 : P --t M2 are any smooth maps with the property h o gl = 12 0 g2, then the map (gl,g2): P --t (fl X 12)-1 (~) given by (gl, g2) (x) = (gl (x), g2 (x)) is smooth and is the unique smooth map such that prl 0 (gl,g2) = gl and pr2 0 (g1,g2) = g2. Proof. We leave the proof as an exercise. Hint: h x 12 is transverse to ~ if and only if h rh 12. D
2.7. The Tangent and Cotangent Bundles
81
2.7. The Tangent and Cotangent Bundles We define the tangent bundle of a manifold M as the (disjoint) union of the tangent spaces; T M = UPEM TpM. We show in Proposition 2.55 below that T 111 is a smooth manifold, but first we introduce a couple of definitions.
Definition 2.50. Given a smooth map J : M -+ N as above, the tangent maps TpJ on the individual tangent spaces combine to give a map
TJ:TM-+TN on the tangent bundle which is linear on each fiber. This map is called the tangent map or sometimes the tangent lift of J. For smooth maps J : M -+ Nand 9 : N -+ M we have the following simple looking version of the chain rule:
T(goJ) = TgoTf.
If U is an open set in a finite-dimensional vector space V, then the tangent space at x E U can be viewed as {x} xV. For example, recall that an element vp = (p, v) corresponds to the derivation J r---+ vpJ := ftlt=o J(p+tv). Thus the tangent bundle of U can be viewed as the product U x V. Let Ul and U2 be open subsets of vector spaces V and W respectively and let J : U1 -+ U2 be smooth (or at least C 1). Then we have the tangent map TJ : TU1 -+ TU2. Viewing TU1 as U1 x V and similarly for TU2, the tangent map T J is given by (p, v) r---+ (J(p), D J(p) . v). Definition 2.51. If J : M -+ V, where V is a finite-dimensional vector space, then we have the differential dJ (p) : TpM -+ V for each p. These maps can be combined to give a single map dJ : T M -+ V (also called the differential) which is defined by dJ(v) = dJ(p)(v) when v E TpM.
If we identify TV with the product V x V, then dJ = pr2 0 TJ, where pr2 : TV = V x V -+ V is the projection onto the second factor. Remark 2.52 (Warning). The notation "df" is subject to interpretation. Besides the map dJ : T M -+ V described above it could also refer to the map dJ : p r---+ dJ(p) or to another map on vector fields which we describe later in this chapter. Definition 2.53. The map 1fTM : TM -+ M defined by 1fTM(V) = p if v E TpM is called the tangent bundle projection map. (The set T AI together with the map 1fTM : T AI -+ M is an example of a vector bundle which is defined later.)
82
2. The Tangent Structure
Whenever possible, we abbreviate JrTM to Jr. For every chart (U, x) on M, we obtain a chart (if, x) on T M by letting
if:= TU = and by defining
Jr- 1 (U)
c TM
x on U by the prescription
x(vp) = (xl(p), ... ,xn(p),v l , ... ,vn ), where vp E TpM, and where v!, ... , v n are the (unique) coefficients in the coordinate expres. a Ip' Thus ~-l( I ... , u n, v I , ... , v n) -_ " a Ix-l(u)" slOn v -_ "i...J v i axi xu, i...J v i axi Recall that if vp = L vi a~i Ip' then vi = dx i (vp). From this we see that x = (xl 0 Jr, ... , xn 0 Jr, dx l , ... , dxn). For any (U, x), we have the tangent lift Tx : TU -+ TV where V = x (U). Since V C IRn, we can identify Tx(p) V with {x (p)} x IRn. Let us invoke this identification. Now let vp E TpU and let, be a curve that represents vp so that ,'(0) = vp. Exercise 2.54. Under the identification of Tx(p) V with {x (p)} x IR n we have Tpx,vp = (x (p), (x 0 , ) ) . [Hint: Interpret both sides as derivations.]
ftlt=o
If vp = a~i Ip' then we can take ,(t) := x-l(x (p) i-th member of the standard basis of IRn. Thus
Tpx,
f)~i Ip =
(x (p) , :t It=O (x (p)
+ tei) )
+ tei), = (x
where ei is the
(p) , ed·
Now suppose that vp = Lvi a~i Ip' Then Tpx . vp
~ Tpx. =
(L {J~i IJ
(x (p),
vi
L viei)
=
(xl(p), ... , xn(p), vI, ... , vn).
x
From this we see that Tx is none other than defined above, and since if = TU, we see that an alternative and suggestive notation for (if , is
x)
(TU, Tx), and we adopt this notation below. This notation reminds one that the charts we have constructed are not just any charts on T M, but are each associated naturally with a chart on M and are essentially the tangent lifts of charts on M. They are called natural charts. Proposition 2.55. For any smooth n-manifold M, the set T M is a smooth 2n-manifold in a natural way and JrTM : TM -+ M is a smooth map. Furthermore, for a smooth map f : M -+ N, the tangent map T f is smooth and
2.7. The Tangent and Cotangent Bundles
83
the following diagram commutes:
TM~TN
~
f
M
~
>N
Proof. For every chart (U, x), let TU = 11"-1 (U) and let Tx be the map Tx: TU ---+ x(U) x jRn. The pair (TU, Tx) is a chart on TM. Suppose that (TU, Tx) and (TV, Ty) are two such charts constructed as above from two charts (U, x) and (V, y) and that U n V i= 0. Then TU n TV i= 0 and on the overlap we have the coordinate transitions Ty 0 Tx- 1 : (x, v) t--t (y, w) where y = yo X-I (x), w
= D(y 0
x- 1 )lx v.
Thus the overlap maps are smooth. It is easy to see that Tx(TU n TV) and Ty(TU n TV) are open. Thus we obtain a smooth atlas on T M from an atlas on M and this generates a topology. It follows from Proposition 1.32 that T M is Hausdorff and paracompact. To test for the smoothness of 11", we look at maps of the form x011"o(Tx) -1. We have X011"O
(Tx)-l (x,v)
=X011"
(vi aail ) =xox-1(x)=x, x x-1(x)
which is just a projection and so clearly smooth. The remainder is left for the exercise below. 0 In the above proof we observed that Tx(TU n TV) and Ty(TU n TV) are open. This must be checked because of (ii) in Definition 1.25 and is the kind of detail we may leave to the reader as we move forward. Exercise 2.56. For a smooth map
f : M ---+ N, the map
Tf: TM ---+ TN is itself a smooth map.
If p E U n V and x(p) = (xl(p), ... , xn(p)), then, as in the proof above, TyoTx- 1 sends (xl(p), ... ,xn(p),v 1, ... ,vn ) to (yl(p), ... ,yn(p), wi, ... , w n ), where
84
2. The Tangent Structure
If we abbreviate the i-th component of yo x-I(XI(p), .. . , xn(p)) to yi = yi (xl (p), ... , xn (p)), then we could express the tangent bundle overlap map by the relations ) " 8yi k Yi = Yi ( x I ( p, ... , x n (p)) an d w i = 'L..J 8x k v .
Since this is true for all p E x(U n V), we can write the very classical looking expressions Yi
= Yi( x I , ... , x n) and
wi
" 8yi k = 'L..J 8x k v ,
where we now can interpret (xl, ... , xn) as an n-tuple of numbers. Once again we note that local expression could either be interpreted as living on the manifold in the chart domain or equally, in Euclidean space on the image of the chart domain. This should not be upsetting since, after all, one could argue that the charts are there to identify chart domains in the manifold with open sets in Euclidean space.
Definition 2.57. The tangent functor is defined by assigning to a manifold M its tangent bundle T M and to any map f : M -t N the tangent map T f : T M -t TN. The chain rule shows that this is a covariant functor (see Appendix A). Recall that we also defined a "pointed" tangent functor. We have seen that if U is an open set in a vector space V, then the tangent bundle is often taken to be U x V. Suppose that for some smooth n-manifold M, there is a diffeomorphism F : TM -t M x V such that the restriction of F to each tangent space is a linear isomorphism TpM -t {p} X V and such that the following diagram commutes: TM--_F_~) MxV
~~ M Then for some purposes, we can identify TM with M x V.
Definition 2.58. A diffeomorphism F : T M -t M x V such that the map FITp M : TpM -t {p} X V is linear for each p and such that the above diagram commutes is called a (global) trivialization of TM. If a (global) trivialization exists, then we say that T M is trivial. For an open set U eM, a trivialization of TU is called a local trivialization of T Mover U. For most manifolds, there does not exist a global trivialization of the tangent bundle. On the other hand, every point p in a manifold M is contained in an open set U so that T M has a local trivialization over U. The
2.7. The Tangent and Cotangent Bundles
85
existence of these local trivializations is quickly deduced from the existence of the special charts which we constructed above for a tangent bundle. Next we introduce the cotangent bundle. Recall that for each P E M, the tangent space TpM has a dual space T; M called the cotangent space at
p. Definition 2.59. Define the cotangent bundle of a manifold M to be the set
T*M:=
U
T;M pEM M to be the obvious projection taking
and define the map 7rT" M : T* M -t elements in each space T; M to the corresponding point p.
Remark 2.60. We will denote both the tangent bundle projection and the cotangent bundle projection simply by 7r whenever no confusion is likely. Remark 2.61. Suppose that J : M -t N is a smooth map. It is important to notice that even though for each P E M, the map TpJ : TpM -t Tf(p)N has a dual map (TpJ)* : (Tf(p)N) * -t (TpM)*, these maps do not generally combine to give a map from T* N to T* M. In general, there is nothing like a "cotangent lift". To see this, just consider the case where J is a constant map. We now show that T* M is also a smooth manifold. Let A be an atlas on M. For each chart (U, x) E A, we obtain a chart (T*U, T*x) for T* M which we now describe. First, T*U = 7rr:M(U) = UPEu T;M. Secondly, T*x is a map which we now define directly and then show that, in some sense, it is dual to the map Tx. For convenience, consider the map Pi : {}p t---+ ~i which just peals off the coefficients in the expansion of any {}p E T; M in the basis
(dxll p "'" dxnl p):
Notice that we have
(}p(a~il) = L~j dxil p (a~il) = L~jol =~j = Pi ({}p) , and so
Pi(Op) = Op (
8~i
IJ .
With this definition of the Pi in hand, we can define
T*x =
(Xl 0 7r, ... ,
xn
0 7r,PI, ...
,Pn)
on T*U. We call (T*U, T*x) a natural chart. If x = (Xl, ... , x n ), then for the natural chart (T*U, T*x), we could use the abbreviation T*x =
2. The Tangent Structure
86
(xl, ... ,Xn,P1, ... ,Pn)' Another common notation is qi := xi notation is very popular in applications to mechanics.
07r.
This
We claim that if we take advantage of the identifications of TxjRn = jRn = (jRn)* = T*jRn where (jRn)* is the dual space of jRn, then T*x acts on each fiber T; M as the dual of the inverse of the map Tpx, i.e. the contragredient of Tpx: ((TpX)-l) * (fJp) . (v)
= fJp ((Tpx)-l . v) .
Let us unravel this. If fJp E T; M for some P E U, then we can write fJp = L~i dxil p
for some numbers We have
~i
depending on fJp which are what we have called Pi (fJp).
((Tpx)-l)* (fJp)' (v)
= fJp ((Tpx)-l. v) = L~i dxil p ' ((Tpx)-l. v)
IJ
~ Lei dxil (~>' a=' ~ Leivi p
Thus, under the identification ofjRn with its dual we see that
((T x)-l) * (fJ p
p)
is just (6, ... ,~n). But recall that T*x(fJp) = (x1(p), ... ,xn(p),6"'.'~n)' Thus for fJp E T; M we have T*x(fJp) = (x(p), ((Tpx)-l)* (fJp)).
Suppose that (T*U, T*x) and (T*V, T*y) are the coordinates constructed as above from two charts (U, x) and (V, y) respectively with unv i- 0. Then on the overlap T* U n T* V we have T*y
0
(T*x)-l : x(U n V) x jRn*
-t
y(U n V) x jRn*.
This last map will send something of the form (x,~) E U x jRn* to (x, () = (yox-1 (x), D(xoy-1)* .~), where D(xoy-1)* is the dual map to D(xoy-1), which is the contragredient of the map D(y 0 x-I). If we identify jRn* with jRn and write ~ = (6, ... , ~n) and ( = ((1, ... , &), then in the classical style we have: . 1 " ox k ft. = yt(x , ... , xn) and ~i = L...J ~k oyi . k
This should be compared to the expression (2.2). It is now clear that we have an atlas on T* M constructed from an atlas on M. The topology of T* M (induced by the above atlas) is easily seen to be paracompact and Hausdorff.
2.8. Vector Fields
87
In summary, both T M and T* M are smooth manifolds whose smooth structure is derived from the smooth structure on M in a natural way. In both cases, the charts are derived from charts on the base M and are given by the n coordinates of the base point together with the n components of the element of T M (or T* M) in the corresponding coordinate frame.
2.8. Vector Fields In this section we introduce vector fields. Roughly, a vector field is a smooth assignment of a tangent vector to each point of a manifold. Definition 2.62. If 1l' : M -+ N is a smooth map, then a (global) section of 1l' is a map (J : N -+ M such that 1l' 0 (J = id. If (J is defined only on an open subset U of Nand 1l' 0 (J = idu, then we call (J a local section. In case the section (J is a smooth (or cr) map, we call (J a smooth (or cr) section. Clearly, if 1l' : M -+ N has a (global) section, then it must be surjective. Definition 2.63. A smooth vector field on M is a smooth map X: M -+ T M such that X (p) E TpM for all p E M. In other words, a vector field on M is a smooth section of the tangent bundle 1l' : T M -+ M. We often write Xp = X(p). Convention: Obviously the notion of a section or field that is not smooth makes sense. Sometimes one is interested in merely continuous sections or measurable sections. In this book, by "vector field" or "section", we will always mean "smooth vector field" or "smooth section" unless otherwise indicated explicitly or by context. A local section of T M defined on an open set U is just the same thing as a vector field on the open manifold U. If (U, x) is a chart on a smooth n-manifold, then writing x = (Xl, .. . ,xn ), we have vector fields defined on Uby
:pi--t aa.j. aa. xt xt p The ordered set of fields (Ixr, ... , 8~n) is called a coordinate frame field (or also "holonomic frame field"). If X is a smooth vector field defined on some set including this chart domain U, then for some smooth functions Xi defined on U we have
X(p) = or in other words
L Xi(p) a~i jp,
2. The Tangent Structure
88
Notation 2.64. In this context, we will not usually bother to distinguish from its restrictions to chart domains and so we just write = L: a~i .
X
X
Xi
Lemma 2.65. If v E TpM then there exists a vector field X such that X(p) = v.
Ip·
Proof. Write v = L: vi a~i Define a field Xu by the formula L: vi a~i where the vi are taken as constant functions on U. Let (3 be a cut-off function with support in U and such that (3(p) = 1. Then let X := (3Xu on U and extended to zero outside of U. 0
Let us unravel what the smoothness condition means for a vector field. Let (TU, Tx) be one of the natural charts that we constructed for T M from a corresponding chart (U, x) on M. To test the smoothness of X, we look at the composition Tx 0 X 0 x-I. For x E x(U), we have Tx 0 X
0 X-I (x)
(I: Xi a~i) x- (x) = Tx(I: X (x- (x)) aa I
= Tx 0
1
0
1
i
= (x,
T
)
i
X
x- 1 (x)
X-l(X)X(I: X i (x- 1 (x)) aaX i Ix-1(x) ))
= (x, X I 0 x-I (x), ... , Xn 0 x-I (x)) . Our chart was arbitrary, and so we see that the smoothness of X is equivalent to the smoothness of the component functions in every chart of an atlas for the smooth structure.
Xi
Exercise 2.66. Show that if X : M -t T M is continuous and 7r 0 X = id, then X is smooth if and only if X f : p H Xpf is a smooth function for every locally defined smooth function f on M. Show that it is enough to consider globally defined smooth functions. Notation 2.67. The set of all smooth vector fields on M is denoted by X(M). Smooth vector fields may at times be defined only on some open set U c M so we also have the notation X(U) = XM(U) for these fields.
We define the addition of vector fields, say X and Y, by
(X
+ Y) (p)
:=
X(p)
+ Y(p),
and scaling by real numbers, by
(cX) (p) := cX(p).
89
2.8. Vector Fields
Then the set X(M) is a real vector space. If we define multiplication of a smooth vector field X by a smooth function f by
(f X) (p)
:=
f(p)X(p),
then the expected algebraic properties hold making X(M) a module over the ring COO(M) (see Appendix D). It should be clear how to define vector fields of class cr on M and the set of these is denoted xr(M) (a module over Cr(M)). The notion of a vector field along a map is often useful.
Definition 2.68. Let f : N ---+ M be a smooth map. A vector field along I is a smooth map X : N ---+ T M such that 1fTM 0 X = f. A vector field along a regular submanifold ScM is a vector field along the inclusion map S,-+ M. (Note that we include the case where S is an open submanifold.) We let Xf denote the space of vector fields along f. It is easy to check that for a smooth map COO(N)-module in a natural way.
f : N ---+
M, the set Xf is a
We have seen how individual tangent vectors in TpM can be identified as derivations at p. The derivation idea can be globalized. We explain how we may view vector fields as derivations.
Definition 2.69. Let M be a smooth manifold. A (global) derivation on COO(M) is a linear map V : COO(M) ---+ COO(M) such that
V(fg)
= V(f)g + fV(g).
We denote the set of all such derivations of COO(M) by Der(COO(M)). Notice the difference between a derivation in this sense and a derivation at a point.
Definition 2.70. To a vector field X on M, we associate the map LX COO(M) ---+ COO(M) defined by
(Lxf)(p)
ex
:=
Xpj.
is called the Lie derivative on functions.
It is important to notice that (Lxf)(p) = Xp' f = df(Xp) for any p and so Lxf = df 0 X. If X is a vector field on an open set U, and if I is a function on a domain V c U, then we take LX f to be the function defined on V by p f---t Xpf for all p E V. It is easy to see that we have CaX+bY = aLx + bLy for a, bE lR and X, Y E X(M).
Lemma 2.71. Let U c M be an open set and X E X(M). If Lxf all IE COO(U), then Xiu = o.
= 0 lor
2. The Tangent Structure
90
Proof. Let p E U be given. Working locally in a chart (V, x), let X = 'LX i8/8xi. We may assume p EVe U. Using a cut-off function we may find functions fi defined on U such that fi coincides with xi on a neighborhood of p. Then we have Xi(p) = Xpx i = Xpfi = (.cxf i ) (p) = O. Thus X (p) = 0 for an arbitrary p E U. 0
The next result is a very important characterization of smooth vector fields. In particular, it paves the way for the definition of the bracket of vector fields which plays a central role in differential geometry. Theorem 2.72. For X E X(M), we have.c x E Der(CC'O(M)), and if V E Der(COO(M)), then V = .cx for a uniquely determined X E X(M). Proof. That .cx is in Der(COO(M)) follows from the Leibniz law, in other words, from the fact that Xp is a derivation at p for each p. If we are given a derivation V, we define a derivation Xp at p (i.e. a tangent vector) by the rule Xpf := (V J) (p). We need to show that the assignment p H Xp is smooth. Recall that any locally defined function can be extended to a global one by using a cut-off function. Because of this, it suffices to show that p H Xpf is smooth for any f E COO(M). But this is clear since Xpf := (VJ) (p) and Vf E COO(M). Suppose now that V = .cXI = .cX2' Notice that .cXI - .cX2 = .cXI-X2 and so .cXI-X2 is the zero derivation on COO(M). By Lemma 2.71, we have Xl - X2 = O. 0
Because of this theorem, we can identify Der(COO(M)) with X(M) and we can and often will write X f in place of .c x f:
Xf:= .cxf. The derivation law (also called the Leibniz law) .cx(Jg) = g.cx f + f.cx 9 becomes simply X(Jg) = gXf + fXg. Another thing worth noting is that if we have a derivation of COO(M), then from our discussion above we know that it corresponds to a vector field. As such, it can be restricted to any open set U c M, and thus we get a derivation of COO(U). If f E COO(U) we write X f instead of the more pedantic Xlu J. While it makes sense to talk of vector fields on M of differentiability r where 0 < r < 00 and these do act as derivations on cr(M), it is only in the smooth case (r = 00) that we can say that vector fields account for all derivations of cr(M). Theorem 2.73. If VI, V2 E Der(COO(M)), then [VI, V2J E Der(COO(M)) where
91
2.8. Vector Fields
Proof. We compute
VI ('0 2 (fg)) = VI (V 2(f)g
+ f V 2(g))
= ('0 1'021) 9 + V2fV 1g + VdV 2g + fV 1V 2g. Writing out the similar expression for '0 2 (VI (f g)) and then subtracting we obtain, after a cancellation,
[VI, '0 2] (fg) = ('0 1'021) 9 + f V 1V 2g - ((V2V d) 9 + f V 2V 1g)
= ([VI, V 2]f) 9 + J[V1' V 2]g.
0
Corollary 2.74. If X, Y E X(M), then there is a unique vector field [X, Y] such that £[X,Yj = £x 0 £y - £y 0 £x.
Since £xf is also written Xf, we have [X, Y]f = X (Yf) - Y (Xf) or
[X,Y] = XY - Yx. Definition 2.75. The vector field [X, Y] from the previous corollary is called the Lie bracket of X and Y. Proposition 2.76. The map (X, Y) X, Y, Z E X(M) we have
f-7
[X, Y] is bilinear over JR, and for
(i) [X, Y] = -[Y, X];
+ [Z, [X, YlJ = 0 = fg[X, Y] + f (Xg) Y - 9 (Y f) X
(ii) [X, [Y, ZlJ+ [Y, [Z, X]]
(Jacobi Identity);
(iii) [f X, gY]
for all f, 9 E COO(M).
Proof. These results follow from direct calculation and the previously mentioned fact that £aX+bY = a£x + bey for a, bE JR and X, Y E X(M). 0
The map (X, Y) f-7 [X, Y] is bilinear over JR, but by (iii) above, it is not bilinear over COO(M). Also notice that in (ii) above, X, Y, Z are permuted cyclically. We ought to see what the local formula for the Lie derivative looks like in conventional "index" notation. Suppose we have X = LXi 8~' and Y = L yi 8~i. Then we have the local formula
Exercise 2.77. Verify this last formula.
The JR-vector space X(M) together with the JR-bilinear map (X, Y) f-7 [X, Y] is an example of an extremely important abstract algebraic structure:
92
2. The Tangent Structure
Definition 2.78 (Lie algebra). A vector space a (over a field IF) is called a Lie algebra if it is equipped with a bilinear map a x a -t a (a multiplication) denoted (v, w) t--t [v, w] such that
[v,w] = -[w,v] and such that we have the Jacobi identity
[x, [y, z]]
+ [y, [z, x]] + [z, [x, y]] = 0
for all x,y,z E a.
Definition 2.79. A Lie algebra a is called abelian (or commutative) if [v, w] = 0 for all v, w E a. A subspace ~ of a is called a Lie subalgebra if it is closed under the bracket operation, and it is called an ideal if [v, w] E ~ for any v E a and w E ~. (We indicate this by writing [a,~] C ~.) Notice that the Jacobi identity may be restated as [x, [y, z]] = [[x, y], z] + [y, [x, z]], which just says that for fixed x the map y t--t [x, y] is a derivation of the Lie algebra a. This is significant mathematically and also an easy way to remember the Jacobi identity. The Lie algebra X(M) is infinite-dimensional (unless M is zero-dimensional), but later we will be very interested in certain finite-dimensional Lie algebras which are subalgebras of X(M). Given a diffeomorphism ¢ : M -t N, we define the pull-back ¢*Y E X(M) for Y E X(N) and the push-forward ¢*X E X(N) of X E X(M) by ¢ by ¢*Y ¢*X
= T¢-l 0 Y 0 ¢ and = T ¢ 0 X 0 ¢ -1.
In other words, (¢*Y)(p) = T¢-l. Y<j>(P) and (¢*X)(p) = T¢·X<j>-l(p). Notice that ¢*Y and ¢*X are both smooth vector fields. Warning: Since many authors use the notation f* for the tangent map T f, the notation f*X might be interpreted to mean T foX, which is actually a vector field along the map f rather than an element of X(N). We shall not use f* as a notation for Tf.
To summarize a bit, if f : M -t N is a smooth map, then for each p we have the tangent map Tpf : TpM -t Tf(p)N, the tangent lift T f : T M -t TN (a "bundle map"), and if f is a diffeomorphism, we have the induced maps on the level of fields f* : X(M) -t X(N) and f* : X(N) -t X(M). Notice that if ¢ : M -t Nand 'ljJ : N -t Pare diffeomorphisms, then we have ('ljJ
0
¢)* = 'ljJ*
0
¢* : X(M)
('ljJ
0
¢)* = ¢*
0
'ljJ* : X(P) -t X(M).
-t
X(P),
We have right and left actions of the diffeomorphism group Diff(M) on the space of vector fields. The left action Diff(M) x X(M) -t X(M) is given by
2.8. Vector Fields
93
(¢, X) t-+ ¢*X, and the right action x(M) x Diff(M) -+ x(M) is given by (X, ¢) t-+ ¢* X.
On functions, the pull-back is defined by ¢*g := go ¢ for any smooth map, but if ¢ is a diffeomorphism, then we can also define a push-forward ¢* := (¢-1)*. With this notation we have the following proposition. Proposition 2.80. The Lie derivative on functions is natural with respect to pull-back and push-forward by diffeomorphisms. In other words, if ¢ : M -+ N is a diffeomorphism and f E COO(M), 9 E COO(N), X E x(M) and Y E x(N), then
and
Proof. We use Definition 2.19. For any p we have
(£<jJ*y¢*g) (p) = (¢*Y)p ¢*g = (T¢-l = (T¢-l . Y<jJ(p») [g
0
0
Y
0
¢)p [g
0
¢]
¢] = T¢ (T¢-lY<jJ(p») 9
= Y<jJ(p)g = (£yg) (¢(p)) = (¢* £yg) (p). The second statement follows from the first since ¢* = (¢ -1) *.
0
Even if f : M -+ N is not a diffeomorphism, it may still be that there is a vector field Y E x(N) such that
TfoX=Yof. In other words, it may happen that Tf· Xp = Yf(p) for all p in M. In this case, we say that Y is f -related to X and write X '" f Y. It is not hard to check that if Xi is f-related to Y; for i = 1,2, then aX l + bX1 is f-related to aYl + bYl .
Example 2.81. Let M and N be smooth manifolds and consider the projections pr1 : M x N -+ M and pr2 : M x N -+ N. Since T(p,q) (M x N) can be identified with TpM x TqN, we see that for X E x(M), Y E x(N) we obtain a vector field X x Y E x(M x N) defined by (X x Y) (p, q) = (X(p), Y(p)). Then one can check that X x Y and X are pr1-related
and
X x Y and Yare prTrelated. Exercise 2.82. Let M, N, X, Y and X x Y be as in the example above. Show that if ~q : M -+ M x N is the insertion map p t-+ (p, q), then X and X x Y are ~q-related if and only if Y(q) = O.
94
2. The Tangent Structure
Lemma 2.83. Suppose that f : M -+ N is a smooth map, X E X(M) and Y E X(N). Then X and Yare f-related if and only if X(g 0 f) = (Y g) 0 f for all 9 E COO(N). Proof. Let p E M and let 9 E Coo (N). Then
X(g
0
f)(p)
= Xp(g 0
f)
= (Tpf . Xp) 9
and (Y go f) (p)
so that X (g 0 f) = (Y g)
0
= Yf(p)g
f for all such 9 if and only if Tpf . Xp = Yf(p).
0
Proposition 2.84. If f : M -+ N is a smooth map and Xi is f -related to Yi for i = 1,2, then [Xl, X2] is f-related to [YI, Y2]. In particular, if ¢ is a diffeomorphism, then [¢*X I , ¢*X2] = ¢*[X I , X 2] for all Xl, X 2 E X(M). Proof. We use the previous lemma: Let 9 E COO(N). Then X I X2(g 0 f) = X I ((Y2g) 0 f) = (YIY2g) 0 f. In the same way, X 2X I (g 0 f) = (Y2Y I g) 0 f and subtracting we obtain
[XI,X2] (g
0
f) = XIX2(g 0 f) - X2 X I(g 0 f) = (YIY2g) 0 f - (Y2YIg) 0 f =
([YI , Y2]g)
0
f.
Using the lemma one more time, we have the result.
o
If S is a submanifold of M and X E X(M), then the restriction Xis E X(S) defined by Xis (p) = X(p) for all pES is i-related to X where i : S y M is the inclusion map. Thus for X, Y E X(M) we always have that [Xis, Yl s] is i-related to [X, Y]. This just means that [X, Y] (p) = [Xis, Yls] (p) for all p. We also have
Proposition 2.85. Let f : M -+ N be a smooth map and suppose that X "'f Y. Then we have £x (f*g) = j*£yg for any 9 E COO(N).
The proof is similar to what we did above and is left to the reader. 2.8.1. Integral curves and flows. Recall that if c : I -+ M is a smooth curve, then the velocity at "time" t is
:t c(t) = c(t) = Ttc· :u It'
Iu
where is the standard field on 1R given at a E 1R as the equivalence class a f = f' (a). of the curve t f-t a + t or by the derivation
Iu I
2.8. Vector Fields
95
Definition 2.86. Let X be a smooth vector field on M. A curve c: I -t M is called an integral curve for X if for all t E I, the velocity of c at time t is equal to X(c(t)), that is, if
c= X
0
c.
Thus if c is an integral curve for X and
f
is a smooth function, then
Xc(t)f = (f 0 c)' (t) for all t in the domain of c. If the image of an integral curve c lies in U for a chart (U, x), and if X = L: Xi 8~i' then c = X 0 c gives the local expressions
d . dt xt 0
.
C
= Xt
0
c for i
= 1, ... ,n,
which constitute a system of ordinary differential equations for the functions xi 0 c. These equations are classically written as d:1ti = Xi.
A (complete) flow is a map
0 such that the composition Xk Xl rptk 0'" 0 rptl is defined on U and maps U into 0 whenever it, ... , tk E (-E, E).
Proof. If the flow box corresponding to rpXi is (Ui , Ei, rpf), then by shrinking UI further we may arrange things so that rp;l maps UI into 0 for all t E (-EI' EI), and then inductively we arrange for rpXi to map Ui into Ui-l for all t E (-Ei' Ei). Now let E = min{ EI,"" Ek}. 0 Remark 2.93. When making compositions of local flows, we will not always make careful statements about domains, but the previous lemma will be invoked implicitly. If Cp(t) is an integral curve of X defined on some interval (a, b) containing
o and Cp(O) = p, then we may consider the limit lim Cp(t).
t~b-
If this limit exists as a point PI EM, then we may consider the integral curve cPl beginning at Pl. One may now use Lemma 2.89 to combine t I-t Cp(t) with t I-t Cpl (t-b) to produce an extended integral curve beginning at p. We may repeat this process as long as the limit exists. We may do a similar thing in the negative direction. This suggests that there is a maximal integral curve defined on a maximal interval := (Tp~x' Tp~x), where Tp~x might be -00 and Tp~x might be +00. We produce this maximal integral curve
J;
2.8. Vector Fields
99
as follows: Consider the collection Jp of all pairs (J, a), where J is an open interval containing 0 and a : J -t M is an integral curve of X with a(O) = p. Then let J; = U(J,a)EJpJ and define cmax(t) := a(t) whenever t E J for (J, a) E Jp. By existence and uniqueness, this definition is unambiguous. -t .M is the desired maximal integral curve and is The curve Cmax : easily seen to be unique.
J;
Definition 2.94. Let X be a Coo vector field on M. For any given p EM, let := (Tp~x' Tp~x) c ~ be the domain of the maximal integral curve c: -t M of X with c(O) = p. The maximal flow cpx is defined on the set (called the maximal flow domain)
J; J;
U J; x {p}
1)x =
pEM
by the prescription that t such that cpx (0, p) = p.
H
cpx (t,p) is the maximal integral curve of X
Thus by definition, X is a complete vector field if and only if 1)x IRxM. We will abbreviate (T;'x,Tp~x) to (Tp-, T p+).
Theorem 2.95. For X E X(M), the set 1)x is an open neighborhood of {O} x M in ~ x M and the map cpx : 1)x -t M is smooth. Furthermore,
(2.4)
cpx (t
+ s,p) = cpx (t, cpx (s,p))
whenever both sides are defined. If the right hand side is defined, then the left hand side is defined. Suppose that t, s ::::: 0 or t, s ::; O. Then if the left hand side is defined so is the right hand side. Proof. Let q = cpx (s,p). If the right hand side is defined, then s E (T;,T:) and t E (Tq-,T:). The curve 7jJ : T H cpx(s + T,p) is defined for T E (Tp- - s, T: - s), and this is the maximal domain for 7jJ. We have
ddT I 7jJ = dd cpx (s 7
+ T,p)
T
=
dd U
I
cpx (u,p)
U=S+7
= X(cpX (s + T,p)) = X(7jJ(T)).
17=0
We also have 7jJ(0) = cpx (T + s, p) = cpX (s, p) so 7jJ is an integral curve starting at q = cpX(s,p). Thus (Tp- - s,Tp+ - s) C (Tq-,Tq+) and 7jJ = !pX(.,q) on (Tp- - s,Tp+ - s). But the maximal domain for 7jJ is (Tp- - s, ~ - s) and so in fact (Tp- - s, T: - s) = (Tq-, T:) for otherwise cpx (., q) would be a proper extension. But then, since t E (Tq-, T:), we have that t E (T; - s, Tp+ - s) and so 7jJ(t) = cpx (t + s,p) is defined and
cpx (t
+ s, p)
=
cpx (t, cpx (s, p)).
2. The Tangent Structure
100
Now let us assume that t, s > 0 and that cpx (s+t,p) is defined (the case of t, s ~ 0 is similar). Then since s, t ~ 0, we have s, t, t + s E
(Tp-, T:) .
Let q = cpx(s,p) as before and let O(u) = cpx(s + u,p) be defined for u with 0 ~ u ~ t. But O(u) is an integral curve with 0(0) = q. Thus we have that cpx (u, q) must also be defined for u = t and O(t) = cpx (t, q). But cpX(t,q) cpx(t,cpX(s,p)), which is thereby defined, and we have
cpx (s + t,p) :_ O(t) _ cpx (t, cpx (s,p)). Now we will show that Vx is open in lR x M and that cpx : Vx --+ M is smooth. We carefully define a subset S C 'Dx by the condition that (t,p) E S exactly if there exists an interval J containing 0 and t and also an open set U C M such that the restriction of cpx to J x U is smooth. Notice that S is open by construction. We intend to show that S = 'Dx. Suppose not. Then let (tQ,po) E 'Dx n Sc. We will assume that to > 0 since the case to < 0 is proved in a similar way. Now let T := sup{t : (t,po) E S}. We know that (O,po) is contained in some flow box and so T > O. We also have T ~ to by the definition of to. Thus T E J:O and we define qo := cpx (T,PO). Now applying the local theory we know that qo is contained in an open set Uo such that cpx is defined and smooth on (-f, f) X Uo for some f > o. We will now show that cpx is actually defined and smooth on a set of the form (-8,r) x 0 where 0 is open, 8,r > 0, and (T,PO) E (-8,r) x O. Since this contradicts the definition of T, we will be done. We may choose tl > 0 so that T E (tl' tl +f) and so that cpX(tl'PO) E Uo. Note that (tl,PO) E S since tl < T. SO on some neighborhood (-8, tl +8) x U1 of (tl,PO) the flow cpx is smooth. By choosing Ul smaller if necessary we can arrange that cpx ({tl} x U1 ) c Uo. Now consider the equation
cpx (t, p) - cpX (t - tl, cpx (tl,p)). If It - tl < f and p E UI, then both sides are defined and the right hand side is smooth near such (t,p). But the right hand side is already known to be smooth on (-8, tl + 8) X UI. We now see that cpx is smooth on (-8, tl + f) X UI, which contains (T,PO) contradicting the definition of T. Thus S Vx. 0 Remark 2.96. In this text, cpx will either refer to the unique maximal flow defined on Vx or to its restriction to the domain of a flow box. In the latter case we call cpx a local flow. We could have introduced notation such as cp!ax, but prefer not to clutter up the notation to that extent unless necessary. We hope that the reader will be able to tell from context what we are referring to when we write cpx.
2.8. Vector Fields
101
If cpx is a flow of X, then we write CPt for the map p t---+ CPt (p). The (maximal) domain of this map is V~ - {p : t E (Tp~x' T:'-x )}' Note that, in general, the domain of CPt depends on t. Also, we have the tangent map Tocp; : ToJR ---t TpM and
:t It=o
cP; (t) = Tocp; :u 10 = X p,
where :'U 10 is the vector at 0 associated to the standard coordinate function on IR (denoted by u here). Exercise 2.97. Let sand t be real numbers. Show that the domain of is contained in V~+t and show that for each t, V~ is open. Show that 0, then at °we have
10
1=
8~t
10
lox
= lim -hl [J(X(O, '" h, .• .) - 1(0)] h-+O
= h-+O lim -hl [1(0, ... , h, ... ) -
1(0)]
= 88ut·1 0 f.
Thus ToX = id and so by the inverse mapping theorem (Theorem 2.25) we see that after restricting X to a smaller neighborhood of zero, the map x := X-I is a chart map. We have already seen that TX I = X 0 X. But then for 1 E Coo (M) we have
Iur
-; 1
8x
p
1=
81 1 1 0 8u x(P)
x-I =
8 1 1 lox 8u x(p)
=Tx 88 1 1 1= (Xox)(x(p))I=Xpl, u x(p) so that
Ixr = x.
o
2.8.2. Lie derivative. We now introduce the important concept of the Lie derivative of a vector field extending the previous definition. The Lie derivative will be extended further to tensor fields.
Definition 2.102. Given a vector field X, we define a map LX : X(M) -+ X(M) by
LXY:= [X, Y]. This map is called the Lie derivative (with respect to X).
2. The Tangent Structure
104
The Jacobi identity for the Lie bracket easily implies the following two identities for any X, Y, Z E X(M) -+ X(M):
..cx[Y, Z] = [..cxY,Z] + [Y,..cxZ], (Le . ..c[X,y] = ..cx o..cy - ..cy 0 ..cx). ..c[X,y] = [..cx,..c y ] We will see below that ..cx Y measures the rate of change of Y in the direction X. To be a bit more specific, (..cx Y) (p) measures how Y changes in comparison with the field obtained by "dragging" Yp along the flow of X. Recall our definition of the Lie derivative of a function (Definition 2.70). The following is an alternative characterization in terms of flows: For a smooth function f : M -+ R and a smooth vector field X E X(M), the Lie derivative ..cx of f with respect to X is given by
..cx f (p)
:t
=
10 f 0 0 such that c( -1) = Po, c(O) = p and d (0) = vp. Then
-d f(C(T)) == - 1 id dT 0 dT 0 =
j
e[
1,7']
a
d~ 10 11[-1,0] a + ddT 10 1 [0,7'] a
= 0+
1
d~ 10 foT c*a - :1' 10 fo7' g(t) dt
= g(O),
2.10. Line Integrals and Conservative Fields
119
where c*o: = 9 dt. On the other hand,
O:(Vp ) = o:(c'(O)) =
= c*o: (:t
0:
IJ
(Toc.
:t IJ
= g(O) dtl o (:t
IJ
= g(O),
iT
where t is the standard coordinate on R Thus 10 f(C(T)) = o:(vp ) for any vp E TpM and any p E M. Now the result follows from the previous lemma. D
It is important to realize that when we say that a form is conservative in this context, we mean that it is globally conservative. It may also be the case that a form is locally conservative. This means that if we restrict the I-form to an open set which is diffeomorphic to a Euclidean ball, then the result is conservative on that ball. The following examples explore in simple terms these issues.
Example 2.126. Let 0: = (x 2 + y2(1 (-y dx + x dy). Consider the small circular path c given by (x, y) = (xo + £ cos t, Yo + £ sin t) with 0 ::; t ::; 27r and £ > O. If (xo, YO) = (0,0), we obtain
1
e 0:
{27r
= Jo
= Thus
0:
1 ( £2
1
27r
o
dx -y(t) dt
dY)
+ x(t) dt
1 2" ( - (£ sin t) (-£ sin t) £
dt
+ (£ cos t) (£ cos t)) dt = 27r.
is not conservative and hence not exact. On the other hand, if
(xo, Yo) f. (0,0), then we pick a ray Ro that does not pass through (xo, Yo) and a function O(x, y) which gives the angle of the ray R passing through (x, y) measured counterclockwise from Ro. This angle function is smooth
Y5, Ie
and defined on U = JR2\Ro. If £ < !Jx~ + then c has image inside the domain of 0 and we have that 0:1 U = dO. Thus 0: = O(c(O)) -O(c(27r)) = O. We see that 0: is locally conservative. Example 2.127. Consider (3 = y dx - x dy on JR 2. If it were the case that for some small open set U C JR2 we had (3lu = df, then for a closed path c with image in that set, we would expect that (3 = f(c(27r)) - f(c(O)) = O. However, if c is the curve going around a circle of radius £ centered at
Ie
120
2. The Tangent Structure
(xo, Yo), then we have
1~ 1(y( = =
t) ~~ - x( t) ~~) dt
1271' (( Xo + c sin t) (-c sin t) -
(Yo + c cos t) (c cos t)) dt
= -2c 2 7l', so we do not get zero no matter what the point (xo, Yo) and no matter how small c. We conclude that ~ is not even locally conservative. The distinction between (globally) conservative and locally conservative is often not made sufficiently clear in the physics and engineering literature. Example 2.128. In classical physics, the static electric field set up by a fixed point charge of magnitude q can be described, with an appropriate choice of units, by the 1-form q
q
q
gxdx+ gydy+ gzdz, p p p where we have imposed Cartesian coordinates centered at the point charge and where p = x 2 + y2 + z2. Notice that the domain of the form is the punctured space ]R3 \ {o}. In spherical coordinates (p, (), ¢), this same form is
J
q
p2dp= d
(-q) p ,
so we see that the form is exact and the field is conservative.
2.11. Moving Frames It is important to realize that it is possible to get a family of locally defined vector (resp. covector) fields that are linearly independent at each point in their mutual domain and yet are not necessarily of the form a~. (resp. dx i ) for any coordinate chart. In fact, this may be achieved by carefully choosing n 2 smooth functions f~ (resp. a~) and then letting Ek := 2:i f~ a~. (resp. ()k
:=
2:i a~dxi).
Definition 2.129. Let E 1, E 2, ... , En be smooth vector fields defined on some open subset U of a smooth n-manifold M. If E1 (P), E2 (P), ... , En (p) form a basis for TpM for each p E U, then we say that (E1' E2,"" En) is a (non-holonomic) moving frame or a frame field over U. If E1, E2, ... ,En is a moving frame over U C M and X is a vector field defined on U, then we may write
X = I::XiEi on U,
121
2.11. Moving Frames
for some functions Xi defined on U. If the moving frame (E1, ... , En) is not identical to some frame field (Ixr, ... , arising from a coordinate chart on U, then we say that the moving frame is non-holonomic. It is often possible to find such moving frame fields with domains that could never be the domain of any chart (consider a torus).
Ixn)
Definition 2.130. If E 1, E2, . .. ,En is a frame field with domain equal to the whole manifold M, then we call it a global frame field. Most manifolds do not have global frame fields. Taking the basis dual to (E1{p), ... , En{P)) in T;M for each P E U we get a moving coframe field (o1, ... , on). The Oi are I-forms defined on U. Any I-form 0: can be expanded in terms of these basic I-forms as 0: = L ad)i. Actually it is the restriction of 0: to U that is being expressed in terms of the 02 , but we shall not be so pedantic as to indicate this in the notation. In a manner similar to the case of a coordinate frame, we have that for a vector field X defined at least on U, the components with respect to (E1, .. . , En) are given by Oi{X):
Let us consider an important special situation. If M x N is a product manifold and (U, x) is a chart on M and (V, y) is a chart on N, then we have a chart (U x V, x x y) on M x N where the individual coordinate functions are xl 0 pr1, ... , xm 0 pr1, y1 0 pr2, ... , yn 0 pr2, which we temporarily denote bY -1 x , ... , x::::Tn , -1 y , ... , ::-:n y. Now we conSl·der wh a t·IS the reIa t·IOn b et ween the coordinate frame fields (Ixr, ... m ), (-/yr, ... -/yn) and the frame field
(Ixr, ... , a~n).
a:
The latter set of n + m vector fields is certainly a linearly independent set at each point (p, q) E U x V. The crucial relations are a~.f = a~i (f 0 prl) and a~' = a~' (f 0 p r 2)· Exercise 2.131. Show that Tpr2
a~' Ip
= Tprl ai?,I(p) and x ,q
l,Y I
q
a~Y I(p,q) .
Remark 2.132. In some circumstances, it is safe to abuse notation and denote Xi 0 pr1 by xi and yi 0 pr2 by yi. Of course we are denoting a~. by a~' and so on.
A warning (The second fundamental confusion of calculus 4 ): For a chart (U, x) with x = (xl, ... , x n ), we have defined for any appropri-
l!r
ately defined smooth (or C1) function
f. However, this notation can be
'In [Pen], Penrose a.ttributes this cute terminology to Nick Woodhouse.
122
2. The Tangent Structure
i!r
ambiguous. For example, the meaning of is not determined by the coordinate function xl alone, but implicitly depends on the rest of the coordinate functions. For example, in thermodynamics we see the following situation. We have three functions P, V and T which are not independent but may be interpreted as functions on some 2-dimensional manifold. Then it may be the case that any two of the three functions can serve as a coordinate depends on whether we are using the coordinate system. The meaning of functions (P, V) or alternatively (P, T). We must know not only which function we are allowing to vary, but also which other functions are held fixed. To get rid of the ambiguity, one can use the notations (Us) V and (U) T" In the first case, the coordinates are (P, V), and V is held fixed, while in the second case, we use coordinates (P, T), and T is held fixed. Another way to avoid ambiguity would be to use different names for the same functions depending on the chart of which they are considered coordinate functions. For example, consider the following change of coordinates:
Us
y2
yl = xl + x 2 , = xl _ x 2 + x 3 , y3 = x3.
Here y3 - x 3 as functions on the underlying manifold, but we use different symbols. Thus may not be the same as The chain rule shows that in fact = + This latter method of destroying ambiguity is not very helpful in our thermodynamic example since the letters P, V and T are chosen to stand for the physical quantities of pressure, volume and temperature. Giving these functions more than one name would only be confusing.
b
b
ib a?
a?
Problems (1) Show that if j : M -7 N is a diffeomorphism, then for each p E M the tangent map Tpj : TpM -7 Tf(p)N is a vector space isomorphism. (2) Let M and N be smooth manifolds, and j : M -7 N a Coo map. Suppose that M is compact and that N is connected. If j is injective and Tpj is an isomorphism for each p EM, then show that j is a diffeomorphism. (Use the inverse mapping theorem.) (3) Find the integral curves in ]R2 of the vector field X = edetermine if X is complete or not.
xtx + ~ and
(4) Which integral curves of the field X = x 2 fx + y /y are defined for all times t?
Problems
123
(5) Find a concrete description of the tangent bundle for each of the following manifolds: (a) Projective space IRpn. (b) The Grassmann manifold G (k, n).
(6) Recall that we have charts on IRP2 given by [x,y,z]1-t (Ul,U2) = (x/z,y/z) on U3 = {z i= O}, [x, y, z]1-t (VI, V2) = (x/y, z/y) on U2 = {y i= O}, [x, y, z] I-t (WI, W2) = (y/x, z/x) on Ul = {x i= O}. Show that there is a vector field on IRP2 which in the last coordinate chart above has the following coordinate expression:
a
a
WI--W2-· aWl aW2
What are the expressions for this vector field in the other two charts? (Caution: Your guess may be wrong!). (7) Show that the graph r(f) = {(p, f(p)) E M x N : p E M} of a smooth map f : M -+ N is a smooth manifold and that we have an isomorphism T(p,/(p)) (M x N) ~ T(p,j(p))r(f) EB Tf(p)N. (8) Prove Theorem 2.112. (9) Show that a manifold supports a frame field defined on the whole of M exactly when there is a trivialization of TM (see Definitions 2.130 and 2.58). (10) Prove Proposition 2.85. (11) Find natural coordinates for the double tangent bundle TTM. Show that there is a nice map s : TT M -+ TT M such that s 0 s = idTTM and such that T1f 0 S = T1fTM and T1fTM 0 s = T7r. Here 1f : TM -+ M and 1fTM : TTM -+ TM are the appropriate tangent bundle projection maps. (12) Let N be the subset of IR n+1 x IRn +l defined by N = {(x, y) : Ilxll = 1 and x . y = O} is a smooth manifold that is diffeomorphic to Tsn. (13) (Hessian) Suppose that f E COO(M) and that dfp = 0 for some p E M. Show that for any smooth vector fields X and Y on M we have that Yp(Xf) = Xp(Yf). Let Hf,p(v,w) := Xp(Yf) , where X and Y are such that Xp = V and Yp = w. Show that Hf,p(v, w) is independent of the extension vector fields X and Y and that the resulting map Hf,p : TpM x TpM -+ IR is bilinear. Hf,p is called the Hessian of f at p. Show that the assumption djp = 0 is needed. (14) Show that for a smooth map F : M -+ N, the (bundle) tangent map T F : T M -+ TN is smooth. Sometimes it is supposed that one can
124
2. The Tangent Structure
obtain a well-defined map F. : X (M) --t X (N) by thinking of vector fields as derivations on functions and then letting (F.X) f = X (f 0 F) for f E COO(N). Show why this is misguided. Recall that the proper definition of F. : X (M) --t X (N) would be F.X := T FoX 0 F-l and is defined in case F is a diffeomorphism. What if F is merely surjective?
(15) Show that if'l/J : M' --t M is a smooth covering map, then so also is T'l/J: TM' --t TM. (16) Define the map f : Mnxn(1R) --t sym(Mnxn(1R)) by f(A) := AT A, where Mnxn(1R) and sym(Mnxn(1R)) are the manifolds of n x n matrices and n x n symmetric matrices respectively. Identify TA (Mnxn(1R)) with Mnxn(lR) and Tf(A)sym(Mnxn(lR)) with sym(Mnxn(lR)) in the natural way for each A. Calculate TJ f : Mnxn(lR) --t sym(Mnxn(lR)) using these identifications.
(17) Let h, ... , fN be a set of smooth functions defined on an open subset of
T;
a smooth manifold. Show that if dh (p), ... , dfN (p) spans M for some p E U, then some ordered subset of {h, ... , fN} provides a coordinate system on some open subset V of U containing p.
(18) Let
~r
be the vector space of derivations on Cr(M) at p E M, where r ~ 00 is a positive integer or 00. Fill in the details in the following outline which studies ~r. It will be shown that ~r is not finitedimensional unless r = 00. (a) We may assume that M = lRn and p = 0 is the origin. Let mr := {f E cr(lRn) : f(O) = O} and let m~ be the subspace spanned by the functions of the form fg for f, 9 E mr . We form the quotient space mr/m~ and consider its vector space dual (mr/m~r. Show that if 8 E ~r, then 8 restricts to a linear functional on mr and is zero on all elements of m~. Conclude that 8 gives a linear functional on mr/m~. Thus we have a linear map ~r --t (mr/m~r. (b) Show that the map ~r --t (mr/m~r given above has an inverse. Hint: For a A E (mr/m~r, consider 8>.(f) := A([J - f(O)]), where f E cr(lRn) and hence [f - f(O)] E mr/m~. Conclude that by taking r - 00 we have ToRn = ~oo ~ (mr/m~)·. The case r < 00 is different as we see next. (c) Let r < 00. The goal from here on is to show that mr/m~ and hence (mr/m~r are infinite-dimensional. We start out with the case lRn = R First show that if f E mr, then f(x) = xg(x) for 9 E Cr-1(lR). Also if f E m~, then f(x) = x 2 g(x) for 9 E Cr-1(lR). (d) For each r E {I, 2, 3, ... } and each c E (0,1), define
o
0, for x ~ O.
125
Problems
Then g; E mr , but g; ~ Cr +1(R). Show that for any fixed r E {1, 2, 3, ... }, the set of elements of the form [g;] := gr + m; for c E (0,1) is linearly independent in the quotient. Hint: Use induction on r. In the case of r = 1, it would suffice to show that if we are given 0 < cl < ... < Cl < 1 and if 2:i-l ajg~J Em;, then aj = 0 for all j. (Thanks to Lance Drager for donating this problem and its solution.) (19) Find the integral curves of the vector field on R2 given by X(x, y) := x
2
a
ax
a + xYay'
(20) Show that it is possible that a vector field defined on an open subset of a smooth manifold M may have no smooth extension to all of M. (21) Find the integral curves (and hence the flow) for the vector field on R2 given by X(x, y) := -yfx + x/y. (22) Let N be a point in the unit sphere 8 2. Find a vector field on 8 2 \ {N} that is not complete and one that is complete. (23) Using the usual spherical coordinates (cp, 8) on 8 2, calculate the bracket [¢,,81¢]. (24) Show that if X and Yare (time independent) vector fields that have flows r.pf and r.pf, then if [X, Y] = 0, the flow of X + Y is r.pf 0 r.pf, (25) Recall that the tangent bundle of the open set GL(n, R) in Mnxn(R) is identified with GL(n, R) x Mnxn(R). Consider the vector field on GL(n,R) given by X: 9 f--t (g,g2). Find the flow of X. (26) Let t f--t Qt =
(~~:: ~~~~t ~)
001 for t E R. Let ¢(t, P) := QtP, where P is a plane in R3. Show that this defines a flow on the Grassmann manifold G(3,2). Find the local expression in some coordinate system of the vector field XQ that gives this flow. Do the same thing for the flow
t
f--t
Rt =
(co;t
~ -~nt)
sin t 0
cos t
and find the vector field XR. Find the bracket [XR,XQ]. (27) Develop definitions for tangent bundle and cotangent bundle for manifolds with corners. (See Problem 21.) [Hint: A curve into an n-manifold with corners should be considered smooth only if, when viewed in a chart, it has an extension to a map into Rn. Similarly, a functions is smooth at a corner (or boundary) point only if its local representative
126
2. The Tangent Structure
in some chart containing the point can be extended to an open set in ]Rn.]
(28) Show that if p(x) for some mEN,
= p(Xl, .. . ,xn) is a homogeneous polynomial, so that
P(tXl, ... , tXn) = tmp(Xl, ... , x n ), then as long as c ifold of ]Rn.
=1=
0, the set p-l(c) is an (n - I)-dimensional subman-
(29) Suppose that 9 : M ~ N is transverse to a submanifold WeN. For another smooth map f : Y ~ M, show that f rh g-l(N) if and only if
(g 0 f) rh W. (30) Let M x N be a product manifold. Show that for each X E X(M) there is a vector field X E X(M x N) t~at is prl-related to X and prTrelated to the zero field on N. We call X the lift of X. Similarly, we may lift a field on N to M x N.
Chapter 3
Immersion and Submersion
Suppose we are given a smooth map J : M ---+ N. Near a point p E M, the tangent map TpJ : TpM ---+ TpN is a linear approximation of J. A very important invariant of a linear map is its rank, which is the dimension of its image. Recall that the rank of a smooth map J at p is defined to be the rank of TpJ. It turns out that under certain conditions on the rank of J at p, or near p, we can draw conclusions about the behavior of J near p. The basic idea is that J behaves very much like TpJ. If L : V ---+ W is a linear map of finite-dimensional vector spaces, then Ker Land L(V) are subspaces (and hence submanifolds). We study the extent to which something similar happens for smooth maps between manifolds. In this chapter we make heavy use of some basic theorems of multivariable calculus such as the implicit and inverse mapping theorems as well as the constant rank theorem. These can be found in Appendix C (see Theorems C.l, C.2 and C.5). More on calculus, including a proof of the constant rank theorem, can be found in the online supplement to this text [Lee, J effj. 3.1. Immersions Definition 3.1. A map J : M ---+ N is called an immersion at p E M if TpJ : TpM ---+ Tf(p)N is an injection. A map J : M ---+ N is called an immersion if it is an immersion at every p EM. Note that TpJ : TpM ---+ Tf(p)N is an injection if and only if its rank is equal to dim(M). Thus an immersion has constant rank equal to the dimension of its domain.
-
127
128
3. Immersion and Submersion
Immersions of open subsets of R2 into R3 appear as surfaces that may self-intersect, or periodically retrace themselves, or approach themselves in various limiting ways. The map R2 -+ R3 given by (u, v) H (cos u, sin u, v) is an immersion as is the map (u,v) H (cosusinv,sinusinv, (1- 2 cos2 v) cos v). The map S2 -+ R3 given by (x, y, z) H (x, y, z - 2z 3 ) is also an immersion. By contrast, the map f : 8 2 -+ R3 given by (x, y, z) H (x, y, 0) is not an immersion at any point on the equator 8 2 n {z = o}.
Example 3.2. We describe an immersion of the torus T2 := 8 1 x 8 1 into R3. We can represent points in T2 as pairs (eilh,ei02). It is easy to see that, for fixed a, b > 0, the following map is well-defined: (e l01 , ei(2 ) H (x(e l01 , ei(2 ), y(el01 , el(2 ), z(ei01 , ei(2 )), where
x(el01 , ei(2 ) = (a + bcos{h) cos 02, y(e~Ol,ei02) = (a z(e~Ol,ei02) =
+ bCOSOl) sin 02,
bsinOl.
Exercise 3.3. Show that the map of the above example is an immersion. Give conditions on a and b that guarantee that the map is a 1-1 immersion. Theorem 3.4. Let f : M -+ N be a smooth map that is an immersion at p. Then for any chart (x, U) centered at p, there is a chart (y, V) centered at f (p) such that f (U) c V and such that the corresponding coordinate expression for f is (xl, ... , xk) H (xl, ... , xk, 0, ... ,0) ERn. Here, n is the dimension of Nand k = dim(M) is the rank of Tpf. Proof. Follows easily from Corollary C.3.
o
Theorem 3.5. If f : M -+ N is an immersion (so an immersion at every point), and if f is a homeomorphism onto its image f(M) (using the relative topology on f(M)), then f(M) is a regular submanifold of N. Proof. Let k be the dimension of M and let n be the dimension of N. Clearly f is injective since it is a homeomorphism. Let f(p) E f(M) for a unique p. By the previous theorem, there are charts (U, x) with p E U and (V, y) with f (p) E V such that the corresponding coordinate expression for f is (xl, ... , xk) H (xl, ... , xk, 0, ... ,0) ERn. We arrange to have f(U) C V. But f(U) is open in the relative topology on f(M), so there is an open set o C V in M such that f(U) - f(M) n O. Now it is clear that (0, yi o ) is a chart with the regular submanifold property, and so since p was arbitrary, we conclude that f(M) is a regular submanifold. 0 If f : M -+ N is an immersion that is a homeomorphism onto its image (as in the theorem above), then we say that f is an embedding.
3.1. Immersions
129
Exercise 3.6. Show that every injective immersion of a compact manifold is an embedding. Exercise 3.7. Show that if I: M -+ N is an immersion and p E M, then there is an open U containing p such that 1Iu is an embedding. Exercise 3.S. Recall the definition of a vector field along a map (Definition 2.68). Let X be a vector field along 1 : N -+ M. Show that if 1 is an embedding, then there is an open neighborhood U of I(N) and a vector field X E X(U) such that X - X 0 I. Recall that a continuous map 1 is said to be proper if 1-1 (K) is compact whenever K is compact. Exercise 3.9. Show that a proper 1-1 immersion is an embedding. [Hint: This is mainly a topological argument. You may assume (without loss of generality) that the spaces involved are Hausdorff and second countable. The slightly more general case of paracompact Hausdorff spaces follows.] Definition 3.10. Let Sand M be smooth manifolds. A smooth map 1 : S -+ M will be called smoothly universal if for any smooth manifold N, a mapping 9 : N -+ S is smooth if and only if log is smooth.
Definition 3.11. A weak embedding is a 1-1 immersion which is smoothly universal. Let 1 : S -+ M be a weak embedding and let A be the maximal atlas that gives the differentiable structure on S. Suppose we consider a different differentiable structure on S given by a maximal atlas A2. Now suppose that f : S -+ M is also a weak embedding with respect to A2. Resorting to seldom used pedantic notation, we are supposing that both 1 : (S, A) -+ M and f : (S, A2) -+ M are weak embeddings. From this it is easy to show that the identity map gives smooth maps (S, A) -+ (S, A2) and (S, A 2) -+ (S, A). This means that in fact A = A2, so that the smooth structure of S is uniquely determined by the fact that 1 is a weak embedding. Exercise 3.12. Show that every embedding is a weak embedding.
3. Immersion and Submersion
130
Figure 3.1. Figure eight immersions
In terms of 1-1 immersions, we have the following inclusions: {proper embeddings}
c c
{embeddings} {weak embeddings}
c
{1-1 immersions} .
3.2. Immersed and Weakly Embedded Submanifolds We have already seen the definition of a regular submanifold. The more general notion of a submanifold is supposed to realize the "subobject" in the category of smooth manifolds and smooth maps. Submanifolds are to manifolds what subsets are to sets in general. However, what exactly should be the definition of a submanifold? The fact is that there is some disagreement on this point. From the category-theoretic point of view it seems natural that a sub manifold of M should be some kind of smooth map I : S ---7 M. This is not quite in line with our definition of regular submanifold, which is, after all, a type of subset of M. There is considerable motivation to define sub manifolds in general as certain subsets; perhaps the images of certain nice smooth maps. We shall follow this route. Definition 3.13. Let S be a subset of a smooth manifold M. If S is a smooth manifold such that the inclusion map L : S ---7 M is an injective immersion, then we call S an immersed submanifold. Notice that in the above definition, S certainly need not have the subspace topology! Its topology is that induced by its own smooth structure. The reader may rightfully wonder just how S could acquire such a smooth structure in the first place. If f : N ---7 M is an injective immersion, then S := f(N) can be given a smooth structure so that it is an immersed submanifold. Indeed, we can simply transfer the structure from N via the bijection f : N ---7 f(N). However, this may not be the only possible smooth structure on f(N) which makes it an immersed submanifold. Thus it is imperative to specify what smooth structure is being used. Simply looking at
3.2. Immersed and Weakly Embedded Submanifolds
131
v
Figure 3.2. Immersions can approach themselves
the set is not enough. For example, in Figure 3.1 we see the same figure eight shaped subset drawn twice, but with arrows suggesting that it is the image of two quite different immersions which provide two quite different smooth structures. Suppose that S is a k-dimensional immersed submanifold of a smooth n-manifold M, and let pES. Then using Theorem 3.4, we see that there is a chart (0, x) on S, and a chart (V, y) on M, with p E 0 C V, such that yo
La
x- 1
-
yo x- 1 : x (0) ---+ Y (V)
has the form (a\ ... , a k ) t--+ (a\ ... , ak , 0, ... ,0). This means that y (0) = yo x l(x (0)) is a relatively open subset of ]Rk x {O}. Thus there is an open subset W of y (V) c ]Rn such that y (0) = W n (]Rk x {O}). Letting Ul := y-l(W), we see that y (U1 nO)
= y(U1 ) n
(]Rk x
{O}).
Thus the chart (ylUl ,Ut) has the submanifold property with respect to 0 (but not necessarily with respect to S). The set 0 has a smooth structure as an open submanifold of S. But this is the same smooth structure 0 has as yk 0 a regular submanifold. To see this note that the restrictions combine to give an admissible chart on S. Indeed, using the functions we obtain a bijection of 0 with an open subset of ]Rk. We only need to show that this bijection is smoothly related to the chart (0, x), and this amounts to showing that yk 0 are smooth. But this follows immediately 1 from the fact that yo x- is smooth. Notice that unlike the case of a regular submanifold, it may be that no matter how small V,
y1lo ' ... , I
y1Io' ... ' I
y(V n 0) # y(V) n (]Rk
x {O}),
as indicated in Figure 3.2. So in summary, each point of an immersed submanifold has a neighborhood that is a regular submanifold. Proposition 3.14. Let ScM be an immersed submanifold of dimension k and let f : N ---+ S be a map. Suppose that L of: N ---+ M is smooth,
132
where /, : S y also smooth.
3. Immersion and Submersion
M is the inclusion. Then ij j : N -+ S is continuous, it is
Proof. We wish to show that j : N -+ S is smooth if j-1(0) is open for every set 0 C S that is open in the manifold topology on S. Let pEN and choose a chart (V, y) for M centered at /, 0 j (p), so that U
= {q E V
: yk+1 (q)
= ... = yn (q) = O}
is an open neighborhood of p in S and such that y1j u ' ... , yk jU are coordinates for S on U. By assumption j-1(U) is open. Thus (/, 0 f) (I-1(U)) C U. In other words, /, 0 j maps an open neighborhood of pinto U. To test for the smoothness of j, we consider the functions (y' j u) 0 j on the set j-1 (U). But (yijU) oj=yio/,oj, and these are clearly smooth by the assumption that /, 0 j is smooth. 0 Sometimes the previous result is stated differently (and somewhat imprecisely): Suppose that j : N -+ M is a smooth map with image inside an immersed submanifold S; then j is smooth as a map into S if it is continuous as a map into S. The lack of a notational distinction between j as a map into Sand j as a map into M is what makes this way of stating things less desirable. Let ScM and suppose that S has a smooth structure. To say that an inclusion S y M is an embedding is easily seen to be the same as saying that S is a regular submanifold, and so we also say that S is embedded in M.
Corollary 3.15. Suppose that ScM is a regular submanijold. Let j : N -+ S be a map such that /, 0 j : N -+ M is smooth. Then j : N -+ S is smooth. Proof. The map f., : S y M is certainly an immersion, and so by the previous theorem we need only check that j : N -+ S is continuous. Let 0 be open in S. Then since S has the relative topology, 0 = un S for some open set U in M. Then 1-1(0) = 1- 1(UnS) = 1-1(/,-1(U)) = (/, 0 f)-1 (U), which is open since /, 0 j is continuous. Thus j is continuous. 0 Definition 3.16. Let S be a subset of a smooth manifold M. If S is a smooth manifold such that the inclusion map /, : S -+ M is a weak embedding, then we say that S is a weakly embedded submanifold. From the properties of weak embeddings we know that for any given subset ScM there is at most one smooth structure on S that makes it a weakly embedded submanifold.
3.2. Immersed and Weakly Embedded Submanifolds
133
Corresponding to each type of injective immersion considered so far we have in their images different notions of submanifold: {proper submanifolds}
c c
{regular submanifolds} {weakly embedded submanifolds}
C {immersed submanifolds}.
We wish to further characterize the weakly embedded submanifolds. Definition 3.17. Let S be any subset of a smooth manifold M. For any XES, denote by Ox(S) the set of all points of S that can be connected to x by a smooth curve with image entirely inside S. It is important to be clear that Ox(S) is not necessarily the connected component of S with its relative topology since, for example, S could be the image of an injective nowhere differentiable curve. In the latter case, Cx(S) = {x} for all XES! Definition 3.18. We say that a subset S of an n-manifold M has property W(k) if for each So E S there exists a chart (U, x) centered at So such that x(Cso(U n S)) = x(U) n (jRk x {O}). Here jRn = jRk X jRn-k. Together, the next two propositions show that weakly embedded submanifolds are exactly those subsets that have property W(k) for some k. Our proof follows that of Michor [Michl, who refers to subsets with property W(k), for some k, as initial submanifolds. With Michor's terminology, the result will be that the initial submanifolds are the same as the weakly embedded submanifolds. Proposition 3.19. If an injective immersion I : S ~ M is smoothly universal, then the image I (S) has property W( k) where k = dim( S). In particular, if SCM is a weakly embedded submanifold of M, then it has property W(k) where k = dim(S). Proof. Let dim(S) - k and dim (M) = n. Choose So E S. Since I is an immersion, we may pick a coordinate chart (W, w) for S centered at So and a chart (V, v) for M centered at I(so) such that
volow I(y) = (y,O) = (yl, ... ,yk,O, ... ,O).
°
Choose an r > small enough that Bk(O, 2r) C w(W) and Bn(O, 2r) C v(V). Let U = v-I(Bn(O,r)) and WI = w-I(Bk(O,r)). Let x := vl u . We show that the coordinate chart (U, x) satisfies the conditions of Definition 3.18:
x-l(x(U) n (jRk x {a})) = x-I{(y,0) : Ilyll < r} =Iow Io(xolow 1)-I({(y,O): Ilyll 0 such that c = 'Y 0 h. In this case, we say that 'Y and c have the same sense and provide the same orientation on the image. We assume that 'Y : I ~ ]Rn has IIT'II > 0, which is the case of interest. Such a curve is called regular, which just means that the curve is an immersion.
Definition 4.3. If 'Y : I ---t]Rn is a regular curve, then T(t) := 'Y'(t)/ 11T'(t)II defines the unit tangent field along 'Y. (Of course, we then have IITII = 1.) We have the familiar notion of the length of a curve defined on a closed interval'Y: [tI, t2J ~ ]Rn:
L=
l
t2
tl
!Ir' (t) II dt.
4. Curves and Hypersurfaces in Euclidean Space
146
One can define an arc length function for a curve 'Y : I --+ lRn by choosing to E I and, then defining s
= h(t)
:-I 1i'Y'(T)\\ t
dT.
to
Notice that s takes on negative values if t < to, so it does not always represent the length in the ordinary sense. If the curve is smooth and regular, then h' - 1i'Y'(T)II > 0, and so by the inverse function theorem, h has a smooth inverse. We then have the familiar fact that if e(s) = 'Y 0 h- 1 (s), then Ile'll (s) :- Ile'(s)11 = 1 for all s. Curves which are parametrized in terms of arc length are referred to as unit speed curves. For a unit speed curve, ~~(s) = T(s). Since parametrization by arc length eliminates any component of acceleration in the direction of the curve, the acceleration must be due only to the shape of the curve. Definition 4.4. Let c : I --+ lRn be a unit speed curve. The vector-valued function dT
K(S)
:=
ds (s)
is called the curvature vector. The function
K,
defined by
is called the curvature function. If K,(s) > 0, then we also define the principal normal dT 11-1 Ts(s), dT N(s):= Il Ts(s)
so that ~'!'
= K,N.
Let 'Y : I --+ lRn be a regular curve. An adapted orthonormal moving frame along 'Y is a list (El," . ,En) of smooth vector fields along 'Y such that El(t) = 'Y'(t)/Ii'Y'(t)11 and such that (E1(t), ... ,En(t)) is a basis of T'Y(t)lR n for each tEl. Identifying El (t), . .. , En(t) with elements of lRn written as column vectors, we say that the orthonormal moving frame is positively oriented if
Q(t) = [E1(t), ... , En(t)] is an orthogonal matrix of determinant one for each t. Definition 4.5. A moving frame El(t), ... , En(t) along a curve 'Y : 1--+ lRn is a Frenet frame for 'Y if 'Y(k) (t) is in the span of El (t), . .. , Ek(t) for all t and 1 ~ k ~ n. As we have defined them, Frenet frames are not unique. However, under certain circumstances we may single out special Frenet frames. For example,
147
4.1. Curves
if c: I -4 lR3 is a unit speed curve with /'i, > 0, then the principal normal N is defined. By letting
B=TxN we obtain a Frenet frame T, N, B. It is an easy exercise to show that we obtain dT ds dN
-
dB ds
=
/'i,N -/'i,T
ds
+
TB
-TN
for some function T called the torsion. This is the familiar form presented in many calculus texts. In matrix notation,
[ °~ -/'i,~ ~T° 1
:8[T,N,B] =[T,N,B]
•
Notice that for a regular curve, /'i, ~ 0, while T may assume any real value. Another special feature of the frame T, N, B is that it is positively oriented. If [ is injective, then we can think of /'i, and T as defined on the geometric image [(1). Thus, if p = ,(80), then /'i,(p) is defined to be equal to /'i,(80)' Exercise 4.6. Let c : I -4 lR3 be a unit speed regular curve with /'i, > 0. Show that (( c' (8) X e" (8)) , e"' (8)) c 11 I () T 8 = /'i, (8) 2 lor a 8 E . Exercise 4.7. Let c: I -4lR3 be a unit speed curve and 80 E I. Show that we have a Taylor expansion of the form
,(8) -,(80) = ((8 - 80) -
~(8 -
80)3/'i,2(8 0)) T(80) 80) 3 d/'i,)) d8 (80 N(80 )
+ ( 12 (8 -
80) 2 /'i,(80)
+ (~(8 -
80 )3/'i,(8 0 )T(80 )) B(8)
1 + 6(8
+ 0((8 -
80)3).
We wish to generalize the special properties of the Frenet frame T, N, B to higher dimensions thereby obtaining a notion of a distinguished Frenet frame. For maximum generality, we do not assume that the curve is unit speed. A curve, in lRn is called k-regular if {['(t), ,"(t), .. . ,[(k)(t)} is a linearly independent set for each t in the domain of the curve. For an (n - 1)regular curve, the existence of a special orthonormal moving frame can be easily proved. One applies the Gram-Schmidt process: If El (t), ... , Ek (t)
4. Curves and Hypersurfaces in Euclidean Space
148
are already defined for some k < n - 1, then
Ew(t)
:~ [-y(W) (t) - ~ U
k +1) (t),
Ck
E,(t)) E,(t)] ,
where Ck is a positive constant chosen so that IIEk+1 (t) II = 1. Inductively, this gives us El(t), ... , En-l(t), and it is clear that the Ek(t) are all smooth. Now we choose En(t) to complete our frame by letting it be of unit length and orthogonal to E1(t), ... ,En-1(t). By making one possible adjustment of sign on En(t) we obtain a moving frame that is positively oriented. In fact, En(t) is given by the construction of Lemma 4.2, from which it follows that En(t) is smooth in t. By construction we have a nice list of properties:
(1) For 1 ~ k ~ n, the vectors El (t), .. . , Ek(t) have the same linear span as -y'(t), ... ,-y(k)(t) so that there is an n x n upper triangular matrix function U (t) such that
b'(t), ... , -y(n) (t)]U(t) = [E1(t), ... , En(t)]. (2) For 1 ~ k ~ n - 1, the vectors E1(t), ... , Ek(t) have the same orientation as -y' (t), ... , -y(k) (t). Thus U (t) has diagonal elements which are all positive except possibly the last one.
(3) (E1(t), ... , En(t)) is positively oriented as a basis of T'Y(t)R n ~ JRn • Exercise 4.8. Show that the moving frame we have constructed is the unique one with these properties. We call a moving frame satisfying the above properties a distinguished Frenet frame along -y. For any orthonormal moving frame, the derivative of each Ei(t) is certainly expressible as a linear combination of the basis E1(t), ... , En(t), and so we may write
d
n
dtEj(t) = ~Wi3(t)Ei(t). ~=l
Of course, Wij(t) = (Ei(t),-9tEj(t)), but since (Ei(t), Ej(t)} = elij, we conclude that Wij(t) = -Wji(t), i.e., the matrix w(t) = [wtj(t)] is antisymmetric. However, for a distinguished Frenet frame, more is true. Indeed, if (E1(t), ... , En(t)) is such a distinguished Frenet frame, then for 1 ~ j < n we have Ej(t) - E{=l Ukj'Y(k) (t), where U(t) = [Uk3 (t)] is the upper triangular matrix mentioned above. Using the fact that U, -9tU, and U- 1 are all upper triangular, we have
149
4.1. Curves
But 'Y(k+1) (t) so that
= "k+l (U- 1) r,k+l E r (t) , and 'Y(k) (t) L.,.,r=l :tEj(t) =
=
t, (! tr (d j
Ukj ) ')Ik)(t)
dt Ukj
j
+
+
t,
= "k (U- 1) rk E r (t) L.,.,r=l
ukd k+1)(t)
) ?; (U-1)rkEr(t) k
k+1
L Uk) L (U- )r,k+1 Er(t). 1
k=l
r=l
From this we see that ;1tEj(t) is in the span of (Er(t)h~r~j+1' Thus w(t) = (Wij(t)) can have no nonzero entries below the sub diagonal. But w is antisymmetric, so we conclude that W has the form
o W(t) =
o
-Wn,n-l(t)
Wn,n-l(t)
0
We define the i-th generalized curvature function by
Wt+1,i(t) () Ki t := 1l'Y'(t) II . Thus if 'Y : I -t
]Rn
is a unit speed curve, we have
o W(s) =
o
-Kn-l(S)
Kn-l(S)
0
Note: Our matrix w is the transpose of the W presented in some other expositions. The source of the difference is that we write a basis as a formal row matrix of vectors. Lemma 4.9. IJ'Y : I -t ]Rn (n ~ 3) is (n - I)-regular, then Jor 1 :::; i :::; n-2, the generalized curvatures Ki are positive.
150
4. Curves and Hypersurfaces in Euclidean Space
Proof. By construction, for 1
~
i
~
=
L Uji(t)-y(J) (t),
n- 1
i
Ei(t)
J-1 i
'Y(i)(t) =
L (U)j/ (t)Ej(t), 3=1
with Un
> 0,
w.+1,.(t)
and hence
(U- 1 )ii > o.
Thus if 1 ~ i ~ n - 2, we have
~ (E,+J, ! E.) ~ ( E.+l, :t ~ Uj;(thu1 (t)) = / E~+l, \
t
J-1
'Y(J)(t) :t
UJ~(t)) + / E~+l, \
t UJ~(t)-y(j+1)(t)) J
1
= Un (E~+1(t)''Y(i+l)(t)) = Un (U- 1)i+1,i+1 > O. In passing from the second to the third line above, we have used the fact that E~+1 is orthogonal to all 'Y(J) for j ~ i since these are in the span of
{EJ}J=l, .... ~.
0
The last generalized curvature function II:n-l is sometimes called the torsion. It may take on negative values. Exercise 4.10. If 'Y : I -+ JRn is (n - l)-regular, show that E1 = T and E2 = N. If 'Y is parametrized by arc length, then fs 'Y - E1 and ~'Y = 1I:1E 2. Conclude that 11:1 II: (the curvature defined earlier). The orthogonal group O(JR n ) is the group of linear transformations A : JRn -+ JRn such that (Av, Aw) = (v, w) for all v, w E JR n . The group O(JRn) is identified with the group of orthogonal n x n matrices denoted O(n). The Euclidean group Euc(JR n ) is generated by translations and elements of O(JRn ). Every element ¢ E Euc(JR n ) can be represented by a pair (A, b), where A E O(JRn ) and bE JRn and where ¢(v) = Av + b. Note that in this case, D¢ - A (the derivative of ¢ is A). The elements of the Euclidean group are called Euclidean motions or isometries of JR n . If ¢ E Euc(JR n ), then for each p E JR n , the tangent map Tp¢ : TpJRn -+ T.p(p)JR n is a linear isometry. In other words, (Tp¢. vp, Tp¢· wp)/(p) - (vp, wp)p for all vp, wp E TpJR n . The group SO(JRn ) is the special linear group on JRn and consists of the elements of O(JR n ) which preserve orientation. The corresponding matrix group is SO(n) and is the subgroup of O(n) consisting of elements of determinant l. The subgroup SEuc(JR n ) c Euc(JR n ) is the group generated by translations and elements of SO(JRn). It is called the special Euclidean group.
4.1. Curves
151
Rn and '1 : I ~ R n be two (n - I)-regular curves with corresponding curvature functions Ki and K.i (1 < i ~ n - 1). If -y'(t) II I'1'(t) II and Ki(t) = Ki(t) for all t E I and 1 ~ i ~ n - 1, then there exists a unique isometry 4> E S Euc(Rn) such that
Theorem 4.11. Let"( : I
~
'1-4>0"(. Proof. Let (E1(t), ... , En(t)) and (E1(t), ... , En(t)) be the distinguished Frenet frames for "( and '1 respectively and let Wtj and i:hJ be the corresponding matrix elements as above. Fix to E I and consider the unique isometry 4> represented by (A, b) such that 4>b(to)) = '1(to) and such that
A(Ei(tO)) = Ei(tO) for 1 ~ i ~ n. Since 1I"f'(t) II = 11'1'(t)II and Ki(t) = K.i(t), we have that Wtj(t) = Wtj(t) for all i,j and t. Thus we have both
and
d
n
dt AEi(t) =
L Wji(t)AEJ (t). J
1
El and AEi satisfy the same linear differential equation, and since A(Ei(tO)) - Ei(to), we conclude that A(Ei(t)) = Ei(t) for all t and 1 ~ i < n. In particular, A-y'(t) = 1I"f'(t)II AEl(t) = 11'1'(t)II E1(t) = '1'(t). Thus Hence
4>b(t)) - 4> b(to)) =
it it
(4) 0 -y)' (T) dT =
to
=
to
A-y'(T) dT =
it
it
D4>· -y'(T) dT
to
'1'(T) dT = '1(t) - '1(to),
to
from which we conclude that 4>b(t)) - '1(t). For uniqueness, we argue as follows. Suppose that 'Ij; 0 -y = '1 for 'Ij; E Euc(Rn) and suppose that 'Ij; is represented by (B, c). The fact that D'Ij; must take the Frenet frame of "( to that of '1 means that A = D'IjJ = D4> = B. The fact that 'lj;b(to)) - '1(to) implies that b c and so 'Ij; 4>. 0 Conversely, we have Theorem 4.12. If Kl,' .. , Kn-l are smooth functions on a neighborhood of E R such that K t > 0 for i < n -1, then there exists an (n -1) -regular unit speed curve -y defined on some interval containing So such that Kl, ... ,Kn 1 are the curvature functions of -y.
So
152
4. Curves and Hypersurfaces in Euclidean Space
Proof. We merely sketch the proof: Let
A(s) :=
a
-KI
KI
a
a
a a
a a
a
a Kn-l
-Kn-l
a
and consider the matrix initial value problem
X'=XA, X(so) = I. This has a unique smooth solution X on some interval I which contains so. The skew-symmetry of A implies that X (s) is orthogonal for all s E I. If we let Xl be the first column of X, then
,(s):=
1 8
xI(t)dt
80
defines a unit speed (n - l)-regular curve with the required curvature func-
0
fuM.
Exercise 4.13. Fill in the details of the previous proof. If n > 3, then a regular curve, need not have a Frenet frame. However, a regular curve still has a curvature function, and if the curve is 2-regular, then we have a principal normal N which is defined so that T and N are the Gram-Schmidt orthogonalization of " and ,". We will sometimes denote this principal normal by E2 in order to avoid confusion with the normal to a hypersurface, which is denoted below by N.
4.2. Hypersurfaces Suppose that Y is vector field on an open set in IRn. For p in the domain of Y, and vp E TplRn, let c be a curve with c(O) = p and c(O) = vp. For any t near 0, we can look at the value of Y at c(t). We let V'vpY := (Y 0 c)' (0),
which is defined since Yo c is a vector field along c. Note that in this context (Y 0 c)' (0) is taken to be based at p = c(O). If X is a vector field, then a vector field V'xY is given by V'xY : pH V'xpY. In fact, it is easy to see that if X = E Xiei and Y = E Y~ei' then V'x Y =
It
L (Xyi)ei
since (Xyi) (p) = XpY~ = 10 yi 0 c. We have presented things as we have because we wish to prime the reader for the general concept of a covariant derivative that we will meet in later chapters. However, it must be confessed
153
4.2. Hypersurfaces
that under the canonical identification of IRn with each tangent space, \7 Xp Y is just the directional derivative of Y in the direction Xp. The map (X, Y) I---t \7 x Y is COO(IRn)-linear in X but not in Y. Rather, it is IR-linear in Y and we have a product rule:
\7xfY = (XI) Y
+ f\7xY.
Because of these properties, the operator \7 x : Y I---t \7 x Y, which is given for any X, is called a covariant derivative (or Koszul connection). In Chapter 12 we study covariant derivatives in a more general context. Since we shall soon consider covariant derivatives on submanifolds, let us refer to 'Vas the ambient covariant derivative. Notice that we have \7 x Y - \7y X -
[X,Yj. There is another property that our ambient covariant derivative \7 satisfies. Namely, it respects the metric:
X (Y, Z) = (\7xY, Z)
+ (Y, \7xZ).
Similarly, if vp E TpIRn, then vp (Y, Z) = (\7 vp Y, Zp) + (Yp, \7 vp Z). Consider a hypersurface M in IRn. By definition, M is a regular (n - 1)dimensional sub manifold of IRn. (If n = 3, then such a submanifold has dimension two and we also just refer to it as a surface in IR3.) A vector field along an open set 0 C M is a map X : 0 -+ TlRn such that the following diagram commutes: TIRn
Y1
o '---
IRn
Here the horizontal map is inclusion of 0 into IRn. If X (p) E TpM for all p EO, then X is nothing more than a tangent vector field on O. If N is a field along 0 such that (N(p), vp) = 0 for all vp E TpM and all p E 0, then we call N a (smooth) normal field. If N is a normal field such that (N(p) , N(p)) = 1 for all p E 0, then N is called a unit normal field. Note that because of examples such as embedded Mobius bands in IR3, it is not always the case that there exists a globally defined smooth unit normal field. Definition 4.14. A hypersurface Min IRn is called orientable if there exists a smooth global unit normal vector field N defined along M. We say that M is oriented by N. We will come to a more general and sophisticated notion of orientable manifold later. That definition will be consistent with the one above. Exercise 4.15. Show that for a connected orientable hypersurface, there are exactly two choices of unit normal vector field.
154
4. Curves and Hypersurfaces in Euclidean Space
In this chapter, we study mainly local geometry. We focus attention near a point p E M. We consider a chart (0, u) for ~n that is a single-slice chart centered at p and adapted to M. Thus if u = CuI, ... ,un), then the restrictions of the functions UI , ... ,un - I give coordinates for M on the set o = 0 n M = {un = o}. We denote these restrictions by ul, ... , un 1. Thus if we write u:= (u I , ..• ,un - I ), then (O,u) is a chart on M. We may further arrange that u( 0) is a cube centered at the origin in ~n and so, in particular, 0 is connected and orientable. Let us temporarily call such charts special. The coordinate vector fields a~. for i = 1, ... , n are defined on 0, while the vector fields a~. are defined on O. We have
r::J~' (p) =
-
uu t
r::J0. (P) for all p E 0 and i = 1, ... , n - 1. uu t
-
-
If X is a vector field on an open set 0, then its restriction to 0 - 0 Mis a vector field along 0, which certainly need not be tangent to M. If X is a vector field along 0, then there must be smooth functions Xi on 0 such that n
X(p) =
.
a
I: xt(P) ou (p) i
1
t
for all p EO. Then X is a tangent vector field on 0 precisely when the last component xn is identically zero on 0 so that n 1
a
X = '"' Xi_. L...J aut t
1
Now if X is a field along M, then it can be extended to a field X on 0 C ~n by considering the component functions Xi as functions on 0 which happen to be constant with respect to the last coordinate variable un. In other words, if 7r : 0 --+ 0 is the map (a I , ... , an) ~ (a I , .. . , an-I, 0), then
-
~
- a
X = L...J x taut' t
1
where it = xiou I 07rou• This last composition makes good sense because (0, u) is a special single-slice chart as described above. We will refer to X as an extension of X, but note that the extension is based on a particular special choice of single-slice chart. The constructions on the hypersurface that we consider below do not depend on the extension. Given a choice of unit normal N along a neighborhood of p EM, V'vpN is defined for any vp E TpM by virtue of the fact that N only needs to be defined along a curve with tangent vp. Alternatively, we can define V'vpN to be equal to V'vp IV for an extension IV of N in a special single-slice chart
155
4.2. Hypersurfaces
N
Figure 4.1. Shape operator
adapted to M and containing p. We note that ('VvpN,N(p)) = O. Indeed, since (N, N) 1, we have
0= vp (N, N) = 2 ('VvpN, N). Thus 'V vpN E TpM since TpM is exactly the set of vectors in TplRn perpendicular to Np = N(p). Exercise 4.16. Suppose we are merely given a unit normal vector at p. Show that we can extend it to a smooth normal field near p. For dimensional reasons, any two such extensions must agree on some neighborhood of p. Definition 4.17. Given a choice of unit normal Np at p E M, the map SNp : TpM ~ TpM defined by
SNp(Vp) := -'VvpN for any local unit normal field N with Np = N(p), is called the shape operator or Weingarten map at p. From the definitions it follows that if c is a curve with c(to) = p and = vp, then SNp(Vp) = - (N 0 c)' (to). If 0 c M is an open set oriented by a choice of unit normal field N along 0, then we obtain a map SN : TO ~ TO by SNITpM := SN p. This map is also called the shape operator (on 0). We have the inner product (" ')p on each tangent space TplRn and TpM C TplRn. For each p E M, the restriction of this inner product to each tangent space TpM is also denoted by (', ')p or just (', .). We also denote this inner product on TpM by gp so that gp("') = (-, ')p. The map p f--t gp is smooth in the sense that p f--t gp (X(p), Y(p)) is smooth whenever X and Yare smooth vector fields on M. Such a smooth assignment of inner product to the tangent spaces of M provides a Riemannian metric on M, a concept
c(to)
156
4. Curves and Hypersurfaces in Euclidean Space
studied in more generality in later chapters. We denote the function p r-+ gp (X(p), Y(p)) simply by 9 (X, Y) or (X, Y). In short, a Riemannian metric is a smooth assignment of an inner product to each tangent space. Definition 4.18. A diffeomorphism f : Ml ---+ M2 between hypersurfaces in JRn is called an isometry if Tpf : TpMl ---+ Tf(p) M2 is an isometry of inner product spaces for all p E MI. In this case, we say that MI is isometric to M 2·
Proposition 4.19. SNp : TpM ---+ TpM is self-adjoint with respect to (', ')p. Proof. Let (0, y) be a special chart centered at p and let 0 = 0 n M as above. Let Xp and Yp be elements of TpM and ex~end th.-:m to .:'ector fields X and Y on O. Then extend X and Y to fields X and Y on O. Similarly, extend Np to N and then fir. Note that Xp(fir, X) = Xp (N, X) = 0 and Yp(fir, Y) = O. Using this, we have
- (SNpXp, Yp)
+ (Xp, SNpYp)
= ('VxpN, Yp) - (Xp, 'VYpN) = ('V xfir, Y)p - (X, 'Vyfir)p
= Xp(fir, Y) - (fir, 'V xY)p - Yp(fir, X) + (fir, 'VyX)p = ('VyX - 'V xY, fir)p = ([Y, Xl, fir)p
= ([Y, Xl (p) , fir (p)) = ([Y, Xl (p) , N (p)) = 0 since [Y, Xl (p) sion map.
= [Y, Xl (p)
E
TpM by Proposition 2.84 applied to the inclu0
Definition 4.20. The symmetric bilinear form IIp on TpM defined by
IIp(vp,wp):= (Vp,SNpWp) = (SNpVp,Wp) is called the second fundamental form at p. If N is a unit normal on an open subset of M, then for smooth tangent vector fields X, Y, the function II(X, Y) defined by p H IIp(Xp, Yp) is smooth. The assignment (X, Y) H II(X, Y) defined on pairs of tangent vector fields is also called the second fundamental form. The form II is bilinear over COO(O), where 0 is the domain of N. As we shall see, the shape operator can be recovered from the second fundamental form II together with the metric 9 = (.,.) (first fundamental form). If the reader keeps the definition in mind, he or she will recognize that the second fundamental form appears implicitly in much that follows. We return to the second fundamental form again explicitly later. Exercise 4.21. Let Xp E TpJRn and let Y be any smooth vector field on JR n . Show that if f : JRn ---+ JRn is a Euclidean motion, then we have
157
4.2. Hypersurfaces
Tf· 'V xpY = 'VTj.xpf.Y. More generally, show that this is true if f is affine (i.e. if f is of the form f{x) - Ax + b for some linear map A and b E ]Rn).
Theorem 4.22. If Ml is a connected hypersurface in]Rn and if f : ]Rn ---t is a Euclidean motion, then M2 - f{M 1 ) is also a hypersurface and
]Rn
(i) the induced map flMl : Ml ---t M2 is an isometry; (ii) if Ml and M2 are oriented by unit normals Nl and N2 respectively, then, after replacing Nl or N2 by its negative if necessary, we have TfoSNl = SN2oTf.
Proof. That f(M I ) is a hypersurface is an easy exercise, which we leave to the reader. After noticing that flMl : Ml ---t M2 is smooth (why?), we argue as follows: Let ¢ := flMl and notice that T¢ . v = T f . v for all v tangent to Ml. Since T f preserves the inner products on T]Rn, we see that Tp¢ is an isometry for each p E MI. Since ¢ is clearly a bijection we conclude that (i) holds. Note that we must have Tpf· NI (p) = ±N2(p) and, since MI is connected, one possible change of sign on unit normals gives Tf
° NI ° f- I
-
N2.
Then by Exercise 4.21 Tf· SN1V
or Tf
0
SNl
= -Tf· 'VvNl = SN2
0
- -'VTj.vf.Nl
=
-'VTf·v N 2 = SN2 (Tf· v),
0
Tf.
Notice that an arbitrary isometry Ml ---t M2 need not be the restriction of an isometry of the ambient ]Rn and need not preserve shape operators. Definition 4.23. Let M be a hypersurface in ]Rn, take a point p E M, and let Np be a unit normal at p. The mean curvature H(p) at p in the direction Np is defined by 1 H(p) := --1 trace(SNp ) '
n-
Notice that changing Np to -Np changes H(p) to -H(P). If N is a unit normal field along an open set U c M, then the function p ~ H(p) is smooth. It is called a mean curvature function. If M is an orient able hypersurface, then it has a global mean curvature function for each unit normal field. Definition 4.24. Let M be a hypersurface in ]Rn. Let a unit normal Np be given at p. The Gauss curvature K(p) at p is defined by K(p) := det(SNp )'
4. Curves and Hypersurfaces in Euclidean Space
158
If N is a unit normal field along an open set U eM, then the function p H K(P) det(SN(p») is smooth. It is called a Gauss curvature function associated to the normal field, and always exists locally. If M is orient able, then it has a global Gauss curvature function for every choice of unit normal field. Notice that changing Np to -Np changes K(p) to (-lr- 1 K(p). It follows that if n - 1 is even, then there is a unique global Gauss curvature function regardless of whether M is orientable or not. In particular, this is the case for surfaces in IR3.
The shape operator S Np encodes the local geometry of the sub manifold at p and measures the way M bends and twists through the ambient Euclidean space. Since SNp is self-adjoint, there is a basis for TpM consisting of eigenvectors of SNp. An eigenvector for SNp is called a principal vector, and a unit principal vector is called a principal direction or a direction of curvature. The eigenvalues are called principal curvatures at p. If k1 , ... ,kn - 1 are the principal curvatures at p, then n-l
H(p)
_1_", k n 1 L...J 1 i
n-l
and
1
K(p) =
II k t
i•
1
If u E TpM is a tangent vector with (u,u) = 1, then k(u) = (SNpU,U) is the normal curvature in the direction u. Of course, if u is a unit length eigenvector (principal direction), then the corresponding principal curvature k is just the normal curvature in that direction. Notice that k( -u) - k(u). A vector v E TpM is called asymptotic if (SNpV,v) = O.
Proposition 4.25. Let M be a hypersurface and N a unit normal field. Let "( : I -+ M c ]Rn be a regular curve with image in the domain of N. Then (SNp "('(t), "('(t)) (N(-y(t)), "("(t)). Proof. Since (N (-y( t)), "(' (t)) - 0, differentiation gives
(N(-y(t)), "("(t))
+ (:t N(-y(t)), :t "((t)) = O.
Thus (SNp"(' (t), "(' (t))
= (-\} -y/(t)N, "(' (t)) = ( - :t N(-y(t)), :t "((t)) - (N(-y(t)) , "("(t)).
0
In particular, if u is a unit vector at p and u = C(O) for some unit speed curve c, then k(u) = (Np, d'(O)). This shows that all unit speed curves with a given velocity u have the same normal component given by the normal curvature in that direction. This curvature is forced by the shape of M and we see how normal curvatures measure the shape of M.
159
4.2. Hypersurfaces
Corollary 4.26. Let c : I --+ M c ]Rn be a unit speed curve. If I'\:(so) = 0 for some So E I, then k(c(so)) = O. If I'\: (s) > 0 and E2 is the principal normal defined near s, then k(c(s)) - ~(s) cosO(s), where O(s) is the angle between N(c(s)) and E2(S). Proof. Suppose ~(O) = O. Then k(d(O)) = (N(c(O)), c"(O)) 0, then E2 is defined for an interval containing s. We have
k(c'(s))
= O.
If ~ (s) >
(N(c(s)), c"(s)) = (N(c(s)), ~E2(C(S))) = ~(s) (N(c(s)), E2(S)) = ~(s) cos O(s).
=
o
If P is a 2-plane containing p E M and ~uch that Np is tan~ent to P, then for a small enough open neighborhood 0 of p, the set C = 0 n P n M is a regular one-dimensional submanifold of ]Rn. We parameterize C by a unit speed curve c with c(O) = p. This curve c is called a normal section at p. Notice that in this case E 2(O) - ±Np. Then k(c(O)) ±~l(O) where the" " sign is chosen in case Np - -E2(O). Thus we see that k(c(O)) is positive if the normal section c bends away from Np •
Definition 4.27 (Curve types). Let M be a hypersurface in I --+ M be a regular curve.
]Rn
and let
"y :
(i) , is called a geodesic if the acceleration ," (t) is normal to M for all t E I. (ii) , is called a principal curve (or line of curvature) if -y(t) is a principal vector for all tEl. (iii) , is called an asymptotic curve if -y(t) is an asymptotic vector for all tEl. Proposition 4.28. Let M be a surface in ]R3. Suppose that a regular curve "y : I --+ M is contained in the intersection of M and a plane P. If the angle between M and P is constant along " then, is a principal curve. Proof. The result is local, and so we assume that M is oriented by a unit normal N. Let v be a unit normal along P. Since P is a plane, v is constant. By assumption, (N, v) is constant along ,. Thus
0= :t (N
0"
v0
,)
= ("V"yN, v 0
,),
so 'V"yN is orthogonal to v along,. By the same token, "V"yN is orthogonal to N since (N, N) = 1. Thus "V"yN must be collinear with -y. In other words, SN-Y - "V"yN = >'-y for some scalar function >.. 0
160
4. Curves and Hypersurfaces in Euclidean Space
Example 4.29 (Surface of revolution). Let t t-+ (9(t), h(t)) be a regular curve in ]R2 defined on an open interval I. Assume that h > O. Call this curve the profile curve. Define x : I x ]R --+ ]R3 by
x(u, v) = (9(U), h(u) cosv, h(u) sin v). This is periodic in v and its image is a surface. The curves of constant u and curves of constant v are contained in planes of the form {x - c} and {z = my} (or {y = mz}) and so they are principal curves. For a surface of revolution, the circles generated by rotating a fixed point of the profile are called parallels. These are the constant u curves in the example above. The curves that are copies of the profile curve are called the meridians. In the example above, the meridians are the constant v curves. We know from standard linear algebra that principal directions at a point in a hypersurface corresponding to distinct principal curvatures are orthogonal. Generically, each eigenspace will be one-dimensional, but in general they may be of higher dimension. In fact, if SNp is a multiple of the identity operator, then there is only one eigenvalue and the eigenspace is all of TpM. In this case, every direction is a principal direction. It may even be the case that S Np = O. If c : I --+ M is a unit speed curve into a hypersurface, then rather than use the distinguished Frenet frame of c, we can use a frame which incorporates the unit normal to the hypersurface: Definition 4.30. Let c : I --+ M be a unit speed curve into a surface in R3 and let N be a unit normal defined at least on an open set containing the image of c. A frame field (Dl' D2, D3) (along c) such that Dl = T (= c), D3 = N 0 c and D2 D3 x Dl is called a Darboux frame. Exercise 4.31. Let c: I --+ M be a unit speed curve into a surface M in ]R3 and let D 1, D2, D3 be an associated Darboux frame. Show that there exist smooth functions 91, 92 and 93 such that 91 D 2 + 92D3, -91 D l + 93D 3, -92Dl 93D2. Show that 91 - 0 along c if and only if c is a geodesic. Show that 92 = 0 along c if and only if c is asymptotic, and 93 = 0 along c if and only if c is principal. dDl/ds dD 2 /ds = dD3/ ds =
The function 91 from the previous exercise is called the geodesic curvature function and is often denoted K- g • Let us obtain a formula for K- g • Define J : TpM --+ TpM by
4.2. Hypersurfaees
161
Figure 4.2. Curvature vectors
Notice that IIJvp 11 = Ilvpll and so J is an isometry of the 2-dimensional inner product space (TpM, (', ')p). But J also satisfies J2 = - id, and in dimension 2 this determines J up to sign since it must be a rotation by ±7l' /2. We now assume that N is globally defined so that the map J extends to a smooth map T M -+ T M. For a unit speed curve e : I -+ M, the geodesic curvature is given by (4.1)
Kg
/I dT JT ) = ( e, J e' ) - \/ ds'
.
Indeed, abbreviating No e to N we have from the first equation in Exercise 4.31 the following: dT
/I
,
,
ds = e = KgN x e + 92 N = KgJe + 92N. Taking inner products with Jd gives formula (4.1). Let e : I -+ M be a curve in a surface that is parametrized by arc length. The curvature vector K.(s) can be decomposed into a component Kg(S) tangent to M and a component K.n(s) normal to M:
K.(s) = K.g(s)
+ K.n(s).
The curve will be a geodesic if and only if K. g (s) = 0 for all s. The vector Kg(S) is the geodesic curvature vector at e(s). It is easy to show that Kg = ± II K.g(s) II. Figure 4.2 depicts a sphere of radius R with a conical "hat". The cone intersects the sphere in a curve of latitude. Since the cone is tangent to the sphere, the vectors K., K.g and K. n apply equally well to both surfaces at least along the curve. Using the Pythagorean theorem and similar triangles, it is possible to show that IIK.gll - 1/a, where a is the distance from the curve to the vertex of the cone. Now imagine cutting the cone along a generating line and unrolling it as shown in Figure 4.3. The result is a planar region with a circular arc of radius a and curvature whose
162
4. Curves and Hypersurfaces in Euclidean Space
magnitude is l/a. The reader might want to try and give a reason why this is to be expected. We will answer this later in this chapter.
Figure 4.3. Unrolling a cone
Definition 4.32. If SNp is a multiple of the identity operator, then p is called an umbilic point. If SNp = 0, then p is called a flat point. If every point of a hypersurface is umbilic, then we say that the hypersurface is totally umbilic. Example 4.33. If aIx l plane P in ]Rn, then
+ a2x2 + ... + anxn = 0 is the equation of a hypern
N = Laiei i-I
is a normal field when restricted to P. Since clearly SNp = 0 for all p E P, we see that every point of P is flat and that the Gauss and mean curvatures are identically zero. Example 4.34. Let sn-I be the unit sphere in ]Rn. Then the map p = (al, . .. ,an) f---t N(p) := ~~=I ai~ is a unit normal field along sn-I. We can calculate SN(V) = -\lvN. Let c be a curve in sn-I with C(O) = v. Then
-
dl
-\lvN = - dt
=-
N(c(t)) t=Q
d I L~ dt n
i=1
i
ei = -v.
t=Q
We are really just using the fact that up to a change in base point we have N (c( t)) = c( t). Thus S N = - id and we see that every point is umbilic and every principal curvature is unity. Also, K = 1 and the mean curvature H = -1 everywhere. Exercise 4.35. What is the shape operator on a sphere of radius r? Composition with a Euclidean motion preserves local geometry so if we want to study a hypersurface near a point p, then we may as well assume
163
4.2. Hypersurfaces
that p is the origin and that Np = en(O) = ~Io' Locally, M is then the graph of a function f : jRn-l ---t jR with ~(O) - 0 for i = 1, ... ,n -1. A normal field N which extends No = en(O) is given by n
1
.
(1 + L (8f /8x~)2)
-1/2
n-l
8t
(en - L 8xiei)' i-I
i=1
Let Z = en - E~:11 -£!'ei so that N = gZ, where 9 is the first factor in the formula above. For v = E~:ll v~ei tangent to M at the origin we have
V'v (gZ)
= (vg) Z + gV'vZ,
where vg means that v acts on 9 as a derivation. But vg vanishes at the origin since 9 takes on a maximum there. Since g(O) = 1, we have V'v (gZ) = V'vZ. Thus we may compute SNo using Z: SNoV
= -V'v Z .
We have n-l
SNoV
=
LV
(81 /8x i ) ~(O)
i=l
n-ln-l
.
81 I
= ~ ~ v 8xi8x j 0 e~(O). J
We conclude that the shape operator at the origin is represented by the (n - 1) x (n - 1) matrix
[D2 t] (0) =
[8:~XJ (0)] l~i,j~n-l
.
This is only valid at the origin. In other words, [D2 t] does not give us a representation of the shape operator except at the origin where M is tangent to jRn-l. We can arrange, by rotating further if necessary, that el, ... ,en 1 are directions of curvature. In this case, the above matrix is diagonal with the principal curvatures kI, . .. , k n - 1 down the diagonal. Example 4.36. Let M be the graph of the function f(x, y) = xy. We look at 0 E M. We have
[D21] = which diagonalizes to
[6 -.?d
[~ ~],
with corresponding eigenvectors ~
(el
+ e2)
and ~ (el - e2). Thus the Gauss curvature at the origin is -1 and the mean curvature is O. We can understand this graph geometrically. The normal sections created by intersecting the graph with the planes y = 0 and x = 0 are straight lines, and so the normal curvatures in those directions
4. Curves and Hypersurfaces in Euclidean Space
164
are zero. However, the normal section given by intersecting with the plane y = x is concave up, while that created by the plane y = -x is concave down. be a surface. If VI, V2 E TpM are linearly independent, then for a given unit normal Np we have
Proposition 4.37. Let M C
]R3
SNpVI x SNpV2 SNpVI x V2
+ VI
X
SNpV2
= K(p) (VI x V2), = 2H(p) (VI x V2) .
Proof. We prove the first equation and leave the second as an easy exercise. Let (s;) be the matrix of SNp with respect to the basis VI, v2. Then SNpVI x SNpV2
= (visi + v2s~) x (vls~ + v2s~) = sis~ - s~s~ = det(SNp)'
o
We remind the reader of the easily checked Lagrange identity:
(v x wax b) = '
I (v, a) (v, b) I (w,a)
(w,~
for any a, b, v, wE ]R3 (or in any Tp]R3). If N is a normal field defined over an open set in the surface M, and if X and Yare linearly independent vector fields over the same domain, then we can apply the Lagrange identity and the above proposition at each point to obtain
and also H=~
I(SNX,X) (Y,X)
(SNX, Y) (Y,Y)
I+ I (X,X)
(SNY,X)
(X, Y) (SNY,Y)
2~------~I-(~X-,X~)~(X-'-Y~)~I------~ (Y,X)
(Y,Y)
These formulas show clearly the smoothness of K and H over any region where a smooth unit normal is defined. Also, the formula k± =H± JH2_K
gives the two functions k+ and k- such that k+(p) and k-(p) are principal curvatures at p. These functions are clearly smooth on any region where k+ > k- and are continuous on all of the surface. Furthermore, if every point of an open set in M is an umbilic point, then k(p) := k+(p) = k-(p) defines a smooth function on this open set. This continuity is important for
165
4.3. The Levi-Civita Covariant Derivative
obtaining some global results on compact surfaces. Notice that for a surface in ]R3 the set of nonumbilic points is exactly the set where k+ > k-, and so this is an open set.
Definition 4.38. A frame field E I , ... , En I on an open region in a hypersurface M is called an orthonormal frame field if EI (p), ... , E n - l (p) is an orthonormal basis for TpM for each p in the region. An orthonormal frame field is called principal frame field if each Ei(P) is a principal vector at each point in the region.
Theorem 4.39. Let M be a surface in lR3 . If p E M is a point that is not umbilic, then there is a principal frame field defined on a neighborhood of p.
Proof. The set of nonumbilic points is open, and so we can start with any frame field (say a coordinate frame field) on a neighborhood of p. We may take this open set to be oriented by a unit normal N. We then apply the Gram-Schmidt orthogonalization process simultaneously over the open set to obtain a frame field FI, F2. Since p is not umbilic, we can multiply by an orthogonal matrix to assure that F I , F2 are not principal at p and hence not principal in a neighborhood of p. On this smaller neighborhood we have
SNFI
= aFI + bF2,
SNF2 = bFI + cF2 for functions a, b, c with b =1= O. Now define GI , G2 by
G1 = bFI + (k+ - a)F2' G2 = (k- - C)Fl
+ bF2
and check by direct computation that SNG I = k+G I and SNG2 = k+G2. We have used a standard linear algebra technique for changing to an eigenbasis. Since b =1= 0, we see that IIGtiland IIG211 are not zero. Finally, let EI = Gd IIGtil and
E2 =
G 2 / IIG211.
0
4.3. The Levi-Civita Covariant Derivative The Levi-Civita covariant derivative is studied here only in the special case of a hypersurface in lRn. The more general case is studied in Chapters 12 and 13. We derive some central equations, which include the Gauss formula, the Gauss curvature equation and the Codazzi-Mainardi equation. Let M be a hypersurface. For Xp E TpM, and Y a tangent vector field on M (or an open subset of M), define V xp Y by Vx"Y:= projT"MVX"Y,
where projTpM : TplRn -+ TpM is orthogonal projection onto TpM. For convenience, let us agree to denote the orthogonal projection of a vector
166
4. Curves and Hypersurfaces in Euclidean Space
v E TplRn onto TpM by v T and the orthogonal projection onto the normal direction by v.i. We call v T the tangent part of v and v.i the normal part. Then V'xl' Y = (V'xl'Y) T. Exercise 4.40. Show that for! a smooth function and Xp and Y as above, we have V'xl'!Y = (Xpl) Y(p) + !(p)V'xl'Y' If N is any unit normal field defined near p, then
V'xl'Y = V'xl'Y + (Np, V'xl'Yp) Np. Since
0= Xp (N, Y) = (-SNXp, Y) we obtain the Gauss formula:
+ (N, V'xl' Y) ,
(4.2) Notice that the right hand side of the above equation is unchanged if N is replaced by -N. It follows that if X and Yare smooth tangent vector fields, then p ~ V' xl' Y is smooth and we may then define the field V' x Y by (V'xY) (p) := V'xl'Y. By construction (V'xY) (p) = (V'zY) (p) if X(p) = Z(p). It is also straightforward to check that the map (X, Y) M V'xY is COO(M)-linear in X, but not in Y. Rather, like V', it is IR-linear in Y and we have the product rule V'x!Y
= (XI) Y + !V'xY,
which follows directly from Exercise 4.40. Thus V' is a covariant derivative on M and V' x Y is defined for X, Y E X(M). We remind the reader that X(M) is the space of tangent vector fields and not to be confused with vector fields along M which may not be tangent to M. Let Y and Z be smooth tangent vector fields on M and take Xp E TpM. We study the situation locally near p. Let (i5, \1) be a special chart ce~tered at p and let (~, u) be_the chart obtained by restrictio~ where 0 = 0 n M as before. If Y and Z are extensions of Y and Z to 0, then since S N IS self-adjoint, we have
(V'yZ - V' zY) (p) = (V'yZ - V' zY) (p)
= (V'yZ - V' zy) (p)
= [Y, Z]p = [Y, Z]p. Thus V' y Z - V' z Y = [Y, Z] for all Y, Z E X (M). This fact is expressed by saying that V' is torsion free. Also,
Xp (Y, Z) = Xp (y, Z) = (V'xl'y, Zp) = (V' xpY, Zp)
+ (¥p, V' xl'Z),
+ (Yp, V'xl'Z)
4.3. The Levi-Civita Covariant Derivative
167
so that if X E X(M), then X (Y, Z) = (V'xY, Z) + (Y, V'xZ). We express this latter fact by saying that V' is a metric covariant derivative on M. There is only one covariant derivative on M that satisfies these last two properties, and it is called the Levi-Civita covariant derivative (or Levi-Civita connection). We prove the uniqueness later in this chapter. With coordinates u\ ... ,un - 1 as above we have functions
If n-l
X
,8
= LX~-8i u
i=l
n-l
and Y
,8
= L Y3 -8" i=l
uJ
then (X, Y) =
L 9ijXi yj. i,j
The length of a curve in M is just the same as its length as a curve in the ambient Euclidean space. One may define a distance function on M by dist(p, q) := inf{L(c)}, where the infimum is over all curves connecting p and q. This gives M a metric space structure whose topology is the same as the underlying topology. This will be proved in more generality in Chapter 13. For now, the point is that metric aspects of M are determined by (" .) and locally by the 9'3 associated to each chart of an atlas for M. For example, the length of a curve c : [a, b] --+ 0 c M is given by
(4.3)
L(c) =
rL
J(1
b n-l
dci dd
9ij (c(t)) dtdt
dt,
a i,j=l
where ci(t) := ui
0 C.
Exercise 4.41. Deduce the above local formula for length from the formula for the length of the curve in the ambient Euclidean space. Returning to our covariant derivative V', we have
8 V'..L 8.. 1 = Bu' ·w
n-l
8
Lr~8U k
k=l
168
4. Curves and Hypersurfaces in Euclidean Space
for smooth functions rt known as the Christoffel symbols of V. For X and Y expressed as above, we have
and so we have
(4.4)
_~ ~ k j) t:i (~8Yk f-t X + /;::l rijX Y i
Vx Y -
Thus the functions
8 8u k '
i
8ui
rt determine V in the coordinate chart.
Also note that
o = [88., 88uJ ] = V...JL.. 8a - V...iL 88. = Lk (rfj - rji) 88u k . u~ au' u) au u~ 1
It follows that
rt =
(4.5)
rj~ for all i, j, k = 1, ... ,n - 1.
n-l
=
L
(r~igsj + r~jgsi) .
s=1
The matrix (gij) is invertible, and it is traditional to denote the components of the inverse by gi j , so that Er gjrgri = 6}. One may solve to obtain (4.6)
r~.~J = ~2 "" ks ~g
(8 9Si _ 8g~j 8ui 8us
+ 88u9 jsk )
.
II
Exercise 4.42. Prove formula (4.6) above. [Hint: First write the formula ~~i = Ell r~igsj + r~jgsi two more times, but cyclically permuting i, j, k to obtain three expressions. Subtract the second expression from the sum of the first and third. Use equality (4.5).] Proposition 4.43. The Levi-Civita connection on a hypersurface is determined uniquely by the properties of being a torsion free metric connection.
169
4.3. The Levi-Civita Covariant Derivative
Proof. In deriving the local formula for the Christoffel symbols, we only used the fact that V is a torsion free metric connection. We do this again in Chapter 13 in a more satisfying way. 0
If an object is determined completely by the metric on M, then we say that the object is intrinsic. Equivalently, if all local coordinate expressions for an object can be written in terms of the metric coefficients gij (and their derivatives, etc.), then that object is intrinsic. We have just seen that the connection V is intrinsic. It follows that if f : Ml ~ M2 is an isome1
2
try of hypersurfaces and V and V are the respective Levi-Civita covariant
derivatives, then 1
2
f*Vx Y = V/.xf*Y
for all vector fields X, Y E I(M). One way to see this is to examine the situation using a chart (V, u) on Ml and the chart (f(V), u 0 f 1) on M2. 2
A better way is to show that (X, Y) t--+ f*V /.xf*Y defines a torsion free metric connection on Ml and then use Proposition 4.43. (Exercise!)
If Y : I define
~
T M is a vector field along a curve c : I
~
M
c
]Rn,
then we
for any to E I and then define V a Y by at
( V ..!!. at
Y) (t) := V Yfor tEl. ..!!.I
at t
Suppose that c : I ~ M is such that c(J) and some chart (V, u). We can then write
Y(t) =
~ yi(t) a~i I i=l
c V for some subinterval J c I
for all t E J.
c(t)
Let us focus attention on a to such that c(to) =1= 0 and let J be an open interval with to E J. By restricting J if necessary ,!e may also assume that c is an embedding. Then there is a smooth field Y on a neighborhood of c(J) such that Yo c = Y. In fact, a simple partition of unity argument shows that we may arrange that Y be defined on all of V (in fact, on all of
170
4. Curves and Hypersurfaces in Euclidean Space
M). For t
E J we have
(V It Y) (t) = y' (t) T = ( (Y
0
c)' (t)) T
= (VC(t)Y) T = Vc(t)Y
n-l (n-l ~
= {;
=
,n-l + ij;l
ayk d aui (c(t)) d~
L (aY m + .L (rfj
n-l
k
n-l
k=l
0
c) (t)
d
rfj(c(t))
i
d~ (t)yj(c(t))
d~ (t)yj(t) auka I i
)
)
.
c(t)
l,j=1
We arrive at
We would like to argue that the above formula holds for general curves. If c(to) f:. 0, then there is an interval around to so that the formula holds as we have just seen. If c(to) = 0, we consider two cases. If there exists a sequence ti converging to to such that C(ti) = 0 for all i, then the formula holds for each ti and hence by continuity at to. Otherwise there must be an interval J containing to such that c is constant on J, say c( t) = p for all t E J, and then in this interval Y is just a map into the vector space TpM. In this case we have
(V:,Y) (t) = Y'W = =
(! ~>'(t) 8~.1J
ayk a I L m(t) auk
But this agrees with the formula since
c(t)
T
.
7t' = O.
Exercise 4.44. Show that for a vector field X along c : I --+ M and smooth function h E Coo (1) we have
I!.
Va/athX = hVa/atX + h'X.
Exercise 4.45. Show that for vector fields X, Y along c we have d dt (X, Y)
= (Va/at X , Y) + (X, Va/atY) = O.
The operator Va/at involves the curve c despite the fact that the latter is not indicated in the notation.
171
4.3. The Levi-Civita Covariant Derivative
We pause to consider again the question posed earlier about the circular arc on the unrolled cone in Figure 4.3. The key lies in the fact that the absolute geodesic curvature IKgl is intrinsic. We remarked earlier that the operator J is intrinsic up to sign (the latter being determined by orientation). On the other hand, in Problem 11 the reader is asked to derive the formula (Til, J,') K
But
g -
"'!"':--'--"';"":-
1Ir'11 3
IK 1- 1(Til, ±J,') 1_ I(V8/ 8 t'Y', ±J,') I g
-
1,'11 3
-
1Ir'11 3
'
and since both V and the pair ±J are intrinsic, we see that 1Kg 1 is intrinsic. A little thought should convince the reader that if a curve in one surface is carried to a curve in another surface by an isometry, then the curves will have equal absolute geodesic curvatures at corresponding points. The unrolling of the cone in Figure 4.3 can be thought of as inducing an isometry between the cone (minus a line segment) and a region in a planar surface in JR3. Thus we expect IKgl to be the same for both curves. But IKgl for a circular arc in a plane is just the reciprocal of the radius. Definition 4.46. A tangent vector field Y along a curve e : I -+ M is said to be parallel (in M) along e if V Jt Y = 0 for all tEl. Ilt
If e is self-parallel, Le. V8/8tC(t)
= 0 for all tEl, then it is easy to see
that e is a geodesic (in M) and in fact, this could serve as an alternative definition of geodesic curve, which will be the basis of later generalizations. Notice that if e is a curve in ]Rn, then Y can be considered as taking values in TJRn. However, Y being parallel in M is not the same as Y being parallel as a TJRn-valued vector field along e. In particular, a geodesic in M certainly need not be a straight line in ]Rn. For example, constant speed parametrizations of great circles on S2 C ]R3 are geodesics. Exercise 4.47. Given a smooth curve e : I -+ M, show that d' = 0 if and only if c is a geodesic in M such that (SNC(t), c(t)) = 0 for all t and choice of unit normal at e(t). Suppose that e" is never zero. Show that e is a geodesic in M if and only if e" is normal to M (Le. d'(t) ~ Tc(t)M for all tEl). The following simple result follows from the preceding exercise: Proposition 4.48. Let Ml and M2 be hypersurfaces in JRn. Suppose that c : I -+ ]Rn is such that e(t) E Ml n M2 for all t. If e" is not zero on any subinterval of I, then Tc(t)Ml = Tc(t)M2 for all tEl. Proof. By Exercise 4.47,
Tc(t)Ml
= d'(t).l = Tc(t)M2
for all t.
D
172
4. Curves and Hypersurfaces in Euclidean Space
~ M are parallel, then (X, Y) (t) := (X(t), Y(t)) is constant in t. In particular, a parallel vector field has constant length.
Proposition 4.49. If vector fields X, Y along a curve c : I
Proof.
ft (X, Y) = ('V 8/fJtX, Y) + (X, 'V 8/fJtY) = o.
o
Corollary 4.50. A geodesic has a velocity vector of constant length. Definition 4.51. Suppose Y E X(M) is a smooth vector field on Mj then Y is called a parallel vector field on M if 'V x Y = 0 at all points of M and for all smooth vector fields X E X(M). Obviously, Y is a parallel field if and only if Y all curves c.
0
c is parallel along c for
We now move on to prove two basic identities and introduce the curvat~r~ t~sor. It is easy to check by direct computation that for vector fields X, Y, Z on an open subset of ]Rn, we have
(4.8) If M is a hypersurface and X, Y, Z are tangent vector fields on a !le~h.E0r hood of an arbit~y p E M, then we may extend these to fields X, Y Z on a neighborhood 0 in ]Rn. Then we have
('Vx'VyZ - 'Vy'VxZ - 'V[X,YjZ) (p)
= ('V x 'VyZ - 'Vy'V xZ - 'V[X,y]Z) (p) = 0, and so (4.9) wherever the fields are all defined on M. Suppose that N is a unit normal field defined on the same domain in M. We apply the Gauss formula (4.2) to equation (4.9) above and then decompose it into tangent and normal parts:
0= 'Vx ('VyZ + (SNY, Z) N) - 'Vy ('VxZ + (SNX, Z) N) - 'V[X,YjZ = 'Vx'VyZ + (SNX, 'VyZ) N + X (SNY, Z) N - (SNY, Z) SNX - 'Vy'VxZ - (SNY, 'VxZ) N - Y (SNX, Z) N + (SNX, Z) SNY - 'V[X,YjZ - (SN[X, Y], Z) N. Equating the tangential parts of the above gives the Gauss curvature equation:
(4.10) 'V x'VyZ - 'Vy'V xZ - 'V[X,YjZ = (SNY, Z) SNX - (SNX, Z) SNY. The normal parts give
0= (SNX, 'VyZ) + X (SNY,Z) - (SNY, 'VxZ) - Y (SNX, Z) - (SN[X, Y], Z),
4.3. The Levi-Civita Covariant Derivative
173
or (VXSNY, Z) - (VySNX, Z) - (SN[X, YJ, Z) = 0 for all Z. From this we obtain the Codazzi-Mainardi equation:
(4.11) Let us give an application of the Codazzi-Mainardi equation and then return to the Gauss curvature equation.
Proposition 4.52. Let M be a connected hypersurface in ]Rn oriented by a unit normal field N. If every point of M is umbilic (i. e. M is totally umbilic), then the normal curvatures are all equal and constant on M. Furthermore, M is an open subset of a hyperplane or a sphere according to whether the normal curvatures are zero or nonzero. In particular, if M is a closed subset of]Rn, then it is a sphere or a hyperplane according to whether it is compact or not. Proof. There is a function k such that S N - kI, where I is the identity on each tangent space. This function is continuous since k = nIl trace S N. Let Xp E TpM and pick Yp E TpM so that Xp and Yp are linearly independent. Extend these to tangent fields X and Y on a neighborhood of p. By the Codazzi-Mainardi equation we have
0= VxkY - VykX - k[X, Yj = (Xk) Y + kVxY - ((Yk)X + kVyX) - k[X, Yj = (Xk) Y - (Yk) X, where we have used VxY - VyX = [X, Yj. In particular, at p we have (Xpk) Yp-(Ypk) Xp = O. Since Xp and Yp are linearly independent, Xpk = O. Since p and Xp were arbitrary and M is connected, we see that k is in fact constant. If the constant is k = 0, then SN = 0 on M, and this means that N = No is constant along M. This implies that M is in a hyperplane normal to No. If k =1= 0, then (changing N to -N if necessary) we may assume that k > O. Define a function f : M --t ]Rn by f(p) - p + tcN(p). We identify tangent spaces with subspaces of]Rn and calculate Df(p). Let v E TpM and choose a curve c: (-a, a) --t M with C(O) = v. Then we have
Df(p)·v= ddl f oc =C(0)+-k1 ddl t t=O t t 1
Noc 0
1
= v - "kSNV = V - "kkv - O. Thus D f (p) = 0 for all p EM, and since M is connected, f is constant (Exercise 2.23). Thus p + tcN(p) = q for some fixed q and all p. In other
174
4. Curves and Hypersurfaces in Euclidean Space
words, all p E M are at a distance 11k from q E ]Rn, and so M is contained 0 in that sphere of radius 11k. The left hand side of the Gauss curvature equation (4.10) is given its own notation
R(X, Y)Z := 'iJ x'iJyZ - 'iJy'iJ xZ - 'iJ[X,YjZ, and looking at the right side of (4.10) we see that (R(X, Y)Z) (p) depends only on the values of X, Y, Z at the point p, which means that we obtain 8 map Rp : TpM x TpM x TpM --t TpM defined by the formula
Rp(Xp, Yp)Zp := (R(X, Y)Z) (p). We say that R is a tensor since it is linear in each variable separately. We study tensors systematically in Chapter 7. The tensor R is called the Riemannian curvature tensor and the map p ~ Rp is smooth in the sense that if X, Y, and Z are smooth tangent vector fields, then p ~ Rp(Xp, Yp)Zp is a smooth vector field. We often omit the subscript p and just write R(Xp, Yp)Zp. (Notice that equation (4.8) just says that the curvature of lRn associated to the ambient covariant derivative 'iJ is identically zero.) Using (4.10) again, it is also easy to check that Rp is linear in each slot separately on TpM. In particular, for fixed X p, Yp E TpM we have a linear map R(Xp, Yp) : TpM --t TpM.
Theorem 4.53. If M is a surface in]R3 and (Xp, Yp) is an orthonormal basis for TpM, then
Proof. Using 4.10, and abbreviating SNp to S, we have
(R(Xp, Yp)Yp , Xp) = (SYp, Yp) (SXp, Xp) - (SXp, Yp) (SYp, Xp)
= det S = K (P).
0
We have shown that 'iJ is intrinsic and thus R is also intrinsic. Thus the previous theorem implies that the Gauss curvature K for a surface in lR3 is intrinsic. This is the content of Gauss's Theorema Egregium, which we prove (again) below using local parametric notation. If M is a k-dimensional submanifold in ]Rn, and (V, u) is a chart on M with U = u (V), then u- 1 : U --t M is a parametrization of a portion of M, and we denote this map by x: U --t M.
This notation is traditional in surface theory. Composing with the inclusion " : M y ]Rn, we obtain an immersion " 0 x : U --t ]Rn, but we normally identify" 0 x and x when possible. One may study immersions that are not
175
4.3. The Levi-Civita Covariant Derivative
necessarily one-to-one. A reason for this extension is that it might be the case that while x : U ---+ ]Rn is not one-to-one, its image is a submanifold M and so we can still study M via such a map. For example, consider the map x: ]R2 ---+ ]R3 given by
x(u, v) = ((a + b cos u) cos v, (a + bcos u) sin v, bsin u)
(4.12)
for 0 < b < a. This map is periodic and its image is an embedded torus. Notice that the restriction of x to sets of the form (uo - 7r /2, Uo + 7r /2) x (vo - 7r /2, Va + 7r /2) are parametrizations whose inverses are charts on the torus. Another example is the map x : ]R2 ---+ S2 C ]R3 given by x( 'P, 0)
= (cos 0 sin 'P, sin 0 sin 'P, cos 'P)'
The restriction of this map to (0,7r) set of measure zero.
X
(0,27r) parametrizes all of S2 but a
If x : U ---+ M is a parametrization of a hypersurface in ]Rn, then for UE U, the vectors (u) are tangent to M at p = x(u) and in fact form the
&:.
coordinate basis at p in somewhat different notation. In fact, if we abuse notation and write ui for ui 0 x-I, we obtain a chart (V, u) with x(U) = V. Then (u) is essentially just a~i ix(u)' We have = 9ij 0 x, but in the current context of viewing things in terms of the parametrization, we just change our notations slightly so that (::.':::,) = 9ij' These 9iJ are the components of the metric with respect to the parametrization. In these terms, some of the calculations look a bit different. For example, if 'Y : [a, b] ---+ M is a curve whose image is in the range of x, then the length of the curve can be computed in terms of the 9ij, which are now functions of the parameters u i . First, 'Y must be of the form t H x(u1(t), ... ,un-1(t)), where u : t H (u 1 (t), ... , Un-l (t)) is a smooth curve in U. Then we have
&:.
(&:., g;)
which is only notationally different from formula (4.3) due to our current parametric viewpoint. Exercise 4.54. Show that there always exists a local parametrization of a hypersurface M C ]Rn around each of its points such that 9~J (0) - 6~J and ~(O) = 0 for all i,j, k. [Hint: Argue that we may assume that M is written as a graph of a function f : ]Rn-l ---+ ]R with f(O) = 0 and Df(O) = O. Then let x: ]Rn-l ---+]Rn be defined by (u\ ... ,un - 1 ) H (U1 , ...
,un-l , f( u 1 , ... ,un-l)) .]
176
4. Curves and Hypersurfaces in Euclidean Space
Theorem 4.55 (Gauss's Theorema Egregium). Let M be a surface in]R3 and let p EM. There exists a parametrization x : U --t M with x(O, 0) = p such that 9ij = dij to first order at 0 and for which we have
K(p) = 82 912 (0) 8u8v
! 82 9222 (0) - ! 829112 (0). 2 8u
2 8v
Proof. In the coordinates of the exercise above, which give the parametrization (u, v) H (u, v, f(u, v)), where p is the origin of IR3 , we have
~f [ 911(U,V) 912(U,V)] _ [ 1 + (m)2 921(U, v) 922(U, v) ~~ 1 + (~)2
],
from which we find, after a bit of straightforward calculation, that
!
!
8 2912 (0) - 8 2922 (0) - 8 2911 (0) 8u8v 2 8u 2 2 8v 2
82 f 82 f
82 f
2
= 8u 2 8v 2 - 8u8v = det D f(O) = det S(p) = K(p).
0
Let us introduce some traditional notation. For x : U --t M c IR3, and denoting coordinates in U again by (u, v), and in IR3 by (x, y, z), we have
Xu=
( 8X 8y 8X) 8u'8u'8u x'
( 8X 8y 8X) 8v'8v'8v x' 82x 8 2y 8 2x) ( Xuv = 8u8v' 8u8v' 8u8v x' Xv
and so on. In this context, we always take the unit normal to be given, as a function of u and v, by
xuxXv x Xv I A careful look at the definitions gives (
)
N u,v =
(based at x(u,v)).
I XU
8N
8N
8u = SN(Xu) and 8v = SN(Xv).
Recall that the second fundamental form is defined by II(v,w) = (SNV,W), and this makes sense if the tangent vectors v, w are replaced by fields along x defined on the domain of N. The traditional notation we wish to introduce is E = (xu, xu) , 1=
(SNX u, xu) ,
F = (Xu,x v ) ,
G = (xv, xv) ,
m= (SNX u, xv) ,
n = (SNX v , xv) .
177
4.3. The Levi-Civita Covariant Derivative
Thus the matrix of the metric (.,.) (sometimes called the first fundamental form) with respect to Xu, Xv is
[:~~ :~:] = [~ ~], while that of the second fundamental form I I is
[! :]. The reader can check that Ilxu x Xv 112 = EG - F2. The formula for the length of a curve written as t r--t x(u(t), v(t)) on the interval [a, b] is
fa b
) a
E
(du)2 dt
dv (dV)2 + 2F du dt dt + F dt dt.
For this reason the classical notation for the metric or first fundamental form is
ds 2 = Edu 2 + 2Fdudv + Fdv 2, where ds is taken to be an "infinitesimal element of arc length" . Consider the map 9 : ]R3 -t (]R3)* given by v r--t (v, .). With respect to the standard basis and its dual basis, the matrix for this map is [~~]. Similarly, we can consider the second fundamental form as a map II: ]R3 -t (JR.3)* given by v r--t II (v, .), and the matrix for this transformation is [; ~]. Then since II (v, w) = (SNV, w), we have II = 9 0 SN and so SN = g-1
0
II.
We conclude that SN is represented by the matrix
This matrix may not be symmetric even though the shape operator is symmetric with respect to the inner products on the tangent spaces. Taking the determinant and half the trace of this matrix we arrive at the formulas nl-m2
K = EG-F2' H _ Gl + En - 2Fm 2 (EG - F2) .
Exercise 4.56. Show that 1 = (N, x uu ), m = (N, xuv) and n = (N, xvv). Exercise 4.57. Consider the surface of revolution given parametrically by
x(u, v) = (g(u), h(u) cosv, h(u) sin v)
178
4. Curves and Hypersurfaces in Euclidean Space
with h > O. Denote the principal curvature for the meridians through a point with parameters (u, v) by kll and that of the parallels by k7r • Show that these are functions of u only given by
k Il -
-
g' I g"
h' h"
I
-----;,------c~
((g,)2 + (h,)2)3/2 '
g' k - ------..----=-----;;-----:7r h((g,)2 + (h,)2)1/2'
4.4. Area and Mean Curvature In this section we give a result that provides more geometric insight into the nature of mean curvature. The basic idea is that we wish to deform a surface and keep track of how the area of the surface changes. Let M C IR3 be a surface and let x : U --+ V C M be a parametrization of a portion V of M. We suppose that V has compact closure. The area of V is defined by A(V):=
L
Ilxu x
Xvii dudv.
The total area of M (if it is finite) can be obtained by breaking M up into pieces of this sort whose closures only overlap in sets of measure zero. However, our current study is local and it suffices to consider the areas of small pieces of M as above. Suppose x : U x (-c, c) --+ ]R3 is a smooth map such that for each fixed t E (-c, c), the partial map x(·, " t) : (u, v) t-+ x(u, v, t) is an embedding and such that is normal to Xu and Xv for all t. For each t, the image \It = x(U, t) is a surface. We have in mind the case where x(',', 0) is a parametrization of a portion of a given surface M so that Vo = V c M (see Figure 4.4). The normal N = Xu x Xvi Ilxu x Xv depends on t and at time t provides a unit normal to the surface \It. Thus \It is a one parameter family of surfaces.
tt:
Theorem 4.58. Let x : U x (-c, c) --+ ]R3 and let \It = x(U, t) be as above so that ~~ is normal to the surface \It. Let H (t) denote the mean curvature of the surface \It. Then
:t
A (\It) = -2
£I ~: I
Proof. Since ~ is parallel to N, we have ~~
axu at
=~ auat
H(t) dA.
=
I tt: II N.
Thus
ax = ~ S (x) +au ~ Ilaxll au (1laxll at N) = -II axil &t at N N
u
179
4.4. Area and Mean Curvature
Figure 4.4. Deformation of a patch
and so
Similarly,
(N,
Xu x
[}~v) = -II~~II (N, XU x SN (Xv)).
Now we calculate using the second formula of Proposition 4.37:
:t A = :t J Ilxu x xvii dudv = ! J (N,xu x xv) dudv = =
JI JI
d
\ dt N, XU
x Xv ) du dv +
[}xu
JI
[}x v )
\ N, Tt x Xv + Xu x [}t
= - J
= -2 J
[}xu \ N, Tt x Xv + Xu x [}Xv) [}t du dv
du dv
II~~II {II SN (xu) x Xv +xu X SN (Xv)II} dudv H(p)
I ~~llllxu x xvii dudv = -2 J I :11 H dA.
0
In particular, we can arrange that II ~ II = 1 at time t = 0, and then we have dd
I
t t=O
A = -2JHdA.
Thus, H is a measure of the rate of change in area under perturbations of the surface. Definition 4.59. A hypersurface in lRn for which H is identically zero is called a minimal hypersurface (or minimal surface if n = 3 so that dimM = 2).
4. Curves and Hypersurfaces in Euclidean Space
180
Example 4.60. For c> 0, the catenoid is parametrized by
x(u,v) = (u, c cosh(u/c)cosv, c cosh u sin v) and is a minimal surface. Indeed, a straightforward calculation gives [ EF FG]
[ c2 cosh2 (u/c)
[~
[~c I~C]'
It follows that H
0 ] cosh2(u/c)'
o
:] =
= O.
Example 4.61. For each b =1= 0, the map x(u,v) = (bv,ucosv,usinv) is a parametrization of a surface called a helicoid and this is also a minimal surface. The reader may enjoy plotting this and other surfaces using a computer algebra system such as Maple or Mathematica.
4.5. More on Gauss Curvature In this section we construct surfaces of revolution with prescribed Gauss curvature and also prove that an oriented compact surface of constant curvature must be a sphere. Let
x(u, v) = (g(u), h(u) cosv, h(u) sin v) be a parametrization of a surface of revolution. By a reparametrization of the profile curve (g(u), h(u)) we may assume that it is a unit speed curve so that (g,)2 + (h,)2 = 1. When this is done, we say that we have a canonical parametrization of the surface of revolution. A straightforward calculation using the results of Exercise 4.57 shows that for any surface of revolution given as above we have
g'
g' K=--
I g"
h ((g,)2
h' h"
I
+ (h,)2)
2'
If we assume that it is a canonical parametrization, then we have
K=
- (g,)2 h" + g' g" h' h
.
On the other hand, differentiation of (g,)2 + (h,)2 = 1 leads to g' g" = -h'h", and so we arrive at -h" K=h' which shows the expected result that K is constant on parallels (curves along which u is constant). Now suppose we are given a smooth function K defined on some interval I, which we may as well assume to contain O.
4.5. More on Gauss Curvature
181
We would like K to be our Gauss curvature and so we wish to solve the differential equation h" + Kh = 0 subject to h(O) > 0 and Ih'(O)1 < 1. The first condition simplifies the analysis, while the second condition allows us to obtain the canonical situation (g,)2 + (h,)2 = 1. In fact, we let
g(u)
=
fou V1 - (h'(t))2 dt,
and this will give the desired solution defined on the largest open subinterval J of I such that h > 0 and Ih'I < 1. With this solution, our surface of revolution will be defined, but only for u E J. Suppose we try to obtain a surface of revolution with constant positive Gauss curvature K = 1/c2 for some constant c. Then a solution of the equation for h will be h(u) = acos(u/c) for an appropriate a > O. Then
g(u)
=
fou
1- :: sin2 (t/c) dt,
and with the resulting profile curve, we obtain a surface of revolution with Gauss curvature 1/c2 at all points. There are three cases to consider. First, if a = c, then the interval J on which h > 0 and Ih'I < 1 is easily seen to be (-rrc/2, rrc/2) and we have
h(u)
=
ccos(u/c) and g(u)
= csin(u/c).
This gives a semicircle which revolves to make a sphere minus the two points on the axis of revolution. We already know that these two points can be added to give the sphere ofradius c = 1/.JK, which is a compact surface of constant positive Gauss curvature K. It turns out that the spheres are the only compact surfaces (without boundary) of constant positive Gauss curvature. Now consider the case 0 < a < c. The interval J is the same as before, but the surface extends in the x direction between Xl = limu -t-7rcj2 g( u) and X2 = limu -t7rc/2 g(u). It is easy to show that Xl < -a and X2 - -Xl> a. Since the maximum value of h is now smaller than c, the profile curve is shallower and wider than the semicircle of the a = c case above. Although liIIlu-t±7rc/2 h( u) = 0 as before, we now have the profile curve tangents at the endpoints given by lim
u-t±7rc/2
(g' (u), h' (u)) =
lim
u-t±7rc/2
(
1-
a:c sin (u/c), -~c Sin(u/C)) 2
( a2 a)
= 1 - c2 ' =t=~ • This shows that the revolved surface is pointed at its extremes and forms an American football shape. In this case, there is no way to add in the missing points on the axis of revolution to obtain a smooth surface. The two principal curvatures are no longer equal, but their product is still1/c2 •
182
4. Curves and Hypersurfaces in Euclidean Space
The third case a with boundary.
> 0 gives a surface that has an extension to a surface
Exercise 4.62. Analyze the case a
> O.
From above we see that there is an infinite family of surfaces with constant curvature (only one extending to a compact surface). The situation for constant negative curvature is similar, but we obtain no compact surfaces. One particular constant negative curvature surface is of special interest: Example 4.63 (Bugle surface). This surface has Gauss curvature and is given by
-lie?
x(u, v) = (u, h(u) cosv, h(u) sin v), where h is the solution of the differential equation h'-
- vc -h h 2-
2'
subject to initial condition limu-+o h( u) = c. The function h is defined on (0,00). Lemma 4.64. Let M be a surface in IR3. Let p E M be a nonumbilic point and EI, E2 a principal frame on a neighborhood of p (oriented by N) so that SNEI = k+ EI and SNE2 = k- E 2. If we define functions
then
Proof. Since (E2, E2) = I, we have ('VEl E2, E2) = 0 and so there is some function hI such that 'VEl E2 = hiEI. Similarly, 'V E2EI = h2E2 for some function h2. We find formulas for each of these functions. We have
and
0= EI (EI,E2) = ('VE 1 EI,E2)
+ (EI, 'VE 1 E2),
from which it follows that 'VEl EI = -hIE2' Similarly, 'V E2E2 = -h2El' We have
4.5. More on Gauss Curvature
183
We now apply the Codazzi-Mainardi equations (4.11):
0= 'VE1SNE2 - 'VE2SNEI - SN[EI,E2] = 'VEl k- E2 - 'V E2 k+EI - SN (hiEI - h2E2)
= (EIk- E2 + k-'VE1E2) - (E2k+ EI + k+'VE2EI) - (hIk+ EI - h2k- E2) = (Elk- E2
+ k-hIEI) -
(E2k+ EI
+ k+h2E2)
- (hlk+ EI - h2k- E2) = (k-h i
-
E2k+ - hIk+) EI
+ (Elk- + h2k- -
k+h2)E2.
Setting the coefficients of EI and E2 equal to zero we obtain
hI
-E2k+ k- and h2
= k+ _
E1kk- .
= k+ -
We compute as follows:
R(EI, E2)E2 = 'VEl 'V E2E2 - 'V E2 'V E1E2 - 'V[El,E2]E2 = 'VEl (-h2EI) - 'VE2 (hIEI) - 'V(hlEl h2E2)E2 = - (EIh2) EI - h2 'V El EI - (E2 hl) EI - hI 'V E2EI - hl'VE1E2 + h2'VE2 E2 = - (EIh2) EI - (E2hl) EI - h2 (-hIE2) - hI (h2E2) - hI (hIEI) + h2 (-h2EI) = - (Elh2) EI - (E2hl) EI - h~EI - h~EI'
Using the Gauss curvature equation (4.10) we arrive at
(4.13)
K
= (R(E I , E2)E2, EI) = -EIh2 -
E2hl - h~ - h~.
0
Corollary 4.65. If p is a nonumbilic point for which both k+ and k- are critical, then
Proof. Using equation (4.13), we have
K = -Elh2 - E2hl - h~ - h~
= -EI (k~~~- )
- E2
(k~E~~~ ) - (k~E~~~)
2-
(k~~~- )
2
At p we have E2k+ = E1k- = 0, and so (suppressing evaluations at p) we obtain -(k+ - k-)E?ko K(p) = (k+ _ k-)2
184
4. Curves and Hypersurfaces in Euclidean Space
Corollary 4.66 (Hilbert). Let M C ]R3 be a surface. Suppose that K is a positive constant on M. Then k+ cannot have a relative maximum at a nonumbilic point. Similarly, k- cannot have a relative minimum at a nonumbilic point. Proof. Since K = k+ k- > 0 is constant, k+ has a relative maximum exactly when k- has a relative minimum, so the second statement follows from the first. For the first statement, suppose that k+ has a relative maximum at p. Notice that X 2k+ ~ 0 and X 2k- ~ 0 for any tangent vector field defined near p. Near p, there is a principal frame as above, and if we use Corollary 4.65, we have E 2k+ - E 2kK (p) = 2 k+ _ k (p) < 0,
2
contradicting our assumption about K.
o
We are now able to use the above technical results to obtain a nice theorem. Theorem 4.67. If M C ]R3 is an oriented connected compact surface w~th constant positive Gauss curvature K, then M is a sphere of radius 1/.fK. Proof. The hypotheses give us the existence of a global unit normal field N. Note that k+ ~ ../K at every point, and since M is compact, k+ must have an absolute maximum at some point p. By Corollary 4.66, p is an umbilic point and so k+(p) = k-(p). But then (k+(p))2 = k+(p)k-(p) = K, and thus the maximum value of k+ is ../K. We have both k+ ~ ../K and k+ < ../K on all of M, and so k+ = k- = ../K everywhere (M is totally umbilic). By Proposition 4.52 we see that M is a sphere of radius 1/../K. 0
We already know from our surface of revolution examples that there are many noncompact surfaces of constant positive Gauss curvature, but now we see that the sphere is the only compact example. Actually, more is true: If a surface of constant positive Gauss curvature is a closed subset of ]R3, then it is a sphere. This follows from a theorem of Myers (see Theorem 13.143 in Chapter 13).
4.6. Gauss Curvature Heuristics The reader may be left wondering about the geometric meaning of the Gauss curvature. We will learn more about the Gauss curvature in Section 9.9 and much more about curvature in general in Chapter 13. For now, we will simply pursue an informal understanding of the Gauss curvature.
185
4.6. Gauss Curvature Heuristics
Negative Curvature
Zero Curvature
Positive Curvature
Figure 4.5. Curvature bends geodesics
On a plane, geodesics are straight lines parametrized with constant speed. If two insects start off in parallel directions and maintain a poliey of not turning either left or right, then they will travel on straight lines, and if their speeds are the same, then the distance between them remains constant. On a sphere or on any surface with positive curvature, the situation is different. In this case, geodesics tend to curve toward each other. Particles (or insects) moving along geodesics that start out near each other and roughly parallel will bend toward each other if they travel at the same speed. The second derivative of the distance between them will be negative. For example, two airplanes traveling due north from the equator at constant speed and altitude will be drawn closer and, if they continue, will eventually meet at the north pole. It is the curvature of the earth that "pulls" them together. On a surface of negative curvature, initially parallel motions along geodesics will bend away from each other. The second derivative of the distance between them will be positive. These three situations are depicted in Figure 4.5. For a more precise formulation of these ideas it is best to consider a parametrized family of curves, and we will do this in Chapter 13. It should be mentioned that according to Einstein's theory, it is the curvature of spacetime that accounts for those aspects of gravity (such as tidal forces) that cannot be nullified by a choice of frame (accelerating frames cause gravity-like effects even in a fiat spacetime). For example, an initially spherical cluster of particles in free fall near the earth will be deformed into an egg shape. For a wonderful popular account of gravity as curvature, see
[Wh]. Another way to see the effects of curvature is by considering triangles on surfaces whose sides are geodesic segments. These geodesic triangles are affected by the Gauss curvature. For instance, consider the geodesic triangle on the sphere in Figure 4.6. The angles shown are actually measured in the tangent spaces at each point based on the tangents of the curves at the endpoints of each segment. We have the following Gauss-Bonnet formula
186
4. Curves and Hypersurfaces in Euclidean Space
Figure 4.6. Curvature and triangle
involving the interior angles:
where D is the region interior to the geodesic triangle. This formula is true for geodesic triangles on any surface M. For a sphere of radius a, this becomes (31 + (32 + (33 = 7r + AIa 2 , where A is the area of the region interior to the geodesic triangle. The fact that the sum of the interior angles for a triangle in a plane is equal to 7r regardless of the area inside the triangle is exactly due to the fact that a plane has zero Gauss curvature. If one starts at p and moves counterclockwise around the triangle, then at the corners one must turn through angles Gl, G2, G3. In terms of these turning angles, the statement is 3
L i=l
Gi
= 27r
-1
K dB.
D
By an argument that involves triangulating a surface, it can be shown that if one integrates the Gauss curvature over a whole surface (without boundary) Me JR3, then something amazing occurs. We obtain
This result is called the Gauss-Bonnet theorem. Here X(M) is a topological invariant called the Euler characteristic of the surface and is equal to 2 - 2g, where g is the genus of the surface (see [Arm]). Thus while the left hand side of the above equation involves the shape dependent curvature K, the right hand side is a purely topological invariant! A presentation of a very general version of the Gauss-Bonnet theorem may be found in [Poor] (also see [Lee, J effj).
187
Problems
Problems (1) Let, : I -+ ]R2 be a regular plane curve. Let J : ]R2 -+ ]R2 be the rotation (x, y) t--+ (-y, x). Show that the signed curvature function K2 (t):=
,"(t) . J,'(t) 1Ir" (t) 11 3
determines, up to reparametrization and Euclidean motion.
+ g2 = 1. Let to E (a,b) and suppose that j(to) - cosfJo and g(to) = sinfJo for some ()o E R Show that there is a unique continuous function () : (a, b) -+ ]R such that ()(to) = fJo and
(2) Let j, 9 : (a, b) -+ ]R be differentiable functions with j2
f(t)
= cosfJ(t),
g(t)
= sinfJ(t)
for t E (a,b).
(3) Let, : (a, b) -+ ]R2 be a regular plane curve. (a) Given to E (a, b) and ()o with
" (to)
II,' (to) I
. = (cos fJo, sm fJo),
show that there is a unique continuous turning angle function ()"( : (a, b) -+ ]R such that ()(to) = fJo and
,'(t)
11r'(t) I
.
= (cos()"((t),smfJ"((t)) for
t E (a,b).
(b) Show that fJ~(t) = 1Ir'(t) I K2(t), where K2 is as in Problem 1. (4) Show that if K2 for a curve, : (a, b) -+ ]R2 is 1/r, then, parametrizes a portion of a circle of radius r. (5) Let,:]R -+]R3 be the elliptical helix given by t t--+ (a cos t, bsint, ct). (a) Calculate the torsion and curvature of ,. (b) Define a map F : ]R -+ SO(3) by letting F have columns given by the Frenet frame of ,. Show that F is a periodic parameterization of a closed curve in SO(3). (6) Calculate the curvature and torsion for the twisted cubic ,(t) := (t, t 2 , t 3 ). Examine the behavior of the curvature, torsion, and Frenet frame as t -+ ±oo. (7) Find a unit speed parametrization of the catenary curve given by c(t) := (a cosh(tla) , t). Revolve the resulting profile curve to obtain a canonical parametrization of a catenoid and find the Gauss curvature in these terms.
4. Curves and Hypersurfaces in Euclidean Space
188
(8) (Four vertex theorem) Show that the signed curvature function K2 for a simple, closed plane curve is either constant or has at least two local maxima and at least two local minima. (9) If 'Y : I ---+ ]R3 is a regular space curve (not necessarily unit speed), then _
show that B(t) -
-y'x-y" "'!'x"'!"II'
_ (-y'x"'!")·"'!'" _ I"'!'x"'!"11 I"'!' x",!" 112 and K 1"'1"113 •
T -
(10) With'Y as in Problem 1, show that 'Y"
=
(dt bill) T + b /11 2 KN.
(11) Let 'Y : 1---+ M C ]R3 be a curve which is not necessarily unit speed and 'Y" .J'Y') suppose M is oriented by a unit normal field N. Show that Kg =
W'
(12) Show that if a surface M c ]R3, either the image by an immersion of an open domain in ]R2 or is the zero set of a real-valued function f (such that df of. 0 on M) then it is orientable.
(13) Calculate the shape operator at a generic point on the cylinder {(x, y, z) : x 2 + y2 = r2}. (14) (Euler's formula) Let p be a point on a surface M in ]R3 and let Ul,U2 be principal directions with corresponding principal curvatures kl,k2 at p. If u = (cosO) Ul + (sinO) U2, show that k(u) = kl cos 2 0 + k2sin20. (15) Show that a point on a hypersurface in ]Rn is umbilic if and only if there is a constant ko such that k(u) = ko for all u E TpM with Ilull = 1. (16) Let Z be a nonvanishing (not necessarily unit) normal field on a surface M c ]R3. If X and Yare tangent fields such that X x Y = Z, then show that (Z,VxZ x VyZ) (Z, VxZ x Y +X x VyZ) 4 K = IIZI1 and H = 211Z11 3 (17) Find the Gauss curvature K at (x, y, z) on the ellipsoid x 2/a 2 + y2/b2+ y2/c 2 = 1. (18) Show that a surface in ]R3 is minimal if and only if there are orthogonal asymptotic vectors at each point. (19) Compute E, F, G, 1, m, n as well as Hand K for the following surfaces: (a) Paraboloid: (u, v) ~ (u, v, au 2 + bv 2) with a, b > O. (b) Monkey Saddle: (u,v) ~ (u,v,u 3 - 3uv 2). (c) Torus: (u,v) ~ ((a+ bcosv) casu, (a+ bcosv) sinu, bsinv) with a> b > O. (20) (Enneper's surface) Show that the following is a minimal but not 1-1 immersion: x(u,v):= (u-
~3 +uv2,
Chapter 5
Lie Groups
One approach to geometry is to view it as the study of invariance and symmetry. In our case, we are interested in studying symmetries of smooth manifolds, Riemannian manifolds, symplectic manifolds, etc. The usual way to deal with symmetry in mathematics is by the use of the notion of a transformation group. The wonderful thing for us is that the groups that arise in the study of geometric symmetries are often themselves smooth manifolds. Such "group manifolds" are called Lie groups. In physics, Lie groups play a big role in connection with physical symmetries and conservation laws (Noether's theorem). Within physics, perhaps the most celebrated role played by Lie groups is in particle physics and gauge theory. In mathematics, Lie groups play a prominent role in harmonic analysis (generalized Fourier theory), group representations, differential equations, and in virtually every branch of geometry including Riemannian geometry, Cartan geometry, algebraic geometry, Kahler geometry, and symplectic geometry.
5.1. Definitions and Examples Definition 5.1. A smooth manifold G is called a Lie group if it is a group (abstract group) such that the multiplication map J.I. : G x G ---t G and the inverse map inv : G ---t G, given respectively by J.I.(g, h) = gh and inv(g) = g-1, are Coo maps. If the group is abelian, we sometimes opt to use the additive notation 9 + h for the group operation. We will usually denote the identity element of any Lie group by the same letter e. Exceptions include the case of matrix or linear groups where we
-
189
190
5. Lie Groups
use the letter I or id. The map inv : G ~ G given by 9 H 9- 1 is called inversion and is easily seen to be a diffeomorphism.
Example 5.2. JR is a one-dimensional (abelian) Lie group, where the group multiplication is the usual addition +. Similarly, any real or complex vector space is a Lie group under vector addition. Example 5.3. The circle 8 1 = {z E C: IzI2 = 1} is a 1-dimensional (abelian) Lie group under complex multiplication. It is also traditional to denote this group by U(l). Example 5.4. Let JR* = JR\ {O}, C* = C\ {O} and IHI* = IHI\ {O} (here JH[ is the quaternion division ring discussed in detail later ). Then, using multipli. cation, JR*, C*, and IHI* are Lie groups. The Lie group IHI* is not abelian. The group of all invertible real n x n matrices is a Lie group denoted GL(n, JR). A global chart on GL(n, JR) is given by the n 2 functions x~, where if A E GL(n, JR) then xj(A) is the ij-th entry of A. We study this group and some of its subgroups below. Showing that GL(n, JR) is a Lie group is straightforward. Multiplication is clearly smooth. For the inversion map one appeals to the usual formula for the inverse of a matrix, A-I = adj(A)j det(A). Here adj(A) is the adjoint matrix (whose entries are the cofactors). This shows that A-I depends smoothly on the entries of A. Similarly, the group GL(n, C) of invertible n x n complex matrices is a Lie group.
Exercise 5.5. Let H be a subgroup of G and consider the cosets gH, g E G. Recall that G is the disjoint union of the cosets of H. Show that if H is open, then so are all the cosets. Conclude that the complement He is also open and hence H is closed. Theorem 5.6. If G is a connected Lie group and U is a neighborhood of the identity element e, then U generates the group. In other words, every element of g is a product of elements of U. Proof. First note that V = inv(U) n U is an open neighborhood of the identity with the property that inv(V) = V. We say that V is symmetric. We show that V generates G. For any open WI and W2 in G, the set WI W2 = {WI W2 : WI E WI and W2 E W2} is an open set being a union of the open sets U9EWlgW2. Thus, in particular, the inductively defined sets
V n = VV n- I , n
= 1,2,3, ... ,
are open. We have e EVe V2
c ... Vn
C ....
5.1. Dennitions and Examples
It is easy to check that each
191
vn is symmetric and so also is the union 00
V oo
:_
UV
n.
n=l
Moreover, V OO is not only closed under inversion, but also obviously closed under multiplication. Thus VOO is an open subgroup. From Exercise 5.5, yeo is also closed, and since G is connected, we obtain V OO = G. 0 In general, the connected component of a Lie group G that contains the identity is a Lie group denoted Go, and it is generated by any open neighborhood of the identity. We call Go the identity component of G. Definition 5.7. For a Lie group G and a fixed element 9 E G, the maps L9 : G --7 G and Rg : G --7 G are defined by
Lgx = gx for x E G, Rgx = xg for x E G, and are called left translation and right translation (by g) respectively. It is easy to see that Lg and Rg are diffeomorphisms with L;l = Lg 1 and Rg 1 = Rg 1. If G and H are Lie groups, then so is the product manifold G x H, where multiplication is (gl, hI) . (g2, h2) = (glg2, h1h2)' The Lie group G x H is called the product Lie group. For example, the product group 8 1 x 8 1 is called the 2-torus group. More generally, the higher torus groups are defined by Tn = 8 1 X ••. X 8 1 (n factors). Definition 5.S. Let H be an abstract subgroup of a Lie group G. If H is a Lie group such that the inclusion map H y G is an immersion, then we say that H is a Lie subgroup of G. Proposition 5.9. If H is an abstract subgroup of a Lie group G that is also a regular submanifold, then H is a closed Lie subgroup. Proof. The multiplication and inversion maps, H x H --7 Hand H --7 H, are the restrictions of the multiplication and inversion maps on G, and since H is a regular submanifold, we obtain the needed smoothness of these maps. The harder part is to show that H is closed. So let Xo E H be arbitrary. Let (U, x) be a single-slice chart adapted to H whose domain contains e. Let 8 : G x G --7 G be the map 8(gl' g2) = g1 1g2, and choose an open set Y such that e EVe V c U. By continuity of the map 8 we can find an open neighborhood 0 of the identity element such that 0 x 0 C 8- 1 (V). Now if {hi} is a sequence in H converging to Xo E H, then xo 1hi --7 e and Xo 1ht E 0 for all sufficiently large i. Since h-;lhi = (xo1h 3 f1 xo1h tl we
192
5. Lie Groups
have that hjl h~ E V for sufficiently large i, j. For any sufficiently large fixed j, we have Since U is the domain of a single-slice chart, un H is closed in U. Thus since each h j 1hi is in un H, we see that h-;lxo E Un H c H for all sufficiently large j. This shows that Xo E H, and since Xo was arbitrary, we are done. 0 By a closed Lie subgroup we shall always mean one that is a regular submanifold as in the previous theorem. It is a nontrivial fact that an abstract subgroup of a Lie group that is also a closed subset is automatically a closed Lie subgroup in this sense (see Theorem 5.81).
Example 5.10. 8 1 embedded as 8 1 x {I} in the torus 8 1 x 8 1 is a closed subgroup. Example 5.11. Let 8 1 be considered as the set of unit modulus complex numbers. The image in the torus T2 = 8 1 X 8 1 of the map JRl ---+ 8 1 X 81 given by t I--t (e~27rt, e~27rat) is a Lie subgroup. This map is a homomorphism. If a is a rational number, then the image is an embedded copy of 8 1 wrapped around the torus several times depending on a. If a is irrational, then the image is still a Lie subgroup but is now dense in T2. The last example is important since it shows that a Lie subgroup might actually be a dense subset of the containing Lie group.
5.2. Linear Lie Groups Let V be an n-dimensional vector space over IF, where IF = JR or C. The space L(V, V) of linear maps from V to V is a vector space and therefore a smooth manifold. A global chart for L(V, V) may be obtained by first choosing a basis for V and then defining n 2 functions {x~ h~~.j 1. Example 5.103. If H is a Lie subgroup of a Lie group G, then we can consider Lh for any h E H and thereby obtain a Lie group action of H onC. Recall that a subgroup H of a group G is called a normal subgroup if gkg- 1 E K for any k E H and all 9 E G. In other words, H is normal if gHg- 1 c H for all 9 E G, and it is easy to see that in this case we always have gHg- 1 = H. Example 5.104. If H is a normal Lie subgroup of G, then G acts on H by conjugation: g. h:= Cgh = ghg- 1 • Notice that the notation g. h cannot reasonably be abbreviated to gh in this example. Suppose that a Lie group G acts on smooth manifolds M and N. For simplicity, we take both actions to be left actions, which we denote by land
230
5. Lie Groups
A, respectively. A map f : M -t N such that f 0 19 = Ag 0 f for all 9 E G, is said to be an equivariant map (equivariant with respect to the given actions). This means that for all 9 the following diagram commutes:
(5.3)
M
-l- N
!
19!
Ag
M-l-N
If f is also a diffeomorphism, then f is an equivalence of Lie group actions.
Example 5.105. If if> : G -t H is a Lie group homomorphism, then we can define an action of G on H by A(g, h) = Ag(h) = L¢(g)h. We leave it to the reader to verify that this is indeed a Lie group action. In this situation, ¢ is equivariant with respect to the actions A and L (left translation). Example 5.106. Let rn = 8 1 x ... X 8 1 be the n-torus, where we identify 8 1 with the complex numbers of unit modulus. Fix k = (k}, ... , k n ) E ]Rn. Then JR acts on JRn by rk(t, x) = t· x := x + tk. On the other hand, JR. acts on Tn by t· (zl, ... , zn) = (eitklzl, ... , eitkn zn). The map JRn -t rn given by (xl, ... , xn) t-+ (e iX1 , ... , eixn ) is equivariant with respect to these actions. Theorem 5.107 (Equivariant rank theorem). Suppose that f : M -t N is smooth and that a Lie group G acts on both M and N with the action on M being transitive. If f is equivariant, then it has constant rank. In particular, each level set of f is a closed regular submanifold. Proof. Let the actions on M and N be denoted by 1 and A respectively as before. Pick any two points PI, P2 EM. Since G acts transitively on M, there is a 9 with 19P1 = P2. By hypothesis, we have f 0 19 = Ag 0 I, which corresponds to the commutative diagram (5.3). Upon application of the tangent functor we have the commutative diagram Tp1J
Tpl M ~ TJ(pI)N TP1lg
1
!
Tf(PIlAg
T p2 J
Tp2M ~ T f (P2)N
Since the maps TP1lg and Tf(Pl)Ag are linear isomorphisms, we see that TpJ must have the same rank as Tp2f. Since PI and P2 were arbitrary, we see that the rank of f is constant on M. Apply Theorem C.5. 0 There are several corollaries of this nice theorem. For example, we know that O(n,JR) is the level set f- 1 (I), where f : GL(n,JR) -t g£(n,JR) =Mnxn. is given by f(A) = AT A. The group O(n,JR) acts on itself via left translation, and we also let O(n,JR) act on g£(n, JR.) by Q . A := QT AQ (adjoint
231
5.7. Lie Group Actions
action). One checks easily that f is equivariant with respect to these actions, and since the first action (left translation) is certainly transitive, we see that O(n, JR.) is a closed regular submanifold of GL(n, JR.). It follows from Proposition 5.9 that O(n, JR.) is a closed Lie subgroup of GL(n, JR.). Similar arguments apply for U(n, q c GL(n, q and other linear Lie groups. In fact, we have the following general corollary to Theorem 5.107 above. Corollary 5.108. If ¢ : G --+ H is a Lie group homomorphism, then the kernel Ker(h) is a closed Lie subgroup of G. Proof. Let G act on itself and on H as in Example 5.105. Then ¢ is equivariant, and ¢-l(e) = Ker(h) is a closed Lie subgroup by Theorem 5.107 and Proposition 5.9. 0 Corollary 5.109. Let 1 : G x M --+ M be a Lie group action, and let Gp be the isotropy subgroup of some p EM. Then Gp is a closed Lie subgroup
a/G. Proof. The orbit map Op : G --+ M given by Op(g) = gp is equivariant with respect to left translation on G and the given action on M. Thus by the equivariant rank theorem, Gp is a regular submanifold of G, and then by Proposition 5.9 it is a closed Lie subgroup. 0 Proper Actions and Quotients. At several points in this section, such as the proof of Proposition 5.111 below, we follow [Lee, John]. Definition 5.110. Let 1 : G x M --+ M be a smooth (or merely continuous) group action. If the map P : G x M --+ M x M given by (g,p) t--+ (lgP,p) is proper, we say that the action is a proper action. It is important to notice that a proper action is not defined to be an action such that the defining map 1 : G x M --+ M is proper. We now give a useful characterization of a proper action. For any subset K eM, let g. K := {gx : x E K}.
Proposition 5.111. Let 1 : GxM --+ M be a smooth (or merely continuous) group action. Then 1 is a proper action if and only if the set G K := {g E G : (g . K)
nK
=1=
0}
is compact whenever K is compact. Proof. Suppose that 1 is proper so that the map P is a proper map. Let 1f'G be the first factor projection G x M --+ G. Then GK
= {g: there exists an x E K such that gx E K} = {g: there exists an x E M such that peg, x) E K x K} = 1rG(P-l(K x K)),
a.nd so GK is compact.
232
5. Lie Groups
Next we assume that GK is compact for all compact K. If C c M x M is compact, then letting K = 7I"1(C) U 71"2 (C), where 71"1 and 71"2 are the first and second factor projections M x M -; M respectively, we have p
1 (C) C
P
1 (K
x K)
c {(g, x) : gx E K
and x E K}
cGKxK. Since p-1(C) is a closed subset of the compact set GK x K, it is compact. This means that P is proper since C was an arbitrary compact subset of MxM. 0 Using this proposition, one can show that Definition 1.106 for discrete actions is consistent with Definition 5.110 above. Proposition 5.112. If G is compact, then any smooth action 1 : G x M -} M is proper. Proof. Let B c M x M be compact. We find a compact subset K such that B c K x K as in the proof of Proposition 5.111. Claim: P-1(B) is compact. Indeed,
P-l(B)
c P-1(K x K)
cM
= UkEKP-1(K x {k})
= UkEK{(g,P) : (gp,p) E K x {k}} = UkEK{(g, k) : gp E K} C U kEK (G x
{k}) = G x K
Thus P-1(B) is a closed subset of the compact set G x K and hence is compact. 0 Exercise 5.113. Prove the following: (i) If 1 : G x M -; M is a proper action and H eGis a closed subgroup, then the restricted action H x M -; M is proper.
(ii) If N is an invariant submanifold for a proper action l : G x M
~
M,
then the restricted action G x N -; N is also proper. Let us consider a Lie group action 1 : G x M -; M that is both proper and free. The orbit map at p is the map Op : G -; M given by Op(g) = g.p. It is easily seen to be smooth, and its image is obviously G . p. In fact, since the action is free, each orbit map is injective. Also, Op is equivariant with respect to the left action of G on itself and the action 1 :
Op(gx) - (gx) . p = g. (x· p) = g. Op(x) for all x, 9 E G. It follows from Theorem 5.107 (the equivariant rank theorem) that Op has constant rank, and since it is injective, it must be an
233
5.7. Lie Group Actions
G·p
Figure 5.1. Action-adapted chart
immersion. Not only that, but it is a proper map. Indeed, for any compact K c M the set O;I(K) is a closed subset of the set GKU{p}, and since the latter set is compact by Theorem 5.111, O;I(K) is compact. By Exercise 3.9, ()p is an embedding, so each orbit is a regular submanifold of M. It will be very convenient to have charts on M which fit the action of G in a nice way. See Figure 5.l. Definition 5.114. Let M be an n-manifold and G a Lie group of dimension k. If 1 : G x M ~ M is a Lie group action, then an action-adapted chart on M is a chart (U, x) such that (i) x(U) is a product open set Vi x
V2 C Rk X Rn-k = ]Rn;
(ii) if an orbit has nonempty intersection with U, then that intersection has the form {X k +l
= el , ... , xn = en- k }
£or some const ant s cI , ... , en-k . Theorem 5.115. If 1 : G x M ~ M is a free and proper Lie group action, then for every p E M there is an action-adapted chart centered at p. Proof. Let p E M be given. Since G . p is a regular submanifold, we may choose a regular submanifold chart (W, y) centered at p so that (G· p) n W is exactly given by yk+l = ... = yn = 0 in W. Let S be the complementary slice in W given by yl = ... = yk = O. Note that S is a regular submanifold. The tangent space TpM decomposes as TpM = Tp (G· p) EEl TpS.
Let t.p : G x S ~ M be the restriction of the action 1 to the set G x S. Also, let ip : G -* G x S be the insertion map 9 ~ (g,p) and let ie : S ~ G x S be the insertion map s ~ (e, s). (See Figure 5.2.) These insertion maps
5. Lie Groups
234
are embeddings, and we have Op - tp 0 ip and also t.p 0 je = t, where t is the inclusion S rphism. One can then use Corollary 3.27 to show that the induced map f is smooth. 0 If a Lie group G acts smoothly and transitively on M (on the right or left), then M is called a homogeneous space with respect to that action. Of course it is possible that a single group G may act on M in more than one way and so M may be a homogeneous space in more than one way. We will give a few concrete examples shortly, but we already have an abstract example on hand.
Theorem 5.125. If H is a closed Lie subgroup of a Lie group G, then the map G x G/H -; G/H, given by I: (g,gIH) -; gglH, is a transitive Lie group action. Thus G / H is a homogeneous space with respect to this action. Proof. The fact that l is well-defined follows since if 91H = g2H, then g2'l g1 E H, and so 992H = 9g292191H = 9glH. We already know that G/H is a smooth manifold and 7r : G -; G/H is a surjective submersion.
5. Lie Groups
242
We can form another submersion idG following diagram commute:
GxG
X7r :
G x G -+ G x G/H making the
) G
idGX~t ~ ~1 GxG/H-G/H Here the upper horizontal map is group multiplication and the lower horizontal map is the action l. Since the diagonal map is smooth, it follows from Proposition 3.26 that I is smooth. We see that 1 is transitive by observing that if 91H,92H E G/H, then 19291 1 (91H) = 92H. 0 It turns out that up to appropriate equivalence, the examples of the above type account for all homogeneous spaces. Before proving this let us look at some concrete examples. Example 5.126. Let M = JRn and let G = Euc(n, JR) be the group of Euclidean motions. We realize Euc(n, JR) as a matrix group
Euc(n, R) = {
[~ ~]: v E JRn and Q E O(n) }
The action of Euc(n, JR) on JRn is given by the rule
[~ ~]. x = Qx + v, where x is written as a column vector. Notice that this action is not given by a matrix multiplication, but one can use the trick of representing the points x of JRn by the (n + 1) x 1 column vectors [i 1, and then we have ~ [i 1= [Qx\v]· The action is easily seen to be transitive.
[ S]
Example 5.127. As in the previous example we take M = Rn, but this time, the group acting is the affine group Aff(n, JR) realized as a matrix group: Aff(n,JR) The action is
= {[
~ ~]: v E JRn and A E GL(n,JR)}.
[~ ~]. x = Ax + v,
and this is again a transitive action. Comparing these first two examples, we see that we have made JRn into a homogeneous space in two different ways. It is sometimes desirable to give different names and/or notations for Rn to distinguish how we are acting on the space. In the first example we might write lEn (Euclidean space), and
5.B. Homogeneous Spaces
243
in the second case we write An and refer to it as affine space. Note that, roughly speaking, the action by Euc(n, ~) preserves all metric properties of figures such as curves defined in En. On the other hand, Aff(n,~) always sends lines to lines, planes to planes, etc. Example 5.128. Let M = H := {z E e : Imz > o}. This is the upper half-plane. The group acting on H will be SL(2, ~), and the action is given by
b). z = cz+d az + b.
( a cd This action is transitive.
Example 5.129. We have already seen in Example 5.102 that both O(n) and SO(n) act transitively on the sphere sn-l C ~n, so sn-l is a homogeneous space in at least two (slightly) different ways. Also, both SU(n) and U(n) act transitively on s2n-l C en. Example 5.130. Let V~,k denote the set of all k-frames for ~n, where by a k-frame we mean an ordered set of k linearly independent vectors. Thus an n-frame is just an ordered basis for ~n. This set can easily be given a smooth manifold structure. This manifold is called the (real) Stiefel manifold of k-frames. The Lie group GL(n,~) acts (smoothly) on V~ k by 9 . (el,"" ek) = (gel, ... , gek). To see that this action is transitiv~, let (el,'''' ek) and (h, ... , ik) be two k-frames. Extend each to n-frames el,"" ek,"" en) and (h, ... , ik, ... , fn)' Since we consider elements of JRn as column vectors, these two n-frames can be viewed as invertible n x n matrices E and F. If we let 9 := EF-l, then gE = F, or g. (el, ... , ek) gel,··., gek) = (h, ... , ik)· Example 5.131. Let Vn,k denote the set of all orthonormal k-frames for ~n, where by an orthonormal k-frame we mean an ordered set of k orthonormal vectors. Thus an orthonormal n-frame is just an orthonormal basis for ~n. This set can easily be given a smooth manifold structure and is called the Stiefel manifold of orthonormal k-frames. The group O(n,~) acts t ansitively on Vn,k for reasons similar to those given in the last example. Theorem 5.132. Let M be a homogeneous space via the transitive action I : G x M --+ M, and let Gp be the isotropy subgroup of a point p EM. Recall that G acts on G/Gp. If G/Gp is second countable (in particular f G is second countable), then there is an equivariant diffeomorphism 1, the groups SO(n), SU(n) and U(n) are connected while the group O(n) has exactly two components: SO(n) and the subset ofO(n) consisting of elements with determinant-1. Proof. The groups SO(l) and SU(l) are both connected since they each contain only one element. The group U(l) is the circle, and so it too is connected. We use induction. Suppose that SO(k), SU(k) are connected for 1 k n - 1. We show that this implies that SO(n), SU(n) and U(n) are connected. From Example 5.138 we know that sn-l = SO(n)/SO(n -1). Since sn 1 and SO(n - 1) are connected (the second by the induction hypothesis), we see that SO(n) is connected. The same argument works for SU(n) and U(n).
:s :s
Every element of O(n) has determinant either 1 or -1. The subset SO(n) C O(n) is closed since it is exactly {g E O(n) : detg = I}. Fix an element ao with det ao = -1. It is easy to show that aoSO( n) is exactly the set of elements of O(n) with determinant -1 so that SO(n) U aoSO(n) - O(n) and SO(n) n aoSO(n) = 0. Indeed, by the multiplicative property of determinants, each element of aoSO(n) has determinant -1. But aoSO(n)
247
5.8. Homogeneous Spaces
also contains every element of determinant -1 since for any such 9 we have 9 - ao (ao1g) and ao1g E SO(n). Since SO(n) and aoSO(n) are complements of each other, they are also both open. Both sets are connected, since 9 t-t aog is a diffeomorphism which maps the first to the second. Thus we see that SO(n) and aoSO(n) are the connected components of O(n). D We close this chapter by relating the notion of a Lie group action with that of a Lie group representation. We give just a few basic definitions, some of which will be used in the next chapter. Definition 5.145. A linear action of a Lie group G on a finite-dimensional vector space V is a left Lie group action A : G x V -+ V such that for each 9 E G the map Ag : v H A(g, v) is linear. The map G -+ GL(V) given by 9 H A(g) := Ag is a Lie group homomorphism and will be denoted by the same letter A as the action so that .\(g)v := A(g, v). A Lie group homomorphism A : G -+ GL(V) is called a representation of G. Given such a representation, we obtain a linear action by letting A(g, v) := A(g)V. Thus a linear action of a Lie group is basically the same as a Lie group representation. The kernel of the action is the kernel of the associated homomorphism (representation). An effective linear action is one such that the associated homomorphism has trivial kernel, which, in turn, is the same as saying that the representation is faithful. Two representations A : G -+ GL(V) and>..' : G -+ GL(V') are equivalent if there exists a linear isomorphism T : V -+ V' such that T 0 Ag = A~ 0 T for all g. Exercise 5.146. Show that if A : G x V -+ V is a map such that Ag : v H ). g, v) is linear for all g, then A is smooth if and only if Ag : G -+ GL(V) is smooth for every 9 E G. (Assume that V is finite-dimensional as usual.) We have already seen one important example of a Lie group representation, namely, the adjoint representation. The adjoint representation came from first considering the action of G on itself given by conjugation which leaves the identity element fixed. The idea can be generalized: Theorem 5.147. Let 1 : G x M -+ M be a (left) Lie group action. Suppose that Po EM is a fixed point of the action (lg(Po) = Po for all g}. The map
l(Po) : G -+ GL(TpoM) gwen by 8
a Lie group representation.
5. Lie Groups
248
Proof. Since
l(Po)(9192) = TPo (lgH12) = TPO(l91 Olg2)
= TpOlgl 0 Tpolg2
= l(po) (91)l(po) (92),
we see that l(Po) is a homomorphism. We must show that l(po) is smooth. It will be enough to show that 9 t-t a(Tpolg·v) is smooth for any v E TPoM and any a E T:aM. This will follow if we can show that for fixed Vo E TPoM, the map G -+ T M given by 9 t-t Tpolg . Vo is smooth. This map is a composition
G -+ TG x T M ~ T (G x M)
!! T M,
where the first map is 9 t-t (Og, vo), which is clearly smooth. By Exercise 5.146 this implies that the map G x TpoM -+ TpoM given by (g, v) t-t TpOlg·v is smooth. 0 Definition 5.148. For a Lie group action 1: G x M -+ M with fixed point Po, the representation Z(po) from the last theorem is called the isotropy
representation for the fixed point. Now let us consider a transitive Lie group action 1 : G x M -+ M and a point Po. For notational convenience, denote the isotropy subgroup GPo by H. Then H acts on M by restriction. We denote this action by A: H x M -+ M, A: (h,p) t-t hp for h E H = Gpo.
Notice that Po is a fixed point of this action, and so we have an isotropy representation ,A(po) : H x TPoM -+ TpoM. On the other hand, we have another action C : H x G -+ G, where Ch : G -+ G is given by 9 t-t h9h 1 for h E H. The Lie differential of Ch is the adjoint map Adh : g -+ g. The map Ch fixes H, and Adh fixes ~. Thus the map Adh : g -+ g descends to a map Adh : g/~ -+ g/~. We are going to show that there is a natural isomorphism TpoM ~ g/~ such that for each h E H the following diagram commutes:
(5.4)
One way to state the meaning of this result is to say that h t-t Adh is a representation of H on the vector space g/~, which is equivalent to the linear isotropy representation. The isomorphism TPoM ~ g/~ is given in the
5.9. Combining Representations
249
~ E ~
following very natural way: Let have
Te7r(~) =
and consider
Te7r(~) E
TpoM. We
dd 7r(exp~t) = 0 t t=O since exp~t E ~ for all t. Thus ~ C Ker(Te 7r). On the other hand, dim~ = dimH = dim(Ker{Te 7r)), so in fact ~ = Ker(Te 7r) and we obtain an isomorphism g/~ ~ TpoM induced from T e7r. Let us see why the diagram (5.4) commutes. First, Lh is well-defined as a map from G / H to itself and the following diagram clearly commutes:
I
Ch
G
.. G
~l
l~
G/H~G/H Using our equivariant diffeomorphism ..g with respect to B by [>..g].13 we obtain a homomorphism G ~ GL(n, IF)
250
5. Lie Groups
given by 9 t--+ [AglB. In general, a Lie group homomorphism of a Lie group G into GL(n, IF) is called a matrix representation of G. We have already seen that any Lie subgroup of GL(JRn ) acts on IRn by matrix multiplication, and the corresponding homomorphism is the inclusion map G c......t GL(JRn ). More generally, a Lie subgroup G of GL(V) acts on V in the obvious way simply by employing the definition of GL(V) as a set of linear transformations of V. We call this the standard action of the linear Lie subgroup of GL(V) on V, and the corresponding homomorphism is just the inclusion map G c......t GL(V). Choosing a basis, the subgroup corresponds to a matrix group, and the standard action becomes matrix multiplication on the left of lFn , where the latter is viewed as a space of column vectors. This action of a matrix group on column vectors is also referred to as a standard action. Given a representation A of G in a vector space V, we have a dual rep~ resentation A* of G in the dual space V* by defining A*(g) := (A(g 1))*: V* -+ V*. Recall that if L : V -+ V is linear, then L * : V* -+ V* is defined by L*(a)(v) = a(Lv) for a E V* and v E V. This dual representation is also sometimes called the contragredient representation (especially when IF = R). Now let AV and AW be representations of a lie group G in IF-vector spaces V and W respectively. We can then form the direct sum repr~ sentation AV E9 AW by (AV E9 AW)g :- A~ E9 A't' for 9 E G, where we have (A~ E9A't') (v,w) (A~V,AWW). We will not pursue a serious study of Lie group representations but simply note that a major goal in the subject is the identification and classification of irreducible representations. A representation A : G -+ GL(V) is said to be irreducible if there is no nonzero proper subspace W of V such that Ag(W) c W for all g. A large class of Lie groups known as semisimple Lie groups have the property that their representations break into direct sums of irreducible representations. Example 5.149. A homogeneous polynomial of degree don ((:2 is a linear combination of monomials of total degree d. Let HJ denote the vector space of homogeneous polynomials of degree 2j, where j is a nonnegative "halfinteger" (j = k/2 for some nonnegative integer k). Define Aj : SU(2) --t GL(HJ ) by (Aj(g)f)(z) := f(g-lz) for z = (Zl,Z2) E ((:2. Then AJ is an irreducible representation called the spin-j representation of SU(2 . The spin-l/2 representation turns out to be equivalent to the standard representation of SU(2) in ((:2. These spin representations play an important role in quantum physics. One can also form the tensor product of representations. The definitions and basic facts about tensor products are given in the more general context of
251
5.9. Combining Representations
module theory in Appendix D. Here we give a quick recounting of the notion of a tensor product of vector spaces, and then we define tensor products of representations. Given two vector spaces V and W over some field IF, consider the class CVxw consisting of all bilinear maps V x W --t X, where X varies over aU IF-vector spaces, but V and Ware fixed. We take members of CYxW as the objects of a category (see Appendix A). A morphism from, say, J.l.l : V x W --t X to J.l.2 : V x W --t Y is defined to be a linear map l : X -t Y such that the diagram
Y
VxW
X
'-
~Y commutes. There exists a vector space Tv,w together with a bilinear map : V x W--t Tv,w that has the following universal property: For every bilinear map J.l. : V x W --t X, there is a unique linear map Ii : Tv,w --t X such that the following diagram commutes:
If such a pair (Tv,w, ®) exists with this property, then it is unique up to isomorphism in CYxw. In other words, if ®: V x W --t Tv,w also has this universal property, then there is a linear isomorphism Ty,w ~ Tv,w such that the following diagram commutes:
y VxW
~
Tv,w
~
Tv,w
We refer to such universal object as a tensor product of V and W. We will indicate the construction of a specific tensor product that we denote by V W with corresponding bilinear map ® : V x W --t V ® W. The idea is
252
5. Lie Groups
simple: We let V ® W be the set of all linear combinations of symbols of the form v ® W for v E V and W E W, subject to the relations (VI
+ V2) ® W = VI ® W + v2 ® W,
V ® (WI + W2) = V ® WI + V ® W2, r (v ® w) = rv ® W = V ® rw, for rEF. The map ® is then simply ®: (v,w) ---+ v®w. Somewhat more pedantically, let F(V x W) denote the free vector space generated by the set V x W (the elements of V x Ware treated as a basis for the space, and so the free space has dimension equal to the cardinality of the set V x W). Next consider the subspace R of F(V x W) generated by the set of all elements of the form
(av,w) - a(v,w), (v, aw) - a(v, w), (VI + V2, w) - (VI. w) + (V2' w), (V, WI + W2) - (V, WI) + (V, W2), for VI, V2, V E V, WI, W2, wE W, and a E F. Then we let V ® W be defined as the quotient vector space F(V x W)/R, and we have a corresponding quotient map F(V x W) ---+ V ® W. The set V x W is contained in F(V x W), and the map ® : V x W ---+ V ® W is then defined to be the restriction of the quotient map to V x W. The image of (v, w) under the quotient map is denoted by V ® w. Tensor products of several vector spaces at a time are constructed similarly to be a universal space in a category of multilinear maps (Definition D.13). We may also form the tensor products two at a time and then use the easily proved fact that (V ® W) ® U ~ V ® (W ® U), which is then denoted by V ® W ® U. Again the reader is referred to Appendix D for more about tensor products. Elements of the form V ® w generate V ® W, and in fact, if (eI,"" er ) is a basis for V and (fI, .•• , fs) is a basis for W, then
{ei ® Ii : 1 ~ i ~ r, 1 ~ j ~ s} is a basis for V ® W, which therefore has dimension rs
= dim V dim W.
One more observation: If A : V ---+ X and B : W ---+ Y are linear maps, then we can define a linear map A ® B : V ® W ---+ X ® Y. First note that the map (v, w) H Av ® Bw is bilinear. Thus, by the universal property, there is a unique map A ® B such that
(A ® B) (v ® w) = Av ® Bw, for all v E V, w E W.
Problems
253
Notice that if A and B are invertible, then A ® B is invertible with (A ® B)-l(v ® w) = A-Iv ® B lw. Pick bases for V and W as above and bases {ei, ... , e~} and {if, ... , I;} for X and Y respectively. For rEV ® W, we can write r = r ij ei ® 13 using the Einstein summation convention. We have
A ® B(r) = A ® B(rije, ® Ij) = r,j Aei ® Blj
= r ij Afe~ ® B~I{ ij Bl. (e'k' : 7r- 1 (U) -t U x F occurring in the definition of a fiber bundle are said to be local trivializations of the bundle. It is easy to see that such a local trivialization must be a map of the form 4> = (7r17l"-1(u) , Ill) where
a = (7r, <pa) : 7r- 1 (Ua) ~ Ua x F
and similarly for 4>fJ = (7r, 0 0 4>-;/ : (Uo n U(:1) x F
~ (Ua n U(:1) x F.
Since .)-bundle it must be the case that both of the co cycles are contained in a larger cocycle that gives the transitions for the atlas obtained as the union of the collection of charts from {(¢et, Uet ), (9et{3) , A} and {(¢~, U;), (9~j)' A}. We handle ineffective actions this way so as to keep aligned with the notion of associated bundle introduced later. If there is no chance of confusion, we will drop the adjective effective. The reader is warned that some standard expositions on fiber bundles allow ineffective actions right from the start, but in some cases assertions are made that would only be true in the effective case! It is interesting to note that in his famous book on the subject [St], Norman Steenrod restricts himself
6.1. General Fiber Bundles
265
to effective actions, although he announces this restriction in one easily overlooked sentence early in the book. Notice that an alternative way to say that A(90.fJ(P), y) = <po.fJlp (y) is ~a 0 '), then there exists a vector bundle over M with a VB-atlas {(Ua, q)a)} satisfying q)a 0 q)~ 1 (p, v) = (p, gaf3 (p) . v)) on nonempty overlaps Ua n Uf3. In other words, there exists a vector bundle with (G, >.)-atlas. Proof. This is essentially a special case of Theorem 6.15. One only needs to check linearity of the q)a on fibers. D Perhaps some clarification is in order. In the case of a vector bundle, the raw transition maps iPa{3 take values in the general linear group GL(V), which is a Lie group. They correspond to a >'o-bundle structure where >'0 is the standard linear action of GL(V) on V (the standard representation), and they automatically satisfy the co cycle condition. The more general transition maps that define a (G, >.)-bundle structure (G-bundle structure) are G-valued. It is important to note that G may be small compared to GL(V) and certainly need not be thought of as a subset of GL(V). For example, the tensor bundles have (possibly ineffective) GL(V)-bundle structures coming from tensor representations, but the tensor bundles themselves generally have rank greater than k = dim (V). Since in the vector bundle case, the iP af3 arise directly from a VB-atlas and act by the standard action, we will call these standard transition maps, and the corresponding GL(V)bundle structure will be called the standard GL(V)-bundle structure. The standard GL(V)-bundle structure is the structure that a vector bundle has simply by virtue of being an IF -vector bundle with typical fiber V. Remark 6.31. We have previously mentioned that the notion of a representation is equivalent to that of a left linear action. When dealing with vector bundles it is perhaps more common to use the representation terminology and notation and this we shall do as convenient. So if>. is a left linear action, then the map G -----t GL(V) given by 9 H >.(g) := >.g is a representation of G. Conversely, if>. is such a representation, we obtain a linear action by letting >.(g, v) := >.(g)v.
6.2. Vector Bundles
275
We already know what it means for two vector bundles over M to be equivalent. Of course any two vector bundles that are equivalent in a natural way can be thought of as the same. Since we can and often do construct our bundles according to the above recipe, it will pay to know something about when two vector bundles over M are isomorphic, based on their respective transition functions. Notice that the standard transition functions are easily recovered from every (G, A)-atlas by the formula A (go.f3{p)) Y = <po.f3lp (Y)· Proposition 6.32. Two vector bundles 7[' : E -t M and 7[" : E' -t M with standard transition maps {
01
0
4>-t(p, S{3(p))
= 4>-pI(P, s.a(P))
= 4>fl1 0l.a(p)·
o
Suppose we have two vector bundles, 11"1 : E1 ~ M and 11"2 : E2 ~ M. We give two constructions of the Whitney sum bundle 11"1 EEl7r2 : E1 EElE2 4 M. This is a globalization of the direct sum construction of vector spaces. In fact, the first construction simply takes E1 EEl E2 = UPEME1p EEl E2p. Now, we have a vector bundle atlas {(4)a, Ua )} for 11"1 and a vector bundle atlas {(1Pa, Ua )} for 11"2. Assume that both atlases have the same family of open sets (we can arrange this by taking a common refinement). Now let 4>a EEl1Pa : (vp, wp) t-+ (P,pr2 0 4>01 (vp) ,pr2 0 1Pa (wp)) for all (vp, wp) E (E1 EEl E2)lu... Then {(4)a EEl 1Pa, Ua )} is a VB-atlas for 11"1 EEl 11"2 : E1 EEl E2 ~ M. Another method of constructing this bundle is to take the co cycle {gaP} for 11"1 and the co cycle {ha.a} for 11"2 and then let ga{3 EEl ha{3 : Ua n U.a 4 GL(IFkl X IFk2) be defined by (ga.a EElhafl) (x) = ga{3(x) EEl ha{3(x) : (v,w) t-+ (ga.a(x)v, ha.a(x)w). The maps 9a.a EEl ha.a form a co cycle which determines a bundle by the construction of Proposition 6.30, which is (isomorphic to) 11"1 EEl 11"2 : El EEl E2 ~ M. The pull-back of a vector bundle 11" : E ~ M by a smooth map f : M is naturally a vector bundle whose linear structure on each fiber (f* E)q = {q} x Ep is the obvious one induced from Ep. Put another way, ';!e give the unique linear structure to each fiber that makes the bundle map f : f* E ~ E linear on fibers. When given this vector bundle structure, we call f* E the pull-back vector bundle.
N
~
Example 6.35. Let 11"1 : E1 ~ M and 11"2 : E2 ~ M be vector bundles and let 6. : M ~ M x M be the diagonal map x t-+ (x, x). From 11"1 and 11"2 one can construct a bundle 1I"E lxE2 : E1 x E2 ~ M x M by 1I"E l XE2 (EI' E2) := (11"1 (El) , 11"2 (E2)). The Whitney sum bundle E1 EElE2 defined previously is naturally isomorphic to the pull-back 6.*11" El X E2 : 6.* (E1 X E 2) ~ M (Problem 10). Exercise 6.36. Recall the space r f(~) from Definition 6.19. Show that if 11" : E ~ M is an IF-vector bundle and f : N ~ M is a smooth map, then both r f(~) and r(f*~) are modules over COO(N; IF), and that the natural correspondence between r f(~) and r(f*~) is a module isomorphism. Every vector bundle has global sections. An obvious example is the zero section which maps each x E M to the zero element Ox of the fiber
277
6.2. Vector Bundles
E3J. The image of the zero section is also referred to as the zero section and is often identified with M. (Of course, the image of any global section is a submanifold diffeomorphic to the base manifold.) We have the following simple analogue of Lemma 2.65: Lemma 6.37. Let 11" : E -+ M be an IB'-vector bundle with typical fiber V. llv E 1I"-1(p) then there exists a global section 0' E r(e) such that O'(p) = v. Furthermore, if s is a local section defined on U, and V is an open set with compact closure with V c V c U, then there is a section 0' E r(e) such that (J = S on V. Proof. Using a local trivialization one can easily get a local section O'loe defined near p such that O'(p) = v. Now just use a cut-off function as in the proof of Lemma 2.65. For the second part we just choose a cut-off function Pwith support in U and such that (3 = 1 on V. Then (3s extends by zero to the desired global section. D
If a section of a vector bundle takes the zero value in some fiber we say that it vanishes at that point. Global smooth sections that never vanish do not always exist; such sections are called nowhere vanishing or nonvanishing. However, there is one case where it is easy to see that nonvanishing smooth sections exist: Proposition 6.38. Any bundle equivalent to a (trivial) product bundle must have a nowhere vanishing smooth global section. It is a fact that the tangent bundle of 8 2 does not have any such nowhere vanishing smooth sections. In other words, all smooth (or even continuous) vector fields on 8 2 must vanish at some point. This is a result from algebraic topology called the "hairy sphere theorem" (see Theorem 10.15). If one fancifully imagines a vector field on a sphere to be hair, then the theorem suggests that one cannot comb the hair neatly "flat" without creating a cowlick somewhere. More generally, the analogous result holds for 8 2n if n,~ 1. Exercise 6.39. Modify either the construction of Example 6.9 or Example 6.16 to obtain a rank one vector bundle version of the Mobius band and give an argument proving that every global continuous section of this bundle must vanish somewhere.
e
Definition 6.40. If = (E, 11", M, V) is a vector bundle and p EM, then a vector space basis for the fiber Ep is called a frame at p. Definition 6.41. Let 11" : E -+ M be a rank k vector bundle. A k-tuple (J = (0'1, •.. , O'k) of sections of E over an open set U is called a (local) frame field over U if for all p E U, (0'1 (p), ... , O'k (p)) is a frame at p.
278
6. Fiber Bundles
If we choose a fixed basis {edi-l, ... ,k for the typical fiber V, then a choice of a local frame field over an open set U c M is equivalent to a local trivialization (a vector bundle chart). Namely, if
0 at p for at least one a: easily gives the result that g is positive definite at each p and so it is a Riemannian metric onE. 0 Example 6.46 (Tautological line bundle). Recall that ]Rpn is the set of all lines through the origin in ]Rn+1. Define the subset IL(]Rpn) of]Rpn x ]Rn+1 consisting of all pairs (1, v) such that vEl (think about this). This set together with the map 7rlRP" : IL(Rpn) --+ ]Rpn given by (l,v) t-+ 1, is a rank one vector bundle. Example 6.47 (Tautological bundle). Let G(n, k) denote the Grassmann manifold of k- planes in ]Rn. Let 'Yn,k be the subset of G (n, k) x ]Rn consisting of pairs (P, v) where P is a k-plane (k-dimensional subspace) and v is a vector in the plane P. The projection 7rn,k : 'Yn,k --+ G(n, k) is simply (P,v) t-+ P. The result is a vector bundle bn,k,7rn,k,G(n,k),Rk ). We leave it to the reader to discover an appropriate VB-atlas (see Problem 12). These tautological vector bundles are not just trivial bundles, and in fact their topology or twistedness (for large n) is of the utmost importance for classifying vector bundles (see [Bo-Tu]). One may take the inclusions Rn C IRn+1 C ... c ]Roo to construct inclusions G(n, k) C G(n + 1, k) c ... and 'Yn,k C 'Yn+l,k. Given a rank k vector bundle 7r : E --+ M, there is an n such that 7r : E --+ M is (isomorphic to) the pull-back of 'Yn,k by some map I: M --+ G(n, k):
E
~
/*'Yn,k
t
M
--~ .. 'Yn,k
f
t
.. G(n,k)
Exercise 6.48. To each point on a unit sphere in IRn, attach the space of all vectors normal to the sphere at that point. Show that this normal bundle is in fact a (smooth) vector bundle. Generalize to define the normal bundle of a hypersurface in IRn. When is such a normal bundle trivial? Exercise 6.49. Fix a nonnegative integer j. Let Y = ]Rx (-1,1) and let Xl,Yl) '" (X2,Y2) if and only if Xl = X2 + jk and Yl = (-1)jk Y2 for some integer k. Show that E := Y/"" is a vector bundle of rank 1 that is trivial if and only if j is even. Prove or at least convince yourself that this is the Mobius band when j is odd.
282
6. Fiber Bundles
6.3. Tensor Products of Vector Bundles Given two vector bundles 11'"1 : El -+ M and 11'"2 : E2 -+ M with respective typical fibers VIand V2, we let
El ® E2 :=
U Elp ® E2p
(a disjoint union).
pEM
Then we have a projection map 11'" : El ® E2 -+ M given by mapping any element in a fiber E 1p ® E2p to the base point p. We show how to construct a VB-atlas for El ® E2 from an atlas on each of El and E2. The smooth structure and topology can be derived from the atlas as usual in such a way as to make all the relevant maps smooth. We leave the verification of this to the reader. The resulting bundle is the tensor product bundle. As usual we can assume that the atlases are based on the same open cover. Thus suppose that {(Ua ,4>a)} is a VB-atlas for El while {(Ua,I/'a)} is a VB-atlas for E 2. Now let -Po ® Wo : (El ® E 2)l u", -+ VI ® V2 be defined by (-Po ® Wa)IElp E2p := -PalElp ® WalE2p for p E Ua . Then let 4>0 ® 1/;0: (El ® E2)l u", -+ Ua
X
(VI ® V2)
be defined by 4>0®1/;0 := (11'", -Po®w o ). To clarify, the map -POIElP ® Wa lE2P : E 1p ® E2p -+ V 1 ® V2 is the tensor product map of two linear maps as described at the end of Chapter 5. To see what the transition maps look like, we compute; (-Pa ® WO)IElP®~P
0
(-P~ ® W~) E~p®E2P
1 = -PalElp ® WoIE2p 0 -P~IE-llp ® W~ E-2p
= (-PaIElP
0
-P~IE~p) ® (WaIElP 0 W~IE~p)
= -Po~(P) ® Wo~(p).
Thus the transition maps are given by p -+ -Pa~(P) ® wa~(P), which is a map from Uo to GL(VI ® V2). The group GL(V1 ® V2) acts on V1 ® V2 in a standard way, and this is the standard effective structure group of the bundle as we have just seen. However, it is also true that the bundle E1 ®Ez has (ineffective) structure group GL(VI) x GL(V2) via a tensor product representation. Indeed, if 1.1 denotes the standard representation of GL(V d in VI and 1.2 denotes the standard representation of GL(V2) in V2, then we have a tensor product representation 1.1 ®1.2 of GL(V1) x GL(V2) in V1 ®V2. This is usually not a faithful representation. Using the GL(Vl) x GL(V2)valued co cycle p r-t ho~(p) := (-Po~(P), Wa~(P)), together with 1.1 ® 1.2, we see that by definition
283
8.4. Smootb Functors
Furthermore, if VI = V2 = V, then the tensor product representation is usually defined as a representation of GL(V) rather than GL(V) x GL(V), and so E1 0 E2 would have a (GL(V), I, 0 I,)-bundle structure where I, is the standard representation. In this case 1,0(, is still not a faithful representation since - idv is in the kernel. We can reconstruct the same vector bundle using any of these representation-co cycle pairs via Lemma 6.30. In fact, it is quite common that we have different representations by one group. Suppose that we have two faithful representations )11 and A2 of a Lie group G acting on VI and V2 respectively. If {ga:,B} is a co cycle of transition maps, then we can use the pair {ga:,B, Ad in Lemma 6.30 to form a vector bundle E1 that has a (G, A1)-bundle structure by construction. Similarly, we can construct a vector bundle E2 with (G, A2)-bundle structure. If we use Al 0 A2 and the same co cycle {ga:,B}, then we obtain a bundle which, as a vector bundle, is E1 ® E 2 • But by construction, it has a (G, Al ® A2)-structure (possibly ineffective). This is the case in the following exercise: Exercise 6.50. Suppose that E is a vector bundle with a (G, A)-bundle structure given by a (G, A)-atlas with a corresponding cocycle of transition functions. Show how one may use Theorem 6.30 to construct bundles isomorphic to E*, E®E and E0E* which will have a (G, A*)-bundle structure, a (G, ). ® A)-bundle structure and a (G, A0 A*)-bundle structure respectively.
6.4. Smooth Functors We have seen that various new vector bundles can be constructed starting with one or more vector bundles. Most of the operations of linear algebra extend to the vector bundle category. We can unify our thinking on these matters by introducing the notion of a Coo functor (or smooth functor). With IF = R or C, the set of alllF-vector spaces together with linear maps is a category that we denote by Lin(lF). The set of morphisms from V to W is the space of IF-linear maps L(V, W) (also denoted Hom(V, W)). Definition 6.51. A covariant Coo functor F of one variable on Lin(lF) consists of a map, denoted again by F, that assigns to every IF-vector space Van IF-vector space FY, and a map, also denoted by F, which assigns to every linear map A E L(V, W), a linear map FA E L(FY,FW) such that (i) F ~ L(V, W) -t L(FY, FW) is smooth; (ii) F(idv)
= idF\'
for alllF-vector spaces V;
(iii) F(A 0 B) = FA 0 FB for all A E L(U, V) and B E L(V, W) and vector spaces U, V and W. As an example we have the Coo functor which assigns to each V the k-fold direct sum ffik V = V E9 ... E9 V and to each linear map A E L(V, W)
6. Fiber Bundles
284
the map
E9A : E9v -t E9w
k k k given by E9kA(Vl,"" Vk) := (AVI,"" AVk). Similarly there is the functor which assigns to each V the k-fold tensor product (j!;lv = V Q9 ... Q9 V and to each A E L(V, W) the map ®k A : ®kV -t ®kW given on homogeneous elements by (®k A)(VI Q9 ... Q9 Vk) := AVI Q9'" Q9 AVk. One can also consider Coo covariant functors of several variables. For example, we may assign to each pair of vector spaces (V, W) the tensor product V Q9 W, and to each pair (A,B) E L(V, V') x L(W, W'), the map A Q9 B : V Q9 W -t V' Q9 W'. There is also a similar notion of contravariant Coo functor: Definition 6.52. A contravariant Coo functor F of one variable on Lin(lF) consists of a map, denoted again by F, which assigns to every JFvector space V an IF-vector space FY, and a map, also denoted by F, which assigns to every linear map A E L(V,W) a linear map FA E L(FW,FV) (notice the reversal) such that
(i) F: L(V, W) -t L(FW, FY) is smoothj (ii) F(idv) = idw for alllF-vector spaces Vj (iii) F(A 0 B) = FB 0 FA for all A E L(U, V) and B
E
L(V, W) and
vector spaces U, V and W. The map that assigns to each vector space its dual and to each map its dual map (transpose) is a contravariant Coo functor F. One may define the notion of a Coo functor of several variables which may be covariant in some variables and contravariant in others. For example, consider the functor of two variables that assigns to each pair (V, W) the space V Q9 W* and to each pair (A, B) E L(Vb V2) X L(WI' W2) the map AQ9B* : VI Q9Wi -t V2Q9Wi. Theorem 6.53. Let F be a Coo functor of m variables on Lin(lF) and let EI, ... ,Em be IF -vector bundles with respective typical fibers V I, ••. , Vm' Then the set
E:= F(E1, ... ,Em):= UF(EIlp, ... , Emlp} p
together with the map 1r : E -t M which takes elements of F(Ellp I • • • , Em Ip) to p is naturally a vector bundle with typical fiber F(V1, .•. , Vm).
Proof. We will only prove the case of m = 2 with covariant first variable and contravariant second variable. This should make it clear how the general case would go while keeping the notational complexity under control.
6.5. Hom
285
Given vector bundles 7r1 : El -+ M and 7r2 : E2 -+ M! the total space of the constructed bundle is UpF( Ellp , E21p) with the obvious projection which we call 7r. Let (4)a! Ua ) be a VB-atlas for El and (1/Ja, Ua) a VBatlas for E2 (we have arranged that both atlases use the same cover by going to a common refinement as usual). For each p, let Ep denote the fiber F(E1I p ' E2Ip). Fix a and for each p E Ua define 8 a lp E L(Ep,F(Vl, V2)) by 8 alp := F( alp, Wal;l), where 4>a = (7rl' a) and 1/Ja = (7r2' wa). Then define 8 a : 7r- 1 (Ua) -+ F(Vl' V2) by 8 a (€) = Sal p (10) whenever 10 E F(Ell p ' E2Ip). Next define Oa = (7r, Sa) : 7r- 1Ua -+ Ua
X
F(Vl' V2).
The family {(Oa, Ua )} is to be a VB-atlas for E. We check the transition maps: 8 al'(P) = Sal p 0 8~11p = F(alp! Wal;l) of(I'l p ' '111'1;1)-1
= F(alp, Wal;l) of(I'lp 1, wpl p ) = F(alp
0
I'1;1! wpl p
0
Wal;l)
= F(ap(P) , wPa(P))'
(Remember that the functor is contravariant in the second variable.) Now we can see from the properties of al" '11 I'a and the definition of Coo functor that F(al'(P) , wI'a(P)) E GL(F(Vl, V2)) and the maps 8 al' : Ua n Up -+ GL(F(Vl, V2)) are smooth. 0
6.5. Hom Let 6 := (E1' 7rt, M, V) and 6 := (E2' 7r2, M, V) be smooth IF-vector bundles. The bundle whose fiber over P E M is L(E1p, E2p ) = Hom(Elp, E2p) is denoted by Hom(6,6) or less precisely, by referring to the total space Hom(E1, E2)' Here Hom(E1p, E2p) denotes IF-linear maps. If I : E1 -+ E2 is vector bundle homomorphism over M, then we may obtain a section s of Hom(El, E2) by defining s : P H IIEl,,' Conversely, given s E r (Hom(El, E 2)) we define f : El -+ E2 by requiring that fiE I" = s(p). Thus every element of Hom(Et, E 2) can be identified with a vector bundle homomorphism over M. Exercise 6.54. Let E1 -+ Ml and E2 -+ M2 be smooth vector bundles. Show that the set of vector bundle homomorphisms along a smooth
286
6. Fiber Bundles
map 9 : Ml -t M2 is in natural bijection with the sections of the bundle Hom(El' g* E2). Since r (El) and r (E2) are COO(M, IF) modules, we can look at the COO(M, IF) module Hom(r (El) ,r (E2))' Then we have Proposition 6.55. Let EI -t M and E2 -t M be smooth IF -vector bundles. Then r (Hom(E!, E2)) and Hom(r (EI) , r (E2)) are naturally isomorphic as COO(M,IF) modules. Proof. To each section s E r (Hom(EI, E2)) we assign a map ¢s ; r (E1 ) r (E2) defined by the formula
¢s (0") (p) = s(P)O"(P) for 0"
E
-t
r (EI) .
Then we obtain a map o:(u)y) in the domain of every bundle chart (uo:, 0:/3} associated to a VB-atlas, then the principal bundle obtained by the above construction is (equivalent to) the linear frame bundle F(E). Letting GL(V) act on V according to the standard action we have F(E) XGL(V) V, which is equivalent to the original bundle (E, 71", M, V). More generally, if
302
6. Fiber Bundles
A : G ~ GL(V) is a Lie group representation, then by treating A as a linear action we can form P x>. V. Proposition 6.89. Let P be a principal G-bundle and let A : G ~ GL{V) be a representation. Then P x>. V has a natural vector bundle structure with typical fiber V. Proof. This follows from Theorem 6.87, but we can argue more directly. Let us denote the total space of P x>. V by B and let Bp be the fiber over some point p E M. Then for each u E Pp there is a map 1/Ju := [u,·J : V ~ Bp given by v t-+ [u, vJ. We compare 1/Ju with 1/Jug for 9 E G and u E Pp • Since lug, vJ = [u, A (g) vJ for all v E V, the following diagram commutes:
V
>.(g)
¢~ Bp h
'V
From this it follows that 'l/Ju transfers the linear structure of V to Bp independently of the choice of u E Pp • We leave it to the reader to show that the local trivializations of P x>. V constructed as in the proof of Theorem 6.87 are linear on each fiber. 0 Example 6.90. Let M be an n-manifold and let F(M) be the frame bundle of M. Then, if AO is the standard action of GL(n, IR) on IRn, we have the following vector bundle isomorphisms:
F(M) x>'O IRn ~ TM, F(M) x>'o IR n ~ T* M, F(M) x>'o®>'o IRn ~ TM ® T* M. If E = P x>. V is an associated vector bundle for A a representation, then we can map P into the frame bundle of E. Indeed, the map is just 1/J : u t-+ 'I/J(u) = 'l/Ju, where 1/Ju := [u,·J as above. Furthermore, 1/J(ug) = 1/J(u) 0 A (g) and so we have a principal bundle morphism with respect to the homomorphism A:
P
¢.
F(B)
~/ M The map 'I/J : P ~ F(E) is only injective if the action A is effective. Based on what we have seen above we can say that the theory of principal bundles and associated bundles is an alternative and "invariant" approach to G-bundles. By "invariant" we mean that the foundations can be laid out
303
Problems
without recourse to strict equivalence classes of G-atlases or the use of cocycles (of course, these notions can be brought in as convenient). According to this approach, the central notion is the principal bundle, and one recovers the other G-bundles of interest as associated bundles. Developing the theory in this way has the advantage that much can be accomplished without the direct need of bundle atlases. It is a more "intrinsic" approach. This approach seems to have originated with Ehresmann and is the approach followed by [H us].
Problems (1) Show that 8 n x JR and 8 n x 8 1 are parallelizable.
(2) Let X := [0,1] x JRn. Fix a linear isomorphism L : JRn -+ JRn and consider the quotient space E - XI"', where the equivalence relation is given by (0, v) rv (1, Lv). Show that E is the total space of a smooth vector bundle over the circle 8 1 . (3) Exhibit the vector bundle charts for the pull-back bundle construction of Definition 6.18. (4) Let ~ = (E, 7[, M, F) be a G-bundle. Let 9afj be cocycles associated to a G-altas {(Ua ,4>a)} for~. Show that ~ is G-equivalent to a product bundle if and only if there exist functions Aa : Ua -+ G such that 9fja(X) = A,s(x)A;;1(x)
and all
O!,
for all x E Ua n Ufj
(3.
(5) Show that the twisted torus of Example 6.17 is trivial as a fiber bundle but not trivial as a Z2-bundle. (Use Problem 4.) (6) Show that the space of sections of a vector bundle over a compact base is a finitely generated module. Show that if the bundle is trivial, then the space of sections is a finitely generated free module. (7) Let El -+ M1 and E2 -+ M2 be smooth vector bundles. Show that if F : E1 -+ E2 is a vector bundle homomorphism along a map f : M1 -+ M2 such that F is an isomorphism of fibers, then E1 -+ M1 is isomorphic to the pull-back bundle f* E2 -+ M1. ~ = (E, 1f, M) is a vector bundle with a positive definite Show that the metric induces a vector bundle isomorphism
(8) Suppose that metric. E~E*
(9) Show that the tangent bundle of the real projective plane is a vector bundle isomorphic to Hom (JL (JRpn) , JL(JRpn).L) , where JL(JRpn) -+ JRpn
6. Fiber Bundles
304
is the tautological line bundle and 1L(lRpn).1 -+ JRpn is the rank n vector bundle whose fiber at l E]R.Pn is {(l,v) E JRpn x JRn+1 : v.il}. (10) Recall Example 6.35. Show that the Whitney sum bundle El EEl E2 is naturally isomorphic to the pull-back l::. *1rE1 XE2 : l::. * (EI x E2) -+ M. (11) (a) Let 7r : E -+ M be an IF-vector bundle. We wish to show that T1r : T E -+ T M is naturally a vector bundle. Consider the maps a : E EEl E -+ E and Its : E -+ E for each sElF given by
a(vp, wp) := vp + wp for vp, wp Its(e p) := se p for ep E Ep
E
Ep
Show that we may identify T (E EEl E) with the submanifold of T Ex TE given by
{(v,w)
E
TE x TE: T7r· v = T1r· w}
Now suppose that for v, wET E with T7r . v = T7r . w we define v EB w := Ta . (v, w) and for sElF and vETE we define 8 • V : Tits· v. Show that with these definitions of addition and scalar multiplication, T1r: TE -+ TM is indeed an IF-vector bundle. (b) Let E be as above but assume for simplicity that IF = JR. Let xl, ... ,xn be coordinates on U eM. Suppose that el, ... , ek is a frame field over U. Let ~n be defined on Elu by y = L: ~i(y)ei(7r(Y)) for any y E E. Then, identifying xi with xi 0 1r, the functions Xl, ... , x n , ,{n are a coordinate system for E defined on Elu and such that the a~. are in the kernel of T7r. Now if v, wET E are such that T1r . v = T1r . w, then we may express v and w as
e, ... , e, ...
and w-
La a~il_ + LbO a~ol_· i
y
~
Y
0
Here y and fi are the base points of v and w, and the fact that the a's are the same for both v and w is a result of the condition T1r . v = T1r . w. Show that
v EB w =
L a a~i I i
~
y+y
+L a
(b
O
+ ba) a~a I ~
y+y
'
where in v EB w, the EB refers to the addition described in part (a). (12) Exhibit a VB-atlas for the tautological bundle of Example 6.47. (13) Show that the tautological bundle over JRpl is a Mobius band.
Problems
305
(14) Let P -+ M be a principal bundle with group G. If H is a Lie subgroup of G, then the quotient P/H is an H-principal bundle. Show that P/H-+ M admits a global section if and only if the structure group of P -+ M is reducible to H. (15) Show that the notions of smooth fiber bundle and vector bundle make sense when the base space is allowed to be a manifold with boundary. What issues arise if one considers allowing both the base space and typical fiber to have boundary?
Chapter 7
Tensors
In this chapter we shall employ the Einstein summation convention. For example, r;kaivk is taken to be shorthand for
where the range of summation is understood from the context. Normally, the repeated indices that are summed over occur once as a subscript and once as a superscript. For example, if A = (a)) is an n x m matrix and B (b;) is an m x k matrix, where in this case we use upper indices to indicate rows and lower indices for columns, then C AB corresponds to
This is reduced by the summation convention to cj = aib~. We will occasionally include the summation symbol l: for emphasis, or to meet the demands of clarity. Tensor fields (often referred to simply as tensors) can be introduced in a rough and ready way by describing their local expressions in charts and then going on to explain how such expressions are transformed under a change of coordinates. With this approach one can gain proficiency with tensor calculations in short order, and this is usually the way physicists and engineers are introduced to tensors. However, since this approach hides much of the underlying algebraic and geometric structure, we will not pursue it here. Instead, we present tensors in terms of multilinear maps.
307
308
7. Tensors
7.1. Some Multilinear Algebra It will be convenient to define the notion of an algebraic tensor on a vector space or module. The reader who has looked over the material in Appendix D will find this chapter easier to understand. In particular, we assume the definition of "multilinear" (Definition D.13). In this chapter, if we say that a module is finite-dimensional, l we mean that it is free and finitely generated and thus has a basis. All modules in this chapter are assumed to be over a commutative ring with unity.
Definition 7.1. Let V and W be modules over a commutative ring R with unity. Then, an algebraic W-valued tensor on V is a multilinear mapping of the form r:V1xV2X"'XVm~W,
where each factor Vi is either V or V*. If the number of V* factors occurring is r and the number of V factors is 8, then we say that the tensor is rcontravariant and 8-covariant. We also say that the tensor is of total type (:). The most common situation is where W is the ring R itself, in which case we often drop the adjective "R-valued". Notice that if r : V* x V x V* x V -+ R is a tensor, then we can define a tensor V* x V x V x V* ~ R by
r:
r(a1' VI, V2, (2) := r(a1' VI, a2, V2)' Although these two tensors clearly contain the same information, they are nevertheless different. We indicate this with a more specific notation. We say that r is a tensor of type (11 11)' while r is of type (1 2 1). More generally, a tensor might, for example, be specified to be of type
(rl 81 r2 82
r a 8b) or (81 r1 82
Sb r a )
The general pattern should be clear. If r = rl + ... + r a and s = 81 + ... +Sb, then the tensor would be of total type (:), which we also write as (r,s). The set of all tensors of fixed type (as above) is easily seen to be an R-module with the scalar multiplication and addition defined as is usual for spaces of functions. As another example, a multilinear map l':VxV*xVxV*xV*~W
is a W -tensor which is of type (1 1 1 2) and total type (~). The set of all Wvalued tensors on V oftype (1 11 2) is denoted TIll 2 (V; W), and we have analogous notations for other types. In many, if not most, circumstances we agree to associate to each tensor of total type (:), a unique element of T"s(V; W) by simply keeping the relative order among the V variables and 1 For
modules, what we mean by dimension is what is usually called the rank.
7.1. Some Multilinear Algebra
309
among the V* variables separately, but shifting all V variables to the right of the V* variables. Following this procedure, we have, for example, the map TIl12 (V; W) -+T32 (V; W). Maps like this will be called consolidation maps or consolidation isomorphisms. Definition 7.2. A tensor
r : V* x V* x ... x V* x V x V x ... x V -+ W ,
v
T
times
'\
Vi
I'
s times
where all the V factors occur last, is said to be in consolidated form. The set of all such (consolidated) W-valued tensors on V will be denoted rs(V; W). As a special case we have rol (V; R) = V*. We will often abbreviate T"s(V; R) to TTS(V). For example, elements of T32 (V; W) are in consolidated form, while tensors from T I I I 2 (V; W) are unconsolidated. Remark 7.3. Some authors consolidate by putting all V arguments first. Also, sometimes it is appropriate to forgo the consolidation especially in connection with the "type changing" operations introduced later. Our policy will be to work with tensors in consolidated form whenever convenient. Example 7.4. One always has the special tensor 8 E T\ (V; R) defined by
8(a, v) = a(v) for a E V* and v E V. This tensor is sometimes referred to as the Kronecker delta tensor.
v:
There is a natural map from V to V** given by v t---+ V, where a: t---+ a(v). If this map is an isomorphism, we say that V is a reflexive module and we identify V with V**. Finite-dimensional vector spaces are reflexive. Exercise 7.5. Show that the COO(M) module of sections of a vector bundle E -+ M is a reflexive module. (It is important here that we are only considering vector bundles with finite-dimensional fibers.) We now consider the relationship between tensors as defined above and the abstract tensor product spaces described in Appendix D. We restrict our discussion to tensors in consolidated form since the implications for the general situation will be obvious. We specialize to the case of R-valued tensors where R is the ring. Recall that the k-th tensor power of an Rmodule V is denoted by ®k V := V ® ... ® V. We always have a module homomorphism
(7.1)
7. Tensors
310
whereby an element UI ® ... ® Ur ® f31 ... ® f3s E (®rV) ® (®8V*) corresponds to the multilinear map given by (a1, ... , ar , VI, •.• ,vs) t-+ a1 (U1) ... ar (u r ) f3I (VI) ... f38 (v s ) . We will identify UI ® ... ® Ur ® f3I ® ... ® f3B with this multilinear map. In particular, this entails identifying V E V with the element E V** where a t-+ a(v). If V is a finite-dimensional vector space, then the map (7.1) is an isomorphism. In fact, it is also true that if V is the space of sections of some vector bundle over M (with finite-dimensional fibers), then V is a COO(M)-module and the map (7.1) is still an isomorphism. A tensor which can be written in the form UI ® ... ® f3s is called a simple or decomposable tensor. Note well that not all tensors are simple.
v
v:
Remark 7.6. The reader should take careful notice of how we treat the orders of the factors: An element of V ® V* ® V* corresponds to a multilinear map V* x V x V ---+ R and not to a map V x V* x V* ---+ R. Since the map (7.1) is not always an isomorphism for general modules, and since no analogous isomorphism exists in the case of tangent spaces to infinite-dimensional manifolds such as those discussed in [Ll], it becomes important to ask to what extent the map (7.1) is needed in differential geometry. Serge Lang has written a very fine differential geometry book for manifolds modeled on Banach spaces [Ll] without the help of such an isomorphism. In any case, we still can and will consider UI ® ... ® U r ® {31 ... ® f38 to be an element of ~(V) as described above. Another thing to notice is that if (7.1) is an isomorphism for all r and s, then in particular V ~ V**, that is, V must be reflexive. Corollary D.33 of Appendix D states that for a finitely generated free module, being reflexive is enough to insure that (7.1) is an isomorphism for all rand s. In the latter case, the consolidation maps introduced earlier can be described in terms of simple tensors. For example, the consolidation map Tl2 2
(V) ---+
T23
(V)
is given on simple tensors by
a ® v ® w ® f3 ® "f ---+ v ® w ® a ® f3 ® "f. Now let us consider the spaces V ® V* and V ® V* ® V*. By a straightforward argument using the universal property of tensor product spaces, one can construct a bilinear map (V ® V*) x (V ® V* ® V*) ---+ V ® V* ® V ® V* ® V* such that (v ® a, w ® f31 ® (32) is mapped to v ® a ® w ® corresponds to a product map ® : TIl (V) x TI 2 (V) ---+ TIl I 2 (V)
f31
® f32. This
7.1. Some Multilinear Algebra
311
such that (8, T) -+ 80 T, where
The general pattern should be clear, but writing down the general case is notationally onerous. This product is the (unconsolidated) tensor product of tensors. Note carefully the order of the factors. To simplify the notation, the tensor product is often defined in a slightly different way when dealing with tensors which are in consolidated form: Definition 7.7. For tensors 8 E Tr~l (V) and T E Tr~2(V)' we define the (consolidated) tensor product 8 ® T E T r 1+r 2 81+ 8 2 (V) by
Whether or not a tensor product is the consolidated version will normally be clear from the context, and so we will drop the word "consolidated". We can also extend to products of several tensors at a time. While it is easy to see that the tensor product defined above is associative, it is not commutative since the order of the slots is an issue. Let T*(V) denote the direct sum of all spaces of the form Tro(V), where we take (V) := R. The tensor product gives T* (V) the structure of an algebra over R as long as we make the definition that r 0 A := r A for r E R.
roo
Proposition 7.8. Let V be a free R-module with basis (el,"" en) and corresponding dual basis (e 1 , ... , en) for V*. Then the indexed set
is a basis for T r8 (V). If 7 E T r8 (V), then
Proof. If 7 E T r8 (V; R) and we define 7 i1 "·}1 ... j. = 7(e~1, ... , e~r, eJ1 , • •• , ej.), then it is easy to check that
and so, in particular, our indexed set spans Trs(V; R). Indeed, if we denote the right hand side of the above equation by 7', we obtain (using the
7. Tensors
312
summation convention throughout) T'( ekl , ... , e kr , eli, ... , el. )
= Ti1 ...ir)1 ...)_. ei1 (e k1 ) ... eir (e kr )e31(el 1 ) ... ej• (ez • ) = Ti1 ... ir .
)1 ... ).
[/1 ... 8kr831 ... 8j8 ~1 ~r II z.
_
Tk1 ... kr
1I ... l.
= T (ekl , ... , e kr , elI , ... , eI. ) . Thus T' and T agree on basis elements, and by multilinearity T' = T. For independence, suppose Til "·~;I ... j. e'l 0 ... 0 e~r 0 ej1 0 ... 0 e)' - 0 for some n T +S elements T i1 ... ir )1 ... j. of R. This is an equality of multilinear maps, and if we apply both sides to (e k1 , ... , ekr , elI, ... , ez.), then we obtain T k1 .. .l_ - O. Since our choices were arbitrary, we see that all n T +S elements T'I ... ~r.)1 ...)_. are equal to O. 0
t.
As a special case we see that if A E T\(VjR), then A = A~) e, ®eJ , where Ai j = A(ei , e)). This theorem is a special case of Theorem D.29 of Appendix D, which we will also invoke below for spaces like W 0 V* . If we are dealing with tensors that are not in consolidated form, it should still be clear how to obtain a basis. For example, {e~ 0 d 0 ek} is a basis for TIll (Vj R), and a typical element A would have an expansion i k
A= Aj
ei
.
0 e3 0 ek·
Notice the purposeful staggered positioning of the indices in Ai) k. Definition 7.9. The elements A i l ...i r jl ... j. from the previous proposition are called the components of T with respect to the basis ell ... , en. Example 7.10. If V = TpM for some smooth manifold M, then we can use any basis of TpM we please. That said, we realize that if p is in a coordinate Ip form a basis for TpM, and chart (U, x), then the vectors ~ Ip ,... , we may form a basis for TTS (TpM) consisting of all tensors of the form
-fxn
--!-I OX'1
p
--!-I
0 .. · 0 ox'r
For example, an element Ap of form as
TIl
Ap = A' j An element Ap of
Ap
0 dxiI p
Ip 0 .. · 0
dxi-I p .
(TpM) can be expressed in coordinate
O~i Ip 0
dx-Jlp .
rrs (TpM) can be written as
--!-I
= A'I ...i r .)1 .. ·).. ox'!
P
--!-I
0 .. · 0 ox'r
P
0 dxi11 P 0 .. · 0 dx-J s ' p'
and this is called the coordinate expression for Ap.
313
7.1. Some Multilinear Algebra
The components of a tensor depend on the basis chosen, and a different choice will give new components related to the first by a transformation law. This is the content of the following exercise: Exercise 7.11. Let eI, ... , en be a basis for V and let e l , ... , en be the corresponding dual basis for V*. IT el, ... ,en is another basis for V with
- = ei
Ckiek,
then the dual basis e1 , ... , en is related to e1 , ... , en by ei = (C- I ) ik e k , where C = (Cj). Show that if ri jk are the components of r with respect to the first basis (and its dual) and if ri jk are the components with respect to the second basis, then
r-i.Jk
_ -
(C-l)ia r a be Cbcc j k
(sum over a, b, c).
This is a transformation law. What is the analogous statement for r E
T"s(V; R)? Example 7.12. It is easy to show that for any basis (with corresponding dual basis) as above, the Kronecker delta tensor 8 has components 8;, where 8~ = 0 if i =1= j and 8\ = 1. It is easy to show that if 8 E T12 (V) and T E T22 (V), then 8 ® T has components given by
(8 ® T)abe defg = Sa de Tbe fg • More generally, if S E TTil (V) and T E rr~2 (V), then
(7.2)
(S ® T)a1 ... ar10 has the form l/> = (-lr, cI», where cI> : 1f- l U -+ IRk and where cI>p := cI>IEp : Ep -+ JRk is a linear isomorphism for each p. We obtain a map cI>~,s : Trs(Ep) -+ TTs(1Rk) by
(cI>;,STp)(al, ... , a r , VI,···, VB) :=
Tp((cI>p)* al, ... , (cI>p)* a r , cI>;lVl,"" cI>;lv s ).
These maps combine to give a map cI>r,s : 1f- l U -+ TTs(JR k ) which is smooth (exercise). Our chart for Tr8(~) is
l/>r,s := (71', cI>r,s) : 1f lU -+ U
X
Trs(lRk).
If desired, one can choose, once and for all, an isomorphism TTs(JR k ) ~ IRk •
A VB-atlas {(Ua , l/>a)} for ~
= (E, 71', M) gives a VB-atlas {(Ua , l/>~/)} for
Tr8(~)'
Exercise 7.21. Show that there is a natural vector bundle isomorphism
We leave it to the interested reader to prove the following useful theorem. Proposition 7.22. Let ~ = (E, 71', M) be a vector bundle as above and let T : M -+ TTs(E) be a map which assigns to each p E M an element of Trs (Ep). Then T is smooth if and only if
p r-+ T(p )(0:1 (P), ... ,00r(P), Xl (p), ... ,Xs(p)) is smooth for all smooth sections p r-+ 0:~(P) and p r-+ X~(P) of E* -+ M and E -+ M respectively. The same statement is true if we use local sections.
7.2. Bottom-Up Approach to Tensor Fields
319
The set of smooth sections of Trs(~)) is denoted r(Trs(~))' If Y E then for XI, ... , Xs any smooth sections of E -+ M and aI, ... , a r smooth sections of E* -+ M, define Y(al,"" ar, Xl, ... , Xs) E COO(M) by r(Trs(~)),
Y(aI, ... , ar, Xl, ... , Xs)(p) :- Yp(a1(p)"", ar(p), X1(P), . .. , Xs(P)). Now we have a map Y : (r E*)k x (r E)l -+ COO(M). This map is clearly multilinear over Coo (M), and we see that we can interpret elements of r(T~ (~)) as such maps when convenient. This extends the idea of thinking of a I-form a as a COO(M) linear map X(M) -+ COO(M). Like most linear algebraic structures existing at the level of a single fiber Ep, the notion of tensor product is easily extended to the level of sections: For T E r(Trs\ (~)) and'f/ E r(Tr~2(~))' we define the (consolidated) tensor product T ® 'f/ E r(r S~~~2(~)) by (T ® 'f/) (P) := Tp ® 'f/p. Thus
(T ® 'f/) (P) (al, ... , a rl +r2 , VI, ... , VSl +S2) _ ( 1 -T a , ... ,arl ,V1,,,,,VSl ) 'f/ ( a r1 +1 , ... ,arl +r2 'VS1+1. .. ·'VS1+S2 ) for all a i E E; and V~ E Ep. Let (Sl, ... , Sk) be a local frame field for ~ over an open set U and let 0'1, .•• , O'k be the dual frame field of the dual bundle E* -+ M so that O'~(sJ) b} Consider the set {cr~1 ® ... ® cr~'" ®
SJ1 ® ... ® sJs : il, ... , iT> iI, ... ,is = 1, ... , k}.
If T E r(Trs(~))' then we have functions T~1 ... ~r jl ... js E COO(U) defined by T~1 ... ~r j1 ... j8 = T(cr i1 , ... , cr ir , Sjll ... , Sjs)' It follows from Proposition 7.8 that T (restricted to U) has the expansion T = Th ... ir J1. .. ·J.. cril ® ... ® crir ® s·J1 ® ... ® s·J. . Also, applying equation (7.2) in each fiber Ep, we see that the component functions for T ® 'f/ are given by (T ® t1 ...ir1+r2. . 'f/ Jl ...J-1 +S2 = Ti1 ...ir1. . 'I'1ir1+1 .. .ir2. 31 .. ·J8 1 ./ J81 +1 ...J8 2 • Here, and wherever convenient, we use the consolidated tensor product. Notation 7.23. Whenever there is no chance of confusion, we will refer to Trs(~) by Trs(E) -+ M or even just Trs(E) (the latter is the notation for the total space of the bundle). In the case of the tangent bundle TM, we have special terminology and notation:
Definition 7.24. The bundle rs(TM) -+ M is called the (r, s)-tensor bundle on M.
320
7. Tensors
By Exercise 7.21, Trs(TM) ~ ((g/TM) ® (®sT*M) and this natural isomorphism is taken as an identification so the latter bundle is also referred to as a tensor bundle. We now restrict ourselves to the case of the tangent bundle of a manifold but note that much of what follows makes sense for general vector bundles. Definition 7.25. The space of sections r(Trs(TM)) is denoted by T;(M) and its elements are referred to as r-contravariant s-covariant tensor fields or just type (r, s)-tensor fields. The space 7O(M) is denoted by P(M) and T;(M) by Ts(M). In summary, a smooth tensor field A is a smooth assignment of a multilinear map on each tangent space of the manifold. Thus for each p, A(P) is a multilinear map
A(p) : (T;Mt x (TpMt -+ JR, or in other words, an element of Trs(TpM). Elements of Trs(TpM) are called tensors at p. We also write Ap for A(p). Example 7.26. In Definition 6.42, we introduced the notion of a Riemannian metric on a real vector bundle. We saw that such metrics always exist. The most important case is where the bundle is the tangent bundle T M of a manifold M. In this case, we say that we have a Riemannian metric on M. Thus a Riemannian metric on M is an element of 72(M) which is symmetric and positive definite at each point. Of course the manifold in question could be an open sub manifold U of M so we have COO(U)-module (r, s)-tensor fields over that set denoted T;(U). The open subsets are partially ordered by inclusion V c U and the tensor fields on these are related by restriction. Let r~ : T;(U) -+ 7;(V) denote the restriction map. The assignment U -+ T;(U) is an example of a presheaf and in fact a sheaf. We will also sometimes deal with tensors with values in T M (or in T* M). First note that the space Trs(TpM; TpM) of all multilinear maps (T;Mr x (TpMt -+ TpM is a vector space. The set Trs(TM;TM) := UpTrs(TpM; TpM) can be given a smooth vector bundle structure in a way that is closely analogous to Trs(TM) -+ M. Definition 7.27. The space of sections r(Trs(TM; TM)) is denoted by T; (M; T M) and its elements are referred to as r-contravariant s-covariant TM-valued tensor fields. Similarly, we may define T* M-valued tensor fields. Note that T M -valued tensor fields can be associated in a natural manner with ordinary tensor fields. For example, if A E T02 (T M; T M), then using
7.2. Bot tom- Up Approach to Tensor Fields
321
the same letter A by abuse of notation, we may define an element A E T~(TM) by
Ap(Op, vp, wp) = Op(Ap(vp,wp)) for Op E r;M and vp,wp E rpM, Many such reinterpretations are possible. For this reason, we shall stick to studying ordinary tensors and tensor fields in what follows. We shall define several operations on spaces of tensor fields. We would like each of these to be natural with respect to restriction. We already have one such operation: the tensor product. If A E T;f(U) and B E T;i(U) and V c U, then r~ (A 0 B) = r~A 0 r~B. A (k, l)-tensor field A may generally be expressed in a chart (U, x) as
..
A=A~1 ...tr
a
a·
®"'®-. ®dx3 1
"'0dx)', ax'" where A i1 .. .i,. )1. .. ·). are functions, 88, E X(U) and dx j E X*(U). Actually, it x is the restriction of T to U that can be written in this way, but because of (7.4)
.. -.
)1 .. ·). ax~1
the naturality of all the operations we introduce, it is generally safe to use the same letter to denote a tensor field and its restriction to an open set such as a chart domain. It is easy to show that
. .
A ~1""r
-
).1 .. ·)'• -
(
A d
X
l1
'''.,
d
x
. a
'r
a)
'-a' Xl1 '''''-a xJ.. ,
and so the components of a smooth tensor field are Coo for every choice of coordinates (U, x) by Proposition 7.22. Conversely, one can obviously define tensors that are not necessarily smooth sections of the appropriate tensor bundle, and then a tensor will be smooth exactly when its components with respect to every chart in an atlas are smooth. Evaluating the expression (7.4) above at a point p E U results in an expression such as that given at the end of Example 7.10. Exercise 7.28 (Transformation laws). Suppose that we have two charts (U,x) and (V, i). If A E Ti(M) has components A;k in the first chart and
A;k in the second chart, then on the overlap Un V we have G axi ax b axe Ak=A -' bc ) axG ax j axk
--.i
where
a axb a . axi = -a. a b' dx' = a x G dx Gand a-' x' x' X
This last exercise reveals the transformation law for tensor fields in r~(M), and there is obviously an analogous law for tensor fields from T's(M)
for any values of rand s. In some presentations, tensor fields are defined in terms of such transformations laws (see [L-R] for this approach). It should be emphasized again that there are two slightly different ways of reading
322
7. Tensors
local expressions like the above. We may think of all of these functions as living on the manifold in the domain U n V. In this interpretation, we read the above as QX a
A;k(P) -
QX b
QX i
A~b(P) QX j (p) QXk (P) QXl (p) for each p E Un V.
This is the default modern viewpoint. Alternatively, we could take g:m to be functions on x(U n V) and write (Xl, ... , xn). Then, ~ would refer to ~ 0 x 0 x-l(xI, ... ,xn ) so that both sides of the equation are functions of variables which we abusively write as (xl, ... , xn). The first version seems theoretically pleasing, but for specific calculations that use familiar coordinates such as polar coordinates, the second version is often convenient. For example, suppose that a tensor 7 has components with respect to rectangular coordinates on JR2 given by AJk and we wish to find the components in polar coordinates. For indexing purposes, we take (x, y) = (u l ,u2 ) and (r,O) = (v I ,v2 ). Then we have
&::..
_
(7.5)
QUa QU b
Ajk = Aab QVJ QV k '
which can be read so that both sides are functions of (vI, V2) by writing u l and u 2 as a function of (vI, v 2 ), etc. Of course, the charts are there to "identify" open sets in Euclidean space with open sets on the manifold so these viewpoints are really somehow the same after all. Using (x, y) and (r, 0), the transformation (7.5) is given in matrix form as [COSO -rsinO] [All A12] [COSO sinO] [ All A12] AI2 A22 sinO rcosO AI2 A22 -rsinO rcosO . We now introduce the pull-back of a covariant tensor field, which will play a big role in the next chapter. Definition 7.29. If f : M --+ N is a smooth map and define the pull-back 1*7 E 7s(M) by
j*7 (Vb
... , vs)(p)
- 7(Tf· VI,
... , Tf·
7 E
7s(N), then we
vs)
for all Vb ... , Vs E TpM and any p E M. Notice the connection of this with the pull-back defined earlier in a purely algebraic context. It is not hard to see that j* : 7s(N) --+ Ts(M) is linear over JR, and for any h E COO(N) and 7 E Is(N) we have j* (h7 = (h 0 J) j*7. If f : M --+ Nand 9 : N --+ P are smooth maps, then of course (g 0 J)* - 1* 0 g*. Let us discover the local expression for pull-back. Choose a chart (UI xl on M and a chart (V, y) on N and assume that f(U) C V. Let us denote
7.3. Top-Down Approach to Tensor Fields
8(Y'of)
ax3
8 '
.
..
323
k 8 I E 7Jfj(p)lfil< f(P)
8 18
by ~ for sImplIcIty. We have Tpf· ax' p =
and
(f*1')~l ...~. (P) = (f*1') (a:tl Ip ,... , a:i.l) =
l'
I
(Tf a:t1 p '''' ,Tf a:ts
I)
aykl a 1 ayks a I ) =1' ( - a i (p) - a k , ... , - a i (P) - a k Y 1 f(P) X • Y S f(p) x 1 a I a I ) aykl ayks =1' ( - k '''''-k (p) ... -(P) ay 1 f(p) ay s f(P) axtl ax tB aykl ayks = 1'kl ... k. (f (p)) axil (p) ... axis (P).
Thus we have
This looks similar to a transformation law for a tensor, but here f is not a change of coordinates and need not even be a diffeomorphism. Pull-back respects tensor products: Exercise 7.30. Let f : M -+ N be as above. Show that for 1'1 E and 1'2 E Ts 2 (N) we have /* (1'1 ® 1'2) = /*1'1 /*1'2.
Ts 1(N)
In the case that f : M -+ N is a diffeomorphism, the notion of pull-back can be extended to contravariant tensors and tensors of mixed covariance. For such a diffeomorphism, let (T f I) * : T; M -+ T; N denote the dual of the map Tf- 1 : TpN -+ TpM. Definition 7.31. If f : M -+ N is a diffeomorphism and l' is an (r, s)-tensor field on N, then define the pull-back /*1' E T r s (M) by f*1'(a1, ... ,ar ,v1, ... ,Vs )(p) :=
1'((Tf-1)* ar, ... , (Tf 1)* a r , Tf· VI, ... , Tf·
Vs)
for all VI, ... , Vs E TpM and aI, ... , a r E T; M and any p E M. The pushforward is then defined for l' E T r s (M) as f*1' := (f 1)*1'.
7.3. Top-Down Approach to Tensor Fields Specializing what we learned from the discussion following Proposition 7.22 to the case of the tangent bundle, we see that a tensor field gives us a COO (M)-multilinear map based on the module X(M). This observation leads
7. Tensors
324
to an alternative definition of a tensor field over M. In this "top-down" view, we simply define an (r, s)-tensor field to be a COO(M)-multilinear map
X*(Mt x X(M)B -+ COO(M). In this view, a tensor field is an element of 'P s (X(M)). For example, a global covariant 2-tensor field on a manifold M is a map r : X(M) xX(M)-t Coo (M) such that
r(hX1 + h X 2, Y) = h r(XI, Y) + hr(X2' Y), r(Y, hXI + hX2) = h r(Y, Xl) + hr(Y, X2) for all h, h E COO(M) and all Xl, X 2 , Y E X(M). As we shall see, it turns out that such COO(M)-multilinear maps determine tensor fields in the sense of the previous section. If we take a top-down approach to tensor fields, then we must work to recover the presheafjsheaf aspects. Indeed, it is not obvious what is the relation between T's(X(M)) and T's(X(U)) for some proper open subset U eM. Indeed, thinking purely in terms of modules makes the issue clear. The module X(M) is not the same module as X(U) unless U = M. A priori, there is no immediate reason to think that a multilinear map with arguments from the module X(M) should be able to take elements of X(U) as arguments! For instance, from the top-down viewpoint, how can we insert coordinate fields a~. and dxi into an element of Trs(X(M)) to get coordinate expressions if the chart domain is not all of M? We address this in the next section indirectly by showing how the top-down approach gives back tensors as sections (the bottom-up approach). Another comment is that both X(U) and 'Ps(X(U)) are finite-dimensional free modules over the ring COO(U) whenever U is a chart domain or, more generally, the domain of a frame field. The reason is that a local frame field and its dual frame field provide a module basis for X(U) and r(U) and the latter really is the dual of the first in the module sense. On the other hand, the COO(M)-modules X(M) and T's(X(M)) are not generally free unless M is parallelizable.
7.4. Matching the Two Approaches to Tensor Fields If we define a tensor field as we first did, that is, as a field of tensors in tangent spaces, then we immediately obtain a tensor as defined in the topdown approach. On the other hand, if r is initially defined as a COO(M)multilinear map, then how should we recover a field of tensors on the tangent spaces?2 Answering this is our next goal. 2This is exactly where things might not go so well if the manifold is not finite-dimensional. What we need is the existence of smooth cut-off functions. Some Banach manifolds support cut-off functions but not all do.
325
7.4. Matcbing tbe Two Approacbes to Tensor Fields
Trs(.X-(M)). Let (h, ... , Or and 1 :::; i :::; r; also let Xl"'" Xs and X!, ... , XS be smooth vector fields such that Xt(p) = Xt(P) for 1 :::; i :::; s. Then we have that Proposition 7.32. Let p E M and
l'
E
(h, ... ,Or be smooth 1-forms such that Oi (P) = Oi (p) for
1'(01, •.. , Or, Xl"'" Xs)(P) = 1'(01, ... , Or, Xl"'" Xs)(P). Proof. The result will follow easily if we can show that
1'(01, •.• ,Or, Xl"'" Xs)(P) = 0 whenever one of OI(P), ... ,Or(P),Xl(P), ... ,Xs(P) is zero. We shall assume for simplicity of notation that r = 1 and s = 2. Now suppose that Xl (p) = O. If (U,x), with x = (xl, ... , x n ), is a chart with p E U, then Xllu = ~~i 8~' for some smooth functions ~i E COO(U). Let 13 be a cut-off function with support in U and j3(P) = 1. Then for any smooth vector field X defined on U we can consider both j3X and 13 2X to be globally defined and zero outside of U. Similarly, if f is a smooth function defined on U, then 13 f can be taken to be globally defined on M and zero outside of U. Now (32 Xl = ~ (j3~i) (13 8~' ). (Notice that in this last expression we have used 13 to extend both the functions ~i and the coordinate fields 8~" which is why we used 13 2 rather than just 13.) Thus
j321'(OI' Xl, X2) = 1'(01, 13 2XI, X2) =
l'
(0 13 ~t 8~i ' X
=
l'
(01,
1,
2
2)
(j3~i) 13 8~i' X 2 )
= j3~i1'(Ol' 13 8~t' X2)' (Notice that the point of the above expression is that, at this moment, l' is defined only for global sections and is linear over global functions. For example, 8~' is not a global section while 13 8~' is a global section.) Since XI(p) - 0, we must have ~i(P) = 0 for all i. Also recall that j3(P) = 1. Plugging P into the formula above we obtain
1'(01 , Xl, X2)(p) =
o.
= 0 or Ol(P) = O. Assume that OI(P) - OI(P), Xl(p) = Xl(p) and X 2(p) = X2(p), Then
A similar argument holds when X2(P) we have
1'(OI' XI, X2) - 1'(01 , Xl, X2)
= 1'(01 - OI, Xl, X2) + 1'(01, Xl - XI, X2) + 1'(01 , XI, X 2 - X 2).
7. Tensors
326
Since 01 - (h, Xl - Xl, and X2 - X2 are all zero at p, we obtain the result that 7(0 1, XI, X2)(P) = 7(0 1, Xl, X2)(p). 0 Thus we have a natural correspondence between Trs(X(M)) and T;(M) (the latter being smooth sections of the bundle T r s (TM) -+ M). For example, if A E TJ(M), then we obtain an element of T13 (X(M)), also denoted by A, by defining a smooth function A(O,X, Y, Z) for given fields (O,X,Y,Z) by
A(O, X, Y, Z)(P)
:=
A(P)(O(P) , X(P), Y(p), Z(P)).
Conversely, if A E T I3 (X(M)), then we can use the above proposition to define an element of (M), which we denote by the same letter. Given A E T 13 (X(M)), define A(p) E T1(TpM) for each p as follows: For X p, Yp, Zp E TpM and Op E T; M we let
TJ
A(p) (Op, X p, Yp, Zp)
:=
A(O, X, Y, Z)(p),
where 0, X, Y, Z are any fields chosen so that O(p) = Op, X(p) = X p, Y(p) Yp, and Z(P) = Zp. By Lemma 6.37 we can always find such extensions, and by Proposition 7.32 above, A is well-defined. That A so defined is smooth follows from Proposition 7.22. The general case should be clear and, all said, we end up with a natural isomorphism of COO(M)-modules:
Similar reasoning shows that there is a correspondence between fields of TM-valued tensors and X(M)-valued tensors on X(M). For example, we have (7.6) Elements of T02(X(M); X(M)) are COO(M)-bilinear maps X(M) x X M
-+ X( M), while elements of 12 (M; T M) are sections of the bundle whose fiber at p is T02(TpMj TpM). So, if A E 12(Mj T M), then for each p, A(p) is an lR-multilinear map TpM x TpM -+ TpM. Similarly, we have (7.7)
T03 (X(M)jX(M)) ~ 73(M;TM).
In fact, later, when we define the curvature tensor on a semi-Riemannian manifold, it will initially be given as a multilinear map on modules of fields with values in X(M). The correspondence is then invoked to get a tenso' field (as defined in the bottom-up approach) with values in TM. It is just as easy to give a similar correspondence between Trs(r(~)) and r (Trs(~)) for some vector bundle ~ = (E, 11", M), where we view r(~) as a COO(M) module. Exercise 7.33. Exhibit the isomorphism (7.7) in detail.
7.5. Tensor Derivations
327
Exercise 7.34. Suppose that Sand T are tensors of the same type and we wish to show that they are equal. Then it is enough to check equality under the assumption that the vector fields inserted into the slots of Sand T are locally defined and have vanishing Lie brackets. Hint: Think about coordinate vector fields. We end this section with some warnings. It may seem that there is a simple way to obtain a pull-back by a smooth map f : M --+ N entirely from the top-down or module-theoretic view. In fact, one often sees expressions like f*T (Xl,'" ,Xs ) = T(f*X I , ... , f*X s )
(problematic expression!).
This looks cute, but invites misunderstanding. The left hand side takes fields Xl"'" Xs as arguments, while on the right hand side, if we consider T as a multilinear map X(N) x ... x X(N) --+ COO(N), then f*X~ must be fields. But the push-forward map f* is generally not defined on fields, and even if it were, the above expression would seem to be an equality of a function on M with a function on N. Note that X(M) is a COO(M)-module, while X(N) is a COO (N)-module. The above expression may be taken to mean something like f*T(X I , ... , Xs)(P) = T{Tf· XI(P), ... , Tf· Xs(P)), but now the right hand side has tangent vectors as arguments, and we are back to the bottom-up approach! A correct statement is the following: Proposition 7.35. Let f : M --+ N be a smooth map. Let T be a (0, s)tensor field. If T and f*T are interpreted as elements of rDs(X(N)) and TOs(X(M)) respectively, then f*T(X I , ... , Xs) whenever ~ is f -related to X t for i
= T(Y1 , .. . , Ys) 0
f
= 1, ... , s.
Of course, we can use Definition 7.31 to make sense of both push-forward and pull-back in the case that f is a diffeomorphism.
7.5. Tensor Derivations We would like to be able to differentiate tensor fields. In particular, we would like to extend the Lie derivative to tensor fields. For this purpose we introduce the following definition, which will be useful not only for extending the Lie derivative, but also in several other contexts. Recall the presheaf of tensor fields U 1-+ T;(U) on a manifold M. Definition 7.36. A tensor derivation is a collection of maps T;(U) --+ T;(U), all denoted by V for convenience, such that
V~ U
7. Tensors
328
(1) V is a presheaf map for 7; considered as a presheaf of vector spaces over R In particular, for all open U and V with V c U we have
'OAlv ='O(Alv) for all A E ~(U), i.e., the restriction of VA to V is just V (Alv).
(2) V commutes with contractions. (3) V satisfies a derivation law. Specifically, for A r{(U) we have
E ~(U) and
BE
'O(A®B) =VA®B+A®VB. For smooth n-manifolds, the conditions (2) and (3) imply that for A E ~(U), aI, ... ,ar E X*(U) and XI, ... ,Xs E X(U), we have
'O(A(al, ... , aT) Xl, ... ,Xs))
= (VA) (al, ... , ar, Xl"'" Xs) r
(7.8)
+ LA(al, ... ,Vai, ... ,ar,Xl, ... ,Xs ) i=l
s
+L
A(al, ... , ar, Xl, ... , VXi, . .. , Xs).
i=l
This follows by noticing that
A(al, ... ,ar,XI, ... ,Xs) = C(A® (al ® ... ® a r ®Xl ® ... ®Xs)) (where C is the repeated contraction) and then applying (2) and (3). Note that V stands for a family of maps whose domains 7;(U) depend not only on rand s, but also on U. The next proposition considers the situation where we only have derivations defined for U = M (the global case). Proposition 7.37. Let M be a smooth manifold and suppose we have a map on globally defined tensor fields V : ~ (M) ---+ 7; (M) for all nonnegative integers r, s such that (2) and (3) above hold for the case U = M. Then there is a unique induced tensor derivation that agrees with V on global sections, that is, on the various T;(M). Proof. We need to define V : ~(U) ---+ ~(U) for arbitrary open U as a derivation. Let 8 be a function in Coo(U) that vanishes on a neighborhood V of p E U. We claim that ('Oo)(p) = O. To see this, let 13 be a cut-off function equal to 1 on a neighborhood of p and zero outside of V. Then = (1 - 13)0 and so
o
'08(P) = '0((1- f3)8)(p) = 8(P)V(1- f3)(P)
+ (1- f3(p))'Oo(p) = O.
329
7.5. Tensor Derivations
Now given r E T;(U), let (3 be a cut-off function with support in U and equal to 1 on a neighborhood of p E U. Then (3r E T;(M) after extending by zero. Define (Vr)(p) := V((3r) (p). To show that this is well-defined let (32 be any other cut-off function with support in U and equal to 1 on a neighborhood of p E U. Then we have
V((3r)(p) - V((32r)(p) = (V((3r) - V((32r))(p) = V(((3 - (32)r)(p) = 0, where the last equality follows from our claim above with
~
= (3 - (32. Thus
V is well-defined on T;(U). We now show that Vr so defined is an element of ~(U). Let (U', x) be a chart with p E U' C U. Then we can write rJul E ~(U') as
We can use this to show that Vr as defined agrees with a global section in a neighborhood 0 of p and so must be a smooth section itself since the choice of p E U was arbitrary. To save on notation, let us take the case r = 1, s = 1. Then Tfl", = rjdxi ® a~" Let (3 be a cut-off function equal to 1 in the neighborhood 0 of p and zero outside of U'. Extend each of the sections (3rJ, f3dx i , and (3 a~' to global sections and apply V to (33r = ((3rj) ((3dxi) ®((3 a~i) to get
By assumption, V takes smooth global sections to smooth global sections, so both sides of the above equation are smooth. On the other hand, we have V(f33r )(q) = V(r)(q) by definition and valid for all q E O. Thus V(r) is smooth and is the restriction of a smooth global section. This gives a unique derivation V : T;(U) -+ T;(U) for all U satisfying the naturality conditions (1), (2) and (3). We leave it to the reader to check this last statement. D Exercise 7.38. Let VI and V2 be two tensor derivations (so satisfying conditions (1), (2) and 3 of Definition 7.36) that agree on functions and vector fields. Then VI = V 2 • [Hint: If a E X*(U) = -rr(U), we must have (V,o:) (X) = Vi (a(X)) - o:(ViX) for i = 1,2. Then both VI and V2 must obey formula (7.8) above.]
7. Tensors
330
Theorem 7.39. If'Du can be defined on Coo(U) and X(U) for each open U c M so that
(1) (2) (3) (4)
'Du(fg) = ('Duf) g + f'Dug for all f, g E Coo(U), ('DMf)lu = 'Du flu for each f E Coo(M), 'Du(fX) = ('Duf) X + f'DuX for all f E COO(U) and X E X(U), ('DMX)lu = 'Du Xlu for each X E X(M),
then there is a unique tensor derivation 'D that is equal to 'Du on Coo (U) and X(U) for all U. Sketch of proof. We wish to define 'D on X* (U) so that
(7.9)
'Du(a
X) - 'Dua ® X + a
'DuX.
By contraction we see that we must have ('Dua)(X) = 'Du(a(X))-a('DuX), which we take as the definition. Then check that (7.9) holds. Now define 'Du by formula (7.8) and verify that we really have a map 7;(U) -+ 7;(U . Check that 'Du commutes with contraction C : /t(U) -+ COO(U) for simple tensors a ® X E If(U). Use the fact that, locally, every element of It can be written as a sum of simple tensors. Next extend to 7; along the lines exemplified by the case of Ii (U) and the contraction CJ as follows: For 7 E Ii (U), we have
('DuCi7) (X) = 'Du ((CJ7) (X)) - (Ci7) 'DuX = 'Du (C(7(" X, .))) - C(7(" 'DuX, .)) = C'Du (7(', X,·) - 7(', 'DuX, .)) = C (('Du7) (', X, .)) = (ci'Du7) (X).
The general case would involve an inconvenient profusion of parentheses. Uniqueness follows from Exercise 7.38. Finally check by direct calculation that (3) of Definition 7.36 holds. Corollary 7.40. The L~e derivative ex can be extended to a tensor derivation for any X E X(M). The last corollary extends the Lie derivative to tensor fields. It follows from formula (7.8) that we have (eXS)(Y1,"" Ys)
(7.10)
X(S(Y1,"" Ys )) s
- L S(Y
1, ...
,1';,-1, ex 1';" Yi+1,"" Ys).
l=l
We now present a different way of extending the Lie derivative to tensor fields that is equivalent to what we have just done. First let A E TJ(M
7.6. Metric Tensors
and recall that if f* A E T;(M) by
f : M -+ M
331
is a diffeomorphism, then we can define
(f* A)(p)(a\ ... , aT, VI, ... , VS)
= A(f(p)) ( (Tpf-I)* (a l ), ... , (Tpf-I) * (aT), Tpf(Vl), ... , Tpf( v s )) for all a 1 , ... , aT E (TpM)* and VI, ... , Vs E TpM. If X is a vector field on M (possibly locally defined), we can define
(7.11) just as we did for vector fields. We leave it as a project for the reader to show that this definition agrees with our first definition of the Lie derivative of a tensor field. The Lie derivative on tensor fields is natural with respect to diffeomorphisms in the sense that for any diffeomorphism f : M -+ N and any vector field X we have (7.12) This property is not shared by some other important derivations such as the covariant derivative, which we define later in this book. Exercise 7.41. Show that the Lie derivative on tensor fields is natural with respect to diffeomorphisms in the above sense of equation (7.12) by using the fact that it is natural on functions and vector fields.
7.6. Metric Tensors We start out again considering some linear algebra that we wish to globalize. Thus the vector space V that we discuss next should be thought of as a tangent space of a manifold or a fiber of some vector bundle. We recall the following definitions: A symmetric bilinear form 9 on a finite-dimensional vector space V is nondegenerate if and only if g( V, w) = o for all w E V implies that V = O. A (real) scalar product on a (real) finite-dimensional vector space V is a nondegenerate symmetric bilinear form 9 : V x V -+ R. A scalar product space is a pair (V,g) where V is a vector space and 9 is a scalar product. We say that 9 is positive (resp. negative) definite if g(v,v) 2: 0 (resp. g(v,v) :S 0) for all V E V and g(v, v) = 0 ===> V = O. In case the scalar product is positive definite, we also refer to it as an inner product and the pair (V,g) as an inner product space. Otherwise we say that the scalar product is indefinite. A scalar product on V is sometimes called a metric tensor on V. We now need to introduce quite a few more definitions.
332
7. Tensors
Definition 7.42. The index of a symmetric bilinear form 9 on V is the dimension of the largest subspace W c V such that the restriction glw is negative definite. The index is denoted ind(g). Definition 7.43. Let (V, g) be a scalar product space. We say that v and ware mutually orthogonal if and only if g(v, w) = O. Furthermore, given two subspaces WI and W 2 of V we say that WI is orthogonal to W 2 and write WI ..1 W 2 if and only if every element of WI is orthogonal to every element of W 2. Since, in general, 9 is not necessarily positive definite or negative definite, there may be nonzero elements that are orthogonal to themselves. Definition 7.44. Given a subspace W of a scalar product space V, we define the orthogonal complement as W1.. = {v E V: g(v,w) = 0 for all
WEW}. Exercise 7.45. We always have dim(W) + dim(W1..) = dim(V), but unless 9 is definite, we may not have wnW1.. = {O}. Definition 7.46. A subspace W of a scalar product space (V, g) is called nondegenerate if glw is nondegenerate. Lemma 7.47. A subspace W W EEl W1.. (inner direct sum).
c (V, g)
is nondegenerate if and only if V =
Proof. This an easy exercise. One uses the standard fact that dim W + dim W1..
= dim(W + W1..) + dim(W n W1..).
0
It is a standard fact from linear algebra, already mentioned in Chapter 5, that if 9 is a scalar product, then there exists a basis eI, ... , en for V such that the matrix representative of 9 with respect to this basis is a diagonal matrix with ones or minus ones along the diagonal. Such a basis is called an orthonormal basis for (V, g). The number of minus ones appearing is the index ind(g) and so is independent of the orthonormal basis chosen. It is easy to see that the index ind(g) is zero if and only if 9 is positive definite. Definition 7.48. For each v E V with (v,v) 1= 0, let €(v):= sgn(v,v). Then if el,"" en are orthonormal, we have €t = €(i) := €(ei)' Thus if el, ... , en is an orthonormal basis for (V, g), then 9 (ei, eJ ) - Et 8tJ , where €~ = g(e t , e~) = ±1 are the entries of the diagonal matrix ind(g) of which are equal to -1 and the remaining are equal to 1. Let us refer to the list of ±1's given by (€I, ... , En) as the signature. We may arrange for the -1's to come first by permuting the elements of the basis. For example, if (-1, -1, 1, 1) is the signature, then the index is 2.
7.6. Metric Tensors
333
Remark 7.49. From now on, whenever context allows, we shall always assume that by "orthonormal basis" we mean an orthonormal basis that is arranged so that the -1 's come first as described above. The convention of putting the minus signs first is not universal, and in fact we used the opposite convention in Chapter 5. The negative signs first convention is popular in relativity theory and semi-Riemannian geometry, but the reverse convention is perhaps more common in Lie group theory and quantum field theory. It makes no difference in the final analysis as long as one is consistent, but it can be confusing when comparing references in the literature. Another difference between the theory of positive definite scalar products and indefinite scalar products is the appearance of the fi'S from the signature in formulas that would be familiar in the positive definite case. For example, we have the following: Proposition 7.50. Let el, ... , en be an orthonormal basis for (V, g). For any v E V, we have a unique expansion given by v = L:i fi(V, ei)ei. Proof. The usual proof works. One just has to notice the appearance of D the €i'S. Definition 7.51. If v E V, then let Ilvll denote the nonnegative number Ig(v, v)1 1 / 2 and call this the (absolute or positive) length or norm of v. Some authors call g(v, v) or g(v, v)I/2 the norm, which would make it possible for the norm to be negative or even complex-valued. We will avoid this. Just as for positive definite inner product spaces, we call a linear isomorphism il> : (VI, gl) -+ (V 2, g2) from one scalar product space to another an isometry if gI(V, w) = g2(il>v, il>w). It is not hard to show that if such an isometry exists, then gl and g2 have the same index and signature. Let (Vi, gi) be scalar product spaces for i = 1, ... ,k. By Corollary D.35 of Appendix D, there is a unique bilinear form
,;(.,.) = (.,.) for all 9 E G. The tangent map TK,: TpM -t T".(p) (MIG) is onto. For x E MIG, let VI,V2 E Tx(M/G). Define hx(VI' V2) = (VI, V2), where VI and V2 are chosen at the same point and such that TK,'Vi = Vi. We wish to show that this is well-defined. Indeed, if Vi E TpM and Wi E TqM are such that TpK, . Vi = TqK, . Wi = Vi for i = 1,2, then there is an isometry >.g with >'gp = q. Furthermore, since >.g is a deck transformation and curves representing Vi and Wi must be related by this deck transformation, we also have Tp>'g . Vi = Wi. Thus
(VI, V2) = (Tp>'gvI, Tp>.gV2) = (WI, W2),
340
7. Tensors
which means hx is well-defined. It is easy to show that x I---t hx is smooth and defines a metric on MIG with the same signature as that of (0,.) and that further, /'i,*h = (', .). In fact, we will use the same notation for either the metric on MIG or on M. Definition 7.61. A lattice of rank k in ]Rn is a set of the form
r
:=
{x
E ]Rn : x =
'Endi where ni
E Z} ,
where h, ... , !k are linearly independent elements of ]Rn. The called the generators of the lattice.
h, ... , !k are
The lattice zn c ]Rn is the standard rank n lattice, and it is generated by the standard basis. A lattice is a subgroup of ]Rn and so acts on ]Rn by a I---t a + v for v E r. This is a discrete, free and proper action, and so the quotient ]Rn Ir provides a simple example of the above construction and so has a metric induced from ]Rn. If the lattice has full rank n, then IRn If is called a flat torus (or flat n-torus) and is diffeomorphic to the product of n copies of the circle 8 1 . Each of these n-dimensional flat tori is locally isometric, but may not be globally isometric. To be more precise, suppose that h, 12, ... , in is a basis for]Rn which is not necessarily orthonormal. Let r f be the lattice consisting of integer linear combinations of h, 12,···, In. Now suppose we have two such lattices r f and r f' When is ]Rn Ir f isometric to ]Rn Ir f? It may seem that, since these are clearly diffeomorphic and since they are locally isometric, they must be (globally) isometric. But this is not the case (see Problem 17). The study of the global geometry of flat tori is quite interesting and even has deep connections with fields outside of geometry such as arithmetic, which we shall not have the space to pursue. We know from Chapter 4 that there are surfaces in ]R3 that are diffeomorphic to a torus 8 1 x 8 1 . Such surfaces inherit a metric from the ambient space, but it turns out that the Riemannian surface obtained in this way cannot be isometric to one of the flat 2-tori introduced here. In Chapter 13 we will see how each metric on a manifold gives rise to an associated curvature tensor. The reason the tori just introduced are referred to as flat is because (being locally isometric to some ]Rn) they have vanishing curvature tensor. If we have semi-Riemannian manifolds (M, g) and (N, h), then we can consider the product manifold M x N and the projections prl : M x N -t M and pr2 : M x N ~ N. The tensor g x h = prig + pr2h provides a semiRiemannian metric on the manifold M x N, which is then called the semiRiemannian product of (M,g) and (N,h). Let (Ul,x) and(U2,Y) denote charts on M and N respectively. Then we may form a product chart for M x N defined on Ul x U2. The coordinate functions of this chart are given by Xi = xi 0 prl and 'if = yi 0 pr2' We have the associated frame fields 88
7.6. Metric Tensors
341
and 8~" The components of 9 x h = prig + pr2h in these coordinates are discovered by choosing a point (pl,P2) E UI X U2 and then calculating. We have
9 x h(
O~i (p,q) , o~j (P,q) I
= prig (
1
~~i I(p,q) ' ~~J I(p,q) ) + pr2h( ~~i I(p,q) ' ~~j I(p,q) )
uX
g(
1
o~t I
p '
= 0+0
uX
O~ix (p,q) ,Tprl o~jY
= 9 (Tprl =
UY
Op)
+ h(Oq,
o~j
I
(p,q)
)
+ h (Tp r2
I)
UY
O~ix
1
(p,q)
,Tpr2
o~jY
I
(p,q)
)
= 0,
and (abbreviating a bit)
gXh(~~tl 'tJ~jl ) =g(tJ°tl ,tJ°il )+h(Oq,Oq)=9ij(P). uX (p,q) uX (p,q) uX p uX p Similarly 9 x h(8~" 8~J )(P, q) = htj(q). In practice, the coordinate functions constructed above are often abusively denoted by (xl, ... , xn1 , yl, ... , yn2) and the frame field,s by -/yr, ... , a:n2' So with respect to these coordinates, the matrix of 9 x h is of the form
-/xr, ... , 8:nl ,
where G
= (gij 0 prl) and H = (hij 0 pr2)'
Notation 7.62. The product metric is often denoted by 9 dinates by ds 2 = gijdxidxj + hk1dykdyl.
+ h or in coor-
Every smooth manifold that admits partitions of unity also admits at least one (in fact infinitely many) Riemannian metric. This includes all (finite-dimensional) paracompact manifolds. The reason for this is that the set of all Riemannian metric tensors is, in an appropriate sense, convex. We record this as a proposition. Proposition 7.63. Every smooth (paracompact) manifold admits a Riemannian metric. Proof. This is a special case of Proposition 6.45.
o
If M is a regular sub manifold of a Riemannian manifold (N, h), then M inherits a Riemannian metric 9 := z* h, where z : M "--+ N is the inclusion map. We have already used this idea for submanifolds of the Euclidean
7. Tensors
342
space lRd • More generally, if f : M -+ N is an immersion, then (M, f* h) is a Riemannian manifold. In particular, if f : M -+ lRd is an immersion, then we obtain a Riemannian metric on M. It turns out that every Riemannian metric on M can be obtained in this way. Actually, more is true! For any Riemannian manifold (M, g) there is an embedding f : M -+ lRd , for some d, such that g := z*gO, where go denotes the standard metric on JRd. What this means is that f(M) is a regular submanifold, and if we give I(M) the metric induced from the ambient space lRd , then f becomes an isometry when viewed as a map into f(M). We say that such an f is an isometric embedding of Minto lRd . In short, the result is that every Riemannian manifold can be isometrically embedded into some Euclidean space of sufficiently high dimension. This difficult theorem is called the Nash embedding theorem and is due to John Forbes Nash (see [Nash!] and [Nash2]). Note that d must be quite large in general (d = (dimM)2 + 5 (dim M) + 3 is sufficient). For an indefinite semi-Riemannian manifold (N, h), the pull-back f*h by a smooth map I: M -+ N may not be a metric because there may be points p E M such that Tpl(TpM) is a degenerate subspace of Tf(p)N. In particular, not every embedding of a manifold M into a semi-Euclidean space 1R~ (with 1 < 1I < d) induces a metric on M. Nevertheless, every metric on M of any index can be obtained using an appropriate embedding into some lR~ (see [Clark]).
Problems
7
(1) Show that if E V ® V* has the same components every basis, then 7] = for some E lR.
ao;
a
7; with respect to
(2) If el, ... , en is a basis for V and iI, ... , 1m is a basis for W, then {Ejh=l .....n j=l .....m is a basis for L(V, W), where Ej(v) :- ei(v)fJ. Show this directly without assuming the isomorphism of W ® V* with L(V, W). (3) Let (Vi, gi) be scalar product spaces for i = 1, ... , k. By Corollary D.35 of Appendix D, there is a unique bilinear form 'P : ®:=l Vi x
such that for
Vi E
Vi and
Wi E
®:=1 Vi -+ lR
Wi,
Show that 'P is nondegenerate and that it is positive definite if each 9 is positive definite.
343
Problems
(4) Define T : X(M) x X(M) -t COO(M) by T(X, Y) = XYf. Show that T does not define a tensor field. (5) Let b~j and b~j be the components of a bilinear form b with respect to bases el,"" en and ei, ... , e~ respectively. Show that in general det(bij ) does not equal det(b~j)' Show that if det(biJ ) is nonzero, then the same is true of det(b~j)' (6) Show that while a single algebraic tensor Tp at a point on a manifold can always be extended to a smooth tensor field, it is not the case that one may always extend a (smooth) tensor field defined on an open subset to a smooth tensor field on the whole manifold. (7) Let 4J : ]R2 -+ ]R2 be defined by (x, y) 1---7- (x + 2y, y). Let T := x ® dy + y ® dy. Compute 4J*T and 4J*T.
tx
t
(8) Prove Proposition 7.35. (9) Let V be a tensor derivation on M and suppose that in a local chart we have V(a~.) = l:Df a~J for smooth functions nf. Show that V(dx j ) = - l: Df dx i . Let X be a fixed vector field with components Xi in our chart. Find the nf in the case that V = ex. (10) Let A E 7(f(M). Show that the component form of the Lie derivative with respect to a chart is given as (£xAtb = aAab Xh _ axa Ahb _ ox bA ah oxh aXh ox h (where we use the Einstein summation convention). Show that if A E TJ! (M), then the formula becomes aAab h axh axh (£x A ) = ox h X + ox a Ahb + ox b A ah .
Ti
Find a formula for A E (M). (11) Show that our two definitions of the Lie derivative of a tensor field agree with each other. (12) In some chart (U, (x, y)) on a 2-manifold, let A = x/y ® dx ® dy + ® dy ® dy and let X = + x/y. Compute the coordinate expression for A. £x 13) Suppose that for every chart (U, x) in an atlas for a smooth n-manifold M we have assigned n 3 smooth functions j , which we call Christoffel symbols. Suppose that rather than obeying the transformation law expected for a tensor, we have the following horrible formula relating the Christoffel symbols r:j on a chart (U', y) to the symbols rfj :
Ix
Ix
r:
'k
r ij
a 2 xl oyk = 0 y~'0 yJ' a XI
t oxr ox s oyk
+ rrs-a'-a yJ'-a Xt y~
(sum).
7. Tensors
344
Assume that such a transformation law holds between the Christoffel symbol functions for all pairs of intersecting charts. For any pair of vector fields X, Y E X(M), consider the functions (DXy)k given in every chart by the formula (Dxyl := ~r;:Xh + rfJxiyi. Show that the local vector fields of the form (Dxy)k~, defined on each chart, are the restrictions of a single global vector field D x y. Show that Dx : Y H DxY is a derivation of X(M) and that with Dxl:= XI for smooth functions, we may extend to a tensor derivation; D x is called a covariant derivative with respect to X. There are many possible covariant derivatives. (14) Continuing on the last problem, show that DfX+gYY = IDxY +gDyT for all I, 9 E COO(M) and X, Y E X(M) and Y E 'Ps(M). (15) Show that if are coordinate vector fields from some chart, then [a~" a~J 1 == O. Consider the vector fields and y arising from standard coordinates on R2 and also the and from polar coordinates. Show that [ix, is not identically zero by explicit computation.
-/t-r, ... ,-Jln
frl
tr
Ix
to
t
(16) Let M -+ M be a semi-Riemannian cover. P~ve that if we have a local isometry ¢> : N -+ M, then any lift ¢> : N -+ M is also a local isometry. (17) Suppose we have two lattices r, and r, in R,n. Let R,n have the standard metric. Describe the induced metrics on R,n jr, and R,n jr, and provide a necessary and sufficient condition for the existence of an isometry R,njr,
-+ R,njr,.
(18) Show that if a, (3 E Tk(V) where V is a scalar product space, then the scalar product on Tk(V) is given in terms of index raising and contraction by
Chapter 8
~ifferential
50rmS
In one guise, a differential form is nothing but an alternating (antisymmetric) tensor field. What is new is the introduction of an antisymmetrized version of the tensor product and also a natural differential operator called the exterior derivative. We start off with some more multilinear algebra. 8.1. More Multilinear Algebra Definition 8.1. Let V and W be real finite-dimensional vector spaces. A kmultilinear map a : V x ... x V ---t W is called alternating if a( VI, ... ,Vk) = o whenever Vi = Vj for some i =1= j. The space of all alternating kmultilinear maps into W will be denoted by L~t(Vj W) or by L:1t(V) if W = lR. By convention, L~lt(Vj W) is taken to be Wand in particular, L~t(V) = R Since we are dealing with the field 1R (which has characteristic zero), it is easy to see that alternating k-multilinear maps are the same as (completely) antisymmetric k-multilinear maps which are defined by the property that for any permutation a of the letters 1,2, ... , k we have
W(Vl' V2,···, Vk)
= sgn(a)W(Vu(l) , Vu(2) , ... , Vu(k))'
Let us denote the group of permutations of the k letters 1,2, ... , k by Sk. In what follows, we will occasionally write ai in place of a(i). Definition 8.2. The antisymmetrization map Alt k : TDk(V) is defined by
---t
L~t(V)
k 1 ~ Alt (W)(Vl' V2, .. ·, Vk) := k! L..J sgn(a)w(vUll VU2 ' " ' ' VUk )· UESk
345
346
8. Differential Forms
Lemma 8.3. For a E TOkl (V) and /3 E yDk2 (V), we have
= Altkl+k2 (a ® /3), Altkl +k2 (a ® Alt k2 /3) = Alt kl +k2 (a ® /3) ,
Altkl+k2(Alt kl a ® /3)
and
Proof. For a permutation a E Sk and any T E TOk(V), let aT denote the element ofTOk(V) given by (aT)(v1, ... ,Vk):= T(V a (l), ... ,Va (k)). We then have Alt k(aT) = sgn( er) Altk (T) as may easily be checked. Also, by definition Altk(T) = Esgn(er)erT. We have Alt kl +k2(Alt kl (a) ® /3)
= Altkl+k2
((k~! L
sgna (aa)) ® /3)
aESkl
= Altkl+k2
(k~! L
sgner (era ® /3))
aESkl
1
= -kI 1·
L
sgn a Alt kl +k2 (aa ® /3) .
aESkl
Let us examine the expression sgn a Alt ki +k2 (era ® /3). If we extend each er E SkI to a corresponding element a' E Skl+k2 by letting a'(i) = a(i) for i ~ kl and er'(i) = i for i > kl' then we have era ® /3 = er'(a ® /3) and also sgn( er) = sgn( a'). Thus sgn a Altkl +k2 (aa ® /3) = sgn a' Alt k1 +k2 er' (a ® !3 and so
We arrive at Altkl+k2(Altkl (a) ® /3) = Altkl+k2 (a ® /3). In a similar way, Altkl+k2(a®Altk2/3) = Altkl+k2 (a ® /3), and so the last part of the theorem follows. 0
347
8.1. More Multilinear Algebra
Given W E L!ft(V) and TJ E L!rt(V), we define their exterior product or wedge product W /\ TJ E L:.ttk2(V) by the formula
w/\TJ:=
(k1 + k2)! k +k k'k' Alt l 2(w01J). 1· 2·
Written out, this is
(8.1)
Warning: The factor in front of Alt in the definition of the exterior product is a convention but not the only convention in use. This choice has an effect on many of the formulas to follow which differ by a factor from the corresponding formulas written by authors following other conventions. It is an exercise in combinatorics that we also have
L
sgn(lT)w(vO"l' ... , VO"I,JTJ(VO"Iol +1' ... , VO"kl +102)'
(kl ,k2 )-shufRes 0"
In the latter formula, we sum over all permutations such that IT (1) < IT(2) < ... < IT(kI) and IT(k1 + 1) < IT(k1 + 2) < ... < IT(k1 + k2)' This kind of permutation is called a (k1' k2 )-shuffie as indicated in the summation. The most important case of (8.1) is for w, TJ E L~t(V), in which case
(w /\ 1J)(v, w) = w(v)TJ(w) - w(w)TJ(v). This clearly defines an antisymmetric bilinear map. Proposition 8.4. For a E L:tt(V), f3 E L:ft(V), and'Y E L:?t(V), we have kl (V) x L k2 (V) ~ L kl +k2 (V) is R-bilinear' (i) /\.. L alt alt alt ' (ii) a /\ f3 = (_1)k l k2 f3 /\ a;
(iii) a /\ (f3 /\ 'Y) = (a /\ f3) /\ 'Y. Proof. We leave the proof of (i) as an easy exercise. For (ii), we consider the special permutation f given by (f(1), f(2), ... , f(k 1+k2)) = (k1 +1, ... , kl +k2, 1, ... , kI). We have that a0f3 = f (f3 0 a). Also sgn(f) = (_1)klk2. So we have Alt kl +k2 (a 0 f3) which gives (ii).
= Alt k1 +k2 (f (f3 0
a))
= (-1 )klk2 Alt k1 +k2 (f3 0
a),
8. Differential Forms
348
For (iii), we compute (kl
0./\
+ k2 + k3)!
(f3 IVy) = k' (k
k )' Alt(a ® (f3/\ ')'))
+ 3· = (kl + k2 + k3)! (k2 + k3)! Al ( 1·
2
k 1·'(k 2 + k)' 3·
=
(kl
k 2·'k 3·,
t a ®
Al (R t
® ')'
f-'
))
+ k2 + k3)! kl!k2!k3! Alt(a ® Alt (f3 ® ')')).
By Lemma 8.3, we know that Alt(a ® Alt (f3 ® ')')) = AIt(a ® (f3 ® ')')), and so we arrive at (kl + k2 + k3)! 0./\ (f3/\ ')') = kl!k2!k3! AIt(a ® (f3 ® ')')). By a symmetric computation, we also have (kl
(a /\ (3) /\ ')' =
+ k2 + k3)! kl!k2!k3! Alt((a ® (3) ® ')'),
and so by the associativity of the tensor product we obtain the result.
0
Example 8.5. Let V have a basis e}, e2, e3 with dual basis el, e2, e3. Let a = 2e1 /\ e2 + e 1 /\ e2 and f3 = e 1 - e3. Then as a sample calculation we have 0./\ f3 = (2e 1 /\ e2 + e 1 /\ e3) /\ (e 1 - e3 )
= 2e 1 /\ e2 /\ e1 + e1 /\ e3 /\ e 1 = -2e1 /\ e2 /\ e3 _ e1 /\ e3 /\ e3 = -2e 1
/\
e2 /\ e3,
where we have used that e 1 /\ e2 /\ e1 = -e 1 /\ e1 /\ e2 = 0, etc.
Lemma 8.6. Let 0. 1 , •.• ,ak be elements of V* - L~t (V) and let VI, ••. ,Vk E V. Then we have 0. 1 /\ ... /\ a k (V}, ... ,Vk)
where A = (aj) is the k
xk
= det A,
matrix whose ij-th entry is
aj =
a~(vj).
Proof. From the proof of the last theorem we have 0./\
(f3 /\ ')') =
(kl
+ k2 + k3)! k1!k2!k3! Alt(a ® (f3 ® ')')).
By inductive application of this we have 0. 1 /\ ••• /\
a k = k! Alt(a 1 ® ... ® a k ).
Thus 0.1
/\ ... /\ ak(vl, ... , Vk)
=
z= 0'
sgn(u)a 1 (vO'l)'" ak(vO'k) = det A.
0
349
B.1. More Multilinear Algebra
Let us define
1
(8.2)
-1
o
if jl, .. " jk is an even permutation of iI, ... , ik, if jI, ... ,jk is an odd permutation of iI, ... ,ik, otherwise.
Then we have Corollary 8.7. Let el, ... ,en be a basis for V and e l , e 2 , .•• ,en the dual basis for V*. Then we have e
il /\
...
/\
e
ik ( ejI' •.. ,ejk )
= EJIil ......ikJk ·
Since any a E L~t(V) is also a member of T>k(V), we may write
where a tI ...tk = a(eill" ., elk)' By Alt (a) = a and the linearity of Alt we have a
= ""' a' . Alt (eil ~ 'I···lk
® ... ® elk)
=~ a . eh k! ""' ~ 'I···'k
/\ ... /\ eik .
We conclude that the set of elements of the form eil /\ ... /\ eik spans L~t(V). Furthermore, if we use the fact that both ai UI ""Uk = sgn (j a'I ... ik and e'ul /\ •.. /\ e'erk = sgn (j eil /\ ... /\ eik for any permutation (j E Sk, we see that we can permute the indices into increasing order and collect terms to get 0: =
~ a eil /\ ... /\ eik k! ""' ~ \I .. ·'k
=
""'
.~ 'I
.
a ' . eil 'I 12,··,'k
/\
ei2
/\ ••. /\
eik ,
0 for some and hence any w E [w]. Equivalently, (el, ... ,en) is positive if ei /\ ... /\ en E [w]. Actually a choice of ordered basis for a real vector space determines an orientation with respect to which it is positive. Indeed, if e1, ... ,en is dual to such a basis, then choose [w] where w = e l /\ ... /\ en. Definition 8.16. Let (Vb [WI]) and (V2' [W2]) be oriented real vector spaces. A linear isomorphism A : V I ---+ V 2 is said to be orientation preserving if )'*W2 = CWI for some c> 0 and some, and hence any, choices WI E [WI] and W2 E [W2]'
8. Differential Forms
354
When one talks about an element A E GL(V) being orientation preserving, one means that det A > 0 and this is tantamount to A : (V, [w]) ~ (V, [w]) being orientation preserving for any choice of orientation [w]. 8.1.1. The abstract Grassmann algebra. We now take a very abstract approach to constructing an algebra that will be seen to be isomorphic to Lalt(V). We wish to construct a space that is universal with respect to alternating multilinear maps. We work in the category of real vector spaces, although much of what we do here makes sense for modules. Consider the tensor space Tk(V) := &;/V (take any realization of the abstract tensor product as in Definition D.17). Let A be the submodule of Tk(V) generated by elements of the form VI ® ... Vi ® ... ® Vi··· ® Vk· In other words, A is generated by simple tensors with two (or more) equal factors. Recall that associated to Tk (V) we have the canonical map I8l : V x··· x V -+ Tk(V) defined so that ®(VI, ... , Vk) = VI ® .. ·®Vk. We define the space of k-vectors to be V A ... A V :=
1\
k
V:= Tk(V)j A.
Let Ak : V x ... x V -+ I\k V be the composition of the canonical map ® with quotient map of Tk (V) onto 1\ k V. This map turns out to be an alternating multilinear map. We will denote Ak(VI, ... , Vk) by VI A ... A Vk. Using the universal property of Tk(V) as described in Appendix D, one can show that the pair (I\k V, Ak) is universal with respect to alternating k-multilinear maps: That is, given any alternating k-multilinear map a : V X ••. x V -+ W, there is a unique linear map a/\ : 1\ k V -+ W such that a = a/\ 0 Ak; that is, the following diagram commutes: Vx···xV~W
/\k!
~
I\kV Notice that we also have that VIA· . ·AVk is the image of VI ® .. ·®Vk under the quotient map Tk(V) -+ I\kV. Next we define 1\ V:= tfJ~ol\kV, which is a direct sum, and we take 1\0 V := R We impose on 1\ V the multiplication generated by the rule
" " (vIA···Avi) x (vIA···Avj) t-tvIA···AviAvIA···Avj E I\i+j V. The resulting graded algebra is called the Grassmann algebra or exterior algebra. (Of course, the definition of A here is different from what we defined previously.) If we need to have a Z grading rather than an fir grading,
8.1. More Multilinear Algebra
355
we may define A. k V := 0 for k < 0 and extend the multiplication in the obvious way. Elements of A. V are called multivectors and specifically, elements of A. k V are called k-multivectors. Notice that since (v+w)/\(w+v) = 0, it follows that v/\w = -w/\v. In fact, any transposition of the factors in a simple element such as VI/\· .. /\ Vk, introduces a change of sign: VI /\ ..• /\ Vi /\ ... /\ Vj /\ ... /\ Vk
=
-VI /\ ... /\ Vj /\ ... /\ Vi /\ ... /\ Vk.
Lemma 8.17. If V has dimension n, then is a basis for V, then the set {eil /\ ... /\ eik : 1 ~ il
is a basis for
A. k V
A. k V = 0 for k > n.
< ... < ik
~
where we agree that eil /\ ... /\ eik
If el, ... , en
n}
=1
if k
= o.
Proof. The first statement is easy and we leave it to the reader. We will show that the set above is indeed a basis. First note that A. n V is spanned by el /\ ... /\ en. To see that el /\ ... /\ en is not zero we let det : V x ... x V -+ lR be the multilinear map given by representing the arguments as column vectors of components with respect to the given basis and then taking the determinant of the n x n matrix built from these column vectors. Then det(eb ... , en) = 1. But by the universal property above there is a linear map det" such that det = det" 0 /\k and so det/\(el/\ ... /\ en) = det/\ 0 /\k (el, ... , en)
= det(eI, ... ,en) = 1; thus, we conclude that el /\ ... /\ en is not zero (and is a basis for A. n V). Now it is easy to see that the elements of the form eil /\ ... /\ eik span A. k V. To see that we have linear independence, suppose that
L
ail ...ikeil /\ ... /\ eik
= o.
l::;il , ¢ll\ or something similar.
8.6. Operator Interactions The Lie derivative acts on differential forms since the latter are, from one viewpoint, alternating tensor fields. When we apply the Lie derivative to a differential form, we get a differential form, so we should think about the Lie derivative in the context of differential forms. Lemma 8.51. For any X E X(M) and any f E OO(M), we have .cxdf
=
dCxf. Proof. For a function
(Cxdf)(Y)
(:tl = Y(:tl
=
f, we compute as
o (cpf)*df)(Y) o (cpf)*f)
=
:tl
o df(Tcpf . Y)
=
:tl
o Y((cpf)*f)
= Y(.cxf) = d(.cxf)(Y),
where Y E X(M) is arbitrary.
D
We now have two ways to differentiate sections in O(M). First, there is the Lie derivative .cx : Oi(M) -+ ni(M), which turns out to be a graded derivation of degree zero, (8.7)
.cx(a 1\ (3) = .cxa 1\ (3 + a 1\ .cx(3.
We may apply .cx to elements of O(U) for U C M, and it is easy to see that we obtain a natural derivation in the sense of Definition 8.33. Exercise 8.52. Prove the above product rule. Second, there is the exterior derivative d which is a graded derivation of degree 1. In order to relate the two operations, we need a third map, which, like the Lie derivative, is taken with respect to a given field X E X (M). This map is defined using the interior product given in Definition 8.12 by letting
(8.8)
ixW(X1, ... , Xi-1)(p) := ixpwp(X1(p), ... , X~ 1(P)).
8. Differential Forms
374
Alternatively, if w E Oi (M) is viewed as a skew-symmetric multilinear map from X (M) x ... x X (M) to Coo (M), then we simply define
ixW(XI , ... , Xi-I) := w(X, Xl, ... ,Xi-I). By convention ixf = 0 for f E Coo(M). We will call this operator the interior product or contraction operator. This operator is clearly linear over R Notice that for any f E Coo (M) we have
ifxw = fixw, and ixdf = df(X) = LXf. Proposition 8.53. ix is a graded derivation of O(M) of degree -1:
ix(a /\ {3) = (ixa) /\ {3 + (-l)ka /\ (ix{3) for a E Ok (M). It is the unique degree -1 graded derivation of O(M) such that ix f = 0 for f E 0° (M) and ixO = O(X) for 0 E 0 1 (M) and X E X (M) .
o
Proof. This follows from Proposition 8.13.
Actually, ix is natural with respect to restriction, so it is a natural graded derivation in the sense of Definition 8.33. Formulas developed for the interior product in the vector space category also hold for vector fields and differential forms. For example, if 01, ... , Ok E 0 1 (M) and X EX (M), then k
. 1,x
(0 1 /\
... /\
Ok)
""" = L...
'""' /\ ... /\ Ok· (-1) k+1 O£(X)Ol /\ ... /\ Of
£=1
Notation 8.54. Other notations for ixw include XJw and (X,w). These notations make the following theorem look more natural: Theorem 8.55. The Lie derivative is a derivation with respect to the pairing (X,w) I-t (X,w). That is,
Lx(iyw) = i,C,xYw + iy,Cxw, or in alternative notations, LX(YJW) = ('cxY)Jw+YJ(,Cxw), 'cx(Y,w) = ('cxY,w)
+ (Y,'cxw).
o
Proof. Exercise.
Now we can relate the Lie derivative, the exterior derivative and the contraction operator.
Theorem 8.56. Let X E XM. Then we have Cartan's formula,
(8.9)
LX = do ix + ix
0
d.
8.7. Orientation
375
Proof. Both sides of the equation define derivations of degree zero (use Proposition 8.35). So by Lemma 8.36 we just have to check that they agree on functions and exact 1-forms. On functions we have ix f = 0 and ixdf = Xf = Cxf so the formula holds. On differentials of functions we have (d 0 ix
+ ix 0 d)df =
(d 0 ix)df = dCxf = Cxdf,
where we have used Lemma 8.51 in the last step.
o
As a corollary, we can extend Lemma 8.51: Corollary 8.57. do Cx
= Cx 0 d.
Proof. We have dCxa
= d(dix + ixd)(a) = dixda = dixda + ixdda = (Cx 0 d) a.
o
Corollary 8.58. We have the following formulas:
(i) i[x,Yj = Cx 0 iy - iy 0 £x; (ii) Cfxw = f Cxw + df 1\ ixw for all wE O(M). Proof. We leave (i) as Problem 9. For (ii), we compute:
+ d(ifxw) = ifxdw + d(f ix w) ixdw + df 1\ ixw + fd (ixw)
Cfxw = ifx dw = f
= f (ix dw + d (ixw)) + df 1\ ixw = f £xw + df 1\ ixw.
0
8.7. Orientation A vector bundle E ---+ M is called oriented if every fiber Ep is given a smooth choice of orientation. There are several equivalent ways to make a rigorous definition: Proposition 8.59. Let E ---+ M be a rank k real vector bundle with typical fiber V. The following are equivalent:
(i) There is a smooth global section w of the bundle /\ k E* ~ L!lt (E) ---+ M such that w is nowhere vanishing. (ii) There is a smooth global section s of the bundle /\ k E ---+ M such that s is nowhere vanishing.
(iii) The vector bundle has an atlas of VB-charts (local trivializations) such that the corresponding transition maps take values in GL+ (V), the group of positive determinant elements of GL(V). This means that the standard GL(V)-structure on E ---+ M can be reduced to a GL + (V)-structure (refer to Chapter 6).
8. Differential Forms
376
Proof. We show that (i) is equivalent to (iii) and leave the rest as an easy exercise. Suppose that (i) holds and that w is a nonvanishing section of L~t(E). Now fix a basis (eI, ... ,ek) on V and recall that with this basis fixed, each VB-chart (U, ¢) for E -+ M corresponds to a local frame field. Indeed, we let ei(p) := ¢-I(p,ei). The transition maps between two charts will have values in GL +(V) exactly when the matrix function that relates the corresponding frame fields has positive determinant (check this). Given a VB-atlas we construct a new atlas. We retain those VB-charts (U,4» whose corresponding frame field eI, ... , ek satisfies w(el, ... , ek) > o. For the charts for which w( eI, ... ,ek) < 0, we replace ell ... ,ek by -el, ... , ek, and the resulting chart will be included in our new atlas. Now if eI, ... , ek and h, ... , fk are two frame fields coming from this atlas, then fi = L: eJ and
C;
w(h, ... ,fk) = (detC)w(el, ... ,ek)· We conclude that det C > o. Conversely, suppose the vector bundle has an atlas {(Uer, ¢er)} taking values in GL +(V). We will use the frame fields coming from this atlas to construct a nowhere vanishing section of L:1t(E) -+ M. If el, ... , ek and h, ... ,fk are two frame fields coming from this atlas, then let fl, ... ,fk be dual to It, ... , fk and consider fl /\ ... /\ fk. We have Ii = L: C; ej and
(II /\ ... /\ fk) (el, ... , ek) = det C > o. For each chart (Uer, ¢er) in our VB-atlas, let ff, ... , fr be the corresponding
frame field and let (J~, ... , f~) be the dual frame field. Then let {Per} be a partition of unity subordinate to the cover {Uer }. Let
w :=
L Perf; /\ ... /\ f~.
Then, w is nowhere vanishing. To see this let P E M and suppose that P E U~ for some chart (U~, ¢~) from the GL + (V)-valued atlas. Then
wuf,···, ff)(p) = L Per(P) det C~er(P) > 0, where C~er is the matrix that relates ff, ... ,ff and
ff, ... , fr.
0
Definition 8.60. If anyone (and hence all) of the conditions in Proposition 8.59 hold, then E -+ M is said to be orientable. A VB-atlas that satisfies (iii) will be called an oriented atlas. If E -+ M is orient able as above, then two nowhere vanishing sections
of L:1t(E), say WI and W2, are said to be equivalent if WI = fW2, where f is a smooth positive function. We denote the equivalence class of such a nowhere vanishing w by [w].
8.7. Orientation
377
Definition 8.61. An orientation for an orient able vector bundle of rank k is an equivalence class [w] of nowhere vanishing sections of /\ k E*. If such an orientation is chosen, then the vector bundle is said to be oriented by
[w]. Notice that if we have two oriented VB-atlases on a vector bundle, then we know what it means for them to determine the same GL + (V)-structure. This was the notion of strict equivalence from Chapter 6. The next exercise shows that the notion of a reduction to a GL + (V)-structure is equivalent to the notion of an orientation as we have defined it. Exercise 8.62. Recall the construction of a nowhere vanishing section win the proof of Proposition 8.59. Show that the class [w] does not depend on the partition of unity used in the construction. Show that if two oriented VBatlases determine the same GL+ (V)-structure, then the constructed sections are equivalent and so determine the same orientation. Conversely, show that an orientation as we have defined it determines a unique reduction to a GL + (V)-structure on the vector bundle. Let E --+ M be oriented by [w]. A frame (VI, ... , Vk) of fiber Ep is positively oriented (or just positive) with respect to [w] if and only if w(P) (Vb" . , Vk) > O. This condition is independent of the choice of representative w for the class [w]. Definition 8.63. Let 7r : E --+ M be an oriented vector bundle. A frame field (h, ... , /k) over an open set U is called a positively oriented frame field if (h (P), ... , fk (P» is a positively oriented basis of Ep for each p E U. Exercise 8.64. Let 7r : E --+ M be an oriented vector bundle. Show that if M is connected, then there are exactly two possible orientations for the vector bundle. Exercise 8.65. If 7r1 : EI --+ M and 7r2 : E2 --+ Mare orientable, then so is the Whitney sum 7r1 EB 7r2 : EI EB E2 --+ M. Definition 8.66. A smooth manifold M is said to be orientable if T M is orientable. An orientation for the vector bundle T M is also called an orientation for M. A manifold M together with an orientation for M is said to be an oriented manifold. Definition 8.67. An atlas {(Ua , xa)} for M is said to be an oriented atlas if the associated frame fields are positively oriented. If this atlas is positively oriented with respect to an orientation on M (an orientation [w] of T M), then we call {(Ua , xa)} a positively oriented atlas. It follows from the definitions that an oriented atlas induces an orientation for which it is a positively oriented atlas. For this reason, a choice of
8. Differential Forms
378
oriented atlas is equivalent to a choice of orientation, and so one often sees an orientation of an orientable manifold defined as simply being given by a choice of oriented atlas. Now let M be an n-manifold. Consider a top form, i.e. an n-form tv E nn(M), and assume that tv is nowhere vanishing. Thus M must be orient able. We call such a nonvanishing tv a volume form for M, and every such volume form obviously determines an orientation for M. If t.p : M -+ M is a diffeomorphism, then we must have that t.p*tv = d'W for some d E COO(M), which we will call the Jacobian determinant of t.p with respect to the volume element tv:
t.p*tv = Jw(t.p)tv. Clearly Jw(t.p) is a nowhere vanishing smooth function. Proposition 8.68. Let (M,
[tvD be an oriented n-manifold. The sign of Jw(t.p) is independent of the choice of volume form tv in the orientation class [tv].
Proof. Let tv' E nn(M). We have never zero on U. Furthermore,
J(t.p)tv and since
7
tv
= atv' for some function a that is
= (t.p*tv) = (a 0 t.p)(t.p*tv') = (a 0 t.p)JWI(t.p)tv' = a 0 t.p tv, a
o
> 0 and tv is nonzero, the conclusion follows.
Let us consider a very important special case of this: Suppose that 'P : ]Rn. Then letting tvo = du 1 /\ ••• /\du fl
U -+ U is a diffeomorphism and U c we have for any x E U,
t.p*tvo(x) = t.p*du 1 /\ •.. /\ t.p*dun(x) =
(~8(~~~t.p)lx du il )
= det
/\ ... /\
(8(~; t.p) (x)) tvo(x) =
(~8(~::t.p)lx du
tn )
Jt.p(x)tvo(x).
So in this case, Jwo(t.p) is just the usual Jacobian determinant of t.p. More generally, let a nonvanishing top form tv be defined on M and let 'W' be another such form defined on N. Then we say that a diffeomorphism t.p : M -+ N is orientation preserving (or positive) with respect to the orientations determined by tv and tv' if the unique function JW,wl such that t.p*tv' = JW,w1tv is strictly positive on M. Exercise 8.69. An open subset of an oriented manifold M inherits the orientation from M since we can just restrict a defining volume form. Show that a chart (U, x) on an oriented manifold is positive if and only if
379
8.7. Orientation
x : U ---+ x (U) is orientation preserving. Here, x (U) inherits its orientation from the ambient Euclidean space with its standard orientation. We now construct a two-fold covering manifold Mor for any manifold M. The orientation cover will itself always be orientable. Recall that the zero section of a vector bundle over M is a submanifold of the total space diffeomorphic to M. Consider the vector bundle whose total space is " n T* M and remove the zero section to obtain (/\n T* M)
x := (/\ n T* M)
\{zero section}.
Define an equivalence relation on (/\n T* M) x by declaring VI '" V2 if and only if VI and V2 are in the same fiber and if VI = aV2 with a > O. The space of equivalence classes is denoted Mo r and we will show that it is a smooth manifold. Let q : (/\n T* M) x ---+ Mo r be the quotient map and give Mor the quotient topology. There is a unique smooth map 7ror making the following diagram commute: (/\nT*M)X _ _ (Mor)
~or! M It is easy to see that for each p E M, the set 7r~I(p) contains exactly two elements, which are the two orientations of TpM. We give the set Mo r a smooth structure. First let [J.Lo] be the standard orientation of]Rn and choose a fixed orientation reversing linear involution TO : ]Rn ---+ ]Rn. Let {(Ua , x a )} be an atlas for M. By composing some of the charts with TO and adding the resulting charts to the atlas, we may suppose that {(Ua , xa )} has the property that for every chart (U, x) in the atlas there is a chart (U, y) in the atlas with the same domain such that yo x-I is orientation reversing. Let us say that such an atlas is "balanced". (The maximal atlas is obviously balanced.) Now for each chart (U, x), where x = (xl, ... , x n ), define a map ¢x : x (U) -+ Mor by
c' then we define Ju a by the same formula. If ~ : V -+ U is an orientation preserving diffeomorphism of open sets in ]Rn, then det Det> > O. Let u l , ... ,un denote standard coordinates on U, and let
391
9. Integration and Stokes' Theorem
392
VI, ... ,vn denote standard coordinates on V. Then by the classical change of variable formula,
fu a = fu a(u) du l ... dun = fv a
0
(v) Idet DI dv l ... dv n
= fv a 0 ( v) det D dv l ... dv n
= fvao(v)detDdvl/\ ... /\dv n = fv*a. So (9.1) Next consider an oriented n-manifold M without boundary and let 0: E nn(M). If a has compact support inside U for some positively oriented chart (U,x), then x-I: x(U) ---+ U and (x-I)*a has compact support in x(U) c lRn. We define
fu
a:=
l(u/ x- )*a. I
The change of variables formula (9.1) shows that this definition is independent of the positively oriented chart chosen. In fact, if (y, V) is another chart and a has support inside Un V, then (x-I)*a has support inside x(U n V) and (y-I)* a has support in y(U n V). Thus since x 0 y-l is orientation preserving, we have
1
(x-I)*a
=
x(U)
1 1
(x-I)*a
=
x(unv)
=
1 1
(x 0 y-I)* (x-I)*a
y(Unv)
(y-l)* a
y(unv)
=
(y-l)*a.
y(U)
The same definition works fine in case M has nonempty boundary, but there is a small technicality. Namely, suppose we wish to work only with positive charts taking values in a fixed half-space lRA>c' Then as long as n ~ 2 there is no problem, but if n = 1, we are faced-with the fact that there may be no positively oriented lRLc-valued atlas at all even if M is orientable. Some authors define manifold with boundary completely in terms of a fixed half-space and seem unaware of this little glitch. Since we allow multiple half-spaces, this is not a problem for us, but in any case, we could modify the definition slightly in a way that works in all dimensions and for any chart. Let M be an oriented manifold with boundary. If a has compact support inside U for some chart (U, x), then (9.2)
( a Ju
= sgn(x)
1
x(U)
(x-I)*a,
9. Integration and Stokes' Theorem
393
where sgn(x) =
{
1 if (U, x) is positively oriented, -1 if (U,x) is not positively oriented.
This could be taken as a definition. Once again, the standard change of variables formula shows that this definition is independent of the chart chosen. Remark 9.2. Because we have used the sgn(x) factor in the definition, we can use arbitrary charts. But the manifold still must be oriented so that sgn(x) makes sense! If a E O~(M) has compact support but does not have support contained in some chart domain, then we choose a positively oriented atlas {(Xi, Ui)} for M and a smooth partition of unity {(pi, Ui)} subordinate to the atlas, and consider the sum
(9.3) Proposition 9.3. In the sum above, only a finite number of terms are nonzero. The sum is independent of the choice of atlas and smooth partition of unity {(pi, Ui)}. Proof. First, for any p E M there is an open set 0 containing p such that only a finite number of Pi are nonzero on O. But a has compact support, and so a finite number of such open sets cover the support. This means that only a finite number of the Pi are nonzero on this support. Now let {(Xi, Vi)} be another positive atlas and Pi a partition of unity subordinate to it. Then we have
o Since the sum (9.3) above is the same independently of the allowed choices, we make the following definition: Definition 9.4. Let (M, [roD be an oriented smooth manifold with or without boundary. Let a E on(M) have compact support. Choose a positively
9. Integration and Stokes' Theorem
394
oriented atlas {(x~, Ui )} and a smooth partition of unity nate to {Ui}. Then we define
f
a:=
i(M,[fIJ])
{(p~,
Ui )} subordi-
L iu,f Pia. i
We usually omit the explicit reference to the orientation [tv] and simply write 1M a. Remark 9.5. Of course if we take (9.2) as a definition, then we may take
and 1(M, fIJI) a = ~t 1u, pta even for an arbitrary unoriented atlas. Once again, we still need M to be oriented. In the online supplement we introduce twisted n-forms, and these may be integrated even on nonorientable manifolds!
In case M is zero-dimensional, and therefore a discrete set of points M = {P!'P2," .}, an orientation [tv] is an assignment of +1 or -1 to each point. Then, if a = f E nO(M), we have
r
i(M,[fIJl)
f =
L ±f(Pt),
where we choose the ± according to the orientation at the point.
9.1. Stokes' Theorem In this section we take up the main theorem of the chapter. It is the fundamental theorem of exterior calculus known as Stokes' theorem. Our definition of integration works for any (positive) atlas, but we can use a specific atlas for theoretical purposes. We will employ lR:1c where c = 0 and>' = -u1 • Let us consider two special cases of integration. Case 1. This is the case of a compactly supported (n - I)-form on IRn , Let Wj = fdu 1 /\ ••• /\ d;;J /\ ... /\ dun be a smooth (n -I)-form with compact support in lRn , where the caret symbol over the du j means that this j-th
9.1. Stokes' Theorem
395
factor is omitted. Then we have
r
1'Ji:
dWj 1
=
0
r
1'Ji:
d(fdu 1
/\ •.• /\
d;J /\ ... /\ dun)
1 1 and i > j. The maps Itk are defined above.
o/Jk 1
It is convenient to define Ck(M, G) to be the trivial group {O} for all k < 0 and define Ok - 0 for all k < O. Then 000 = 0 remains true. The sequence of spaces and maps
... ~ Ck+1(M, G) ~ Ck(M, G) ~ Ck I(M,G) ~ ... is called the singular chain complex with coefficients in G. Let Zk(M, G) Kerok and Bk(M, G) :- 1m Ok-I. Then because of equation (9.5) we have Bk(M, G) c Zk(M, G). The k-th singular homology group Hk(M; G) with coefficients in G is defined by
Zk(M,G) Hk(M; G) = Bk(M, G)' If C E Zk(M, G), then the equivalence class of C is denoted [c]. If CI and C2 are k-chains in the same class, then CI = C2 + OC for some C E Bk(M, G , and in this case we say that q and C2 are homologous (or in the same homology class). The most important choices for G are JR and Z.
Exercise 9.14. Check that if G = JR, then Ck(M, JR), Zk(M, JR), Bk(M,lR and Hk(M; JR) are all vector spaces in a natural way and the boundary maps o extend to linear maps. We define the integral of a k-form as
1
0::=
a
0:
over a smooth singular k-simplex (J
r
JAk
4;>*0:,
9.3. Stokes' Theorem for Chains
and then for a chain c =
403
I:a: CuU E Ck(M, R) we can define
j a=L cu l a . u
c
u
Ju 1 = 1(0). We state
If U is a O-simplex and if 1 E nO(M) = COO(M), then without proof the following version of Stokes' theorem:
Theorem 9.15 (Stokes' theorem for chains). Let M be a smooth manilold. For c E Ck+1(M,R), and a a k-Iorm on M, we have
1 = lac da
a.
For a proof see [War]. Notice that in this version, M is not assumed to be orient able. Also, a need not have compact support since the chain c has finite support. If u is a I-simplex and 1 E COO(M), then the above reduces to
1=
f(u (1)) - I(u (0)).
df
Now we define the de Rham map. First, if a E Zk(M) c nk(M), then we can define an element la of Hk(MjR), the dual space of Hk(MjR): For [c] E Hk(Mj R) represented by c E Zk(Mj R) we define
/a([c]) =
1
a.
This is well-defined since if c+ Bd E [c], then
1 l
la(c+Bc') =
c+ac'
a
a + { a = ja+
=
Jac'
c
=
1
a
c
(since da
1
da
(Stokes)
c'
= 0).
This gives a linear map Zk(M) ---+ Hk(Mj R). Now if a E Bk(M) (image of d), then a - d{3 and
la([eD = ja = jd{3 - { {3 = 0 (since Be = 0). c
c
Jac
Thus we obtain a linear map called the de Rham map H~eR(M) ---+ Hk(Mj R),
where H~eR (M) = Zk (M) / Bk (M) is the de Rham cohomology defined earlier. The content of the celebrated de Rham theorem is in part that this map is an isomorphism. The theorem is fairly difficult to prove.
9. Integration and Stokes' Theorem
404
Theorem 9.16 (de Rham). The de Rham map defined above is an isomorphism, H~eR(M) ~ Hk(M; IR).
o
Proof. See [Bo-Tu] or [War].
We have defined Hk(M; IR) using smooth singular chains, but we could have used continuous chains. The result is isomorphic to Hk(M;IR) as we have defined it (see [War] or [Bo-Tu]).
9.4. Differential Forms and Metrics Let (V,g) be a real scalar product space (not necessarily positive definite). We wish to induce a scalar product on L~t(V) ~ I\k(V*). Even though elements of L!lt(V) ~ I\k(V*) can be thought of as tensors of type (O,k) that just happen to be alternating, we will give a scalar product to this space in such a way that the basis
(9.6) is orthonormal if e1 , •.• ,en is orthonormal. Recall that if a, 13 E V*, then by definition (a,f3) = g(a d, f3d). Now suppose that a = a 1 1\ a 2 1\ •.. 1\ ci and 13 = 13 1 1\ 13 2 1\ ... 1\ 13 k , where the a i and f3i are I-forms. Then we want to have
(alf3) = (a 1 1\ a 2 1\ .. . 1\ ak
I 13 1 1\ 13 2 1\ ... 1\ 13 k )
= det [(a i ,f3i )] ,
where [( ai, f3i)] is an n x n matrix. Notice that we use (a I13) rather than (a,f3) since the latter could be taken to be the natural inner product of a and 13 as tensors-the latter differs from the first by a factor. We want to extend this bilinearly to all k-forms. We could just declare the basis (9.6 above to be orthonormal and thus define a scalar product. Of course, one must then show that this scalar product does not depend on the choice of basis. For completeness, we now show how to arrive at the appropriate scalar product using universal mapping properties and obtain some formulas. Fix 13 1 ,132 , •.• ,13k E V* and consider the map J-Lpl,p2, ... ,pk : V* x ... x V* -+ IR given by J-Lpl,p2, ... ,pk : (al, a 2 , .•• , a k ) t-+ det [(a i ,f3i )] . Since this is an alternating multilinear map, we can use the universal property of 1\ k (V*) to see that this map defines a unique linear map Jipl,p'l, ... ,pk : /\ k (V*) -+ IR
405
9.4. Differential Forms and Metrics
such that Ji{jI,/32, ... ,/3k
.. ] . (a1 /\a2 /\ ..k ·/\a) =det [ (a\/31)
Similarly, for fixed a E Ak(V*), the map ma:: (131, 13 2 , ... ,13k ) I--t ilf31 ,13 2 ,... ,f3k (a) is alternating multilinear and so gives rise to a linear map
_ : /\k (V*) --+ IR ma: such that
ma:(f31/\ 132 /\
.•. /\
13 k ) = il(31,f32, ... ,(3k (a) .
Lemma 9.17. The map k
k
(·1·) : /\ (V*) x /\ (V*) --+ lR defined by
(alf3) 2S
:=
ma:(f3)
symmetric and bilinear. We have
(a 1 /\ a 2 /\ ... /\ a k 1 13 1 /\ 13 2 /\ ... /\ 13 k ) = det [(a i ,f3i )]
Proof. By construction, the map is linear in the second slot for each fixed Fix 13 and write 13 as a sum 13 = 2: bi1 ... ik f3'1 /\ ... /\ f3'k in any way. Then
0:.
ma: (13) =
~ . ma: L..J b·'l""k
(f3i 1 /\ ••• /\ f3i k)
- L..J ~ b·~l'''~kr'f3 . Ii 1 ,(3 2 , ... k ,(3 (a)
•
But each map a I--t Ji(31,(32, ... ,(3k (a) is linear and so a I--t ma:(f3) is also linear. Now let cp(.,.) : W x W --+ lR be any bilinear map on a real vector space W. If SeW spans Wand if cp(81' 82) = cp(82' 81) for all 81, 82 E S, then cp is symmetric. The set of all elements 13 E Ak (V*) of the form 13 = 131 /\ •.• /\ 13 k for I-forms 131, •.• , 13 k , is a spanning set. Since
(a 1 /\ a 2 /\ ... /\ a k 1 13 1 /\ 132 /\ ... /\ 13 k ) = det [(a i ,f3J )] = det [(f3i ,ai )] =
(13 1 /\ 13 2 /\ .•. /\ 13 k a 1 /\ a 2 /\ ••. /\ a k ),
we conclude that (,1,) is symmetric.
1
o
The bilinear map defined in the previous lemma is a scalar product for the vector space Ak(V*). Notice the vertical bar rather than a comma in
9. Integration and Stokes' Theorem
406
the notation. If e1 , ... ,en is an orthonormal basis for V*, then {eiI /\ ... /\ e~k hl 0 we have
for all i. Then {ai}i has a Cauchy subsequence.
Let 1£k = {w E nk(M) : ~w = O} and let (1£k).1 denote {f3 (f3lw) = 0 for all w E 1£k}. Lemma 9.52. There exists a constant C all f3 E (1£k).1.
> 0 such that 11f311
~
E
nk(M):
C II~.B for
421
9.6. The Hodge Decomposition
Proof. Suppose there is no such constant C. Then we may find a sequence {,8l}f C (1-£k).1 such that II,8ill = 1 while liIDi-+oo 1I~,8ill = O. By Proposition 9.51, {,8l}r has a Cauchy subsequence, which we may as well assume to be {,8i}i. Hence we have that for any fixed e E nk(M), the sequence {(,8ile)} is Cauchy in 1R and so converges to some number. Now we define l: nk(M) -t 1R by
£(e) :- lim (,8lle). '-+00
Then,
and since £ is clearly bounded, it is a weak solution to ~w = 0. By Theorem 9.50, there must exist ,8 E nk(M) such that £(e) = (,8le) for all e E nk(M). It follows that (,8le) = limi-+oo (,8i 10) for all 0 and so ,8i -t ,8 since
II,8i - ,8J12 = (,8i - ,81,8i - ,8) = (,8il,8i) - 2(,8il,8) + (,81,8) -t O. Since II,8ill = 1, we must have 11,8J1 = 1 and of course ,8 E (1-£k).1. But by Theorem 9.50, ~,8 = 0 so ,8 E 1-£k n (1-£k) .1 = 0, which contradicts 11,811 - 1. We conclude that C exists after all. 0 Theorem 9.53 (Hodge decomposition). Let (M, g) be compact and oriented (and without boundary). For each k with 0 ::; k ::; n = dim M, the space
of harmonic k-forms 1-£k is finite-dimensional. Furthermore, we have an orthogonal decomposition of nk(M), nk(M) = ~ (nk(M)) ffi 1-£k = d8 (nk(M)) ffi 8d (nk(M)) ffi 1-£k
=d(n k- 1 (M)) ffi8(nk+l(M)) ffi1-£k. Proof. If 1-£k were infinite-dimensional, then it would contain an infinite orthonormal sequence {wi}i. In this case, we would have
t-
for all i,j with i j. By Proposition 9.51, this sequence would contain a Cauchy subsequence. But this contradicts the above equation. Thus 1-£k must be finite-dimensional. We now prove the orthogonal decomposition nk(M) = ~ (nk(M)) ffi1-£k. The other two decompositions can be derived from the first, and we leave
9. Integration and Stokes' Theorem
422
this as a problem for the reader (Problem 9). Choose an orthonormal basis WI, ••• ,Wd for l£k. If a E nk(M), then we may write d
a = 13 + L(Wila)Wi i=l
where 13 E (l£k) 1.. It is easy to show that this decomposition is unique, so we have the orthogonal decomposition nk(M) = (l£k) 1. EB l£k and our task is to show that (l£k) 1. = A (nk(M)). Since (Aalw) = (aIAw) - 0 whenever W E l£k, we see that A (nk(M)) C (l£k) 1.. Now let a E (llk).i and define a linear functional f on A (nk (M)) by
f(AO)
:=
(aIO).
This is well-defined since if AOI = A02, then 01 - 02 E l£k and so (allh) (aI02) = (alOl - (2) = o. We show that f is bounded. Let := 0 - H(O). where Hk : nk(M) --t l£k is the orthogonal projection. Then using Lemma 9.52 we have If(~O)1
= If(~¢)1 = l(al¢)1 ~ lIallll¢11 ~ C IlalIIlA¢1I = C lIa1i1lA011 .
By the Hahn-Banach theorem, the functional f extends to a bounded functionall defined on all ofnk(M), which is then a weak solution of Aw = a. By Theorem 9.50, there is an wE nk(M) with Aw = a so (1I. k ) 1. C A (nk(M)) and so (l£k) 1. = A (nk(M)). 0 In order to take full advantage of the Hodge decomposition, we now introduce a so-called "Green's operator" Gk : nk(M) -+ (l£k) 1.. We simply define Gk(a) to be the unique solution of Aw = a - Hk(a) where Hk : nk(M) --t l£k is the orthogonal projection as above. The Gk combine to give a map G : n(M) --t EBk (l£k) 1. also called the Green's operator. Lemma 9.54. Let Gk : nk(M) --t (l£k)
1.
be the Green's operator defined
above for each k. Then (i) Gk is formally self-adjoint; (ii) if L : nk(M) --t nr(M) is linear and commutes with A, then G commutes with L (that is, L 0 Gk = Gr 0 L). In particular, G commutes with d and 8. Proof. We have
(Gk(a)lf3) = (Gk (a) 113 - Hk(f3)) = (G k(a)I AGk(f3)) = = (a - Hk(a)IG k (13)) = (alGk (13)),
(~Gk(a)IGk
(13))
423
9.6. The Hodge Decomposition
so G is self-adjoint. For each j, let 1rj : OJ(M) -+ (1I.i).1 denote orthogonal projection (thus 1rJ + HJ = id{l.1). Now suppose that L : Ok(M) -+ or(M) is linear and commutes with Ll. Notice that by definition we have Gle = (AI (1-£k) .1) 01rk. The fact that LLl = LlL implies that L(1-£k) C 1-£r. Also, since Ll (Ok(M)) = (1-£k).1, we have L((1-£k).1) C (1-£r).1. Thus
L 0 1rk =
1rr 0
L,
L 0 (LlI (1-£k).1) = (LlI (1-£r).1) 0 LI(,w).L , and (LlI (1-£r) .1)-10 L = LI(,w).L 0 (LlI (l£k) .1)-1. It follows that G commutes with L.
o
We note in passing that G maps bounded sequences into sequences that have Cauchy subsequences. Indeed, suppose that {ai}f C Ok(M) is a sequence with lIaill ~ O. If f3i := G(ai), then using Lemma 9.52 we have lIf3ill ~ II Af3i II = Ilai - H(ai)11 ~ Ilaill ~ 0,
and so by Proposition 9.51, {f3i}f has a Cauchy subsequence. Theorem 9.55. Let (M,g) be a compact Riemannian manifold (without boundary). Then each de Rham cohomology class contains a unique harmonic representative. Proof. First assume that M is orient able and fix an orientation. Let a E
nk(M) and use the Hodge decomposition and the definition of G to obtain a
= LlGk(a) + Hk(a) = d8G k(a) + 8dGk(a) + Hk(a).
Then since G commutes with d, we have a = d8Gk(a)
+ 8Gk+1(da) + Hk(a).
So if da = 0, then a - Hk(a) = d8Gk(a), and so Hk(a) represents the same cohomology class as a. To show uniqueness, suppose that a1 and a2 are both harmonic and in the same class so that a2 - a1 = df3 for some f3. Then we have 0 = df3 + (a1 - a2). But a1 - a2 is orthogonal to df3 since by Proposition 9.48
(df31 a1 - a2) = (f3 I8a1 - 8a2) = (f310) = o. Next suppose that M is nonorientable. If 1r : Mor -+ M is the 2fold orientation cover of M (Section 8.7), then there is a unique metric on Mor such that 1r restricts to an isometry on sufficiently small open sets (1r is a Riemannian covering). Now suppose that c E Hk(M) is a de Rham cohomology class and consider the class 1r"'C E Hk(Mor) (see the discussion after Definition 8.41). Since Mor is orient able , 1r"'C is uniquely represented
9. Integration and Stokes' Theorem
424
by a harmonic form "ij on Mor. If r : Mo r -+ Mo r is the involution which transposes the two points in each fiber, then it is easy to see that r is an isometry so that r*"ij is also harmonic. It follows from the identity 11" 0 r 7r that r*"ij is also a representative of the cohomology class 11"* c. Indeed, [r*17] r* [1j] = r*1I"*c = 1I"*c. So by uniqueness, we must have r*"ij = "ij, which is exactly the condition that guarantees that "ij = 1I"*rJ for some rJ E nk(M). But 11" is a local isometry and so by Problem 5, rJ must be harmonic. We show that rJ represents c. Suppose that c = [JL] for JL E nk(M). Then both 11"* JL and 1I"*rJ represent the class 11"* c, so 11"* (rJ - JL) = df3 for some f3. But then, as for r*"ij and "ij above, r* f3 is in the same class as f3, which means that df3 = d r* f3. Then 11"*
(rJ - JL) = df3 =
~ (df3 + d r* f3) = d (~(f3 + r* f3)) .
Since r is an involution, f3 + r*f3 is r-invariant, that is, r* (f3 + r*{3) (f3 + r* f3), so there exists a E Ok (M) with 11"* a = (f3 + r* f3). Thus
!
11"*
(rJ - JL) = d 11"* a =
=
11"* d a,
and since 1T"* is a local diffeomorphism, we must have rJ - JL = da, which means that [rJ] = [JL] = c. The uniqueness of the harmonic representative TJ is clear. 0 Notice that if r is the involution of Mor introduced in the above proof, then 1T"*a = r*1I"*a for any a E nk(M). Corollary 9.56. If M is compact, then its cohomology spaces Hk(M) a e finite-dimensional. Proof. Any manifold can be given a Riemannian metric, and so if M is compact and oriented, then Theorems 9.53 and 9.55 combine to give the result. Now suppose that M is not orientable. Let a E nk(M) and suppose that 1T"*a = df3 so that 1T"*[a] = O. Then 1I"*a = r*1T"*a = r*df3 = dr*{3. Now 11"*0 = !(f3 + r*f3) for some 0 and so
1I"*a = d (~(f3 + r* f3)) = d1l"*0 = 1I"*dO. Since 11" is a local diffeomorphism, we have a = dO or [a] = O. Thus the map 11"* : Hk(M) -+ Hk(Mor) has trivial kernel and is injective. The finite-dimensionality of Hk(M) now follows from that of Hk(Mor). 0 The results above allow us to give a quick proof of Poincare duality for de Rham cohomology. Choose an orientation for M. We define a bilinear pairing Hk (M) x Hn-k (M) -+ lR as follows:
(([w], [rJ))) =
1M w 1\ rJ·
9.7. Vector Analysis on ]R3
This is well-defined since if WI = then by Stokes' theorem
1M
WI
1\ "71
425
W
+ dO' and "71 = "7 + d(3 (with w, "7 closed),
= 1M W 1\ "7 + 1M da 1\ "7 + 1M W 1\ d(3 +
L
dal\ d(3
1M W1\ "7 + 1M d (a 1\ "7) - 1M d (W 1\ (3) + 1M d (a 1\ d(3) = 1M wl\"7. =
We wish to show that the pairing defined above is nondegenerate. Thus, given any nonzero [w] E Hk(M) we wish to produce a ["7] E Hn-k(M) such that (([w], ["7])) =I O. Choose a Riemannian metric and metric volume element for M. We may assume that W is the harmonic representative for [w]. But it is easily checked that ~ commutes with * and so *w is also harmonic. In particular, *w is closed and so represents a cohomology class [*w]. Then [*w] is the desired class since
(([w], [*w]))
= 1M w 1\ *w =
1M (wi * w) vol = IIwl1
2
> O.
For each fixed [w] E Hk(M), we have the linear map t[w] E (Hn-k(M))* given by t[w] (["7D := (([w], ["7])). Since the pairing is nondegenerate, the map [w] t-+ t[w] defines an isomorphism from Hk(M) to (Hn-k(M))*. Thus we have proved the following: Theorem 9.57 (Poincare duality). If M is an orientable compact n-manifold without boundary, then we have an isomorphism
coming from the pairing defined above.
We will take up Poincare duality again in Chapter 10.
9.7. Vector Analysis on IR3 In ]R3, the I-forms may all be written (even globally) in the form 8 = hdx+ hdy+ !adz for some smooth functions h , h and !a and all 2-forms (3 may be written (3 = gldyl\dz+g2dzl\dx+g3dxl\dy. The forms dyl\dz, dzl\dx, dxl\dy constitute a basis (in the module sense) for the space of 2-forms on]R3 just as dx, dy, dz form a basis for the I-forms. The single form dx 1\ dy 1\ dz provides a module basis for the 3-forms in ]R3. Suppose that x( u, v) parametrizes a surface S C ]R3 so that we have a map x : U ~ ]R3. Then the surface is
9. Integration and Stokes' Tbeorem
426
oriented by this parametrization, and the integral of f3 over S is
if3=i~~A&+~&A~+~~A~ {( 8(y,z) = Ju gl(X(U, v)) 8 (u, v)
8(z,x)
+ g2 (x(u, v)) 8 (u, v)
8(x,y)) + g3 (x(u, v)) 8 (u, v) dudv.
Here and in the following we disregard technical issues about integration (but recall Theorem 9.7). Exercise 9.58. Find the integral of f3 = x dy A dz + dz A dx + xz dx 1\ dy over the sphere oriented by the parametrization given by the usual spherical coordinates ¢, f}, p. If w = h dx Ady Adz has support in a bounded open subset U c 1R3 which we may take to be given the usual orientation implied by the rectangular coordinates x, y, z, then
fu w = fu
hdx A dy A dz
=
fu
hdxdydz.
In order to relate differential forms on 1R3 to vector calculus on ]R3, we will need some ways to relate forms to vector fields. To a l-form f} = hdx+ hdy + hdz, we can obviously associate the vector field ~f} = hi + f2,j + 13k. But recall that this association depends on the notion of orthonormality prov~ed by_the d~t product. If f} is expressed in sa~ sph~ica15oordinates f} = hdp+ hdf}+ hd¢, then it is not true that ~f} = hi+ f2,j + 13k. Neither is it generally true that ~f} = hp + lif + h'¢, where p, 0, '¢ are unit vector fields in the coordinate directions 1 and where the are just the fi expressed in polar coordinates. Rather, our general formulas give -1.- 1 ..... ~f} = fIp + h-f} + 13-.-f}¢. p psm In rectangular coordinates x, y, z, we have
h
~:
~: ~:
dx dy dz
-+ i, -+ j, -+ k,
while in spherical coordinates we have ~:
dp ~ : pdf} ~: psin f} d¢ 1 Here
() is the polar angle ranging from 0 to
'7r.
t-+
p,
t-+
0,.....
t-+
¢.
427
9.7. Vector Analysis on JR3
As an example, we can derive the familiar formula for the gradient in spherical coordinates by first just writing t in the new coordinates t(p, 0, ¢) := !(x(p, 0, ¢), y(p, 0, ¢), z(p, 0, ¢)) and then sharping the differential
at at at df = ap dp + ao dO + a¢ d¢ to get
af a
gradf = ~df = ap ap
1 af a
1
af a
+ pao ao + psinO a¢ a¢'
where we have used
1 0
10 [
[
o P o 0
o o
o 1.
1p
0
psfn8
In order to proceed to the point of including the curl and divergence of traditional vector calculus, we need a way to relate 2-forms with vector fields. This part definitely depends on the fact that we are talking about forms in JR3. We associate to a 2-form 7] the vector fields H*7]). Thus 91dy /\ dz + g2dz /\ dx + ggdx /\ dy gives the vector field X = gli +g2j + ggk. Now we can see how the usual divergence of a vector field comes about. First fiat the vector field, say X = Ili+ hj+ fgk, to obtain pX = /ldx + /2dy + fgdz and then apply the star operator to obtain /ldy /\ dz + hdz /\ dx + fgdx /\ dy. Finally, we apply exterior differentiation to obtain
d(/ldy /\ dz + hdz /\ dx + fgdx /\ dy) =~/\~/\~+~/\~/\~+~/\~/\~
=
( all ax dx + all ay dy + all) az dz /\ dy /\ dz + the other two terms all
= ax dx /\ dy /\ dz
all = ( ax
+
ah afg ax dx /\ dy /\ dz + ax dx /\ dy /\ dz
afg) + ah ax + ax dx /\ dy /\ dz.
Now we see the divergence appearing. In fact, if we apply the star operator one more time, we get the function div X = ~ + ~ +~. We are thus led to *d* (pX) = div X which agrees with the definition of divergence given earlier for a general semi-Riemannian manifold.
9. Integration and Stokes' Theorem
428
What about the curl? For this, we just take d (~X) to get
d (IIdx + hdy + hdz) = dII =
1\
(a;: dx + 0:: dy + 0:: dZ)
= (a h _ a h ) dy 1\ dz
ay
az
dx + dh 1\
+ (a h
az
1\
dy + dh 1\ dz
dx + the obvious other two terms _ alI) dz 1\ dx + (alI _ a h ) dx 1\ dy ax ax ay
and then apply the star operator and sharping to get back to vector fields obtaining
In short, we have ~
* d (~X) =
curlX.
Exercise 9.59. Show that the fact that dd = 0 leads to both of the following familiar facts:
curl (grad f) = 0, div (curl X) = O. The 3-form dx 1\ dy 1\ dz is the (oriented) volume element of ]R3. Every 3-form is a function times this volume form, and integration of a 3-form over a sufficiently nice subset (say a compact region) is given by JD W= JD f dx 1\ dy 1\ dz = JD f dx dy dz (usual Riemann integral). Let us denote dx 1\ dy 1\ dz by dV. Of course, dV is not to be considered as the exterior derivative of some object V. Let u 1 , u 2, u 3 be curvilinear coordinates on an open set U C ]R3 and let 9 denote the determinant of the matrix [gij 1where gij = (a~" a':v ). Then d V = J9 du 1 1\ du 2 1\ du 3 • A familiar example is the case when (u 1 , u 2, u 3 ) are spherical coordinates p, 0, ¢>, in which case dV = p2 sin 0 dp 1\ dO and if D
c
]R3
1\
d¢>,
is parametrized by these coordinates, then
in
in = in
fdV =
f(p, 0, ¢»p2 sinO dp 1\ dO 1\ d¢> f(p, 0, ¢»p2 sinO dpdO d¢>.
If we go to the trouble of writing Stokes' theorem for curves, surfaces and domains in ]R3 in terms of vector fields associated to the forms in an appropriate way, we obtain the following familiar theorems (using standard
9.B. Electromagnetism
notation):
429
1
Vf· dr = f(r(b)) - f(r(a)),
JIs
curl (X) x dS =
JJl
div (X) dV =
t
X . dr (Stokes' theorem),
JIs
X . dS (Divergence theorem).
Similar and simpler things can be done in R2 leading for example to the following version of Green's theorem for a planar domain D with (oriented) boundary c = aD:
l (~~ -~~)
dx/\dy
=
l
d(Mdx+Ndy)
=
1
Mdx+Ndy.
All of the standard integral theorems from vector calculus are special cases of the general Stokes' theorem.
9.8. Electromagnetism In this subsection we take a short trip into physics. Consider Maxwell's equations2 :
V·B=O, V x
aB
E+Ft =0, V· E =
VxB_
aE
at
(!,
=j.
Here E and B, the electric and magnetic fields, are functions of space and time. We write E = E(t,x), B = B(t,x). The notation suggests that we have conceptually separated space and time as if we were stuck in the conceptual framework of the Galilean spacetime. Our purpose is to slowly discover how much better the theory becomes when we combine space and time in Minkowski spacetime Rt. Recall that Rt is treated as a semi-Riemannian manifold, which is ]R4 endowed with the indefinite metric (x, y)1I = -xOyo + E~=l xiy~. Here the standard coordinates are conventionally denoted by (xO, x \ x 2 , x 3 ), and xO is to be thought of as a time coordinate and is also denoted by t (we take units so that the speed of light c is unity). In what follows, we let r = (xl, x 2 , x 3 ), so that (xO,x 1 ,X2 ,x3 )
= (t,r).
2 Actually, thIS is the form of Maxwell's equations after a certain convenient choice of units, and we are ignoring the somewhat subtle distinction between the two types of electric fields E and D and the two types of magnetic fields B and H and their relation in terms of dielectric constants.
9. Integration and Stokes' Theorem
430
The electric field E is produced by the presence of charged particles. Under normal conditions a generic material is composed of a large number of atoms. To simplify matters, we will think of the atoms as being composed of just three types of particle; electrons, protons and neutrons. Protons carry a positive charge, electrons carry a negative charge and neutrons carry no charge. Normally, each atom will have a zero net charge since it will have an equal number of electrons and protons. If a relatively small percent of the electrons in a material body are stripped from their atoms and conducted away, then there will be a net positive charge on the body. In the vicinity of the body, there will be an electric field which exerts a force on charged bodies. Let us assume for simplicity that the charged body which has the larger, positive charge, is a point particle and stationary at ro with respect to a rigid rectangular coordinate system that is stationary with respect to the laboratory. We must assume that our test particle carries a sufficiently small charge, so that the electric field that it creates contributes negligibly to the field we are trying to detect (think of a single electron). Let the test particle be located at r. Careful experiments show that when both charges are positive, the force experienced by the test particle is directly away from the charged body located at ro and has magnitude proportional to qe r2, where r = Ir - ro I is the distance between the charged body and the test particle, and where q and e are positive numbers which represent the amount of charge carried by the stationary body and the test particle respectively. If the units are chosen in an appropriate way, we can say that the force F is given by r-ro F = qe 3' Ir- rol By definition, the electric field at the location r of the test particle is
(9.12)
E=q
r-ro Ir - rol
3'
If the test particle has charge opposite to that of the source body, then one of q or e will be negative and the force is directed toward the source. The test particle could have been placed anywhere in space, and so the electric field is implicitly defined at each point in space and so gives a vector field on ]R3. If the charge is modeled as a smoothly distributed charge density p which is nonzero in some region U C ]R3, then the total charge is given by integration Q = p(t, r) dVr and the field at r is now given by E(t, r) = p(t, y) 1:~~3 dV Since the source particle is stationary at rOt the electric field will be independent of time t. A magnetic field is produced by circulating charge (a current). If charge e is located at r and moving with velocity v in a magnetic field B(t, r), then the force felt by the charge is F = eE + ~v x B, where v is the velocity of the test particle. The test
Ju
Ju y.
9.B. Electromagnetism
431
particle has to be moving to feel the magnetic part of the field! At this point it is worth pointing out that from the point of view of spacetime, we are not staying true to the spirit of differential geometry since a vector field should have a geometric reality that is independent of its expression in a coordinate system. But a change of inertial frame can make B zero. Only by treating E and B together as aspects of a single field can we obtain the proper view. Our next task is to write Maxwell's equations in terms of differential forms. We already have a way to convert (time dependent) vector fields E and B on JR3 into (time dependent) differential forms on JR3. Namely, we use the flatting operation with respect to the standard metric on JR3. For the electric field we have
For the magnetic field we do something a bit different. Namely, we flat and then apply the star operator. In rectangular coordinates, we have
If we stick to rectangular coordinates (as we have been), the matrix of the standard metric is just 1= (dij), and so we see that the above operations do not numerically change the components of the fields. Thus in any rectangular coordinate system we have
and similarly for the B's. It is not hard to check that in the static case where and B are time independent, the first pair of (static) Maxwell's equations are equivalent to
e
dE
= 0 and dB = O.
This is nice, but if we put time dependence back into the picture, we need to do a couple more things to get a nice viewpoint. So assume now that E and B and hence the forms E and B are time dependent, and let us view these as differential forms on spacetime JRt. In fact, let us combine E and B into a single 2-form on JRt by setting
F= B+E 1\ dt.
!
Since F is a 2-form, it can be written in the form F = Fp,vdxP, 1\ dx v , where Fp,v - -Fvp, and where the Greek indices are summed over {O, 1,2, 3}. It is traditional in physics to let the Greek indices run over this set and to let Latin indices run over just the "space indices" 1, 2, 3. We will follow this convention for a while. If we compare F = B + E 1\ dt with ~ Fp,vdxP, 1\ dx v ,
9. Integration and Stokes' Theorem
432
we see that the FJ-LIJ form an antisymmetric matrix which is none other than
[
~x -~x Ey Ez
-Bz By
-t
y
=~:]
0
Bx
-Bx
0
.
Our goal now is to show that the first pair of Maxwell's equations are equivalent to the single differential form equation
dF=O. Let N be an n- manifold and let M = (a, b) x N for some interval (a, b). Let the coordinate on (a, b) be t = xo (time). Let (xl, ... , xn) be a coordinate system on N. With the usual abuse of notation, (xO, xl, ... ,xn) is a coordinate system on (a, b) xN. One can easily show that the local expression dw = BJ-LfJ-Ll ... J-Lk /\ dxJ-L /\ dx/1 1 /\ ••• /\ dXJ-Lk for the exterior derivative of a form w = f /11···J.l.k dXJ-Ll /\ ... /\ dX/1k can be written as 3
dw
(9.13)
=~ /\ dx i L...J {Jw t J-Ll···J-Lk
/\
dXJ-Ll /\ ... /\ dXJ-Lk
i=l
+ BOWJ-Ll ... J-Lk /\ dxO /\ dXJ-Ll /\ ... /\ dXJ-Lk, where the /li sum over {O, 1, 2, ... , n}. Thus we may consider the spatial part ds of the exterior derivative operator d on (a, b) x S = M. That is, we think of a given form w on (a, b) x S as a time dependent form on N so that dsw is exactly the first term in the expression (9.13) above. Then we may write dw = dsw + dt /\ Btw as a compact version of the expression (9.13). The part dsw contains no dt's. By definition F = B + & /\ dt on lR x lR3 = lRt, and so
dF = dB + d(& /\ dt) = dsB + dt /\ BtB + (ds& + dt /\ Bt&) /\ dt = dsB + (BtB + ds&) /\ dt. The part dsB is the spatial part and contains no dt's. It follows that dF is zero if and only if both dsB and BtB + ds& are zero. Unraveling the definitions shows that the pair of equations dsB = 0 and BtB + ds& = 0 (which we just showed to be equivalent to dF = 0) are Maxwell's first two equations disguised in a new notation. In summary, we have
dF = 0
¢:::::?
dsB=O BtB+ds& = 0
V'·B=O V'xE+~~=O
Below we rewrite the last pair of Maxwell's equations, where the advantage of combining time and space together manifests itself to an even greater degree. Let us first pause to notice an interesting aspect of the
9.B. Electromagnetism
433
first pair. Suppose that the electric and magnetic fields were really all along most properly thought of as differential forms. Then we see that the equation dF = 0 has nothing to do with the metric on Minkowski space at all. In fact, if ¢ : IRf -+ IRf is any diffeomorphism at all, we have dF = 0 if and only if d( ¢* F) = 0, and so the truth of the equation dF = 0 is really a differential topological fact; a certain form F is closed. The metric structure of Minkowski space is irrelevant. The same will not be true for the second pair. Even if we start out with the form F on spacetime it will turn out that the metric will necessarily be implicit in the differential forms version of the second pair of Maxwell's equations. In fact, what we will show is that if we use the star operator for the Minkowski metric, then the second pair can be rewritten as the single equation *d * F = J, where J is formed from j = (jl, j2, j3) and p as follows: First we form the 4-vector field J = pat + lax +j 2 ay + j 3 az (called the 4-current) and then using the flatting operation we produce .J - -pdt+j 1 dx+j 2 dy+j 3 dz = JO dt+J1 dx+J2 dy+Jadz, which is the covariant form of the 4-current. We will only outline the passage from *d * F = J to the pair V . E = {! and V x B = j. Let *s be the operator one gets by viewing differential forms on IRt as time dependent forms on IR3 and then acting by the star operator with respect to the standard metric on IR3. The first step is to verify that the pair V . E = {! and V x B - ~~ = j is equivalent to the pair *sds *s E = {! and -atE + *sds *s B = J, where J := j 1dx + j 2 dy + j 3 dz and Band E are as before. Next we verify that
cg;
*F = *s£ - *sB 1\ dt. So the next goal is to get from *d * F = *J to the pair *sds *s E = -atE + *sds *s B = J. The following exercise finishes things off.
{!
and
Exercise 9.60. Show that *d * F = -atE - *sds *s £ 1\ dt + *sds *s Band then use this and what we did above to show that *d * F = J is equivalent to the pair *sds *s £ = {! and -ate + *sds *s B = J. We have arrived at the following formulation of Maxwell's equations:
dF=O, *d*F= J. If we just think of this as a pair of equations to be satisfied by a 2-form F where the I-form J is given, then this last version of Maxwell's equations makes sense on any semi-Riemannian manifold. In fact, on a Lorentz manifold that can be written as (a, b) x S = M with the product metric -dt®dt x g, for some Riemannian metric 9 on S, we can write F = B+E I\dt,
9. Integration and Stokes' Theorem
434
which allows us to identify the electric and magnetic fields in their covariant form.
9.9. Surface Theory Redux Consider a surface M c ]R3. In what follows, we take advantage of the natural identifications of the tangent spaces of ]R3 with ]R3 itself. Let el, e2 be an oriented orthonormal frame field defined on some open subset U of M and let e3 - el x e2. We can think of each e}, e2 and e3 as vector fields along U. Using the identifications mentioned above, we also think of them as ]R3- valued O-forms. Note that the identity map idJR3 may be considered as an ]R3-valued 0form on ]R3. Let I := t* (idJR3) where t : U c M "---+ ]R3 is inclusion. Then dI := t *d (idJR3) = 0 since idJR3 is constant. Thus, I is an ]R3-valued O-form, and we may write 1= el e 1 + e2e2, where (19 1 ,19 2) is the frame field dual to (el' e2). Note that volM = 19 1 /\ 02• We have the ]R3-valued I-forms dej and 3
dej = Lekwj k 1
for some matrix of I-forms (w}). Note that if v E TpM, then dej(v) = 'Vue], where V' is the flat Levi-Civita derivative on ]R3. (In fact, we should point out that if i,j, k is the standard basis on ]R3, then any field X along U may be written as X = hi+ h,j+ 13k for some smooth functions h, 12, 13 defined on U and may be considered as an ]R3-valued I-form. Then dX(v) = V'vX = dh{v)i + dh(v)j + dh(v)k.) For an arbitrary tangent vector v we have
wj(v) = (dej(v),ei) = - (ej,dei(V)) = -w{(v), and so it follows that the matrix (w}) is antisymmetric. We write ]R3-valued forms on U as E~-l ekrl where 17k E O(U). This is to conform with the order of matrix multiplication. Theorem 9.61. Let Me ]R3, e1, e2, 191, 19 2 and (w}) be as above. Then the following structure equations hold: 2
de i
=-
wr /\
Lwl /\ 19 k , k=l
19 1
+ w~ /\ 19 2 =
""'i
0,
3
k dWji = - L..Jwk 1\ Wj' k=l
9.9. Surface Theory Redux
435
Proof. We calculate as follows: 2
o=
dI = d L k 2
2
ek(i - L dek A Ok 1 k-l
3
2
= L ( L wte,) k=l , 2
+L
A Ok
ek A dO k
k=l
1
2
=L (
?= ejwt + e3w~)
k=l
,
2
2
+ ek A dO k
2
A
+
Ok
1
?= ek
A
dO k
,-I
2
2
= L L e,wt A Ok + L ej A dO' + L e3w~ A Ok j=lk=l
=
t
ej
j=l
(dO j
+
j=l
t k
and it follows that dOt = - 2:~ Also,
k-l
wt A Ok)
+ e3
1 wi
(t w~ k
1
A Ok and
A Ok) ,
1
wf A 01 + w~ A 02 = O.
3
0= ddej = dL ekwj k-l 3
3
= L dek A wj + L ek A dw; I
k
k-l
3 3 3
= L(Leiwi ) Awj+ Lei Adw; k-l
i
I
i-I
i)
~ ei (~i = L..J L..Jwk AWjk + dw, , and so
.
dwj = -
i-I 3· 2:k=I A
wl:
k I k wj .
o
Notice that for v E TpM we have de3(V) = Vve3 = -S(v) and so IJ(v,w) = (-de3(V),W) (recall Definition 4.20). Therefore, II(ei,ej)
= (-de3(ei),e,) = / - tekwhei),ej) = \
-w~(ei)
k=1
and so [w1(el) w5(el)] [ II(eI' et} II(el,e2)] _ II(e2' eI,) II(e2' e2) - wl(e2) w5(e2) . Since (el' e2) is an orthonormal frame field, the matrix of the first fundamental form is the identity matrix. It follows that the matrix which represents
9. Integration and Stokes' Theorem
436
the shape operator is the same as that which represents the second fundamental form. Thus
K
= det [
II(el, e1) II(e 1,e2)] II(e2' e1,) II(e2' e2)
Notice that de3 = We have
2:%=1 ekw~
= _ det [w§(e 1) w~(e1)]. w§(e2) w~(e2)
reduces to de3 = e1w§
+ e2w~ since wl = O.
w§ = w§(edO l + w§(e2)02, w~ = w~(edOl +w~(e2)02,
so that
w§ "w~ = - K 01 " 02 , w§ " w~ = K 01 " 02, dw~ = K 01 " 02 • Recall that for v, w E Tx S2 we have vols2 (v, w) = (v x w, x). If we consider e3 as a map e3 : U C M -+ S2, then it is called the Gauss map. If M is assumed oriented, then we may take e3 to be equal to a global normal field. If v, wE TpM, then we have ea vols2(v, w) = (de3(V) x de3(w),e3)
+ w~(v)e2) x (w§(w)e1 + w~(w)e2) ,e3) w§ "w~(v, w) = K 01 " 02(v, w) = KvolM(v, w).
= ((w§(v)e1 =
Thus we obtain ea volS2 = K volM . This shows that the Gauss curvature is a measure of distortion of the signed volume under the Gauss map. In particular, if e3 : U C M -+ S2 is orientation reversing at p, then the curvature at p is negative. Let p E M with p in the domain U of e3. If A is a nice domain in U, then e3 maps A to a set in S2. Without worrying about measure-theoretic technicalities, we have vol(e3 (A)) = =
r
i e3(A)
L
volS2
ea vols2 =
L
K volM .
One can get an idea of the curvature near a point on a surface by visualizing the Gauss map (see Figure 9.2). For example, it is clear that the right circular cylinder has Gauss curvature zero since it maps every region of the cylinder onto a set with zero area. It is also fairly clear that the saddle surface in the diagram has negative curvature.
Problems
437
Figure 9.2. Gauss map
Problems (1) Let /, : 8 2
Y
lR3 \{0} be the inclusion map. Let
7"
= /'*w where
x dy 1\ dz + y dz 1\ dx + z dx 1\ dy (x2 + y2 + z2)3/2
W=-~-""""""''''';;'''''---=--''''''''''''''''''''''''''--~
Compute
fS27"
where 8 2 is given the orientation induced by 7" itself.
(2) Let M be an oriented smooth compact manifold with boundary 8M and suppose that 8M has two connected components No and Nl' Let Zi : Ni Y M be the inclusion map for i = 0, 1. Suppose that a is a = 0 and f3 an (n - p -I)-form with l,if3 = O. Prove p-form with that in this case
zoa
1M da 1\ f3 = (_l)P+l 1M a 1\ df3. (3) Consider the set up in the proof of Theorem 9.22 where'Y E /\ n-k and L-y : /\ k V* -+ lR is defined so that L-y (a) vol = a 1\ 'Y. Show that 'Y H- L-y E (/\ k V*)* is linear. Show that if L-y(a) = 0 for all a E /\ k V*, then 'Y = O. Thus 'Y H- L-y is injective (and hence an isomorphism). See the related Problem 4.
9. Integration and Stokes' Theorem
438
(4) Recall the notion of a manifold with corners as in Problem 21 in Chapter 1. Define orientation on manifolds with corners. Let M be an nmanifold with corners. Show that if eM is the set of corner points of M, then M\ C M is a manifold with boundary. Develop integration theory and Stokes' theorem for manifolds with corners. [Hint: If W is an (n - 1)-form with compact support in the domain of a chart (U, x), then define
where Fi := {x E ]R~ : xi = O}
is given the induced orientation as a subset of the boundary of {x E lRn xi 20}.]
:
(5) Show that if I: (M, g) -+ (N, h) is an isometry or a local isometry and w is a k-form on N, then f*w is harmonic if and only if w is harmonic. (6) Let M be a connected oriented compact Riemannian manifold with Laplace operator~. A smooth nonzero function I is called an eigenfunction for ~ with eigenvalue Aif ~I = AI. (a) Show that zero is an eigenvalue and that all other eigenvalues are strictly positive. (b) Show that if ~11 = Adl and ~h = A2!2 for Al =1= A2, then (11112) = 1M 1112 volM = o. (a) Show that if p : M -+ M is a smooth covering space of multiplicity m, then for any compactly supported wE nn(M) we have fitP*w=m fM w .
(7) Let Me ]Rn+1 be an oriented hypersurface. If N is a positively oriented normal field along M with N = 2: N i 8/8x i , then the following formula gives the volume form corresponding to the induced metric on M: volM :=
2:( _l)i-l Nidxl 1\ ... 1\ ;J;i 1\ ... 1\ dx n +1, i
where xl, ... ,xn +1 are the standard coordinates on are restricted to M. (8) Let U be a starshaped open set in coordinates. Given w=
""" ~
h< .. - 0 and f > 0 at some point. Then w=
f
dx 1 1\ ... 1\ dx n
is exact since every closed form on JRn is exact. However, w cannot be da for an a with compact support, since then we would have
{ w = { da = ( lR~
lRn
loRn
a = 0,
which contradicts the assumption that f is nonnegative with f > 0 at some point. This already shows that H~(JRn) 1= O. Using a bump function with support inside a chart, one can similarly show that H~(M) 1= 0 for any orient able manifold M. Exercise 10.18. Let M be an oriented n-manifold. If w E w 1= 0, then [w] 1= 0 in H~(M).
n~(M)
Exercise 10.19. Show that H;(JR) ~ JR and that H~(M) dim M > 0 and M is connected.
0 whenever
IM
=
and
10.3. Compactly Supported Cohomology
457
If we look at the behavior of differential forms under the operation of pull-back, we immediately realize that the pull-back of a differential form with compact support may not have compact support. In order to get desired functorial properties, we consider the class of smooth proper maps. Recall that a smooth map f : P -+ M is called a proper map if f-l(K) is compact whenever K is compact. It is easy to verify that the set of all smooth manifolds together with proper smooth maps is a category and the assignments M M nc(M) and f M {a M f*a} give a contravariant functor. In plain language, this means that if f : P -+ M is a proper map, then f* : Oc(M) -+ Oc(P) and for two such maps we have (J 0 g). = g. 0 f* as before, but now the assumption that f and 9 are proper maps is essential. We will use a different functorial behavior associated with forms of compact support. The first thing we need is a new category (which is fortunately easy to describe). The category we have in mind has as objects the set of all open subsets of a fixed manifold M. The morphisms are the inclusion maps jv,u : V y U, which are only defined in case V cU. For any such inclusion jv,u, we define a map (jv,u). : Oc(V) -+ Oc(U) according to the following simple prescription: For a E nc(V), let (jv,u). a be the form in Oc(U) which is equal to a at all points in V and equal to zero otherwise (extension by zero). Since the support of a is neatly inside the open set V, the extension (jv,u). a is smooth. In what follows, we take this category whenever we employ the functor nco Let U and V be open subsets which together cover M. Recall the commutative diagram of inclusion maps (10.2). Now for each k, let us define a map (-ih' i2.) : O~(U n V) -+ O~(U) E9 n~(V) by a M (-iha, i2.a). Now we also have the map jh + h. : n~(V) E9 nc(U) -+ O~(M) given by (O:l,a2) M jhal +l2.a2. Notice that if {¢u,¢v} is a partition of unity subordinate to {U, V}, then for any w E O~(M) we can define Wu := ¢uwiu and Wv := ¢vwiv so that we have
(jh + h.) (wu,wv) = jhWU + j2.WV = ¢uw + ¢vw = w. Notice that Wu and wv have compact support. For example, Supp(wu)
= Supp(¢uw)
C Supp(¢u)
n Supp(w).
Since Supp(¢u) n Supp(w) is compact, Supp(wu) is also compact. Thus jh + h. is surjective. We can associate to the diagram (10.2), the new sequence (10.4)
0 -+ n~(V n U)
(-ih ,:2*)
O~(V) E9 O~(U) 31*+,2* O~(M) -+
o.
This is the short Mayer-Vietoris sequence for differential forms with compact support. Theorem 10.20. The sequence (10.4) is exact.
458
10. De Rham Cohomology
We have shown the surjectivity of ih + i2", above. The rest of the proof is also easy and left as Problem 4.
Corollary 10.21. There is a long exact sequence
which is called the (long) Mayer- Vietoris sequence for cohomology with compact supports. Notice that we have denoted the connecting homomorphism by d*. We will need to have a more explicit understanding of d",: If [w] E H~(M), then using a partition of unity {pu, PV} as above we write w = ihWU + h*wv and then d ihWU = -d i2 .. wv on U n V. Then
(10.5) Next we prove a version of the Poincare lemma for compactly supported cohomology. For a given n-manifold M, we consider the projection 1r : M x R -+ M. We immediately notice that 71'* does not map n~ (M) into n~(M x R). What we need is a map 71' .. : n~(M x JR) -+ n~-l(M) called integration along the fiber. Before giving the definition of 71'.. we first note that every element of n~(M x JR) is locally a sum of forms of the following types: Type I: f 71''' A, Type II: f 71''''t.p /\ dt, where A E n~(M), t.p E n~-l(M) and f is a smooth function with compact support on M x R. By definition 71'.. sends all forms of Type I to zero, and for Type II forms, we define 71'* (7I''''t.p •
f /\ dt) - t.p
L:
f(" t) dt.
By linearity this defines 71'", on all forms. In Problem 2 we ask the reader to show that 71'.. 0 d = d 0 71'111 so that 71'111 is a chain map. Thus we get a map on
10.3. Compactly Supported Cohomology
459
cohomology: 71'* : H~(M x
Next choose e E n~(I~.) with n~+1(M x 1R) by
Je -
e* :
1R) -+ H~-l(M). 1 and introduce the map
e* : n~(M)
-+
w H 71'*w /\ 71'2'e,
where 71'2 : M x lR -+ lR is the projection on the second factor. It is easy to check that e* commutes with d, so once again we get a map on the level of cohomology: e* : H~(M) -+ H~+1(M x lR) for all k. Our immediate goal is to show that e* 0 71'* and 71'* 0 e* are both identity operators on the level of cohomology. In fact, it is not hard to see that 71'* 0 e* = id already on n~(M). We need to construct a homotopy operator K between e* 071'* and id. The map K : n~(M x lR) -+ n~-l(M x 1R) is given by requiring that K is linear, maps Type I forms to zero, and if W = 71'*
has kernel Ep at each pEN, and since a~3lp never lies in E, we see that
a«I> -a . =1= 0 on N n U' for all J.. z3
In particular, this holds at (xo, zo), and the implicit mapping theorem tells us that a neighborhood of (xo, zo) in a plaque of N is the graph of a function f : V --7 ]Rm with f(xo) = zoo Define a function F : V --7 ]Rk X ]Rm by F(x) := (x, f(x)). Writing p = F(x), we see that for each i the vector
TF.
~I axi
x
= ~I + " aaxi r (x) ~I axi P ~ az r P r
is a linear combination of vectors Xj defined above:
~I ax'
f(x)
=
+ 2: af~ (x) ~I r r ax'
tc: 8=1
(aa 8 x
az
f(x)
I(x,/(x)) + 2:A~(P) aa I ). r zr (x,f(x))
Collecting terms and comparing we see that
c: = cSt and
ar _ r ax j (x) - A3 (x, f(x)). It follows from Corollary 11.23 that f is uniquely determined on a sufficiently 0 small connected neighborhood of (x, y). It is often convenient to be able to come up with the integrability conditions for a given application without trying to match indexing and notation with the above theorem. The basis of the procedure is to set mixed partials equal to each other. We demonstrate this using the notation of the theorem. We start with
a .
a .
axk Aj (x, f(x)) = ax j Ak(x, f(x)). Apply the chain rule:
aA~ axk (x, f(x))
a
ar
+ 2: azrAj(x, f(x)) axk (x) i
r
_ aAt - - x3 a. (x, f(x))
a i ar +" - a Ak(x, f(x))-a ~ Z8 x3. (x). 8
Finally, substitute back using the original equations (11.5) and replace all occurrences of f(x) in the arguments with the independent variable z. We arrive at the integrability conditions:
"I
"I
8Aj aA~I (x,z) = aAi 8Ai a j (x,z) + ~Aj(x,z)-aI (x,z). 8 x k (x,z) + ~Ak(X,Z)-a I Z x I Z
11. Distributions and Frobenius' Theorem
488
The convenience of this may not be clear yet, but we shall shortly demonstrate the usefulness of this method. Proposition 11.37. Let U be open in}Rk x}Rm and (A~) an m x k matm of smooth functions on U. Let (xo, zo) E U and suppose that for som connected open set V, both h : V ~ U C }Rm and /2 : V ~ U C }Rm are solutions of
;~; (x) =
A; (x, f(x)) for all i,j,
f(xo) = zoo Thenh
=/2.
Proof. This follows from Corollary 11.23 and the considerations in the proof of the previous proposition. 0 Lemma 11.38. Let V be an open set in}Rk and let (A~) be an m x k matm of smooth functions on V x }Rm that are linear in the second argument and satisfy the integrability conditions (11.6) on V x }Rm. Then for any Xo E V there is a ball Bxo C V such that for any (a, b) E Bxo x}Rm there is a solution defined on Bxo with f(a) = b. Proof. Let fi be the solution with fi(XO) = ei, where ei is the i-th standard basis vector of }Rm. Then h, ... , fm are defined and linearly independent on some ball Bxo containing Xo and contained in the intersection of the domains of the ft. Choose (a, b) E Bxo x}Rm and note that b = Ebrfr( for some uniquely defined numbers bi • Now define f = E bi f, on Bxo' Then writing f = (fl, ... , f m ), we have for any x E V,
and f(a)
=
L b fr(a) = b.
o
r
Corollary 11.39. Let V be a simply connected open set in IRk and let (A ) be an m x k matrix of smooth functions on U = V x }Rm. Suppose that ea h is linear in its second argument. If
A;
oA;+ L AIk oA~= oA1 L I oAk ·oxk Ozl ax]+ A ] Ozl I
I
for all i, j, k
11.6. Fundamental Theorem of Surface Theory
489
on V x ]Rm, then given any (xo, zo) E V x ]Rm, there exists a unique smooth map f : V -+ ]Rm such that
afi i ax j (x) = Aj(x, f(x))
for all i,j,
f(xo) - zoo Proof. Let Xj := 8~' Ip + Er Aj(p) 8~r Ip be the fields that span an integrable distribution on V x ]Rm as in Proposition 11.36. Let L(xO,%o) be the maximal integral manifold through the point (xo, zo). Let go denote the restriction of the projection prl : V x ]Rm -+ V to L(xo,zo). Let (aI, bl ) E £(xo,zo) and consider the set Fal = p-l(al).
By Lemma 11.38 above, there is a fixed open set U containing al such that for every (all b) E Fal there is a solution fb : U -+ lRm with f(al) = b. By Corollary 11.23, the graphs of these solutions are all disjoint open sets in L(xo,zo) and p restricts to a diffeomorphism on each such graph. Thus p : £(xo,zo) -+ V is a smooth covering map. The local solutions guaranteed to exist by Proposition 11.36 are local sections of this covering. Thus since V is simply connected, we know from Theorem 1.95 that there is a smooth lift p of idv : V -+ V such that p{xo) = (xo, zo), which in this case means that we have a global section: go 0 p = idv. Now let f := pr2 0 p: V -+ ]Rm. Then p{x) = (x, f(x)) and f(x) must be smooth and for every a E V the function f must agree with the unique local solution which takes the value f(a) at a. 0 We return to the situation studied in Chapter 4. Consider an immersion x : V -+ ]R3, where V is an open set in ]R2 whose standard coordinates will be denoted by ul,u 2 • Let (/I,/2,fg) be the frame fields along x defined by fl :=xul
f2 := Xu2,
,
fg :=N = Xul x xu2/ IIXul x Xu211 , where Xul = entries
ax/ au l , etc.
The first fundamental form is given by the matrix
gij = (xu" x u1 ) for 1 ::; i, j ::; 2, while the second fundamental form is given by the matrix entries iij
= -
(Nul, Xu3)
= (N, X U'u 3) = (/3, X u'u1) .
Let us consider f = (/I,/2,fg) as a matrix function of values in GL(3). We have 3
(fi)ul
-
L P; fr r=1
3
and (fi)u2 =
(U l
L Qi fT r=1
,u2 ) that takes
11. Distributions and Frobenius' Tbeorem
490
for some matrix functions P and Q. In matrix notation, we have
fu1 = fP, fu2 = fQ,
(11.7)
and these are called the frame equations. For convenience, we define a 3 x 3 matrix function G by Gij := (Ii, fJ) for 1 ~ i,j ~ 3 so that
(11.8)
G=
911 921
(
o
0)
912 922
0
0
.
1
For any given x = ~t xi Ii, we have
ei = (x, Ii) = (L xk Ik' Ii) = xk L and so
ei = L
Uk, Ii) = Gki Xk ,
L Gik Xk .
(at)ik xk =
Now let x = (fi)Ul = ~ fkPik. Then xk = Pik, so if we define Bt3 .(Ii, (/j )u1), then we have
L GikPf, = Bi2 = L Gik P;, = Bi3 = L GikP;,
(fi, (h)u 1 ) = Bi1 =
(11.9)
(fi, (h)u 1 )
(Ii, (h)u 1 )
or B = GP. Similarly, if G = (Gij) = (Ii, (fJ)u2), then
(fi, (fdu 2 ) = Gil = (11.10)
L GikQ~,
(fi, (h)u 2 )
= Gi2 = L
(Ii, (f3)u 2 )
= Gi3 =
GikQ~,
L GikQ~,
or G = GQ. We arrive at P = G- 1B,
Q = G- 1G. We denote the entries of G- 1 by 9ij so that
(11.11)
G- 1 :=
(
gIl
912
9 21
g22
o
0 ~ (911)u 2
! (922)u £12
1
11.6. Fundamental Theorem of Surface Theory
491
and
~ (gU)U2 ( 0= (g22)u1 - ~ (gll)u2
(11.13)
~ (g12)u2 - ~ (g22)u1 -£12) ~ (g22)u2 £22
£12
-£22,
0
In particular, P and Q can be written in terms of the matrix entries of the first and second fundamental forms.
Proof. The proof is just a calculation, and we only do part. For example, if i is either 1 or 2, then
Bii
1
1
= (Ii, (fi)u 1) = (Xu·,Xu•u1) = 2' (Xu"XU')u1 = 2' (gii)u1'
Similarly, for i = 1 or 2 we have 1
= (Ii, (fi)u 2) = (Xu"Xu' u2) = 2' (gu)u2' (Xu1,Xu2)u1 = (Xul u1,Xu2) + ~ (gll)u2 from above, and so
Oil Now, ~ (g12)u1 =
B21 =
(12, (l1)u1) = (Xu1u1,Xu2 )
=
1
1
2' (g12)u1 - 2' (gll)u2.
The entries B12, 012, 021 are calculated similarly. Next consider Bi3 for i = 1 or 2. We have and so
B3i
= -Bi3.
The entries 03i and Oi3 are obtained in the same way. (13, fa) = 1, 0=
1
2' (13, fa)u/c =
((f3)u/C , fa)
=
{ B33 if k = 1, 033 if k = 2.
Lastly, since
o
We record an observation to be used later: (11.14)
B +Bt = Gu 1, 0+ ot = Gu 2.
The frame equations (11.7) are a system to which Proposition 11.36 applies. Rather than trying to rewrite the equations in a form that matches that proposition we obtain the integrability conditions by setting
(fu1)u2 and then
= (fu2)u1
11. Distributions and Frobenius' Theorem
492
Substituting from the frame equations we obtain f (Pu 2
-
QuI - (PQ - QP)) = O.
Now f is a nonsingular matrix, so we have the equivalent integrability equation (11.15)
Pu2 - QuI - (PQ - QP)
= O.
At this point we pause to appreciate an important fact. Namely, direct calculation reveals that these equations are equivalent to the combination of the Codazzi-Mainardi equation and the Gauss curvature equation, which we now see are integrability conditions (see Problem 7). We thus refer to the above integrability equations (11.15) as the Gauss-Codazzi equations with apologies to Gaspare Mainardi (1800 1879). We now tum things around. Rather than assuming that we have a surface, we take the (g'3) and (i.,) as some given symmetric smooth matrixvalued functions defined on a connected open V C R.2 with the assumption that (g,,) is positive definite. Furthermore, we now assume that G, P, and Q are actually defined in terms of these by the formulas above, which we found to be true in the case where we started with a surface. We will show that we can obtain a surface with these as first and second fundamental form. Theorem 11.41 (Fundamental existence theorem for surfaces). The following assertions hold:
(i) Let V be an open set in R.2 diffeomorphic to an open disk and let x : V
-+ R.3 be an immersion with the corresponding first and
second fundamental forms given in matrix form as (gi;) and (i.3). Let y : V -+ R.3 be another immersion with the corresponding forms (gi;) and (~j). If
y=fox for some proper Euclidean motion f : R.3 -+ R.3, then
g'j = g",
l'j
= li,.
Conversely, if the last equations hold, then y = fox for som Euclidean motion f. (ii) Suppose that (g,,) and (li,) are given symmetric matrix-valuedfunctions defined on V with (gij) positive definite and suppose that G, Band C are defined in terms of the entries of (gij) and (lij) as in
11.6. Fundamental Theorem of Surface Theory
493
formulas (11.8), (11.11), (11.12) and (11.13). Then if
= G-IB, Q = G-le, P
and if Pu2 - Qul - (PQ - QP) = 0, then there exists an embedding x : V -+ ]R3 such that (gij) and (iij) are the corresponding first and second fundamental forms.
Proof. We leave the proof of the first part of (i) to the reader, but note that it can be proved using direct calculation or it can be derived from Theorem 4.22. For the rest of (i), note that by composing with a translation we may assume that both x and y map some fixed point U E V to the origin in ]R3. Let (II, h, h) be the natural frame for x as above and let (fr, be that of y. By making a rotation we may assume tha~ these two frames agree at p. But since we are assuming that g~} = 9ij and i~j = it}, it follows that both frames satisfy the same frame equations and so by Proposition 11.37 they must agree on the connected set V. In particular, xu. = Yu' for i = 1,2. Thus x and y only differ by a constant, which must be zero since x(u) = y(u).
h, h)
Next we consider (ii) where (9tj) and (iij) are given. We want to construct a surface, but first we construct the frame for the desired surface. Since it is assumed that 9 = (gt}) is positive definite, g and the extended matrix G are both invertible and positive definite. Thus P and Q are well-defined. Since we assume that the integrability equations Pu2 - QuI (PQ - QP) = 0 hold, Theorem 11.36 tells us that we can solve the frame equations locally, near any point U E V and with any initial conditions f(u) = fo holding as desired. But the system is linear and our domain is diffeomorphic to a disk so we can use Corollary 11.39 to obtain a solution on all of V. Since G is positive definite, we may choose these initial conditions so that (ft(u) , fj(u)) = Gtj(u) (ij-th entry of G at u). Having obtained the fi near u, we now wish to obtain a surface. This means solving the system (11.16) and this time the integrability conditions are derived from
(lI)u 2 = (f2)u l
•
Using the frame equations, we obtain integrability conditions
LQ{1i - LP~fj.
11. Distributions and Frobenius' Theorem
494
This just says that the second column of P is equal to the first column of Q, which is true. Thus we can find x : V ~ R3 with x(u) = 0 so that (11.16) holds. Next we show that (Ii, i,) = Gij on all of V. We compute as follows:
(Ii, ij)u 1 = ((fi)u1 ,Ii) + (Ii, (li)u 1 ) = ~ Qi (lr,
Ii) + ~ Qj (Ii, i8) =
r
(QG + (QG)t)ij
8
= (GQ
+ (GQ)t)ij
= (B
+ Bt)ij = (Gul)ij = (Gij)u1
by equations (11.14). Similarly for u 2 • Thus (Ii, Ii) - G" is a constant, which must be zero since it is zero at u. From (Ii, Ii) = Gij it follows that (13, fa) = 1 and that It, 12 are independent and orthogonal to fa. It remains to show that ((f3)u' ,i,) = -iij. We compute as follows: - ((f3)'11.1 ,i,) = (gl1i11 + g12 i 12 ) (It, Ii) + (g 12 i 11 + g22i12) (12, Ii)
+ l 2i12)glj + (g 12i11 + g22i12)g2j ill (gl1 g1j + g12g2j) + i12(g21 g1j + g22g2j) i116j + i 126;.
= (gl1i11 = =
This shows that ((f3)u 1 , Ii) = -iIj for j = 1,2. The computation of - ((f3)u2 ,Ii) is similar and left for the reader. 0
11. 7. Local Fundamental Theorem of Calculus Recall the structure equations (8.15) satisfied by the Maurer-Cartan form WG for a Lie group G:
dwG =
-~ [wG,wG]".
If VI = Xl (e), ... ,Vn = Xn (e) is a basis for the Lie algebra 9 which extends to left invariant vector fields Xl, ... , X n , then the above equation is equivalent to i 1\ wj = dw k = - 'L" ck,w 1\ wj , ~, 2 L- c~,wi ~,
-! '"
ct
i<j
i,j
where the are the structure constants associated to wI, ... , wn , which is the left invariant frame field dual to Xl"'.' X n . If M is some m-manifold and i : M ~ G is a smooth map, then wJ = j*wG is a g-valued I-form on M. By naturality we have
or equivalently
11.7. Local Fundamental Theorem of Calculus
495
where w} = f*w i for i = 1, ... , n. The (I-valued I-form wI is sometimes called the (left) Darboux derivative of f. The right Darboux derivative is defined similarly using the right Maurer-Cartan form.
If we think of a {I-valued I-form on a manifold M as a map TM --+ (I, then wI = f*we = We 0 T f. From this point of view we can understand why wI is a kind of derivative of f by considering the special case where G is a vector space V with its abelian (additive) Lie group structure. In this case, the Lie algebra is V itself and the Maurer-Cartan form is just the canonical map pr2 : TV = V x V -+ V, and so for a smooth map f : M --+ V, the Darboux derivative is the differential df = pr2oTf. Just as for the differential, the Darboux derivative embodies less information than the tangent map since the values that the map takes are "forgotten" and only tangential information is retained. Indeed, notice that if Lg : G --+ G is a left translation and F := Lg 0 f, then WF
= F*we = f* L;we = f*we = WI
since We is left invariant. Hence two smooth maps into G that differ by left translation have the same (left) Darboux derivative. This generalizes the fact that two functions that differ by an additive constant have the same differential. For a smooth I-form {) = gdt on JR, we can always find a smooth function with df = {) since by the Fundamental theorem of calculus one need only choose f(t) = J~ g(T)dT. More generally, if M is simply connected and Gis a vector space V, then the fact that Hl(M) = 0 means that every V-valued 1-form is the differential of some smooth f : M -+ V. For a general G, if {) is a {I-valued I-form on M, then we may ask for an f such that {) = WI' But there is no reason to expect a general {) to satisfy the above structure equation that WI satisfies. Now if we choose a basis {Vi} for g, then there must be I-forms {)1, ... , {)n E 0 1 (M) such that
f
n
{) =
L
Vi{)i.
i=l
Then d{) =
-! [{),{)]/\ is equivalent to d{)k =
_~2 "L...J c~t} {)i A {ji , i,j
where cfj are the structure constants. As we said, this mayor may not hold. These equations are the integrability conditions for the existence of an f such that {) = wI' More precisely, we have the following theorem.
11. Distributions and Frobenius' Tbeorem
496
Theorem 11.42. Let M be an m-manifold and G an n-dimensional Lze group. If iJ is a g-valued 1-form on M that satisfies the structural equation diJ = -~ [iJ, iJ]''', then for every Po E M there is a neighborhood Upo of Po such that given any (a, b) E Upo x G there is a smooth function f : UXo ~ G with f(a) = band iJ wf' Proof. Let prl : M x G -+ M and pr2 : M x G -+ G be the canonical projections and define a g-valued 1-form on M x G by 0:= priiJ - pr:iwG. For each (p, g) E M x G, let E(p,g) = Ker O(p,g). Now define a vector bundle homomorphism T (M x G) ~ (M x G) x 9 by v(p,g) 1---+ ((p, g), O(p,g) (V(p,g»))' By Proposition 6.28, if this homomorphism has constant rank, then the kernel is a subbundle which clearly has fiber E(p,g) at (p, g). By linear algebra, this is equivalent to showing that the dimension of E(p,g) is independent of (p, g). This will follow if we show that Tpr 11 E(p,g) : E(p , g) -+ TpM is an isomorphism for all (p,g). If we identify T(p,g) (M x G) with TpM x TgG, then Tprl is just the projection (v, w) 1---+ v and similarly for Tpr2' Now if (v, w) E E(p,g) and Tprl . (v, w) = 0 then v = O. But, since iJ(v) wa(w), we have w - 0 also. Thus TprllE (p,g) is injective. It is also surjective since for any v E TpM, we clearly have (v, TeLg(iJ(v)) E E(p,g) and this has v as its image. Now we use Proposition 11.19 to show that E is integrable:
dO = pridiJ - pr:idwG = pri ( -~ [iJ, iJ]") - pr2 ( -~ [wG, wG]" ) *w] -"21 [prl*.Qv ,p r *]" l'1? +"21 [pr2*wG,p r2 G" . But priiJ = 0 + pr;wG, so =
dO = =
-~ [(0 + pr2wG) , (0 + pr2wG)]" - ~ [pr;wG, pr2W G] " -~[O,O]" - ~[O,pr:iwGl" - ~[pr:iwG,Ol'"
which makes it clear that dO(X, Y) = 0 whenever 0 (X) = 0 and 0 (Y) O. Now we use the leaves of the distribution to construct the solution. Let Xo E M and fix go E G. Then let L(xO,go) be the maximal integral manifold through (xo, go). The map Tprll E(PO,go) : E(po,go) -+ TpM is an isomorphism so the inverse mapping theorem tells us that prll L(pO,gO restricts to a diffeomorphism on some neighborhood 0 of (Po, go) in L(po,go)' Let
E nk(M, End(E)) and wE nk(M, E), we have
d'il (q>!\ w) = d'il !\ w + (_l)kq>!\ d'il w. Proof. We have
dE((A 0 a)!\ (a 0 13)) := dE(A(a) 0 (a!\ 13)) = \7 E(A(a))!\ (a!\ 13) + A(a) 0 d (a!\ 13) = (_l)k (a!\ \7 E(A(a)) !\ 13) + A(a) 0 d (a!\ 13)
= (_l)k (a!\ {\7End(E) A!\ a + A!\ \7 Ea}!\ 13)
+ A(a) 0
da!\ 13 + (_l)k A(a) 0 a!\ d13
= (\7End(E) A) !\ a!\
+ A(a) 0 =
a!\ (3 + (_l)k A!\ a!\ (\7 Ea) !\ (3
da!\ (3 + (_l)k A(a) 0 a!\ d(3
(\7End(E) A!\ a
+ (_l)k (A 0
+A 0
da) !\ (a 0 (3)
a) !\ (\7 Ea!\ (3 + a 0 d(3)
= dEnd(E) (A 0 a)!\ (a 0 (3)
+ (-ll (A 0 a)!\ dE (a 0
(3).
By linearity we conclude that for q> E nk(M, End(E)) and wE n(M, E) we have
dE(q>!\ w)
= dEnd(E) !\ w + (_l)kq>!\
So in light of our notational conventions we are done.
dEw.
o
12. Connections and Covariant Derivatives
540
Remark 12.63. In the literature, it seems that the different natures of (d'V) k and ('V)k are not always appreciated. For example, the higher derivatives given by (d'V) k are not appropriate for defining k-th order Sobolev spaces since (d'V) 2 is zero for any flat connection.
12.10. Curvature Again The space o'(M, End(E)) is an algebra over COO(M) where the multiplication is according to (Ll ® WI) 1\ (L2 ® W2) = (Ll 0 L 2) ® WI 1\ W2. Now o'(M, E) is a module over this algebra because we can multiply (using the symbol 1\) as follows:
(L ® 0:)
1\
(0' ® ;3) = LO' ® (0: 1\ ;3) .
As usual, this definition on simple elements is sufficient. If X, Y E X (M) and \II E 0,2(M, End(E)), then using 0,2(M; E) ~ L~lt(X(M), r(E)) we have
(\II
1\
0') (X, Y) = \II (X, Y) 0'
for any 0' E o'(M, E).
Proposition 12.64. The map d'V 0 d'V : O,k(M, E) ---+ 0,k+2(M, E) is given by the action of F, the curvature 2-form of'V E , d'V
0
d'V Il
=F
Il for Il E O,k(M, E).
1\
Proof. Let us check the case where k = 0 first. From formula (12.10) above we have for 0' E O,O(M, E), (d'V
0
d'V 0') (X, Y) = 'Vx (d'V O'(Y)) - 'Vy (d'V O'(X)) - O'([X, Y])
= 'Vx'VYO' - 'Vx'VYO' - O'([X, Y]) = F(X, Y)O' = (F 1\ O')(X, Y). More generally, we just check d'V d'V
0
d'V Il = d'V
0
0
d'V on elements of the form Il = 0' ® fJ:
d'V (0' ® fJ)
= d'V (d'V 0' 1\ fJ
= (d'V d'V 0') =
1\
+ 0' ® dfJ)
e - d'V 0' 1\ de
+ d'V 0' 1\ dfJ + 0
(F 1\ 0') 1\ e = F 1\ (0' 1\ e)
= F 1\ Il.
=
F 1\ (0' ® e)
o
Let us take a look at how curvature appears in a local frame field. As before restrict to an open set U on which eu = (el,"" er ) is a given local frame field and then write a typical element s E O,k(M, E) as s = e'r/, where
541
12.11. The Bianchi Identity
'r/ = ('r/1, ... , 'r/T)t is a column vector of smooth k-forms. With Wu connection forms we have d'il
i" s =
d'il d'il (eu'r/u) = d'il (eud'r/u
=
/\ 'r/u)
+ Wu /\ 'r/u) + eu /\ d'il (d'r/u + Wu /\ 'r/u) euwu /\ (d'r/u + Wu /\ 'r/u) + eu /\ dwu /\ 'r/U - eu /\ Wu /\ d'r/u eudwu /\ 'r/U + euwu /\ Wu /\ 'r/U eu (dwu + Wu /\ wu) /\ 'r/u·
= euwu /\ =
(w}) the
+ d'il eu /\ 'r/u)
= d'il (eu (d'r/u + Wu /\ 'r/u)) = d'il eu /\ (d'r/u + Wu /\ 'r/u) + eu /\ d'il (d'r/u + Wu =
=
(d'r/u
The matrix dwu+wu/\wu represents a section ofEnd(E)u@/\2T *U. In fact, we will now check that these local sections paste together to give a global section of End(E) @ /\ 2T* M, or in other words, an element of n2(M, End(E)), which is clearly the curvature form: F : (X, Y) f--7 F(X, Y) E r(End(E)). Let Fu = dwu +wu /\wu and let Fv = dwv +wv /\wv be the corresponding form for a different moving frame eu := evg, where g : Un V -+ GL(JFT), and r is the rank of E. What we need to verify is the transformation law Fv = g-lFug,
which we met earlier in equation (6.2). Recall that Wv = g-lwUg + g-ldg. Using d (g-l) = _g-ldgg- 1 , we have Fv
= dwv + Wv /\ Wv = d (g-lwUg + g-ldg)
+ (g-lwUg + g-ldg) /\ (g-lwUg + g-ldg) = d (g-l) /\ wug + g-ldwUg - g-lwU /\ dg + d(g-l) /\ dg + g-lwU /\ wug + g-ldgg- 1 /\ wug + g-lwU /\ dg + g-ldg /\ g-ldg = g-ldwUg + g-lwU /\ wug = g-l Fug, where we have used that g-ldg /\ g-ldg = d(g-lg) = O.
12.11. The Bianchi Identity In this section we give several versions of the so-called Bianchi identity for a connection \7 on a vector bundle E -+ M. Perhaps the simplest version to understand is the following: If U, V, WE X(M), then [\7u, [\7v, \7wll
+ [\7v, [\7w, \7ull + [\7w, [\7u, \7vll
= 0 (Bianchi identity),
12. Connections and Covariant Derivatives
542
where [\7u, \7v] := \7u 0 \7v - \7v 0 \7u. This identity follows trivially once we observe that the set of linear operators on any vector space is a Lie algebra under the commutator bracket operation [A, B] := A 0 B - BoA. So, in this form, the Bianchi identity is just an instance of the Jacobi identity. Let (U,x) be a chart on M, and let F/1-v : f(E)lu -+ f(E)lu be the local curvature operator defined by
F/1-v where 0/1identity:
= [\7al" \7aJ = F(0/1-,ov),
= 0/ ox/1-, etc. Then we have the following version of the Bianchi
[\7 a!', Fv'\] + [\7 av , F'\/1-]
+ [\7 a>. v , F/1-v]
= 0
(Bianchi identity).
Another revealing form of the Bianchi identity depends on our discussions in the last section and in particular on Proposition 12.62. We have, for any E-valued form T/,
(d'V) 3 T/
= d'V ( (d'V) 2 T/) = d'V (F /\ T/) = d'V F /\ T/ + F /\ d'V T/.
But equally,
(d'V) 3 T/ = (d'V) 2 (d'V T/) = F /\ d'V T/ so it must be that d'V F /\ T/ = 0 for any T/ and so we obtain
=0
d'V F
(Bianchi identity).
Exercise 12.65. Use a calculation in local coordinates and with respect to a local frame el, ... , ek to show that the above versions of the Bianchi identity are equivalent. 12.12. G-Connections If the vector bundle 1f : E -+ M has a G-bundle structure, then there should be certain connections that respect this structure. These are the G-connections. It is time to confess that we have come face to face with a weakness of our approach to connections. It is much easier and more natural to define the notion of G-connection if one first defines connections in terms of horizontal distributions on principal bundles. We treat this in the online supplement [Lee, J efl1. However, we can still say what a G-connection should be from the current point of view without too much trouble.
Recall that if 1f : E -+ M is a rank k vector bundle with typical fiber V that has a G-bundle structure where G acts on V by the standard action as a subgroup of GL(V), then we have the bundle of G-frames
Fc(E) := UPEMFC(Ep), where Fc(Ep) := {u E GL(V,Ep) : U = c/J-l(p,.) for some (U,c/J) E Ac} and Ac is the maximal G-atlas defining the structure. The elements of
12.12. G-Connections
543
FG (Ep) are called G-frames. Having fixed a basis (e1, ... , ek) for V, each element of FG(Ep) can be identified as a basis (U 1, ... , Uk) for Ep according to U : x r---+ L xiUi' If F(Ep) is the set of all frames at p, then FG(Ep) C F(Ep) and FG(E) is a subbundle of F(E). Now let c : [a, b] -+ M be a smooth curve. If (U1, ... , Uk) is a frame, then (PeU1, ... ,PeUk) is also a frame. Thus we get a map Pc : F(Ee(a)) -+ F (Ee(a)). The following is then a workable definition of G-connection: Definition 12.66. Let 11' : E -+ M be a vector bundle with a G-bundle structure as above. A connection on the bundle is called a G-connection if parallel transport Pc takes G-frames to G-frames for all piecewise smooth curves c.
A G-frame field is a frame field which is a G-frame at each point of its domain. Exercise 12.67. Show that if ~ is the covariant derivative associated to a G-connection on E, then for each G-frame field the associated connection forms take values in the Lie algebra of G thought of as a matrix subgroup of GL(n, IF). Let 11' : E -+ M be a vector bundle with metric h = (., .) and standard fiber IRk. The metric h gives a reduction of the structure group to O( n). In this case, an O(n)-frame is an orthonormal frame. A connection on E is an O(n)-connection if and only if the associated covariant derivative \7 satisfies
(12.11) for all 81,82 E r(E) and all v E TM. Exercise 12.68. Prove this last assertion. One may also consider metrics which are nondegenerate but not necessarily positive definite. In this case, the structure group reduces to one of the semiorthogonal groups O(k, n - k). Definition 12.69. A covariant derivative satisfying (12.11) above for some metric h on a vector bundle E -+ M is called a metric covariant derivative (or metric connection). A simple partition of unity argument shows that if h is a given metric, then there exists a (nonunique) metric connection for h (Problem 11). For a metric connection, parallel transport is an isometry (Problem 12). Furthermore if h is the metric and \7 is metric with respect to h, then it is easy to check that \7h = O.
12. Connections and Covariant Derivatives
544
Proposition 12.70. Suppose that h = (-,.) is a metric on a vector bundle E -+ M. If \7 is a metric covariant derivative, then the corresponding curvature satisfies
for all X, Y E X(M) and all 0-1,0-2 E r(E).
Proof. Let X and Y be fixed vector fields. Without loss of generality we may assume that [X, Y] = 0 (recall Exercise 7.34). It is also enough to show that (F(X, Y)o-, 0-) = 0 for all 0-. We have
(F(X, Y)o-,o-) = (\7x\7yo-,o-) - (0-, \7x\7yo-) = X (\7yo-, 0-) - (\7yo-, \7 xo-) - Y (\7 xo-, 0-) - (\7 xo-, \7yo-) 1
= 2" (XY (0-,0-) - Y X (0-,0-)) = 0
o
since [X, Y] = O.
We shall study metric connections on tangent bundles in the next chapter.
Problems (1) Prove Theorem 12.23.
(2) Let M have a linear connection \7 and let T(X, Y) := \7xY - \7yX. Show that T is COO(M)-bilinear (tensorial). The resulting tensor is called the torsion tensor for the connection. (3) Show that a holonomy group is indeed a group. Show that the holonomy at any point of a sphere is isomorphic to the special orthogonal group SO(2). Second order differential equations and sprays (4) A second order differential equation on a smooth manifold M is a vector field on TM, that is, a section X of the bundle TTM (second tangent bundle) such that every integral curve 0: of X is the velocity curve of its projection on M. In other words, 0: = /y where., := 7rTM 00:. A solution curve., : I -+ M for a second order differential equation X is, by definition, a curve with &(t) = X(a(t)) for all T E I.
Problems
545
In the case M = ~n, show that this concept corresponds to the usual system of equations of the form y' = v,
v' = f(y, v), which is the reduction to a first order system of the second order equation y" = f(y, y'). What is the vector field on T~n = ~n X ~n which corresponds to this system? Notation: For a second order differential equation X, the maximal integral curve through vET M will be denoted by O:v and its projection will be denoted by 'Yv := 7rTM 00:.
(5) A spray on M is a second order differential equation, that is, a section X of TT M as in the previous problem, such that for vET M and s E ~, a number t E ~ belongs to the domain of 'Ysv if and only if st belongs to the domain of 'Yv, and in this case
'Ysv(t) = 'Yv(st). Show that there are infinitely many sprays on any smooth manifold M. [Hint: (i) Show that a vector field X E X(T M) is a spray if and only if T7r 0 X = idTM, where 7r = 7rTM:
TTM
71
TM~d 1 TM
TM
(ii) Show that X E X(T M) is a spray if and only if for any s E ~ and any v E TM, Xsv = TJ1s(sXv), where J1s : v t---+ sv is the multiplication map. (iii) Show the existence of a spray on an open ball in a Euclidean space. (iv) Show that if Xl and X 2 both satisfy one of the two characterizations of a spray above, then so does any convex combination of Xl and X2.] (6) Show that if one has a linear connection on M, then there is a spray whose solutions are the geodesics of the connection. (7) Show that given a spray on a manifold there is a (not unique) linear connection V' on the manifold such that 'Y : I ---+ M is a solution curve of the spray if and only if V' 8/Y = 0 for all tEl. Note: 'Y is called a geodesic for the linear connection or the spray. Does the stipulation that the connection be torsion free force uniqueness?
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12. Connections and Covariant Derivatives
(8) Let X be a spray on M. Equivalently, we may start with a connection on M (i.e. a connection on T M) which induces a spray. Show that the set Ox: = {v E T M : ')'V (1) is defined} is an open neighbor hood of the zero section of T M. (9) Finish the proof of Theorem 12.32. (10) Prove formula (12.9). (11) Let h be a metric on a vector bundle E ~ M. Show that there exists a metric connection for h. [Hint: Use orthonormal frames defined on each open set of a locally finite cover.] (12) Show that parallel transport along curves with respect to a metric connection in a vector bundle with metric is an isometry of scalar product spaces.
Chapter 13
Riemannian and Semi-Riemannian Geometry
"The most beautiful thing we can experience is the mysterious. It is the source of all true art and science."
- Albert Einstein
In this chapter we take up the subject of semi-Riemannian geometry, which includes Riemannian geometry and Lorentz geometry as important special cases. The exposition is inspired by [ON1], which we follow quite closely in some places (also see [L1]). Recall that by definition, a semiRiemannian manifold (M, g) has a well-defined index denoted ind (M) or ind(g). In the case of an indefinite metric (ind(M) > 0), we will need a classification: Definition 13.1. Let (V, (-,.)) be a scalar product space. A nonzero vector v E V is called (1) spacelike if (v, v) > 0; (2) lightlike or null if (v, v) = 0;
(3) timelike if (v, v) < 0; (4) nonnull if v is either timelike or spacelike. The terms spacelike, null (lightlike), and timelike indicate the causal character of a vector. The word causal comes from relativity theory and is most apropos in the context of Lorentz manifolds defined below. If (M, g) is a
-
547
548
13. Riemannian and Semi-Riemannian Geometry
TIMELIKE
LlGHTLIKE
SPACELIKE
Figure 13.1. Lightcone
semi-Riemannian manifold, then each tangent space is a scalar product space and the above definition applies. Recall that we define Ilvll = l(v,v)1 1/ 2 , which we call the length of v.
Note: We have so far left the causal character of the zero vector undefined. It may seem reasonable that it should be considered null. A second possibility is that the zero vector should have all three causal characters. Actually, we shall see that if the index of the scalar product is one, then it is convenient to consider the zero vector as being spacelike. Definition 13.2. The set of all null vectors in a scalar product space is called the nullcone or lightcone. If (M, g) is a semi-Riemannian manifold, then the null cone in TpM is called the nullcone at p. Definition 13.3. Let I c lR be some interval. A curve c : I -t (M, g) is called spacelike, null, timelike, or nonnull, according as c(t) E Tc(t)M is spacelike, null, timelike, or nonnull, respectively, for all tEl. While every smooth manifold supports Riemannian metrics by Proposition 6.45, the existence of an indefinite metric on a given smooth manifold has an obstruction:
Theorem 13.4. A compact smooth manifold admits a continuous CO indefinite metric of index k if and only if its tangent bundle has a CO rank k sub bundle. This result is Theorem 40.11 of [St].
Definition 13.5. Let (M, g) be semi-Riemannian. If c : [a, b] piecewise smooth curve, then Lc(a),c(b) (c)
=
lb
l(c(t),c(t))1 1 / 2 dt
is called the arc length or simply length of the curve.
-t
M is a
13. Riemannian and Semi-Riemannian Geometry
549
The word length could cause confusion since the length of a null curve is zero. Thus for indefinite metrics, arc length can have some properties that are decidedly not like our ordinary notion of length. In particular, a curve may connect two different points and the arc length might still be zero! The word length is therefore sometimes reserved for timelike or spacelike curves.
Definition 13.6. A positive reparametrization of a smooth curve c ; [a, b] -t M is a curve defined by composition co h ; [a', b'] -t M, where h; [a', b'] -t [a, b] is a smooth monotonically increasing bijection. Similarly, a negative reparametrization is given by composition with a smooth monotonically decreasing bijection h ; [a', b'] -t [a, b]. By a reparametrization we shall mean either a positive or negative reparametrization. The above definition can be extended to piecewise smooth curves. Suppose c ; [a, b] -t M is a continuous curve such that, for some partition a = to < tl < ... < tk = b, we have that c is smooth on each [ti-l, til. A positive reparametrization of c is a curve co h ; [a', b'] -t M, where h ; [a', b'] -t [a, b] is a monotonically increasing continuous bijection that is smooth on each interval h- 1 ([ti-l, til). Negative reparametrization is defined similarly.
Remark 13.7 (Important fact). The integrals above are well-defined since e(t) is defined and continuous except for a finite number of points in [a, b]. Also, it is important to notice that by standard change of variable arguments, a reparametrization , = c 0 h does not change the arc length of the curve;
J b
h-l(b)
l(e(t),e(t))llj2dt =
a
r Jh-
Ih(u),--y(u))1 1 / 2 du. 1 (a)
Thus the arc length of a piecewise smooth curve is a geometric property of the curve; i.e. a semi-Riemannian invariant.
Definition 13.8. Let (M,g) be semi-Riemannian. Let I be an interval (possibly infinite). If c ; I -t M is a smooth curve with Ilell = 1, then we say that c is a unit speed curve. If c ; I -t M is a curve such that Ilell is never zero, then choosing a reference to E I, we may define an arc length function £ ; I -t I' c lR by
(13.1)
£(t)
;=
it
I(e(t), e(t))1 1 / 2 dt.
to
For a finite interval of definition [a, b]' the reference to is most often taken to be the left endpoint a. Since d£/dt = l(e(t),e(t))1 1 / 2 > 0, we may invert to find £-1 and then reparametrize;
,(s)
;=
c(£-l(s)).
550
13. Riemannian and Semi-Riemannian Geometry
It is easy to see from the chain rule that the resulting curve is a unit speed curve. Conversely, if, is a unit speed curve, then the arc length from ,(Sl) to ,(S2) is S2 - Sl. Often one abuses notation by writing s = £(t) and then ds/dt instead of d£/dt. The use of the letter s for the parameter of a unit speed curve is traditional, and we say that the curve is parametrized by arc length. In the case of timelike curves, people sometimes use the letter T instead of s and refer to it as a proper time parameter.
13.1. Levi-Civita Connection In this chapter we will use the term "connection" to be synonymous with covariant derivative. Let (M, g) be a semi-Riemannian manifold and '\7 a metric connection for M (see Definition 12.11). By definition, we have
x (Y, Z) = (\7 x Y, Z) + (Y, \7 x Z) for all X, Y, Z E X(M). It is easy to show that the same formula holds for locally defined fields. Recall that the operator T : X(M) x X(M) ---+ X(M) defined by T(X, Y) = \7 x Y - \7y X - [X, Y] is a tensor called the torsion tensor of \7. From the previous chapter we know that (X, Y) M T(X, Y) defines a COO(M)-bilinear map X(M) x X(M) ---+ X(M). The isomorphism (7.6) implies that T gives a section of Tg(TM;TM). That is, T can be thought of as defining a TpM-valued 2-tensor field at each p, so if X p, Yp E TpM, then T (Xp, Yp) is a well-defined element of TpM. Recall from our study of tensor fields that T (Xp, Yp) is defined to be T(X, Y)(p) for any fields X, Y such that X(p) = Xp and Y(p) = Yp. Requiring that a connection be both metric and torsion free, pins down the metric completely.
Theorem 13.9. For a given semi-Riemannian manifold (M,g), there is a unique metric connection '\7 such that its torsion is zero, T == O. This unique connection is called the Levi- Civita connection for (M, g). Proof. We will derive a formula that must be satisfied by \7 and that can be used to actually define \7. Let X, Y, Z, W be arbitrary vector fields on M. If \7 exists as stated, then we must have X (Y, Z)
= (\7 x Y, Z) + (Y, \7 x Z),
Y(Z,X)
=
Z(X, Y) =
+ (Z, \7yX), (\7zX, Y) + (-- - Y).
(\7yZ,X)
551
13.1. Levi-Civita Connection
Now add the first two equations and subtract the third to get
+ Y(Z, X) - Z(X, Y) (VxY,Z) + (Y, VxZ) + (VyZ,X) + (Z, VyX)
X(Y, Z) =
- (VzX, Y) - (X, VzY).
If we assume the torsion zero hypothesis, then this reduces to
+ Y(Z, X) - Z(X, Y) (Y, [X, Z]) + (X, [Y, Z]) - (Z, [X, Y]) + 2(V xY, Z).
X(Y, Z) =
Solving, we see that V X Y must satisfy (13.2)
2(VxY, Z) = X(Y, Z)
+ Y(Z, X) -
+ (Z, [X, Y]) -
Z(X, Y)
(Y, [X, Z]) - (X, [Y, Z]).
Since knowing (V X Y, Z) for all Z is tantamount to knowing V X Y, we conclude that if V exists, then it is unique. On the other hand, the patient reader can check that if we actually define (V X Y, Z) and hence V X Y by this equation, then all of the defining properties of a covariant derivative are 0 satisfied and furthermore T will be zero. Formula (13.2) above, which serves to determine the Levi-Civita connection, is called the Koszul formula. It is easy to see that the restriction of a Levi-Civita connection to any open submanifold is just the Levi-Civita connection on that open sub manifold with the induced metric. It is a straightforward matter to show that the Christoffel symbols for the Levi-Civita connection in some chart are given by
rk, = ~ll lJ
where 9jkgki = (12.6).)
63,
2
(09jl ox i
+ ogli ox j
_ 09i j ) ox 1
(Recall that gij = (ai, OJ) and
rfj
'
are given by formula
We know from the study in the last chapter that we may take the covariant derivative of vector fields along maps. The most important cases for this chapter are fields along curves and fields along maps of the form h: (a, b) x (c, d) ---+ M. Exercise 13.10. Show that if 0 : I ---+ M is a smooth curve and X, Yare vector fields along 0, then (X, Y) = (VajatX, Y) + (X, Va/atY).
1t
Exercise 13.11. If h : (a, b) x (c, d) ---+ M is smooth, then oh/ot and oh/os are vector fields along h. Show that VB/Btoh/os = VB/Bsoh/ot. [Hint: Use local coordinates and the fact that V is torsion free.]
13. Riemannian and Semi-Riemannian Geometry
552
13.1.1. Covariant differentiation of tensor fields. Let V' be any natural covariant derivative on M. It is a consequence of Proposition 7.37 that for each X E X(U) there is a unique tensor derivation V' x on T;(U) such that V' x commutes with contraction and coincides with the given covariant derivative on X(U) (also denoted V' x) and with Lx f on Coo(U), Recall that if T E
T}, then s
(V' xT)(Y1 , ... , Ys)
= V' x(T(Y1 , ... , Ys )) -
L T( ... , V' xYi, .. ,), i=l
If Z E
Trl,
we apply this to V' Z E
Tl
and get
(V'xV'Z)(Y) = X(V'Z(Y)) - V'Z(V'xY) = V'x(V'yZ) - V'V'xyZ,
from which we get the following definition: Definition 13.12. The second covariant derivative of a vector field Z E Tr} is Definition 13.13. A tensor field T is said to be parallel if V' ~ T = 0 for all ~ E T M. Similarly, if a : I -t T; (M) is a tensor field along a curve e : I -t M that satisfies V' fit a = 0 on I, then we say that a is parallel along e. Just as in the case of a general connection on a vector bundle we then have a parallel transport map P(en o : T;(M)c(to) -t T;(M)c(t). From the previous chapter we know that V'ata (t )
.
= f-tO hm
+ E) -
P(eH+fa(t
E
a(t)
,
T;, and if eX is the curve t H rpf (p), then I' P(exn+f(T 0 rpf (p)) - Y 0 rpf (p)
It is also true that if T E n
v X
'Y'() 1.
P
=
1m
f-tO
E
.
I t is easy to see that the space of parallel tensor fields of type (r, s) is a vector space over ~. Exercise 13.14. Show that if T is parallel, then for any smooth curve e : [a, b] -t M such that e(a) = p and e(b) = q we have P(e)~T p = T q• Deduce that if M is connected, then the dimension of the space of parallel tensor fields of type (r,s) has dimension less than or equal to dimT;(M)p for any fixed p. The map V' x : T; M -t T; M just defined commutes with contraction by construction. Furthermore, if the connection we are extending is the Levi-Civita connection for a semi-Riemannian manifold (M, g), then V'~g =
0 for all
~ E
TM.
13.2. Riemann Curvature Tensor
553
To see this, recall that V'~(g @ Y @ W) = V'~g @X @Y
+ 9 @ V'~X @Y + g@X @ V'~Y,
which upon contraction yields V'~(g(X, Y)) ~(X, Y)
+ g(V'~X, Y) + g(X, V'~Y), = (V'~g)(X, Y) + (V'~X, Y) + (X, V'~Y). =
(V'~g)(X, Y)
We see that V' ~g == 0 for all ~ if and only if (X, Y) = (V' ~X, Y)+(X, V' ~Y) for all ~,X, Y. In other words, the statement that the metric tensor is parallel (constant) with respect to V' is the same as saying that the connection is a metric connection. Exercise 13.15. Let V' be the Levi-Civita connection for a semi-Riemannian manifold (M, g). Prove the formula
(13.3)
(eXg)(Y, Z)
= g(V' x Y, Z) + g(Y, V' xZ)
for vector fields X, Y, Z E X(M).
13.2. Riemann Curvature Tensor For (M, g) a Riemannian manifold with associated Levi-Civita connection V', the associated curvature tensor field is called the Riemann curvature tensor: For X, Y E X(M) we have the map R (X, Y) : X(M) -t X(M) defined by R(X, Y) Z:= Rx,yZ:= V'xV'yZ - V'yV'xZ - V'[X,YJZ.
Exercise 13.16. Show that V'i- y(Z) - V'~x(Z) = R (X, Y) Z (recall Definition 13.12). "
By direct calculation, or by appealing to Theorem 12.44 and Exercise 12.45 from the previous chapter, we find that (X, Y, Z) H R (X, Y) Z is COO(M)-multilinear (tensorial). Appealing to the isomorphism (7.7), we conclude that R gives a section of T~ (T M; T M). That is, R can be thought of as defining a TpM-valued tensor field at each p. In other words, if X p, Yp, Zp E TpM, then R (Xp, Yp) Zp is a well-defined element of TpM and (Xp, Yp, Zp) H R (Xp, Yp) Zp gives a multilinear map. Here, R (Xp, Yp) Zp is defined to be (R (X, Y) Z) (p) for any fields X, Y, Z such that X(p) = X p, Y(p) = Yp, and Z(p) = Zp- Many interpretations of R arise. From the previous chapter we know that R is also to be thought of as a TM-valued 2-form. From this point on we will freely interpret elements of r;(M) as elements of TT s (X(M)) when convenient. Notice that we will use both the notation R (X, Y) as well as Rx,y. Definition 13.17. A semi-Riemannian manifold (M, g) is called fiat if the curvature tensor is identically zero.
13. Riemannian and Semi-Riemannian Geometry
554
Recall that if f : (M, g) --7 (N, h) is a local diffeomorphism between semi-Riemannian manifolds such that f*h = g, then f is called a local isometry and we say that the manifolds are locally isometric. Theorem 13.18. Let (M,g) be a semi-Riemannian manifold of dimension n and index 1/. If (M, g) is fiat, that is, if the curvature tensor is identically zero, then (M, g) is locally isometric to the semi-Euclidean space IR~. Proof. Let p E M given. If the curvature tensor vanishes, then by Theorem 12.51 we can find local parallel vector fields defined in a neighborhood of p with prescribed values at p. So we may find parallel fields Xl, ... ,Xn such that XI(p), ... , Xn(P) is an orthonormal basis. But since parallel translation preserves the various scalar products (Xi, X j ), we see that we actually have an orthonormal frame field in a neighborhood of p. Next we use the fact that the Levi-Civita connection is symmetric (torsion zero). We have V' XiXj V'xjXi - [Xi,Xj] = 0 for all i,j. But since the Xi are parallel, this means that [Xi, Xj] == O. Therefore there exist coordinates xl, ... ,xn on a possibly smaller open set such that = Xi for all i. 00. xt
The result is that these coordinates give a chart which is an isometry of a neighborhood of p with an open subset of the semi-Riemannian space IR~.
0
For another proof of the previous theorem see the online supplement [Lee, J eflj. The next theorem exhibits the symmetries of the Riemann curvature tensor: Theorem 13.19. The map X, Y, Z, W t--+ (Rx,YZ, W) is tensorial in all variables. Furthermore, the following identities hold for all X, Y, Z, W E
X(M):
(i) Rx,y = -Ry,x. (ii) (Rx'yZ, W) = -(Rx'yW, Z). (iii) Rx'yZ
+ Ry,zX + Rz,xY = 0
(iv) (Rx'yZ, W)
=
(First Bianchi identity).
(Rz,wX, Y).
Proof. Tensorality is immediate from our previous observations. Also, (i) is immediate from the definition of Rand (ii) is just a special case of Proposition 12.70 from the previous chapter. For (iii) we calculate: Rx'yZ + Ry,zX
= V' xV'yZ -
+ Rz,xY V'yV'xZ + V'yV' zX -
V' zV'yX
+ V' zV' xY -
V' xV' zY
= O.
13.2. Riemann Curvature Tensor
555
The proof of (iv) is rather unenlightening and is just some combinatorics. Since R is a tensor, we may assume without loss of generality that [X, Y] = O. For any X, Y, Z, let (CRh,y Z be defined by (CRh,y Z:= Rx,YZ
+ Ry,zX + Rz,xY.
By (iii) we have ((CR)yZ X, W) = 0 for any W. Summing over all cyclic permutations of Y, Z, X,'W, we obtain 0= ((CR)y,z X, W)
+ ((CR)w,y z, X) + ((CR)x,w Y, Z) + ((CR)z,x w, Y).
Expand this expression using the definition of CR, and we have twelve terms. Four pairs of terms cancel due to (i) and (ii) resulting in
2 (Rx,YZ, W)
+ 2 (Rw,zX, Y) = O. D
Using (i) we obtain the result.
Theorem 13.20 (Second Bianchi identity). For X, Y, Z E X(M), we have (\7 zR)(X, Y)
+ \7 x R(Y, Z) + \7y R(Z, X) =
O.
Proof. This is the Bianchi identity for the Levi-Civita connection and in this context is also called the second Bianchi identity. We give an independent proof here. Since this is a tensor equation, we only need to prove it under the assumption that all brackets among the X, Y, Z are zero (recall Exercise 7.34). First we have (\7 zR)(X, Y)W
= \7 z(Rx,YW) - R(\l zX, Y)W - R(X, \7zY)W - RX'y\7zW = [\7z, Rx,Y]W - R(\7zX, Y)W - R(X, \7zY)W.
Using this, we calculate as follows:
+ (\7 xR)(Y, Z)W + (\7y R)(Z, X)W [\7z, Rx,Y]W + [\7x, Ry,z]W + [\7y,Rz,x]W
(\7 zR)(X, Y)W =
- R(\7zX, Y)W - R(X, \7zY)W - R(\7 x Y, Z) W - R(Y, \7 x Z) W - R(\7yZ, X)W - R(Z, \7yX)W
+ [\7x, Ry,z]W + [\7y,Rz,x]W + R([X, Zl, Y)W + R([Z, Yl, X)W + R([Y, X], Z)W [\7 z, [\7 x, \7 y]] + [\7 x, [\7 y, \7 z]] + [\7 y, [\7 z, \7 x]] =
= [\7z, Rx,Y]W
=
O.
The last identity is the Jacobi identity for commutators and is true for purely D algebraic reasons (see the next exercise).
13. Riemannian and Semi-Riemannian Geometry
556
Note: Given a semi-Riemannian manifold (M,g), the tensor R defined by R(X, Y, Z, W) := (Rx,Y Z, W) for all X, Y, Z, W, is also called the Riemann curvature tensor. Often this tensor is defined with a different ordering of the slots and one should always check which conventions are in use. One traditional ordering is R(W, Z, X, Y) := (Rx,y Z, W). Exercise 13.21. Show that if Li, i = 1,2,3, are linear operators, and the commutator is defined as usual ([A, B] = AB - BA), then we always have the Jacobi identity [Ll' [L2' L3ll + [L2' [L3, Llll + [L3, [Ll' L2ll = o.
We now introduce several objects which hold all or part of the information in the curvature tensor in different forms. First we mention that the reader should keep an eye out for expressions of the form (R(v, w)v, w) or (R( v, w)w, v) for v, w in the tangent space of a point on the semi-Riemannian manifold (M, g) under study.
It will be convenient to introduce a little linear algebra at this point. Recall that if (V, (., .)) is a finite-dimensional scalar product space, then there is an associated natural scalar product on 1\ 2V defined so that for an orthonormal basis {ed, the basis {ei 1\ ej h,j for 1\ 2V is also orthonormal. On simple elements we have
(Vl,V3) (V1,V4)) g(Vll\ v 2,v3I\ v4)=det ( ( ) ( ). V2,V3 V2,V4 We will use the angle brackets (.,.) for this scalar product also. The quantity (v 1\ W, v 1\ w) is important and needs special attention in the case of an indefinite scalar product. If (-, .) on V is indefinite, then the induced scalar product is also indefinite and (v 1\ w, v 1\ w) may be zero, even when v and W are linearly independent. For the next lemma, recall that a subspace W of a scalar product space (V, (., .)) is called nondegenerate if (., .) restricted to W is nondegenerate. Lemma 13.22. Let (V, (-, .)) be a finite-dimensional scalar product space.
Let P be a plane spanned by v, w. Then, (i) P is nondegenerate if and only if (v 1\ w, v 1\ w) i- o. (ii) (v 1\ w, v 1\ w) > 0 if and only if (., .) restricted to P is definite. (iii) (v 1\ w, v 1\ w) < 0 if and only if (-,.) restricted to P is indefinite.
o
Proof. Exercise. If v and w span a nondegenerate plane, then I(v squared area of the parallelogram spanned by v and w.
1\
w, v 1\ w) I is the
Lemma 13.23. Let (V, (-, .)) be a finite-dimensional scalar product space. If v, w E V are any two vectors, then there exist vectors v', w' E V arbitrarily
close to v and w respectively such that v', w'span a nondegenerate plane.
13.2. Riemann Curvature Tensor
557
Proof. Assume that (v 1\ w, v 1\ W) = 0, or there is nothing to prove. Any pair of vectors is close to a pair of linearly independent vectors, so we may assume that v and ware linearly independent. There exists a vector x such that (v 1\ x, v 1\ x) < 0. Indeed, if v is null, then we can pick x so that (v, x) of- 0, which means that (v 1\ x, v 1\ x) < 0. If v is not null, then pick x of opposite causal character. That is, pick x to be spacelike if v is timelike and vice versa. Now if WE := W + EX for small E > 0, then (v 1\ WE, V 1\ WE) = 2Eb + E2(v 1\ x, v 1\ x) for some number b independent of E. If b = 0, then (v 1\ WE' V 1\ WE) < 0, and we are done by Lemma 13.22. If b of- 0, then (v 1\ WE' V 1\ WE) is nonzero in case E is sufficiently small, and then v, WE span a nondegenerate plane by Lemma 13.22 again. 0 Note that in the previous lemma, closeness is measured in the standard topology of a finite-dimensional vectors space. One does not try to use an indefinite scalar product to define the topology! The symmetry properties for the Riemann curvature tensor allow that we have a well-defined map (13.4) which is symmetric with respect to the natural extension of 9 to The map 9t is defined implicitly as follows:
g(9t( VI
1\ V2), v3 1\ V4) :=
1\ 2(T M).
(R( VI, V2)V4, V3).
Notice the switch in the indices 3 and 4. This hides a sign and must be remembered to avoid confusion later. Another commonly used quantity is the sectional curvature K. If v and W span a non degenerate plane in TpM, then define
K(v
1\
w) :=
(R(v,w)w,v) (v, v) (w, w) - (v, w)2 (9t(vl\w),vl\w) (vl\w,vl\w)
The value K (v 1\ w) only depends on the oriented plane spanned by the vectors v and w; therefore if P = span{ v, w} is such a nondegenerate plane, we also write K(P) instead of K(v 1\ w). The set of all planes in TpM is denoted Grp (2). We remark that if M is 2-dimensional, then K is a scalar function on M. It turns out that this function is exactly the Gauss curvature introduced in Chapter 4. There we showed that the Gauss curvature is intrinsic and we found an expression for it in terms of the metric, which is still valid in this situation.
In the following definition, V is an R-module. The two cases we have in mind are (1) V is X(M), R = COO(M), and (2) V is TpM, R = R
13. Riemannian and Semi-Riemannian Geometry
558
Definition 13.24. A multilinear function F : V x V x V x V --t R is said to be curvature-like if it satisfies the symmetries proved for the curvature R above; namely, if for all x, y, z, wE V we have
(i) F(x, y, z, w) = -F(y, x, z, w); (ii) F(x,y,z,w) = -F(x,y,w,z);
+ F(y, z, x, w) + F(z, x, y, w)
(iii) F(x, y, z, w)
= 0;
(iv) F(x, y, z, w) = F(w, z, x, y). Exercise 13.25. Define the tensor Cg by Cg(X, Y, Z, W) := g(Y, Z)g(X, W) - g(X, Z)g(Y, W).
Show that Cg is curvature-like. Proposition 13.26. If F is curvature-like and F(v,w,v,w) v, wE V, then F == o.
o for
all
Proof. From (iv) it follows that for each v, the bilinear map (w, z) t-+ F(v,w,v,z) is symmetric, and so if F(v,w,v,w) = 0 for all v,w E V, then F(v,w,v,z) = 0 for all v,w,z E V. Now it is a simple matter to show that (i) and (ii) imply that F == o. 0 Proposition 13.27. If (Rv,wv, w) is known for all v, w E TpM, then R itself is determined at p. If K(P) is known for all nondegenerate planes in TpM, then R itself is determined at p. Proof. Let R2(V, w) := (Rv,wv, w) for v, w E TpM. Using an orthonormal basis for TpM, we see that K and R2 contain the same information, so we will just show that R2 determines R:
~8:
usut =
(R2(v+tz,w+su)-R2(v+tu,w+sz))
I 0,0
~8:
usut
I
{g(R(v+tz,w+su)v+tz,w+su)
0,0
- g(R(v
+ tu, w + sz)v + tu, w + sz)}
= 6R(v,w,z,u).
The second part follows by continuity and the fact that (R(v, w)w, v) (v 1\ W, v 1\ w) K (v 1\ w) for v, w spanning a non degenerate plane P. 0 For each vET M, the tidal operator Rv : TpM --t TpM is defined by Rv(w) := Rv,wv.
We are now in a position to prove the following important theorem.
13.2. Riemann Curvature Tensor
559
Theorem 13.28. The following assertions are all equivalent stant):
(Ii
is a con-
(i) K(P) = Ii for all P E Grp(2), P nondegenerate. (ii) (Rv1,v2V3,V4) = liCg(VI,V2,V3,V4) for all VI,V2,V3,V4 E TpM. (iii) -Rv(w) = Ii(W - (w, v) v) for all w, v E TpM with Ilvll = l. (iv) !.R(~) = Ii~ for all ~ E 1\ 2TpM. Proof. Let p EM. The proof that (ii)=}(iii) and that (iii)=}(i) is left as an easy exercise. We prove that (i) =}(ii)=}(iv)=}(i). (i)=}(ii): Let R be defined by R(VI,V2,V3,V4) := (RV1,V2V3,V4) and let Tg := R - liCg. Then Tg is curvature-like and Tg(v, w, v, w) = 0 for all v, w E TpM by assumption. It follows from Proposition 13.26 that Tg == O. (ii)=}(iv): Let {el,"" en} be an orthonormal basis for TpM. Then {ei 1\ ej h<j is an orthonormal basis for 1\ 2TpM. Using (ii), we see that
(!.R(ei
1\
ej), ek
1\
q)
=
\Rei,ejek, q)
=
(R(ei,ej)ek,el)
=
liCg(VI, V2, v3, V4) Ii (ei 1\ ej, ek 1\ q) for all k, l.
=
This implies that !.R( ei 1\ ej) Ii
liei 1\ ej. (iv)=}(i): This follows because if v, ware orthonormal, then we have = (!.R(v 1\ w), v 1\ w) = K(v 1\ w). 0 =
Definition 13.29. Let (M, g) be a semi-Riemannian manifold. The Ricci curvature is the (1, I)-tensor Ric defined by n
i=l
where (el,"" en) is any orthonormal basis of TpM and Ei := (ei, ei). We say that the Ricci curvature Ric is bounded from below by Ii and write Ric 2:: k if Ric(v,w) 2:: k(v,w) for all v,w E TM. Similar and obvious definitions can be given for Ric :::; k and the strict bounds Ric > k and Ric < k. Actually, it is often the case that the bound on Ricci curvature is given in the form Ric 2:: Ii(n - 1), where n = dim(M). In passing, let us mention that there is a very important and interesting class of manifolds called Einstein manifolds. A semi-Riemannian manifold (M, g) is called an Einstein manifold with Einstein constant k if and only if Ric(v,w) = k(v,w) for all v,w E TM. We write this as Ric = kg or even Ric = k. For example, if (M, g) has constant sectional curvature Ii, then it is an Einstein manifold with Einstein constant k = Ii( n - 1). The effect of this
13. Riemannian and Semi-Riemannian Geometry
560
condition depends on the signature of the metric. Particularly interesting is the case where the index is 0 (Riemannian) and also the case where the index is 1 (Lorentz manifold). Perhaps the first question one should ask is whether there exist any Einstein manifolds that do not have constant sectional curvature. It turns out that there are many interesting Einstein manifolds that do not have constant sectional curvature. For manifolds of dimension > 2 the Einstein manifold condition is natural and fruitful. Unfortunately, we do not have space to explore this fascinating topic (but see [Be]).
Exercise 13.30. Show that if M is connected and dim(M) > 2, and Ric (.,.) = I(-' .), where I E COO(M), then I == k for some k E lR (so (M, g) is Einstein). 13.3. Semi-Riemannian Submanifolds Let M be ad-dimensional submanifold of a semi-Riemannian manifold M of dimension n, where d < n. The metric g(-,.) = (.,.) on M restricts to a tensor on M, which we denote by h. Since h is a restriction of g, we shall also use the notation (.,.) for h. If the restriction h is nondegenerate on each space TpM and has the same index for all p, then h is a metric tensor on M and we say that M is a semi-Riemannian submanifold of M. If M is Riemannian, then this nondegeneracy condition is automatic and the metric h is automatically Riemannian. More generally, if ¢ : N -+ (M, g) is an immersion, we can consider the pull-back tensor ¢* 9 defined by ¢*g(X, Y) = g(T¢· X, T¢· Y).
If ¢* 9 is non degenerate on each tangent space, then it is a metric on N called the pull-back metric and we call ¢ a semi-Riemannian immersion. If N is already endowed with a metric gN, and if ¢*g = gN, then we say that ¢ : (N, 9 N) -+ (.!v!, g) is an isometric immersion. Of course, if ¢* 9 is a metric at all, as it always is if (M, g) is Riemannian, then the map ¢ : (N,¢*g) -+ (M,g) is automatically an isometric immersion. Every immersion restricts locally to an embedding, and for the questions we study here there is not much loss in focusing on the case of a submanifold .!v! C M.
There is an obvious bundle on M which is the restriction of T M to M. This is the bundle TMIAf = UpEMTpM. Recalling Lemma 7.47, we see that each tangent space TpM decomposes as --
TpM ~
--
~
= TpM EB (TpM) , ~
where (TpM) = {v E TpM : (v, w) = 0 for all W E TpM}. Then TM = UPEM (TpM)~, with its natural structure as a smooth vector bundle, is called
13.3. Semi-Riemannian Submanifolds
561
the normal bundle to M in M. The smooth sections of the normal bundle will be denoted by r (TMl..) or X(M)l... The orthogonal decomposition above is globalized as
A vector field on M is always the restriction of some (not unique) vector field on a neighborhood of M. The same is true of any, not necessarily tangent, vector field along M. The set of all vector fields along M will be denoted by X( M) 1M' Since any function on M is also the restriction of some function on M, we may consider X(M) as a submodule of X(M)IM' If X E X(M), then we denote its restriction to M by XIM or sometimes just X. Notice that X(M)l.. is a submodule of X(M) 1M' We have two projection maps, nor: TpM -+ TpM l.. and tan: TpM -+ TpM which in turn give module projections nor: X(M)IM -+ X(M)l.. and tan: X(M)IM -+ X(M). We also have the pair of naturally related restrictions Coo (M) res~ion Coo (M) and X(M) res~ion X(M) 1M' Note that X(M) is a Coo(M)-module, while X(M)IM is a Coo(M)-module. We have an exact sequence of modules:
Now we shall obtain a sort of splitting of the Levi-Civita connection of M along the submanifold M. First we notice that the Levi-Civita connection 'V on M restricts nicely to a connection on the bundle T M IM -+ M. The reader should be sure to realize that the space of sections of this bundle is exactly X(M)IM' We wish to obtain a restricted covariant derivative 'VIM: X(M) x X(M)IM -+ X(M)IM' If X E X(M) and W E X(M)IM' then 'VxW does not seem to be defined since X and Ware not elements of X(M). But we may extend X and W to elements of X(M), use 'V, and then restrict again to get an element of X (M) 1M' Then recalling the local properties of a connection we see that the result does not depend on the extension.
Exercise 13.31. Use local coordinates to prove that 'V x W does not depend on the extensions used. It is also important to observe that the restricted covariant derivative is exactly the covariant derivative obtained by the methods of Section 12.3 of the previous chapter. Namely, it is the covariant derivative along the inclusion map M '---7 M. Thus, it is defined even without an appeal to the process of extending fields described above.
13. Riemannian and Semi-Riemannian Geometry
562
We shall write simply V7 in place of V71 M since the context will make it clear when the latter is meant. Thus for all p EM, we have V7 x W (p) = V7 x W (p), where X and Ware any extensions of X and W respectively. Clearly we have V7 x (Y1 , Y2 ) = (V7 x Y1 , Y2 ) + (YI, V7 x Y2 ) and so V7 is a metric connection on T J\;[IM' For fixed X, Y E X(M), we have the decomposition of V7 x Y into tangent and normal parts. Similarly, for V E X( M)l., we can consider the decomposition of V7 x V into tangent and normal parts. Thus we have
+ (V7xy)l., V7 X V = (V7 X v)tan + (V7 X V)l..
V7x Y = (V7x y )tan
Proposition 13.32. For a semi-Riemannian submanifold Me M, we have
V7 x Y = (V7 X y)tan for all X, Y
E
X(M),
where V7 is the Levi-Civita covariant derivative on M with its induced metric. Proof. It is straightforward to show that if we extend fields X, Y, Z to X, Y, Z, then the Koszul formula (13.2) for V7 xY implies that
(13.5)
2((V7xy)tan,Z) = X(Y,Z)
+ Y(Z,X)
+ (Z, [X, Y]) -
- Z(X, Y)
(Y, [X, Z]) - (X, [Y, Z])
for all Z. But the Koszul formula which determines V7 x Y shows that (V7 x Y) tan = V7 X Y. 0 Definition 13.33. Define maps II: X(M)xX(M) ----7 X(M)l., II: X(M)x X(M)l. ----7 X(M) and V71. : X(M) x X(M)l. ----7 X(M)l. according to
II(X, Y)
:=
(V7 X y)l. for all X, Y E X(M),
II(X, V)
:=
(V7xV)tan for all X E X(M), V E X(M)l.,
V7iv
:=
(V7xV)l. for all X E X(M), V E X(M)l..
It is easy to show that V71. defines a metric covariant derivative on the normal bundle TMl.. The map (X, Y) f---t II(X, Y) is clearly COO(M)-linear in the first slot. If X, Y E X(M) and V E X(M)l., then 0 = (Y, V) and we have
o = V7 x (Y, V) =
(V7 x Y, V)
+ (Y, V7 x V)
= ((V7xy)l. ,V) + (Y, (V7xV)tan) = (II(X, Y), V) + (Y, II(X, V)). It follows that
(13.6)
(II(X, Y), V) = -(Y, II(X, V)).
13.3. Semi-Riemannian Submanifolds
563
From this we see that II(X, Y) is not only COO (M)-linear in X, but also in Y. This means that II is tensorial, and so II(Xp, Yp) is a well-defined element of TpMl.. for each Xp, Yp E TpJovf. Thus II is a TMl..-valued tensor field and for each p we have an IR-bilinear map IIp : TpM x TpM -t TpM1.. (We often suppress the subscript p.) Proposition 13.34. I I is symmetric. Proof. For any X, Y E X(M) we have II(X, Y) - II(Y,Xt)
= (V'xY - V'yX) 1. = ([X, Y])1. = o.
D
We can also easily deduce that II is a symmetric COO(M)-bilinear form with values in X(M) and is similarly tensorial. So II(Xp, Vp) is a welldefined element of TpM for each fixed Xp E TpM and Vp E TpM 1.. We thus obtain a bilinear form 1. IIp: TpM x TpM -t TpM. In summary, I I and I I are tensorial, V'1. is a metric covariant derivative and we have the following formulas: V'xY = V'xY + II(X, Y), 1.V'xV = V'xV + II(X, V)
for X, Y E X(M) and V E X(M)l... Recall that if (V, (., ·)1) and (W, (., ·)2) are scalar product spaces, then a linear map A : V -t W has a metric transpose At : W -t V uniquely defined by the requirement that
(Av,w)2 = (v, Atw\ for all v E V and w E W. Definition 13.35. For v E TpM, we define the linear map BvO := II(v, .). Formula (13.6) shows that the map II(v,·) : TpM1. -t TpM is equal to -Bt : TpM -t TpMl... Writing any Y E X(M)IM as a column vector (ytan, y1.)t, we can write the map V'x : X(M)IM -t X(M)IM as a matrix of operators:
[~;
-:t]·
Next we define the shape operator, also called the Weingarten map. We have already met a special case of the shape operator in Chapter 4. The shape operator is sometimes defined with the opposite sign.
13. Riemannian and Semi-Riemannian Geometry
564
Definition 13.36. Let p E M. For each unit vector u normal to M at p, we have a map called the shape operator Su associated to u defined by
Su(v) := - (V' v U ) tan, where U is any unit normal field defined near p such that U (p) = u. Exercise 13.37. Show that the definition is independent of the choice of normal field U that extends u.
The family of shape operators {Su : u a unit normal} contains essentially the same information as the second fundamental tensor I I or the associated map B. This is because for any X, Y E X(M) and U E X(M)l., we have (SuX,Y) = ((_V'xU)tan ,Y) = (U,-V'xY) = (U,(-V'xy)l.)
=
(U,-IJ(X,Y)).
In the case of a hypersurface, we have (locally) only two choices of unit normal. Once we have chosen a unit normal u, the shape operator is denoted simply by S rather than Su. Theorem 13.38. Let M be a semi-Riemannian submanifold of M. For any V, W,X, Y E X(M), we have
(RvwX, Y) =(RvwX, Y) - (IJ(V, X), IJ(W, Y))
+ (IJ(V, Y), IJ(W, X)).
This equation is called the Gauss equation or Gauss curvature equation.
Proof. Since this is clearly a tensor equation, we may assume that [V, Wl = o (see Exercise 7.34). We have (RvwX, Y) = (V'vV'wX, Y)-(V'wV'vX, Y). We calculate:
+ (V'v(IJ(W, X)), Y) (V'vV'wX, Y) + (V'v(IJ(W, X)), Y) (V'vV'wX, Y) + V(IJ(W, X), Y) - (IJ(W, X), V'vY)
(V'vV'wX, Y) = (V'vV'wX, Y) = =
= (V'vV'wX, Y) - (IJ(W, X), V'vY). Since (IJ(W, X), V'vY)
=
(IJ(W, X), (V'vy)l.)
= (IJ(W, X), IJ(V, Y)), we have (V'vV'wX, Y) = (V'vV'wX, Y) - (IJ(W, X), IJ(V, Y)). Interchanging the roles of V and W and subtracting we get the desired conclusion. 0
13.3. Semi-Riemannian Submanifolds
565
The second fundamental form contains information about how the semiRiemannian sub manifold M bends in M. Definition 13.39. Let M be a semi-Riemannian submanifold of M and N a semi-Riemannian sub manifold of N. A pair isometry : (M, M) -+ (N, N) consists of an isometry : M -+ N such that (M) = N and such that IM : M -+ N is an isometry. Proposition 13.40. A pair isometry : (M, M) -+ (N, N) preserves the
second fundamental tensor: Tp . II( v, w) = II(Tp . v, Tp . w) for all v, wE TpM and all p EM. Proof. Let p E M and extend v, w E TpM to smooth vector fields V and W. Since isometries respect Levi-Civita connections, we have * 'VvW = 'Vcp.v*W. Since is a pair isometry, we have Tp(TpM) C Tcp(p)N and
Tp(TpM.l) C (Tcp(p)N).l. This means that * : X(M)IM -+ X(N)IN preserves normal and tangential components *(X(M)) C X(N) and *(X(M).l) C X(N).l. We have Tp· II(v, w) = *II(V, W)( (p)) = * ('VvW).l ( (p))
= (*'VvW).l((p)) = ('Vcp.v*W).l((p)) II (* V, * W)( (p)) = II (* V, * W)( (p)) = II(Tp . v, Tp . w).
=
o
The following exercise gives a simple but conceptually important example. Exercise 13.41. Let M be the 2-dimensional strip {(x, y, 0) : -11" < X < 11"} considered as a sub manifold of ]R3. Let N be the subset of ]R3 given by { (x, y, J 1 - x 2 ) : -1 < x < 1}. Show that M is isometric to N. Show that there is no pair isometry (]R3, M) -+ (]R3, N). Definition 13.42. Let M be a semi-Riemannian submanifold of a semiRiemannian manifold M. Then for any V, W, Z E X(M) define ('VvII) by
('VvII) (W, Z)
:=
'V& (II(W, Z)) - II('VvX, Y) - II(X, 'VvY).
Theorem 13.43. With M, M, and V, W, Z E X(M) as in the previous
definition we have the following identity: (RvwZ).l = ('VvII)(W, Z) - ('VwII) (V, Z)
(Codazzi equation).
13. Riemannian and Semi-Riemannian Geometry
566
Proof. Since both sides are tensorial, we may assume that [V, W] Then (RvwZ)-L = (V'vV'wZ)-L - (V'wV'vZ)-L. We have (V'vV'wZ)-L
O.
= (V'v (V'wZ))-L - (V'v (II(W, Z)))-L = II(V, V'wZ) - (V'v (II(W, Z)))-L.
Now recall the definition of (V'vII) (W, Z) and find that (V'vV'wZ)-L = II(V, V'wZ) + (V'vII) (W, Z) +II(V'vW, Z) +II(W, V'vZ).
Now compute (V'vV'wZ)-L - (V'wV'vZ)-L and use the fact that V'vWV'wV= [V,W] =0. 0 The Gauss equation and the Codazzi equation belong together. If we have an isometric embedding f : N -+ M, then the Gauss and Codazzi equations on f(N) c M pull back to equations on N and the resulting equations are still called the Gauss and Codazzi equations. Obviously, these two equations simplify if the ambient manifold M is a Euclidean space. For a hypersurface existence theorem featuring these equations as integrability conditions see [Pe]. 13.3.1. Semi-Riemannian hypersurfaces. A semi-Riemannian submanifold of codimension one is called a semi-Riemannian hypersurface. Let M be a semi-Riemannian hypersurface in M. By definition, each tangent space TpM is a nondegenerate subspace of TpM. The complementary spaces (TpM)-L are easily seen to be nondegenerate, and ind (TpM-L) is constant on M since we assume that ind (TpM) is constant on M. The number ind (TpM-L) called the co-index of M. Exercise 13.44. Show that the co-index of a semi-Riemannian hypersurface must be either 0 or 1. Definition 13.45. The sign E of a hypersurface M is defined to be +1 if the co-index of M is 0 and is defined to be -1 if the co-index is 1. We denote it by sgn M. Notice that if E = 1, then ind( M) ind(M) = ind(M) - 1.
= ind( M), while if
E
= -1, then
Proposition 13.46. Let f E COO(M) and M := f-l(c) for some c E JR. Suppose that M of- 0 and that grad f of- 0 on M. Then M is a semiRiemannian submanifold if and only if either (grad f, grad f) > 0 on M, or (grad f, grad f) < 0 on M. The sign of M is the sign of (grad f, grad f), and grad f / Ilgrad fll restricts to a unit normal field along M.
13.4. Geodesics
567
Proof. The relation (grad f, grad f) of. 0 on M ensures that df of. 0 on M, and it follows that M is a regular submanifold of codimension one. Now if v E TM, then (gradf,v) = df(v) = v(f)
= v(fIM) = 0, so grad f is normal to M. Thus for any p E M the space (TpM)l.. is nondegenerate, and so the orthogonal complement TpM is also nondegenerate. The rest is clear. 0 We now consider certain exemplary hypersurfaces in 1R~+l. 1R~+l -+ IR be the quadratic form defined by v
q(x)
:=
(x, x)
= -
2)xi)2 + i=l
Let q :
n+l
L
(x i )2
i=v+l
n i=l
where the reader will recall that Ei = -1 or Ei = 1 as 1 ::; i ::; v or 1 + v ::; i ::; n + 1. Hypersurfaces in 1R~+1 defined by Q(n,r,E) := {x E 1R~+l : q(x) = Er2}, where E = -1 or E = 1, are called hyperquadrics. Exercise 13.47. Let Q(n, r, E) be a hyperquadric as defined above. Let P := 2: XiOi be the position vector field in 1R~+l. Show that the restriction of Plr to Q(n, r, E) is a unit normal field along Q(n, r, E). Exercise 13.48. Show that a hyperquadric Q(n, r, E) as defined above is a semi-Riemannian hypersurface with sign E.
13.4. Geodesics In this section, I will denote a nonempty interval assumed to be open unless otherwise indicated by the context. Usually, it would be enough to assume that I has nonempty interior. We also allow I to be infinite or "half-infinite". Let (M, g) be a semi-Riemannian manifold. Suppose that, : I -+ M is a smooth curve that is self-parallel in the sense that 'iJa/y = 0
along ,. We call, a geodesic. To be precise, one should distinguish various cases as follows: If, : [a, b] -+ M is a curve which is the restriction of a geodesic defined on an open interval containing [a, b], then we call, a (parametrized) closed geodesic segment or just a geodesic for short. If , : [a, (0) -+ M (resp. , : (-00, a] -+ M) is the restriction of a geodesic then we call, a positive (resp. negative) geodesic ray.
13. Riemannian and Semi-Riemannian Geometry
568
If the domain of a geodesic is JR, then we call 'Y a complete geodesic. If M is an n- manifold and the image of a geodesic 'Y is contained in the domain
of some chart with coordinate functions xl, ... , x n , then the condition for 'Y to be a geodesic is
(13.7)
d2 x i 0'Y dt 2 (t)
.
+ L rjkb(t))
dxio'Y dxk0'Y dt (t) dt (t) = 0
for all tEl and 1 :S i :S n. This follows from formula (12.7) and this is a system of n second order equations often abbreviated to f + L ~t = 0, 1 :S i :S n. These are the local geodesic equations. Now consider a smooth curve 'Y whose image is not necessarily contained in the domain of a chart. For every to E I, there is an E > 0 such that 'Yi(to-€,to+€) is contained in the domain of a chart, and thus it is not hard to see that 'Y is a geodesic if and only if each such restriction satisfies the corresponding local geodesic equations for each chart which meets the image of 'Y. We can convert the local geodesic equations (13.7) into a system of 2n first order equations by the usual reduction of order trick. We let v denote a new dependent variable and then we get
d;t:
fjk d::
We can think of xi and vi as coordinates on T M. Once we do this, we recognize that the first order system above is the local expression of the equations for the integral curves of a vector field on TM. Exercise 13.49. Show that there is a vector field G E X(T M) such that a is an integral curve of G if and only if'Y := 7fTM 0 a is a geodesic. Show that the local expression for G is
The vector field G from this exercise is an example of a spray (see Problems 4-8 from Chapter 12). The flow of G in the manifold T M is called the geodesic flow. Lemma 13.50. For each v E TpM, there is an open interval I containing 0 and a unique geodesic 'Y : I --+ M, such that -r (0) = v (and hence 'Y (0) = p). Proof. This follows from standard existence and uniqueness results for differential equations. One may also deduce this result from the facts about flows since, as the exercise above shows, geodesics are projections of integral curves of the vector field G. The reader who did not do the problems
13.4. Geodesics
569
on sprays in Chapter 12 would do well to look at those problems at this time. 0 Lemma 13.51. Let 1'1 and 1'2 be geodesics I ---t M. If'h(to) = 'Y2(tO) for
some to
E
I, then 1'1 = 1'2·
Proof. If not, there must be t' E I such that 1'1 (t') # 1'2 (t'). Let us assume that t' > to since the proof of the other case is similar. The set A = {t E I : t > to and 1'1 (t) # I'2(t)} has an infimum b = inf A. Note that b ~ to. Claim: 'Y1(b) = 'Y2(b). Indeed, if b = to, there is nothing to prove. If b > to, then 'Y1 (t) = 'Y2(t) on the interval (to, b). By continuity 'Y1 (b) = 'Y2(b). Now t H 1'1 (b + t) and t H 1'2 (b + t) are clearly geodesics with initial velocity 'Y1(b) = 'Y2(b). Thus by Lemma 13.50, 1'1 = 1'2 for some open interval containing b. But this contradicts the definition of b as the infimum of A.
0
A geodesic I' : I ---t M is called maximal if there is no other geodesic with open interval domain J strictly containing I that agrees with I' on I. Theorem 13.52. For any v E TM, there is a unique maximal geodesic
I'v with 'Yv(O) = v. Proof. Take the class Yv of all geodesics with initial velocity v. This is not empty by Lemma 13.50. If (x, fJ E Yv and the respective domains Ia and I f3 have nonempty intersection, then (X and fJ agree on this intersection by Lemma 13.51. From this we see that the geodesics in Yv fit together to form a manifestly maximal geodesic with domain I = U')'E9J,. Obviously this geodesic has initial velocity v. 0 Definition 13.53. If the domain of every maximal geodesic emanating from a point p E TpM is all of~, then we say that M is geodesically complete at p. A semi-Riemannian manifold is said to be geodesically complete if and only if it is geodesically complete at each of its points. Exercise 13.54. Let ~~ be the semi-Euclidean space of index v. Show that all geodesics are of the form t H Xo + tw for w E ~~. Definition 13.55. A continuous curve I' : [a, b] ---t M is called a broken geodesic segment if it is a piecewise smooth curve whose smooth segments are geodesic segments. If t* is a point of [a, b] where I' is not smooth, we call I'(t*) a break point. (A smooth geodesic segment is considered a special case.) Exercise 13.56. Prove that a semi-Riemannian manifold is connected if and only if every pair of its points can be joined by a broken geodesic I' : [a, b] ---t M.
570
13. Riemannian and Semi-Riemannian Geometry
Exercise 13.57. Show that if '"Y is a geodesic, then a reparametrization c := '"Y 0 f is a geodesic if and only if f(t) := at + b for some a, b E JR and a =I o. Show that if ~ is never null, then we may choose a, b so that the geodesic is unit speed and hence parametrized by arc length.
The existence of geodesics passing through a point p E M at parameter value zero ~ith any specified velocity allows us to define a very important map. Let Dp denote the set of all v E TpM such that the geodesic '"Yv is defined at least on the interval [0, 1]. The exponential map, exp p : i5p -+ M, is defined by expp v := '"Yv(l). Lemma 13.58. If '"Yv is the maximal geodesic with ~v(O) = v E TpM, then for any c, t E JR, we have that '"Ycv(t) is defined if and only if '"Yv(ct) is defined. When either side is defined, we have
'"Ycv(t) = '"Yv(ct). Proof. Let Jv,c be the maximal interval for which '"Yv(ct) is defined for all t E Jv,c. Certainly 0 E J. Use the chain rule for covariant derivatives or calculate locally to see that t f--7 '"Yv(ct) is a geodesic with initial velocity cv. But then by uniqueness and the maximality of '"Ycv, the interval Jv,c must be contained in the domain of '"Ycv and for t E Jv,c we must have
'"Ycv(t) = '"Yv(ct). In other words, if the right hand side is defined, then so is the left and we have equality. Now let u = cv, s = ct and b = 1/c. Then we just as well have that
'"Ybu(S) = '"Yu(bs), where if the right hand side is defined, then so is the left. But this is just '"Yv(ct) = '"Ycv(t). So left and right have reversed and we conclude that if either side is defined, then so is the other. 0 Corollary 13.59. If '"Yv is the maximal geodesic with then
~v(O) =
v E TpM,
(i) t is in the domain of '"Yv if and only if tv is in the domain of expp; (ii) '"Yv(t) = expp(tv) for all t in the domain of '"Yv. Proof. Suppose that tv is in the domain of expp- Then '"Ytv (1) is defined. But '"Ytv (1) = '"Yv (t) and t is in the domain of '"Yv by the previous lemma. The converse is proved similarly and so we obtain expp(tv) = '"Ytv(1) = '"Yv(t). 0
13.4. Geodesics
571
Now we have a very convenient situation. The maximal geodesic through p with initial velocity v can always be written in the form t f-t expp tv. Straight lines through Op E TpM are mapped by expp onto geodesics which we sometimes refer to as radial geodesics through p. Similarly, we have radial geodesic segments and radial geodesic rays emanating from p. The result of the following exercise is a fundamental observation. Exercise 13.60. Show that (i'v(t),'Yv(t))
= (v,v) for all t in the domain of
"(V·
The exponential map has many uses. For example, it is used in comparing semi-Riemannian manifolds with each other. Also, it provides special coordinate charts. The basic theorem is the following: Theorem 13.61. Let (M, g) be a semi-Riemannian manifold and p E M. There exists an open neighborhood Up C Vp containing Op such that expp IfJ: p is a diffeomorphism onto its image Up. Proof. The tangent space TpM is a vector space, which is isomorphic to IR n and so has a standard differentiable structure. Using the results about smooth dependence on initial conditions for differential equations, we can easily see that exp p is well-defined and smooth in some neighborhood of Op E TpM. The main point is that the tangent map Texp p : TOp(TpM) --+ TpM is an isomorphism and so the inverse mapping theorem gives the result. To see that T expp is an isomorphism, let vOp E Top (TpM) be the velocity of the curve t f-t tv in TpM. Then, unraveling definitions, we have T expp vOp = fA 10 expp tv = v. Thus T expp is just the canonical map vOp f-t v. 0 Definition 13.62. A subset C of a vector space V that contains 0 is called star-shaped about 0 if whenever VEe, tv E C for all t E [0,1]. Definition 13.63. If U c Vp is a star-shaped open set about Op in TpM such that expp IfJ is a diffeomorphism as in the theorem above, then the
image expp(U) = U is called a normal neighborhood of p. In this case, U is also referred to as star-shaped.
Ii
Theorem 13.64. U c M is a normal neighborhood about p with corresponding preimage U C TpM, then for every point q E U there is a unique geodesic"( : [0,1] --+ U c M such that "((0) = p, "((1) = q, '1'(0) E U and expp'Y(O) = q. (Note that uniqueness here means unique among geodesics with image in U.) Proof. The preimage U corresponds diffeomorphic ally to U under expp. Let -1
-
-
(q) so that v E U. By assumption U is star-shaped and so the map p : [0,1] --+ TpM given by t f-t tv has image in U. But then, the
V
= expplfJ
13. Riemannian and Semi-Riemannian Geometry
572
geodesic segment 'Y : t f--7 expp tv, t E [0,1] has its image inside U. Clearly, 'Y(O) = P and 'Y(1) = q. Since p = v, we get
1'(0) = Texppp(O) = Texppv = v under the usual identifications in TpM. Now assume that 'Y1 : [0,1] ---+ U c M is some geodesic with 'Y1(0) and 'Y1 (1) = q. If 1'1 (0) = w, then 'Y1 (t) = expp two
=
P
Claim: The ray PI : t f--7 tw (t E [0,1]) stays inside fJ. If not, then the set A = {t : tw rJ- fJ} is nonempty. Let t* = inf A and consider the set C := {tw : t E (0, t*)}. Then C c fJ and fJ\i5 is contractible (check this). But its image expplif (fJ\C) is U\C, where C is the image of (0, t*) under 'Y1. Since U\C is certainly not contractible, we have contradiction. Thus the claim~is true, and in particular w = P1(1) E fJ. Therefore, both wand v are in U. On the other hand,
exp p w = 'Y1 (1) = q = expp V. Thus, since exp pIif is a diffeomorphism and hence injective, we conclude that w = V. By the basic uniqueness theorem for geodesics, the segments 'Y and 'Y1 are equal and both given by t f--7 exp p tv. D Let (M,g) be a semi-Riemannian manifold of dimension n. Let Po E M and pick any orthonormal basis (e1,"" en) for the semi-Euclidean scalar product space (TpoM, (-, ·)po). This basis induces an isometry I : lR~ ---+ TpoM by (xi) f--7 L; xiei' If U is a normal neighborhood centered at Po EM, then X norm := 10 exp po 10 1 : U ---+ lR~ = lR n is a coordinate chart with domain U. These coordinates are referred to as normal coordinates centered at Po. Normal coordinates have some very nice properties: Theorem 13.65. Ifx norm = (xl, ... , xn) are normal coordinates defined on U and centered at Po, then
gij (po) = \ r~k(pO)
a~i ' a~j ) PO =
Ei 0, then the set B R (p) (also denoted B (p, R)) defined by B R (p) = {q EM: dist(p, q) < R} is called an open geodesic ball centered at p with radius R. It is important to notice that unless R is small enough, BR(p) may not be homeomorphic to a ball in a Euclidean space. To see this just consider a ball of large radius on a circular cylinder of small diameter. Proposition 13.85. Let U be a normal neighborhood of a point p in a Riemannian manifold (M, g). If q E U and if "( : [0, 1] --7 M is the radial geodesic such that "((0) = p and "((1) = q, then"( is the unique shortest curve in U (up to reparametrization) connecting p to q. Proof. Let 0: be a curve connecting p to q (refer to Figure 13.6).Without be the loss of generality we may take the domain of 0: to be [0, b]. Let radial unit vector field in U. Then if we define the vector field R along 0: by t I---t la(t)' we may write 0: = (R, o:)R + N for some field N normal to
tr
Ir
1 Recall
that by our conventions, a piecewise smooth curve is assumed to be continuous.
13. Riemannian and Semi-Riemannian Geometry
586
R (but note that N(O) = 0). We now have
L(a)
=
fob(a, a)1/2 dt
=
fob [(R,a)2
+ (N,N)]1/2
dt
2: fo bl (R,a)ldt2: fob(R,a)dt= fob :t(roa)dt = r(a(b)) = r(q).
On the other hand, if v = '1'(0), then r(q) = f01 Ilvll dt = f01('1', '1')1/2 dt so L(a) 2: L(-y). Now we show that if L(a) = L(-y), then a is a reparametrization of ,. Indeed, if L(a) = L(-y), then all of the above inequalities must be equalities so that N must be identically zero and ft (r 0 a) = (R, a) = I(R,a)l. It follows that a = (R,a)R = (ft(roa))R, and so a travels 0 radially from p to q and must be a reparametrization of ,.
~R
p
Figure 13.6. Normal neighborhood of p
It is important to notice that the uniqueness assertion of Theorem 13.85 only refers to curves with image in U. This is in contrast to the proposition below. Proposition 13.86. Let Po be a point in a Riemannian manifold M. There exists a number EO (p) > 0 such that for all E, 0 < E ::; EO (p) we have the following:
(i) The open geodesic ball B(po, E) is normal and has the form B (po,
(ii) For any p
E) = exp po {v
I I < E} .
E Tpo M : v
E B(Po,E), the radial geodesic segment connecting Po to
p is the shortest curve in M, up to parametrization, from Po to p. (Note carefully that we now mean the shortest curve among curves into M rather than just the shortest among curves with image in B(po, E).)
13.5. Riemannian Manifolds and Distance
587
-
-
Proof. Let U c TpoM be chosen so that U = exp po U is a normal neighborhood of PO. Then for sufficiently small c > the ball
°
B(O,c)
:=
{v E TpoM: Ilvll < c}
is a starshaped open set in fj, and so Apo,c = exppo (B (0, c)) is a normal neighborhood of PO. From Proposition 13.85 we know that the radial geodesic segment (J from Po to P is the shortest curve in Apo,c from Po to p. This curve has length less than c. We claim that any curve from Po to P whose image leaves Apo,c must have length greater than c. Once this claim is proved, it is easy to see that
Apo,c = B(po, c)
= {p EM:
dist(po,p) < c}
and that (ii) holds. Now suppose that a : [a, b] -+ M is a curve from Po to p which leaves Apo,c' Then for any r > with r < c, the curve a must meet the set S(r) := exppo({v E TpoM: Ilvll = r}) at some first parameter value tl E [a, b]. Then al := al[a,tl] lies in Apo,c, and Proposition 13.85 tells us that L(a) ~ L( al[ah]) ~ r. Since this is true for all r < c, we have L(a) ~ c, which is what was claimed. 0
°
Theorem 13.87 (Distance topology). Given a Riemannian manifold, define the distance function dist as before. Then (M, dist) is a metric space, and the metric topology coincides with the manifold topology on M. Proof. To show that dist is a true distance function (metric) we must prove that (1) dist is symmetric; dist(p, q) = dist(q,p); (2) dist satisfies the triangle inequality dist( q, p) :'S dist(p, x) + dist(x, q); (3) dist(p, q)
~
0; and
(4) dist(p, q) = 0 if and only if p = q. Now, (1) is obvious, and (2) and (3) are clear from the properties of the integral and the metric tensor. For (4) suppose that p =1= q. Then since M is Hausdorff, we can find a normal neighborhood U of p that does not contain q. In fact, by the previous proposition, we may take U to be of the form B (p, c). Since (by the proof of the previous proposition) every curve starting at p and leaving B(p, c) must have length at least c/2, we see that dist(p, q) ~ c/2. 0 By definition a curve segment in a Riemannian manifold, say c : [a, b] -+ M, is a shortest curve if L(c) = dist(c(a), c(b)). We say that such a curve is (absolutely) length minimizing. Such curves must be geodesics.
Proposition 13.88. Let M be a Riemannian manifold. A length minimizing curve c : [a, b] -+ M must be an (unbroken) geodesic.
588
13. Riemannian and Semi-Riemannian Geometry
Proof. There exist numbers ti with a = to < t1 < ... < tk = b such that for each subinterval [ti' ti+1], the restricted curve cl[t.z,1+1 t. J has image in a totally star-shaped open set. Thus since cl [t.z, t.t+1 J is minimizing, it must be a reparametrization of a unit speed geodesic (use the uniqueness part of Proposition 13.85). Thus there is a reparametrization of c that is a broken geodesic. But this new reparametrized curve is also length minimizing, and so by Problem 1 it is smooth. D
13.6. Lorentz Geometry In this section we define and discuss a few aspects of Lorentz manifolds. Lorentz manifolds play a prominent role in physics and are often singled out for special study. We discuss the local length maximizing property of timelike geodesics in a Lorentz manifold and derive the Lorentzian analogue of Proposition 13.85. Definition 13.89. A Lorentz vector space is a scalar product space with index equal to one and dimension greater than or equal to 2. A Lorentz manifold is a semi-Riemannian manifold such that each tangent space is a Lorentz space with the scalar product given by the metric tensor. Under our conventions, the signature of a Lorentz manifold is of the form (-1,1" ... ,1,1).2 Each tangent space of a Lorentz manifold is a Lorentz vector space, and so we first take a closer look at some of the distinctive features of Lorentz vector spaces. Let us now agree to classify the zero vector in a Lorentz space as spacelike. For Lorentz spaces, we may classify subspaces into three categories: Definition 13.90. Let V be a Lorentz vector space such as a tangent space of a Lorentz manifold. A subspace W c V is called (1) spacelike if glw is positive definite (or if W is the zero subspace); (2) timelike if glw nondegenerate with index 1; (3) lightlike if glw is degenerate. Thus a subspace falls into one of the three types, which we refer to as its causal character. If we take a timelike vector v in a Lorentz space V, then lRv, the space spanned by v, is nondegenerate and has index 1. By Lemma 7.47, v.l. is nondegenerate and V = lRv EEl v.l.. Since 1 = ind (V) = ind (lRv) + ind (v.l.) , it follows that ind (v.l.) = 0, so that v.l. is spacelike. This little observation is useful enough to set out as a proposition. 2Some authors use (1, -1, ... , -1, -1), but this does not really change the geometry.
13.6. Lorentz Geometry
589
Figure 13.7. Causal character of a subspace
Proposition 13.91. If V is a Lorentz vector space and v is a timelike element, then v~ is spacelike, and we have the orthogonal direct sum V = lRv EEl v~. Exercise 13.92. Show that if W is a subspace of a Lorentz space, then W is timelike if and only if W ~ is spacelike. Exercise 13.93. Suppose that v, ware linearly independent null vectors in a Lorentz space V. Show that (v, w) =I- O. [Hint: Use an orthonormal basis to orthogonally decompose; V = lReo EEl P, where eo is timelike and where (-,.) is positive definite on P. Suppose (v, w) = 0; write v = aeo + PI and w = /3eo + P2. Then show that (PI,P2) = a/3, (Pi, Pi) = a 2 = /3 2 and I(PI,P2)1
= IlpIilllp211·]
Lemma 13.94. Let W be a subspace of a Lorentz space. Then the following conditions are equivalent:
(i) W is timelike and so a Lorentz space in its own right. (ii) There exist null vectors v, w E W that are linearly independent. (iii) W contains a timelike vector. Proof. Suppose (i) holds. Let el, ... , em be an orthonormal basis for W with el timelike. Then el + e2 and el - e2 are both null and, taken together, are a linearly independent pair so that (ii) holds. Now suppose that (ii) holds and let v, w be a linearly independent pair of null vectors. By Exercise 13.93 above, either v+w or v-w must be timelike so we have (iii). Finally, suppose (iii) holds and v E W is timelike. Since v~ is spacelike and W~ C v~, we
13. Riemannian and Semi-Riemannian Geometry
590
see that W~ is spacelike. But then W is timelike by Exercise 13.92 so that (i) holds. 0 Exercise 13.95. Use the above lemma to prove that if W is a nontrivial subspace of a Lorentz space, then the following three conditions are equivalent: (i) W contains a nonzero null vector but no timelike vector.
(ii) W is lightlike. (iii) The intersection of W with the nullcone is one-dimensional. Definition 13.96. The timecone determined by a timelike vector v is the set C (v) := {w E V : (v, w) < o}. In Problem 6 we ask the reader to show that timelike vectors v and w in a Lorentz space V are in the same timecone if and only if (v, w) < o. Exercise 13.97. Show that there are exactly two timecones in a Lorentz vector space whose union is the set of all nonzero timelike vectors. Describe the relation of the nullcone to the timecones. Now we come to an aspect of Lorentz spaces that underlies the twins paradox of special relativity. Proposition 13.98. If v, ware timelike elements of a Lorentz vector space then we have the backward Schwartz inequality
l(v,w)l2: Ilvllllwll, with equality only if v is a scalar multiple of w. Also, if v and ware in the same timecone, then there is a uniquely determined number a 2: 0, called the hyperbolic angle between v and w, such that
(v, w) =
-llvllllwll cosha.
Note: The minus sign appears because of our convention that (v, v) = -1 for timelike vectors. Proof. We may write w = av
a2 (v, v)
+z
where z E v~. We have
+ (z, z)
=
(w, w)
0 such that ,(t) is in C(p), the vector P(r(t)), and hence P(a(t)), must be timelike. For such t, (a, P) < 0 which implies that (1, P) < 0 and hence iq < O. So q starts out negative and goes down hill as long as ,(t) is in C(p). Since, can only exit C(p) by reaching the nullcone (or 0) where q vanishes, we see that, must remain inside C(p).
0,
0,
Now we consider what happens if , is timelike but merely piecewise smooth. The first segment remains in C(p), and at the first parameter value tl where, fails to be smooth, we must have (1(f1), P) < O. But then by the Gauss lemma again (a(tl)' P) < O. The technical restriction of Definition 13.104 forces (a(ti), p) < 0 so that a(ti) E C(p). Applying the Gauss lemma gives (1(ti), P) < 0 and so cannot change sign at tl. We are now set up to repeat the argument for the next segment. The result follows 0 inductively.
iq 0,
The following proposition for Lorentz manifolds should be compared to Proposition 13.85 proved for Riemannian manifolds. In this proposition we find that the geodesics are locally longer than nearby curves. Proposition 13.106. Let U be a normal neighborhood of a point p in a Lorentz manifold. If the radial geodesic, connecting p to q E U is timelike, then it is the unique longest geodesic segment in U that connects p to q. Once again, uniqueness is up to reparametrization. Proof. Let if be related to U as usual. Take any timelike curve a : [0, b] -7 U segment in U that connects p to q. By the previous lemma, j3 := exp;l 0 a
stays inside a single timecone C(p) and so also inside c(p)nif. Thus a stays inside expp(C(p)nif) where it is timelike and where the field R = (P/r)oa is
13. Riemannian and Semi-Riemannian Geometry
594
a unit timelike field along a. We now seek to imitate the proof of Proposition 13.85. We may decompose a as
a=
-
(R,a)R+N,
where N is a spacelike field along a that is orthogonal to R. We have
Iiall =
J- (a,a) = V(R,a)2 -
(N,N)
~ I(R,a)l·
Recall that q(.) = (., .) and q := q 0 exp; 1. Since r = R and so grad r = -P/r, we have (grad r)oa = -R. By the previous lemma (a, R) is negative. Then . . d(roa) l(a,R)I=-(R,a)= dt . Thus we have
L(a) =
fob Ila(t)11 dt ~ r(q) = Lb).
If L(a) = Lb), then N = 0 and we argue as in the Riemannian case to conclude that a is the same as I up to reparametrization. 0
Recall that the arc length of a timelike curve is often called the curve's proper time and is thought of as the intrinsic duration of the curve. We may reparametrize a timelike geodesic to have unit speed. The parameter is then an arc length parameter, which is often referred to as a proper time parameter. We may restate the previous theorem to say that the unit speed geodesic connecting p to q in U is the unique curve of maximum proper time among curves connecting p to q in U.
13.7. Jacobi Fields Once again we consider a semi-Riemannian manifold (M, g) of arbitrary index. We shall be dealing with smooth two-parameter maps h : (-E, E) X [a, b] ~ M. The partial maps t H- hs(t) = h(s, t) are called the longitudinal curves, and the curves s H- h(s, t) are called the transverse curves. Let a be the center longitudinal curve t H- ho(t). The vector field along a defined by V(t) = hs(t) is called the variation vector field along a. We will use the following important result more than once:
1s Is=o
Lemma 13.107. Let Y be a vector field along the smooth map h : (-E, E) X
[a,b]
~
M. Then
Proof. If one computes in a local chart, the result falls out after a mildly tedious computation, which we leave to the curious reader. 0
13.7. Jacobi Fields
595
Suppose we have the special situation that, for each s, the partial maps t f---t hs (t) are geodesics. In this case, let us denote the center geodesic t f---t ho(t) by I. We call h a variation of I through geodesics. Let h be such a special variation and V the variation vector field. Using Lemma 13.107 and the result of Exercise 13.11 we compute
V' EJt V' at V = V' at V' aJ3s h = V' at V' a/3th = V' as V' at 8th + R( 8th, 8 s h )8t h = R(8t h, 8 s h)8th and evaluating at s = 0 we get V'atV'atV(t) = Rb(t), V(t))i'(t). This equation is important and shows that V is a Jacobi field as per the folowing definition:
Definition 13.108. Let I : [a, b] -+ M be a geodesic and let J E X')'(M) be a vector field along I. The field J is called a Jacobi field if
V' at V' aJ = Rb(t), J(t))i'(t) for all t E [a, b].
In local coordinates, we recognize the above as a second order system of linear differential equations and we easily arrive at the following Theorem 13.109. Let (M, g) and the geodesic I : [a, b] -+ M be as above. Given WI, W2 E T')'(a)M, there is a unique Jacobi field J W l,W2 E X')'(M) such that J(a) = WI and V' aJ(a) = W2. The set Jac (r) of all Jacobi fields along I is a vector space isomorphic to T')'(a)M x T')'(a)M. We now examine the more general case of a Jacobi field J W l,W2 along a geodesic I : [a, b] -+ M. First notice that for any curve 0: : [a, b] -+ M with I(a( t), a( t)) I > 0 for all t E [a, b], any vector field Y along 0: decomposes into an orthogonal sum yT + Y l.. This means that yT is a multiple of a and that Y l. is normal to a. If I : [a, b] -+ M is a geodesic, then V' at y l. is also normal to ly since 0 = -9t(yl.,ly) = (V'atYl.,ly) + (yl., V'atly) = (V'atYl.,ly). Similarly, V' at yT is parallel to ly all along I.
Theorem 13.110. Let I : [a, b] -+ M be a geodesic segment.
(i) If Y E X')'(M) is tangent to I, then Y is a Jacobi field if and only ifV'~tY = 0 along I. In this case, Y(t) = (at + b)i'(t). (ii) If J is a Jacobi field along I and there are some distinct tl, t2 E [a, b] with J(tl)..ily(tl) and J(t2)..ily(t2), then J(t)..ily(t) for all t E [a, b]. (iii) If J is a Jacobi field along I and there is some to E [a, b] with J(to)..ily(to) and V'aJ(to)..ily(to), then J(t)..ily(t) for all t E [a,b]. (iv) If I is not a null geodesic, then Y is a Jacobi field if and only if both yT and yl. are Jacobi fields.
13. Riemannian and Semi-Riemannian Geometry
596
Proof. (i) Let Y
= f"r. Then the Jacobi equation reads 'V~J"r(t) = R("f(t) , f"r(t)h(t) = O.
Since I is a geodesic, this implies that
1" = 0 and
(i) follows.
(ii) and (iii) We have ~(J,"r) = (R("f(t),J(t)h(t),"r(t)) = 0 (from the symmetries of the curvature tensor). Thus (J(t), "r(t)) = at + b for some a, bE R The reader can now easily deduce both (ii) and (iii). (iv) The operator 'V~t preserves the normal and tangential parts of Y. We now show that the same is true of the map Y H R("f(t) , Yh(t). Since we assume that I is not null, we have yT = f"r for some "r. Thus R("f(t) , yTh(t) = R("f(t) , f"r(t)h(t) = 0, which is trivially tangent to "r(t). On the other hand, (R("f(t) , y..L(t) h(t), "r(t)) = 0 by symmetries of the curvature tensor. We have
('V~tY) T + ('V~tY)..L = 'V~tY = R("f(t) , Y(t)h(t)
+ R("f(t) , y..L(t)h(t)
=
R("f(t) , yT (t)h(t)
=
0 + R("f(t) , y..L(t)h(t).
So the Jacobi equation 'V~t Y(t) = R("f(t), Y(t)h(t) splits into two equations 'V~tyT (t)
= 0,
'V~ty..L(t) = R"y(t),Y.L(t)"r(t), and the result follows from this.
o
Corollary 13.111. Let I = IV and J~'w be as above. Then J~,rv(t) rt"rv(t). If w.l..v, then (J0,W(t),"rv(t)) = 0 for all t E [0, b]. Furthermore, every Jacobi field JO,w along expv tv with JO,W(O) = 0 has the form JO,w := rt"rv + JO,Wl, where w = 'VaJ(O) = rv + WI, wI.l..v and JO,Wl (t).l.. "rv(t) for all t E [0, b].
The proof of the last result shows that a Jacobi field decomposes into a parallel vector field along I, which is just a multiple of the velocity "r, and a "normal Jacobi field" J..L, which is normal to I at each of its points. Of course, the important part is the normal part since the tangential part is merely the infinitesimal model for a variation through geodesics which are merely reparametrizations of the s = 0 geodesic. Thus we focus attention on the Jacobi fields that are normal to the geodesics along which they are defined. Thus we consider the Jacobi equation 'V~J(t) = R("f(t), J(t)) "r(t) with initial conditions such as in (ii) or (iii) of Theorem 13.110. Exercise 13.112. For v E TpM, let v..L := {w E TpM: (w,v) = O}. Prove that the tidal operator Rv : w H Rv,wv maps v..L to itself.
In light of this exercise, we make the following definition.
13.7. Jacobi Fields
597
Definition 13.113. For v E TpM, the (restricted) tidal force operator Fv : v.l ---+ v.l is the restriction of Rv to v.l C TpM.
Notice that in terms of the tidal force operator the Jacobi equation for normal Jacobi fields is 'l~J(t)
= F"yCt) (J(t)) for all t.
If J is the variation vector field of a geodesic variation, then it is an infinitesimal model of the separation of nearby geodesics. In general relativity, one thinks of a one-parameter family of freely falling particles. Then '1 at J is the relative velocity field and 'l~t J is the relative acceleration. Thus the Jacobi equation can be thought of as a version of Newton's second law with the curvature term playing the role of a force. Proposition 13.114. For v E TpM, the tidal force operator Fv : v.l ---+ v.l is self-adjoint and Trace (Fv) = - Ric(v, v). Proof. First, (FvWl' W2) = (Rv,Wl v, W2) = (Rv,W2 v, WI) = (FvW2' WI) by (iv) of Theorem 13.19. The proof that Trace Fv = - Ric(v, v) is easy for definite metrics but for indefinite metrics the possibility that v may be a null vector involves a little extra work. If v is not null, then letting e2, ... ,en be an orthonormal basis for v.l we have
Ric(v,v) = -
I:>i
(Rv,eiv,ei) = - LEi (Fvei,ei) = -TraceFv .
If v is null, then we can find a vector W such that (w, v) = -1 and w, v span a Lorentz plane L in TpM. Define el := (v + w) /..;2 and e2 := (v - w) /..;2. One checks that el is timelike while e2 is spacelike. Now choose an orthonormal basis e3, . .. ,en for L.l C v.l so that el, ... ,en is an orthonormal basis for TpM. Then we have
Ric(v,v) = (Rv,elv,el) - (Rv,e2v,e2) - LEi (Rv,eiv,ei). i>2
But (Rv,el v, el)
= ! (Rv,wv, w) = (Rv,e2V, e2) and so we are left with
Ric(v,v) = - LEi (Rv,eiv,ei) = - LEi (Fvei,ei). i>2 i>2 Since v, e3, . .. ,en is an orthonormal basis for v.l and Fvv = 0, we have Ric(v, v)
= - LEi (Fvei,ei) i>2
= -
(Fvv, v) - LEi (Fvei' ei) = - Trace Fv.
0
i>2
Definition 13.115. Let, : [a, b] ---+ M be a geodesic. Let the set of all Jacobi fields J such that J(a) = J(b) = o.
Job, a, b) denote
13. Riemannian and Semi-Riemannian Geometry
598
Definition 13.116. Let, : [a, b] -+ M be a geodesic. If there exists a nonzero Jacobi field J E Joh, a, b), then we say that ,(a) is conjugate to ,(b) along ,.
From standard considerations in the theory of linear differential equations it follows that the set Joh, a, b) is a vector space. The dimension of the vector space Joh, a, b) is the order of the conjugacy. Since the Jacobi fields in Joh, a, b) vanish twice, and, as we have seen, this means that such fields are normal to "I all along" it follows that the dimension of Joh, a, b) is at most n - 1, where n = dim M. We have seen that a variation through geodesics is a Jacobi field; so if we can find a nontrivial variation h of a geodesic, such that all of the longitudinal curves t t-7 h8 (t) begin and end at the same points ,(a) and ,(b), then the variation vector field will be a nonzero element of Joh, a, b). Thus we conclude that ,(a) is conjugate to ,(b). We will see that we may obtain a Jacobi field by more general variations, where the endpoints of the curves meet at time b only to first order. Let us bring the exponential map into play. Let, : [0, b] -+ M be a geodesic as above. Let v = "1(0) E TpM. Then,: t t-7 expp tv is exactly our geodesic, which begins at p and ends at q at t = b. Now we create a variation of , by h(s, t) = expp t(v + sw), where W E TpM and s ranges in (-E, E) for some sufficiently small E. We know that J(t) = %818=0 h(s, t) is a Jacobi field, and it is clear that J(O) := %818=0 h(s, 0) = O. If Wbv is the vector tangent in Tbv(TpM) which canonically corresponds to w, in other words, if Wbv is the velocity vector at s = 0 for the curve s t-7 b (v + sw) in TpM, then
J(b) =
~!
uS
8=0
h(s, b) =
~!
uS 8=0
expp b(v + sw)
= Tbv expp(wbv).
(We have just calculated the tangent map of expp at x
= bv!) Also,
'VaJ(O) = 'Vat%s eXPpt(v+SW)!s=o,t=o = 'Va5
Is=0 ~! ut
exppt(v+sw). t=O
But X (s) := ~ 1t=o expp t( v+sw) = v+SW is a vector field along the constant curve t t-7 p, and so by Exercise 12.42 we have 'Va.ls=oX(s) = X'(O) = w. The equality J(b) = Tbv expp(vbv) is important because it shows that if Tbv expp : nv(TpM) -+ Ty(b)M is not an isomorphism, then we can find a vector Wbv E Tbv(TpM) such that Tbv expp (Wbv) = O. But then if W is the vector in TpM which corresponds to Wbv as above, then for this choice of w,
13.8. First and Second Variation of Arc Length
599
the Jacobi field constructed above is such that J(O) = J(b) = 0 so that ')'(0) is conjugate to ')'(b) along ')'. Also, if J is a Jacobi field with J(O) = 0 and \7 aJ(O) = w, then this uniquely determines J and it must have the form %818=0 expp t(v + sw) as above. Theorem 13.117. Let')': [0, b] --+ M be a geodesic. Then the following are equivalent:
(i) ')' (0) is conjugate to ')' (b) along ')'. (ii) There is a nontrivial variation h of ')' through geodesics that all start at p = ')'(0) such that J(b) := ~~ (0, b) = O. (iii) If v = 1 (0), then T bv expp is singular. Proof. (ii)===}(i): We have already seen that a variation through geodesics is a Jacobi field J and that if (ii) holds, then by assumption J(O) = J(b) = 0, and so we have (i). (i)===}(iii): If (i) is true, then there is a nonzero Jacobi field J with J(O) = J(b) = O. Now let w = \7 aJ(O) and h(s, t) = expp t(v + sw). Then h (s, t) is a variation through geodesics and 0 = J (b) = %slo exp p b( v + sw) = Tbv exp p(Wbv) so that Tbv expp is singular. (iii)===}(ii): Let v = 1 (0). If nvexpp is singular, then there is a w with nv expp Wbv = O. Thus the variation h(s, t) = exp p t(v + sw) does the job. 0
13.8. First and Second Variation of Arc Length Let us restrict attention to the case where 0 is either spacelike or timelike (not necessarily geodesic). This is just the condition that 1(6:(t),6:(t))1 > O. Let E = +1 if 0 is spacelike and E = -1 if 0 is timelike. We call E the sign of 0 and write E = sgn o. Consider the arc length functional defined by
L(o) =
lb
(E(6:(t),6:(t)))1/2 dt
=
lb
1(6:(t), 6:(t)) 11/2 dt.
If h : (-c, c) x [a, b] --+ M is a variation of 0 as above with variation vector field V, then formally V is a tangent vector at 0 in the space of curves [a, b] --+ M. By a simple continuity and compactness argument we may
choose a real number c > 0 small enough that I(hs(t), hs(t)) I > 0 for all s E (-c, c). Then we have the variation of the arc length functional defined by
d id I 8=0 L(h s ) := ds 8=0
8Lla (V) := ds
Ib (. . ) a
E(hs(t), hs(t))
1/2
dt.
600
13. Riemannian and Semi-Riemannian Geometry
Thus, we are interested in studying the critical points of L(s) := L(h s ), and so we need to find L'(O) and L"(O). For the proof of the following proposition we use the result of Exercise 13.11 to the effect that 'V' a.8th = 'V' at8sh. Proposition 13.118. Let h : (-E, E) X [a, b] ~ M be a variation of a curve 0::= ho such that l(hs(t),hs(t))1 > 0 for all s E (-E,E). Then
L'(s)
lb
=
E('V' as8th(s, t), 8t h(s, t)) (E(8t h(s, t), 8th(s, t)) )-1/2 dt.
Proof. We have
L'(s)
=
=
:s lb I lb lb lb b
a
I
Ilhs(t) dt
d (.
.
ds E(hs(t), hs(t))
) 1/2
dt
2E('V' ashs(t), hs(t))~ (E(hs(t), hs(t))) -1/2 dt
=
=
E('V' a.8th(s, t), 8t h(s, t)) (E(8t h(s, t), 8t h(s, t)) )-1/2 dt
=E
('V' at8sh(s, t), 8t h(s, t)) (E(8t h(s, t), 8t h(s, t)) )-1/2 dt.
0
Corollary 13.119. We have
oLla (V)
=
L'(O)
=
lb
E
('V' at V(t), a(t)) (E(a(t), a(t)) )-1/2 dt.
Let us now consider a more general situation where 0: : [a, b] ~ M is only piecewise smooth (but still continuous). Let us be specific by saying that there is a partition a = to < tl < ... < tk < tk+l = b so that 0: is smooth on each [ti, ti+l]. A variation appropriate to this situation is a continuous map h : (-E, E) X [a, b] ~ M with h(O, t) = o:(t) such that h is smooth on each set of the form (-E, E) X [ti, ti+1]. This is what we mean by a piecewise smooth variation of a piecewise smooth curve. The velocity a and the variation vector field V(t) := ah~~,t) are only piecewise smooth. At each "kink" point ti we have the jump vector 6a(ti) := a(ti+) - a(ti-), which measures the discontinuity of a at k Using this notation, we have the following theorem which gives the first variation formula: ~ M be a piecewise smooth variation of a piecewise smooth curve 0: : [a, b] ~ M with variation vector field V. If 0: has constant speed c= (E(a,a))1/2, then
Theorem 13.120. Let h : (-E, E) X [a, b]
oLla (V)
=
L'(O)
=
-~
I
b
('V' ata, V) dt - ~ cae
k
b
L (6a(ti), V(ti)) + ~(a, V)I a i=l
C
601
13.8. First and Second Variation of Arc Length
Proof. Since c = (E(6:,6:))1/2, Proposition 13.119 gives
L'(O)
r
ti + 1 iti (\7 8t V(t),
E
~L
=
6:(t)) dt.
-it
Since we have (6:, \7 8t V) = (6:, V) - (\7 8t6:, V), we can employ integration by parts: On each interval [ti, ti+l] we have E lti+l
-
C
E
(\7 8t V, 6:) dt = - (6:, V) c k to get
ti
We sum from i
= 0 to i =
b
Iti+l ti
E lti+l
-
C
ti
L'(O) = ~(6:, V)I - ~ L(66:(ti), V(ti)) - ~ c
k
c
a
lb (\78t6:, V) dt,
C
i =l
(\7 8t 6:, V) dt.
a
o
which is the required result.
A variation h : (-to, to) X [a, b] ---+ M of a is called a fixed endpoint variation if h(s, a) = a(a) and h(s, b) = a(b) for all s E (-to, to). In this situation, the variation vector field V is zero at a and b.
Corollary 13.121. A piecewise smooth curve a : [a, b] ---+ M with constant speed c > 0 on each subinterval where a is smooth is a (nonnull) geodesic if and only if 8Lla (V) = 0 for all fixed endpoint variations of a. In particular, if M is a Riemannian manifold and a: [a, b] ---+ M minimizes length among nearby curves, then a is an (unbroken) geodesic. Proof. If a is a geodesic, then it is smooth and so 66:(ti) = 0 for all ti (even though a is smooth, the variation still only needs to be piecewise smooth). It follows that L'(O) = o. Now if we suppose that a is a piecewise smooth curve and that L' (0) = 0 for any variation, then we can conclude that a is a geodesic by picking some clever variations. As a first step we show that al[t. t.'1.+1 1 is a geodesic for each segment [ti, ti+1]' Let t E (ti' ti+l) be arbitrary and let v be any nonzero vector in Ta(t)M. Let f3 be a cut-off function on [a, b] with support in (t - 8, t + 8) and 8 chosen sufficiently small. Then let V(t) := f3(t)Y(t), where Y is the parallel translation of y along a. We can now easily produce a fixed endpoint variation with variation vector field V by the formula '1"
h(s, t)
:= eXPa(t)
sV(t).
With this variation the last theorem gives
lb (\78t 6:, V) dt = -- I + (\78t6:, f3(t)Y(t)) dt, c
L'(O) = --E
E
a
C
t 8
t-8
which must hold no matter what our choice of y and for any 8> O. From this it is straightforward to show that \7 8t 6:(t) = 0, and since t was an arbitrary
13. Riemannian and Semi-Riemannian Geometry
602
element of (ti' ti+1), we conclude that al[t. t.1.+1 J is a geodesic. All that is left is to show that there can be no discontinuities of a.:. Once again we choose a vector y, but this time y E TCX(ti)M, where ti is a potential kink point. Take another cut-off function [3 with supp [3 C [ti-l,ti+1l = [ti-l, til u [ti' ti+1], [3(ti) = 1, and i a fixed but arbitrary element of {I, 2, ... , k}. Extend y to a field Y as before and let V = [3Y. Since we now have that a is a geodesic on each segment, and we are assuming that the variation is zero, the first variation formula for any variation with variation vector field V reduces to 1.,
for all y. This means that 6a.:(ti) = 0, and since i was arbitrary, we are ~~.
0
°
We now see that, for fixed endpoint variations, L'(O) = implies that a is a geodesic. The geodesics are the critical "points" (or curves) of the arc length functional restricted to all curves with fixed endpoints. In order to classify the critical curves, we look at the second variation but we only need the formula for variations of geodesics. For a variation h of a geodesic ,,(, we have the variation vector field V as before, but now we also consider the transverse acceleration vector field A(t) := \7 8sosh(0, t). Recall that for a curve "( with I(i', 1') I > 0, a vector field Y along "( has an orthogonal decomposition Y = yT + y.l (tangent and normal to "(). Also we have (\7 8t y).l = \7 8t y.l, and so we can use \7 8t y.l to denote either of these without ambiguity. We now have the second variation formula of Synge: Theorem 13.122. Let "( : [a, bl ---+ M be a (nonnull) geodesic of speed c > 0. Let c be the sign of"( as before. If h : (-10, E) X [a, bj is a variation
of"( with variation vector field V and acceleration vector field A, then the second variation of L(8) := L(hs(t)) at 8 = 0 is
._ 1(8s 8h (8, t), 8h Proof. Let H (8, t).8s (8, t)) 11/2 -_ (8h c( 8s (8, t), 8h 8s (8, t)) ) 1/2 . We have L'(8) = iH(8, t) dt. Computing as before, we see that
J:
OH(8, t) c / oh Oh) 08 = H\08(8,t),\78s ot (8,t) .
13.8. First and Second Variation of Arc Length
603
Taking another derivative, we have
8 2H(s, t) = ~ (H~/8h V 8h) _ /8h V 8h)8H) 8s 2 H2 8s \ 8t ' as 8t \ 8t ' as 8t 8s
+ \ 8t ' V 0 at -
1/8h 8h)8H) H \ at ' Va 8t 8s
~ ( \ V as ~~ , Va ~~) + \ ~~ ,V~s ~~) -
;2 (~~ ,Va ~~) 2) .
C
=H
=
8h 8h) \ Vas 8t ' Vas 8t
/8h
(/
8h)
2
8
8
Using Vat8sh
8
8
= Vas8th and Lemma 13.107, we obtain
8h Va Vas 8t 8
8h
= Vas Vat 8s = R
(8h 8h) 8h 8s' at 8s
8h
+ Vat Vas 8s '
and then
8 2H 8s2
8h
C {
8h
/8h
(8h 8h) 8h)
= H (Vat 8s ' Vat 8s) + \ 8t ,R 8s' 8t 8s 8h
+ (8t' Vat Va
8
8h C /8h 8h)2} 8s) - H2 \ at ' Vat 8s .
Now we let s = 0 and get
82 H
8s 2 (0, t)
=
C{
~ (Vat V, V Ot V)
+ b, R(V, 1')V)
+ b, Vat A ) -
:2 b, Vat V)2}.
Before we integrate the above expression, we use the fact that (1', VatA) = (1', A) ('Y is a geodesic) and the fact that the orthogonal decomposition of Vat Vis Vat V = c2 b, Vat V)'Y + Vat V -1,
ft
C
so that (Vat V, Vat V) = 2x b, Vat V)2 + (Vat V -1, Vat V -1). Plugging these identities in, observing the cancellation, and integrating, we get
L"(O)
=
lb ~s~
(O,t)dt =
~
lb
((Vat V-1 , VatV-1)
C Ib . + -b,A) C a
+ (1',R(V,1')V)) dt D
The right hand side of the main equation of the second variation formula just proved depends only on V except for the last term. But if the variation is a fixed endpoint variation, then this dependence drops out. It is traditional to think of the set S1 a ,b(p, q) of all piecewise smooth curves 0: : [a, b] -+ M from p to q as an infinite-dimensional manifold. Then a variation vector field V along a curve 0: E S1(p, q) which is zero at the endpoints is the "tangent vector" at 0: to the curve in S1 a ,b(P, q) given by the corresponding fixed endpoint variation h. Thus the "tangent space"
604
13. Riemannian and Semi-Riemannian Geometry
TaO = Ta (Oa,b(P, q)) at Q is the set of all piecewise smooth vector fields V along Q such that V(a) = V(b) = O. We then think of L as being a function on Oa,b(P, q) whose constant speed and nonnull critical points we have discovered to be nonnull geodesics beginning at P and ending at q at times a and b respectively. Further thinking along these lines leads to the idea of the index form. Let us abbreviate Oa,b(P, q) to f2 a,b or even to 0. For our present purposes, we will not lose anything by assuming that a = 0 whenever convenient. On the other hand, it will pay to refrain from assuming that b = 1. Definition 13.123. For a given nonnull geodesic T [0, b] ----7 M, the index form 1"(: T,,(O x T,,(O ----7 IR is defined by I"((V, V) = L~(O), where L"((s) =
Ji I(hs(t), hs(t))ll/2 dt and V' ash(O, t) = V.
Of course this definition makes sense because L~(O) only depends on V and not on h itself. Also, we have defined the quadratic form I"((V, V), but not directly I"((V, W). Of course, polarization gives I"((V, W), but if V, W E T,,(O, then it is not hard to see from the second variation formula that (13.9) It is important to notice that the right hand side of the above equation is in fact symmetric in V and W.
It is important to remember that the variations and variation vector fields we are dealing with are allowed to be only piecewise smooth even if the center curve is smooth. So let 0 = to < tl < ... < tk < tk+l = b as before and let V and W be vector fields along a geodesic 'Y. We now derive another formula for I"((V, W). Rewrite formula (13.9) as I"((V, W) =
~ 2: k
lti+l . {(V'at V~, V' at W~) + (R(i',
i=O
Vh, W) } dt.
t,
On each interval [ti, ti+l] we have
(V' at V~, V' at W~) = V' at (V' at V~, W~) - (V'~t V~, W~), and substituting this into the above formula we obtain E~ k
I
t,+1 {
I"((V, W) = ~ ~. i=O
~
~
2
~
~
V' at (V' at V ,W ) - (V' at V ,W )
t,
+(R(i', Vh, W) } dt. As for the last term, we use symmetries of the curvature tensor to see that
(R(i', Vh, W) = (R(i', V~h, W~).
605
13.8. First and Second Variation of Arc Length
Substituting we get I, (V, W)
=
f:o' I + t'
C,"", k
~
1
ti
{
-.l
2
-.l
-.l
-.l
\7 8t (\7 8t V ,W ) - (\7 8t V ,W )
+(R('Y, V-.l)"Y, W-.l)} dt. Using the fundamental theorem of calculus on each interval [ti' ti+1], and the fact that W vanishes at a and b, we obtain the following alternative formula:
Proposition 13.124 (Formula for index form). Let, : [0, bJ ----t M be a nonnull geodesic. Then for V, W E T,Oa,b, I, (V, W)
=
r (\7~t c io
-~
b
V-.l
+ R('Y, V-.l)"Y, W-.l) dt
k
C,"", -.l - - ~ (~\7 8t V (ti),
c
-.l
W (ti)),
i=1
Letting V = W we have I, (V, V) =
-~ c
b
k
r (\7~t V-.l+R('Y, V-.l)"Y, V-.l) dt-~ L(~ \7 io c
8t V-.l(ti),
V-.l(ti)) ,
i=1
and the presence of the term (R(~, V-.l)"Y, V-.l) indicates a connection with Jacobi fields.
Definition 13.125. A geodesic segment, : [a, bJ ----t M is said to be relatively length minimizing (resp. relatively length maximizing) if for all piecewise smooth fixed endpoint variations h of , the function L(s) := I(hs(t), hs(t))11/2 dt has a local minimum (resp. local maximum) at s = (where, = ho(t) := h(O, t)).
J: °
If , : [a, bJ ----t M is a relatively length minimizing nonnull geodesic, then L"(O) = 0, which means that I, (V, V) = for any V E T,Oa,b. The adverb "relatively" is included in the terminology because of the possibility that there may be curves in Oa,b which are "far away" from , and which have smaller length than,. A simple example of this is depicted in Figure has greater length than, even though is relatively length 13.8, where minimizing. We assume that the metric on (0,1) x 8 1 is the usual definite metric dx 2+dy2 induced from that on lRx (0,1), where we identify 8 1 x (0, 1) with the quotient lR x (0, l)/((x, y) '" (x + 27r, y)). On the other hand, one sometimes hears the statement that geodesics in a Riemannian manifold are locally length minimizing. This means that for any geodesic , :
°
,2
,2
606
13. Riemannian and Semi-Riemannian Geometry
S
1
X(O,l)
Figure 13.8. Geodesic segments on a cylinder
[a, b] -+ M, the restrictions to small intervals are always relatively length minimizing. But note that this is only true for Riemannian manifolds. For a semi-Riemannian manifold with indefinite metric, a small geodesic segment can have nearby curves that decrease the length. For example, consider the metric -dx 2 + dy2 on lR x (0,1) and the induced metric on the quotient Sl x (0,1) = lR x (0,1)/,,-,. In this case, the geodesic "( in the figure has length greater than all nearby geodesics; the index form 1'1 is now negative semidefinite. Exercise 13.126. Prove the above statement concerning 1'1 for 8 1 x (0,1) with the index 1 metric -dx 2 + dy2. It is not hard to see that if even one of V or W is tangent to -y, then 1'1 (V, W) = and so 1'1 (V, W) = 1'1 (V.l , W.l). Thus, we may as well restrict 1'1 to T¢n = {V E T'Yn : V l..-Y}.
°
Notation 13.127. The restriction of 1'1 to T.:j-n will be called the restricted index and will be denoted by 1~. The nullspace N(I~) is then defined by
N(I~)
:=
{V E T¢n: 1¢(V, W) =
°
for all WE T¢n}.
Theorem 13.128. Let "( : [0, b] -+ M be a nonnull geodesic. The nullspace T.:j-n -+ R is exactly the space Jo(,,(,O,b) of Jacobi fields vanishing at "((0) and "((b).
N(I~) of 1~ :
Proof. The formula of Proposition 13.124 makes it clear that JO("(, 0, b) N(I~).
c
Suppose that V E N(I~). Let t E (ti' ti+l), where the ti determine a partition of [0, b] such that V is potentially nonsmooth at the ti as before. Pick an arbitrary nonzero element y E ("((t)).l c T'Y(t)M and let Y be the unique parallel field along "(I[t.t,t+1 t. 1 such that Y(t) = y. Picking a cut-off
13.8. First and Second Variation of Arc Length
607
function (3 with support in [t + 8, t - 8] C (ti' ti+l) as before we extend (3Y to a field W along 'Y with W(t) = y. Now V is normal to the geodesic and so Ly(V, W) = I~(V, W) and
1
t 8 + (V7~t V I, (V, W) = --E c t-8
+ R('Y, Vh, (3Y) dt.
For small 8, (3Y~ is approximately the arbitrary nonzero y and it follows that V7~t V + R('Y, Vh is zero at t. Since t was arbitrary, V7~t V + R('Y, Vh is identically zero on (ti' ti+l). Thus V is a Jacobi field on each interval (ti' ti+ 1)' and since V is continuous on [0, b], it follows from the standard theory of differential equations that V is a smooth Jacobi field along all of 'Y. Since V E T,n, we already have V(O) = V(b) = O. We conclude that V E Joh, 0, b). 0 Proposition 13.129. Let (M,g) be a semi-Riemannian manifold of index ind(g) and'Y : [a, b] -+ M a nonnull geodesic. If the index form I, is positive semidefinite, then ind(g) = 0 or n (thus the metric is definite and so, up to the sign of g, the manifold is Riemannian). On the other hand, if I, is negative semidefinite, then ind(g) = 1 or n - 1 (so that up to the sign convention, M is a Lorentz manifold). Proof. For simplicity we assume that a = 0 so that 'Y : [0, b] -+ M. Let I, be positive semi-definite and assume that 0 < /I < n (/I = ind( M)). In this case, there must be a unit vector u in T,(o)M which is normal to -y(0) and has the opposite causal character of -Y(O). This means that if E = b(O), -y(0)) / 1h(0)11, then E(U, u) = -1. Let U be the field along 'Y which is the parallel translation of u. By choosing 8 > 0 appropriately we can arrange that 8 is as small as we like and simultaneously that sin(t/8) is zero at t = 0 and t = b. Let V := 8 sin(t/8)U and make the harmless assumption that II'YII = 1. Notice that by construction V ..1 -y. We compute: I, (V, V)
= E fob {(V7 Ot V, V7 Ot V) + (R( -y, Vh, V)} dt
= E fob {(V7 Ot V, V7 Ot V) - (R('Y, V)V, -y)} dt = E fob {(V7 Ot V, V7 Ot V) - K(V A -y)(V A -y, V A -y)} dt = Efob {(V7Ot V, V7 Ot V) - K(V A -y)(V, V)E} dt, where
K(V A -y) := (9t(V A -y), V A -y) = (9t(V A -y), V A -y) (V A -y, V A -y) E(V, V)
13. Riemannian and Semi-Riemannian Geometry
608
as defined earlier. Continuing the computation we have If'(V, V) = c =
fob {(U, U) cos 2 (t/8) + K(V 1\ '1')8 2 sin2 (t/8)} dt
fob {- cos 2 (t/8) + cK(V 1\ '1')8 2 sin2 (t/8)} dt
= -b/2 + 82 fob cK(V 1\ '1') sin2 (t/8) dt. Now as we said, we can choose 8 as small as we like, and since K(V(t) 1\ '1'( t)) is bounded on the (compact) interval [0, b], this clearly means that If'(V, V) < 0, which contradicts the fact that If' is positive semidefinite. Thus our assumption that 0 < v < n is impossible. Now let If' be negative semidefinite. Suppose that we assume that contrary to what we wish to show, v is not 1 or n - 1. In this case, one can find a unit vector U E Tf'(o)M normal to '1'(0) such that c(u, u) = +1. The same sort of calculation as we just did shows that If' cannot be semidefinite; 0 again a contradiction. By changing the sign of the metric the cases handled by this last theorem boil down to the two important cases: 1) where (M, g) is Riemannian, 'Y is arbitrary, and 2) where (M, g) is Lorentz and 'Y is timelike. We consolidate these two cases by a definition: Definition 13.130. A geodesic 'Y : [a, b] -+ M is cospacelike if the subspace '1'(8)1.. C Tf'(s)M is spacelike for some (and consequently all) 8 E [a, b]. Exercise 13.131. Show that if 'Y : [a, b] -+ M is cospacelike, then 'Y is nonnull, '1'(8)1.. C Tf'(s)M is spacelike for all 8 E [a, b], and also show that (M, g) is either Riemannian or Lorentz.
A useful observation about Jacobi fields along a geodesic is the following: Lemma 13.132. If we have two Jacobi fields Jl and J2 along a geodesic 'Y, then (V' oJl' J2) - (J1 , V' oJ2) is constant along ,. Proof. To see this, we note that
V' at (V' oJl' J2)
= (V'~Jl' J2) + (V' oJl' V' at h) = (R('Y, hh, h) + (V'oJl, V'oth) =
(R('Y, hh, h)
+ (V'oJ2' V'oJl)
= V'Ot(V'oJ2' J 1 ).
Similarly, we compute V' at (11, V' oJ2) and subtract the result from the above to obtain the conclusion. 0
609
13.8. First and Second Variation of Arc Length
In particular, if (\7 aJ1' h) = (11, \7 aJ2) at t = 0, then (\7 aJ1, h) (11, \7 at h) = 0 for all t. We need another simple technical lemma: Lemma 13.133. If h, ... , Jk are Jacobi fields along a geodesic 'Y such that (\7 aJi' Jj) = (1i, \7 aJj) for all i, j E {I, ... , k}, then any field Y which can
be written as Y = (\7 at Y, \7 at Y)
L: rpi Ji
has the property that
+ (R(Y, i')Y, 1') = ((8t rpi) Ji, (8t rpi) Ji) + 8t (Y, rpr (\7 aJr)).
Proof. We have \7 at Y = (8t rpi) Ji + rpr (\7 at Jr ) and so using the summation convention,
+ (Y, \7 at [rpr (\7 aJr)]) (( 8t rpi) Ji , rpr (\7 aJr)) + (rpr (\7 aJr) , rpk (\7 aJk)) + (Y, 8t rpr \7 at J r ) + (Y, rpr\7~t Jr ).
8t (Y, rpr (\7 aJr)) = ((\7 at Y ) , rpr (\7 aJr)) =
The last term (Y, rpr\7~Jr) equals (R(Y, i')Y, 1') by the Jacobi equation. Using this and the fact that (Y, 8t rpr \7 at Jr ) = (( 8t rpi) Ji, rpr (\7 at J r )), which follows from a short calculation using the hypotheses on the Ji, we arrive at
8t (Y, rpr (\7 aJr)) = 2((8t rpi) Ji, rpr (\7 aJr))
+ (rpr (\7 aJr) , rpr (\7 aJr))
+ (R(Y, i')Y, 1')' Using the last equation together with \7 at Y = (8t rpi) Ji the result (check it!).
+ rpr (\7 aJr)
gives 0
Exercise 13.134. Work through the details of the proof of the lemma above.
Throughout the following discussion, 'Y : [0, b] -+ M will be a cospacelike geodesic with sign E and speed c. Suppose that there are no conjugate points of p = 'Y(O) along 'Y. There exist Jacobi fields J1,.'" I n- 1 along 'Y which vanish at t = 0 and are such that the vectors \7 aJ1 (0), ... , \7 aJn-1 (0) E TpM are a basis for the space 1'(0)1- c T-y(o)M. Claim: J 1 (t), ... , I n -1(t) are linearly independent for each t > O. Indeed, suppose that C1J1(t) + ... + C2Jn-1(t) = 0 for some t. Then, Z := L::-11 CiJi is a normal Jacobi field with Z(O) = Z(t) = O. But then, since there are no conjugate points, Z = 0 identically and so 0 = \7 atZ(O) := L:~:11 ci\7aJi(O). Since the \7aJi(O) are linearly independent, we conclude that Ci = 0 for all i and the claim is proved. It follows that at each t with 0 < t :::; b the vectors h(t), ... , I n -1(t) form a basis of i'(t)1- C T-y(t)M. Now let Y E T-y(O) be a piecewise smooth
13. Riemannian and Semi-Riemannian Geometry
610
variation vector field along , and write Y = 2:: 'Pi Ji for some piecewise smooth functions 'Pi on (0, b], which can be shown to extend continuously to [O,b] (see Problem 3). Since ('\7aJi' Jj ) = (Ji, '\7aJj) = 0 at t = 0, we have ('\7aJi' Jj ) - (Ji, '\7aJj) = 0 for all t by Lemma 13.132. This allows for the use of Lemma 13.133 to arrive at
('\7at Y , '\7atY)
+ (R(Y,-y)Y,-y) =
(L (8 'Pi) Ji, L t
(8t 'Pi)
Ji )
+ 8t (y, L
'P r ('\7 at J r ) )
and then (13.10) cIl'(Y' Y)
=
~ fob (L (8t 'Pi) Ji, L
(8t 'Pi) J i )dt + ~
(Y, 'P r ('\7 aJr))lg.
On the other hand, Y is zero at a and b and so the last term above vanishes. Now we notice that since, is cospacelike and the Ji are normal to the geodesic, we must have that the integrand in equation (13.10) above is nonnegative. We conclude that cIl'(Y' Y) ~ o. On the other hand, if II'(Y' Y) = 0 identically, then J~ (2:: (8t 'Pi) Ji, 2:: (8t 'Pi) Ji) dt = 0 and (2:: (8t 'Pi) Ji , 2:: (8t 'Pi) Ji ) = O. In turn, this implies that 2:: (8t 'Pi) Ji == 0 and that each 'Pi is constant, in fact zero, and finally that Y itself is identically zero along,. All we have assumed about Y is that it is in the domain of the restricted index I¢ and so we have proved the following: Proposition 13.135. If, E n is cospacelike and there is no conjugate points to p = ,(0) along" then cI¢(Y, Y) ~ 0 and Y = 0 along, if and only if I¢(Y, Y) = o.
We may paraphrase the above result as follows: For a cospacelike geodesic , without conjugate points, the restricted index I¢ is definite; it is positive definite if E = +1 and negative definite if E = -1. The first case (E = +1) is exactly the case where (M,g) is Riemannian (Exercise 13.131). Next we consider the situation where the cospacelike geodesic, : [0, b] -+ M is such that ,(b) is the only point conjugate to p = ,(0) along ,. In this case, Theorem 13.128 tells us that I¢ has a nontrivial nullspace and so II' cannot be definite. Claim: II' is semidefinite. To see this, let Y E Tl'n and write Y in the form (b - t) Z (t) for some (continuous) piecewise smooth Z. Let bi -+ b and define Yi to be (bi - t)Z(t) on [0, bJ Our last proposition applied to := ,1[O,bi] shows that cII'JYi, Yi) ~ O. Now cIl'i (Yi, Yi) -+ cIl'(Y' Y) (some uninteresting details are omitted) and so the claim is true.
,i
Now we consider the case where there is a conjugate point to p before ,I [O,r] with 0 < r < b
,(b). Suppose that J is a nonzero Jacobi field along
13.8. First and Second Variation of Arc Length
611
such that J(O) = J(r) = O. We can extend J to a field Jext on [0, b] by defining it to be 0 on [r, b]. Notice that '\l aJext(r-) is equal to '\l aJ(r), which is not 0 since otherwise J would be identically zero (over determination). On the other hand, '\laJext(r+) = 0 and so the "kink" l::.J~xt(r) := '\l aJext(r+) - '\l aJext(r-) is not zero. Notice that l::.J~xt(r) is normal to "I (why?). We will now show that if W E T)'(O) is such that W(r) = l::.J~xt(r) (and there are plenty of such fields), then cI)'(Jext + £5W, Jext + £5W) < 0 for small enough £5 > O. This will allow us to conclude that I)' cannot be definite since by Proposition 13.135 we can always find a Z with cI)'(Z, Z) > O. We have
cI)'(Jext + £5W, Jext + £5W) = cI)'(Jext , Jext} + 26cI), (Jext, W)
+ c:£5 2 I)'(W, W).
It is not hard to see from the formula of Theorem 13.124 that I)'(Jext , Jexd is zero since it is piecewise Jacobi and is zero at the single kink point r. But using the formula again, cI),(Jext(r), W(r)) reduces to
' ) , W(r)) = --1 1l::.Jext(r) I 12 < 0, --1 (l::.Jext(r
c c and so taking £5 small enough gives the desired conclusion.
Summarizing the conclusions of the above discussion (together with the result of Proposition 13.135) yields the following nice theorem: Theorem 13.136. If'Y : [0, b] -+ M is a cospacelike geodesic of sign c:, then (M, g) is either Riemannian or Lorentz and we have the following three
cases:
(i) If there are no points conjugate to "1(0) along "I, thencI~ is positive definite. (ii) If 'Y(b) is the only conjugate point to "1(0) along "I, then I)' is not
definite, but must be semidefinite. (iii) If there is a point 'Y(r) conjugate to "1(0) with 0 < r
< b, then I)' is
not semidefinite (or definite). As we mentioned the Jacobi equation can be written in terms of the tidal force operator Rv : TpM -+ TpM as
'\l~J(t)
= R.y(t)(J(t)).
The meaning of the term force here is that R.y(t) controls the way nearby families of geodesics attract or repel each other. Attraction tends to create conjugate points, while repulsion tends to prevent conjugate points. If "I is cospacelike, then we take any unit vector u normal to 1'(t) and look at the component of R.y(t)(u) in the u direction. Up to sign this is
(R.y(t) (u), u)u = (R.y(t),u(i'(t)), u)u = -(9\(1'(t) Au), 1'(t) A u)u.
612
13. Riemannian and Semi-Riemannian Geometry
In terms of sectional curvature, (R"y(t) (u), u)u = K('Y(t) 1\ u) ('Y(t) , 'Y(t)) . It follows from the Jacobi equation that if (R"y(t) (u), u) ~ 0, i.e., if K('Y(t) 1\ u)('Y(t), 'Y(t)) ~ 0, then we have repulsion, and if this always happens anywhere along " we expect that ,(0) has no conjugate point along ,. This intuition is indeed correct.
Proposition 13.137. Let, : [0, b] -+ M be a cospacelike geodesic. If for every t and every vector v E ,(t)..L we have (R"y(t) (v), v) ~ 0 (i.e. if K('Y(t) 1\ v)('Y(t), 'Y(t)) ~ 0), then ,(0) has no conjugate point along ,. In particular, a Riemannian manifold with sectional curvature K ~ 0 has no conjugate pairs of points. Similarly, a Lorentz manifold with sectional curvature K ~ 0 has no conjugate pairs along any timelike geodesics.
Proof. Take J to be a Jacobi field along, such that J(O) We have (J, J) = 2 (\7 aJ, J) and
1t
=
0 and J..l'Y.
d2 dt 2 (J, J) = 2(\7 at J, \7 at J)
+ 2(\7~t J, J) = 2(\7aJ, \7aJ) + 2(R"y(t),J('Y(t)), J) = 2 (\7aJ, \7aJ) + 2 (R"y(t) (J), J),
and by the hypotheses ~(J, J) ~ O. On the other hand, (J(O), J(O)) = 0 and 10 (J, J) = O. It follows that since (J, J) is not identically zero we must have (J, J) > 0 for all t E (0, b] and the result follows. 0
1t
13.9. More Riemannian Geometry Recall that a manifold is geodesic ally complete at p if and only if expp is defined on all of TpM. The following lemma is the essential ingredient in the proof of the Hopf-Rinow theorem stated and proved below. Note that this is a theorem about Riemannian manifolds.
Lemma 13.138. Let (M,g) be a connected Riemannian manifold. Suppose that expp is defined on the ball of radius p > 0 centered at 0 E TpM. Let Bp(p):= {x: dist(p,x) < pl. Then each point q E Bp(p) can be connected to p by an absolutely minimizing geodesic. In particular, if M is geodesically complete at p EM, then each point q E M can be connected to p by an absolutely minimizing geodesic.
i=
q and let R = dist(p, q). Choose E > 0 small enough that B2f (p) is the domain of a normal coordinate system. (Refer to Figure 13.9.) By Lemma 13.85, we already know the theorem is true if Bp(p) c Bf(p), so we will assume that E < R < p. Because 8Bf(p) is
Proof. Let q
E Bp(p) with p
613
13.9. More Riemannian Geometry
diffeomorphic to sn-l C jRn, it is compact and so there is a point PE E aBE(p) such that x 1---7 dist(x, q) achieves its minimum at PE' This means that dist(p, q) = dist(p,PE) + dist(PE, q) = E + dist(PE, q). Let "1: [O,p]--+ M be the geodesic with
Ii'I = 1, "1(0) = p,
and 'Y(E)
= PE'
~(p) 1
q
y
Figure 13.9
It is not difficult to see that the set T = {t E [0, R] : dist(p, 'Y(t)) + dist("((t), q) = dist(p, q)}
is closed in [0, R] and is nonempty since E E T. Let tsup = supT > O. We will show that tsup = R from which it will follow that "11 [O,R] is a minimizing geodesic from P to q since then dist("((R), q) = 0 and so 'Y(R) = q. With an eye toward a contradiction, assume that tsup < R. Let x := 'Y(tsup ) and choose El with 0 < El < R - tsup and small enough that B2El (x) c Bp(p) is the domain of normal coordinates about x. Arguing as before we see that there must be a point X E1 E aBE! (x) such that dist(x, q) = dist(x, xE!) + dist(x q , q) = El + dist(x Ell q). Now let "11 be the unit speed geodesic such that "Y1 (0) = x and "11 (El) = x q But since tsup E T and x = 'Y(tsup), we also have dist(p, x)
+ dist(x, q) = dist(p, q).
Combining, we now have dist(p, q) = dist(p, x) + dist(x, X E1 ) + dist(xq , q). By the triangle inequality, dist(p, q)
:s; dist(p, Xq) + dist(x Ell q) and so
dist(p, x) +dist(x,xE1):S; dist(p,xq).
.
614
13. Riemannian and Semi-Riemannian Geometry
+ dist(x, Xq) and so dist(p,xq ) = dist(p,x) + dist(x,xq).
But also dist(p, Xq) :S dist(p, x)
Examining the implications of this last equality, we see that the concatenation of 1'1 [O,tsup] with 1'1 forms a curve from p to Xq of length dist(p, Xq), which must therefore be a minimizing curve. By Problem 1, this potentially broken geodesic must in fact be smooth and so must actually be the geodesic I'I[o,t sup +q]. Then, tsup+El E T which contradicts the definition oftsup . This 0 contradiction forces us to conclude that tsup = R and we are done. Theorem 13.139 (Hopf-Rinow). If(M,g) is a connected Riemannian manifold, then the following statements are equivalent:
(i) The metric space (M, dist) is complete. sequence is convergent.
That is, every Cauchy
(ii) There is a point p E M such that M is geodesically complete at p. (iii) M is geodesically complete. (iv) Every closed and bounded subset of M is compact. Proof. (iv)===>(i): The set of points of a Cauchy sequence is bounded and so has compact closure. Thus there is a subsequence converging to some point. Since the original sequence was Cauchy, it must converge to this point.
(i)===>(iii): Let p be arbitrary and let I'v(t) be the geodesic with 'Yv(O) = v and J its maximal domain of definition. We can assume without loss of generality that (v, v) = 1 so that L( I'vl[tl,t2]) = t2 - tl for all relevant tl, t2. We want to show that there can be no upper bound for the set J. We argue by contradiction: Assume that t+ = sup J is finite. Let {t n } C J be a Cauchy sequence such that tn -7 t+ < 00. Since dist hv (t), I'v (s )) :S 1t - s I, it follows that I'v(tn) is a Cauchy sequence in M, which by assumption must converge. Let q := limn-too I'v (tn) and choose a small ball Bi (q) which is small enough to be a normal neighborhood. Take it with 0 < t+ - tl < E/2 and let 1'1 be the (maximal) geodesic with initial velocity 'Yv(tt). Then in fact 1'1 (t) = I'v (tl + t) and so 1'1 is defined for tl + E/2 > t+ and this is a contradiction. (iii)===>(ii) is a trivial implication. (ii)===>(iv): Let K be a closed and bounded subset of M. For x E M, Lemma 13.138 tells us that there is a minimizing geodesic ax : [0,1] -7 M connecting p to x. Then Ilax(O)11 = dist(p, x) and exp(ax(O)) = x. Using the triangle inequality, one sees that sup{llax(O)II} :S r xEK
13.9. More Riemannian Geometry
615
for some r < 00. From this we obtain {ax(O) : x E K} c Br := {v E TpM : Ilvll :::; r}. The set Br is compact. Now exp(Br) is compact and contains the closed set K, so K is also compact. (ii)===*(i): Suppose M is geodesically complete at p. Now let {xn} be any Cauchy sequence in M. For each X n , there is (by assumption) a minimizing geodesic from p to X n , which we denote by 'YPXn. We may assume that each 'Ypx n is unit speed. It is easy to see that the sequence {In}, where In := Lbpxn) = dist(p, xn), is a Cauchy sequence in IR with some limit, say l. The key fact is that the vectors 'Ypx n are all unit vectors in TpM and so form a sequence in the (compact) unit sphere in TpM. Replacing bpxn} by a subsequence if necessary we have 'Ypx n -+ U E TpM for some unit vector u. Continuous dependence on initial velocities implies that {xn} = {')'PXn (In)} has the limit 'Yu (l). D Let (M, g) be a complete connected Riemannian manifold with sectional curvature K :::; O. By Proposition 13.137, for each point p EM, the geodesics emanating from p have no conjugate points and so Tvp exp p : TvpTpM -+ M is nonsingular for each vp E TpM. This means that exp p is a local diffeomorphism. If we give TpM the metric exp;(g), then exp p is a local isometry. It now follows from Theorem 13.76 that exp p : TpM -+ M is a Riemannian covering. Thus we arrive at the Hadamard theorem. Theorem 13.140 (Hadamard). If (M, g) is a complete simply connected Riemannian manifold with sectional curvature K :::; 0, then exp p : TpM -+ M is a diffeomorphism and each two points of M can be connected by a unique geodesic segment. Definition 13.141. If (M, g) is a Riemannian manifold, then the diameter of M is defined to be
diam(M) := sup{dist(p, q) : p, q EM}. The injectivity radius at p E M, denoted inj(p), is the supremum over all E > 0 such that exp p : B(Op, E) -+ B(p, E) is a diffeomorphism. The injectivity radius of Mis inj(M) := infpEM{inj(p)}. The Hadamard theorem above has as a hypothesis that the sectional curvature is nonpositive. A bound on the sectional curvature is stronger than a bound on Ricci curvature since the latter is a sort of average sectional curvature. In the sequel, statements like Ric 2: C should be interpreted to mean Ric(v, v) 2: C(v, v) for all v E TM. Lemma 13.142. Let (M, g) be an n-dimensional Riemannian manifold and let'Y : [0, L] -+ M be a unit speed geodesic. Suppose that Ric 2: (n - 1) K, > 0 for some constant K, > 0 (at least along 'Y). If the length L of 'Y is greater than or equal to 1r / ",;K" then there is a point conjugate to 'Y (0) along 'Y.
13. Riemannian and Semi-Riemannian Geometry
616
°
Proof. Suppose < 7r / y'K, :s; L. If we can show that 1':;- is not positive definite, then Theorem 13.136 implies the result. To show that 1':;- is not positive definite, we find an appropriate vector field V i- along 'Y such that I(V, V) :s; 0. Choose orthonormal fields E 2, ... ,En so that E 2,·· . ,En is an orthonormal frame along 'Y. For a function f : [0, 7r / yK] -+ lR that vanishes at endpoints, we form the fields f E i . Using (13.9), we have
° -r,
r 1fi {!'(s)2 + f(s)2(REj,"y(Ej(s)), -r(s))} ds,
I-y(f Ej, f Ej ) = io and then n
L I-y(f Ej,J Ej ) = j=2
rrr /fi io {(n - 1) (1,)2 -
f2 Rich,
-r)} ds
0
r
:s; (n - 1) io 1fi ((1,)2 - K,f2) ds. Letting f (s)
= sin( y'K,s),
we get
n
r 1fi
LI(fEj,fEj):S; (n -1) io j=2 0 and so I(f Ej,J Ej)
:s;
K,
(cos 2(y'K,s) - sin2(~s)) ds
°
for some j.
= 0, D
The next theorem also assumes only a bound on the Ricci curvature and is one of the most celebrated theorems of Riemannian geometry. A weaker version involving sectional curvature was first proved by Ossian Bonnet (see [Hicks], page 165).
Theorem 13.143 (Myers). Let (M,g) be a complete connected Riemannian manifold of dimension n. If Ric 2: (n - 1) K, > 0, then (i) diam(M)
(ii)
7rl (M)
:s; 7r / y'K"
M is compact, and
is finite.
Proof. Since M is complete, there is always a shortest geodesic 'Ypq between any two given points p and q. We can assume that 'Ypq is parametrized by arc length: 'Ypq : [0, dist(p, q)] -+ M. It follows that 'Ypq I[O,aj is arc length minimizing for all a E [0, dist(p, q)]. From Proposition 13.129 we see that the only possible conjugate to p along 'Ypq is q. The preceding lemma shows that 7r / y'K, > dist(p, q) is impossible. Since the points p and q were arbitrary, we must have diam( M) It follows from the Hopf-Rinow theorem that M is compact.
:s; 7r / y'K,.
For (ii) we consider the simply connected covering ~ : M -+ M (which is a local isometry). Since ~ is a local diffeomorphism, it follows that ~-l(p)
617
13.10. Cut Locus
has no accumulation points for any p EM. But also, because M is complete and has the same Ricci curvature bound as M, it is compact. It follows that p-1(p) is finite for any p EM, which implies (ii). 0 The reader may check that if sn(R) is a sphere of radius R in JRn+1, then sn(R) has constant sectional curvature K, = 1/ R2 and the distance from any point to its cut locus (defined below) is 7r /..jK,. A result of S. Y. Cheng states that with the curvature bound of the theorem above, if diam(M) = 7r /..jK" then M is a sphere of constant sectional curvature K,. See [Cheng].
13.10. Cut Locus In this section we consider Riemannian manifolds. Related to the notion of conjugate point is the notion of a cut point. For a point p E M and a geodesic, emanating from p = ,(0), a cut point of p along, is the first point q = ,( tt) along , such that for any point r = ,(t") beyond p (Le. t" > tt) there is a geodesic shorter than ,1[O,t'l which connects p with r. To see the difference between this notion and that of a point conjugate to p, it suffices to consider the example of a cylinder Sl x JR with the obvious flat metric. If p = (1,0) E Sl X JR, then for any x E JR, the point (e i7f , x) is a cut point of p along the geodesic ,(t) := (e it7f , tx). We know that beyond a conjugate point, a geodesic is not (locally) minimizing. In our cylinder example, for any f > 0, the point q = ,(1 + f) can be reached by the geodesic segment ,2 : [0,1 - f] -+ Sl X JR given by ,(t) := (e it7f , ax), where a = (1 + f)/(1 - f). It can be checked that is shorter than ,I [0,1 + fl· However, the last example shows that a cut point need not be a conjugate point. In fact, Sl x JR has no conjugate points along any geodesic. Let us agree that all geodesics referred to in this section are parametrized by arc length unless otherwise indicated.
,2
Definition 13.144. Let (M,g) be a complete Riemannian manifold and let p EM. The set C (p) of all cut points to p along geodesics emanating from p is called the cut locus of p.
For a point p EM, the situation is summarized by the fact that if q = ,(tt) is a cut point of p along a geodesic " then for any t" > t' there is a geodesic connecting p with q which is shorter than ,1[O,t'l' while if t" < tt, then not only is there no geodesic connecting p and ,( t") with shorter length but there is no geodesic connecting p and ,(t") whose length is even equal to that of ,1[O,t"l. (Why?) Consider the following two conditions:
(C1): ,(to) is the first conjugate point of p = ,(0) along ,.
618
13. Riemannian and Semi-Riemannian Geometry
(C2): There is a unit speed geodesic a from ')'(0) to ')'(to) that is different from ')'1 [O,to] such that L(a) = L( ')'I[o,to])' Proposition 13.145. Let M be a complete Riemannian manifold.
(i) If for a given unit speed geodesic ,)" either condition (C1) or (C2) holds, then there is a t1 E (0, to] such that ')'(t1) is the cut point of p along ')'. (ii) If ')'(to) is the cut point of p = ')'(0) along the unit speed geodesic ray')', then either condition (C1) or (C2) holds. Proof. (i) This is already clear from our discussion: For suppose (C1) holds, then ')'1 [O,t'] cannot minimize for t' > to and so the cut point must be ')'(t1) for some t1 E (0, to]. Now if (C2) holds, then choose E > 0 small enough that a(to - E) and ')'(to + E) are both contained in a convex neighborhood of ')'(to). The concatenation of al[o,to] and ')'1 [to,to+£J is a curve, say c, that has a kink at ')'(to). But there is a unique minimizing geodesic T joining a(to-E) to ')'(to + E), and we can concatenate the geodesic al[o,to-£] with T to get a curve with arc length strictly less than L(c) = to + E. It follows that the cut point to p along,), must occur at ')'(t') for some t' :S to + Eo But E can be taken arbitrarily small and so the result (i) follows.
(ii) Suppose that ')'(to) is the cut point of p = ')'(0) along a unit speed geodesic ray')'. We let Ei --+ 0 and consider a sequence {aJ of minimizing geodesics with ai connecting p to ')'(to + Ed. We have a corresponding sequence of initial velocities Ui := O:i(O) E 8 1 C TpM. The unit sphere in TpM is compact, so replacing Ui by a subsequence we may assume that Ui --+ U E 8 1 C TpM. Let a be the unit speed segment joining p to ')'(to + Ei) with initial velocity u. Arguing from continuity, we see that a is also minimizing and L(a) = L( ')'I[o,to])' If a i=- ')'1 [O,to] , then we are done. If a = ')'1 [O,toJ, then since ')'1 [O,to] is minimizing, it will suffice to show that Tto"l(o) exp p is singular because that would imply that condition (C1) holds. The proof of this last statement is by contradiction: Suppose that a = ')'1 [O,to] (so that ...y(0) = u) and that Tto"l(o) exp p is not singular. Take U to be an open neighborhood of to...y(O) in TpM such that expplu is a diffeomorphism. Now ai(to + E~) = ')'(to + Ei) for 0 < E~ :S Ei since the ai are minimizing. We now restrict attention to i such that Ei is small enough that (to + EDui and (to + Ei)U are in U. Then we have expp(to + Ei)U = ')'(to =
+ Ei) ai(to + ED =
expp(to
+ EDui,
and so (to + Edu = (to + EDui, and then, since Ei --+ 0 and both U and Ui are unit vectors, we have ...y(0) = U = Ui for sufficiently large i. But then
619
13.11. Rauch's Comparison Theorem
for such i, we have Qi = 'Y on [0, to 'YI [O,tO+EiJ is not minimizing.
+ ti],
which contradicts the fact that 0
Exercise 13.146. Show that if q is the cut point of p along 'Y, then p is the cut point of q along 'Y~ (where 'Y~(t) := 'Y(L - t) and L = Lh)).
It follows from the development so far that if q E M\ C (p), then there is a unique minimizing geodesic joining p to q, and that if B(p, R) is the ball of radius R centered at p, then expp is a diffeomorphism on B(p, R) provided R:S: d(p, C(p)). In fact, an alternative definition of the injectivity radius at pis d(p, C(p)) and the injectivity radius of M is inj(M) = inf {d(p, C(p))}. pEM
Intuitively, the complexities of the topology of M begin at the cut locus of a given point. Let TI M denote the unit tangent bundle of the Riemannian manifold:
TIM = {u E TM:
Ilull = 1}.
Define a function CM : TI M ~ (0,00] by ( ) '=
CM u.
{to if 'Yu(to) is the cut point of 7rTM(U) along 'Yu, 00 if there is no cut point in the direction u.
Recall that the topology on (0,00] is such that a sequence tk converges to the point 00 if limk-+oo tk = 00 in the usual sense. It can be shown that if (M, g) is a complete Riemannian manifold, then the function eM : TI M ~ (0,00] is continuous (see [Kobl).
13.11. Rauch's Comparison Theorem In this section we deal strictly with Riemannian manifolds. Definition 13.147. Let 'Y : [a, b] ~ M be a smooth curve. piecewise smooth vector fields along 'Y, define
I,,(X, Y) := l\"V at X, "Vat Y)
For X, Y
+ (R.y,x"!, Y) dt.
The map X, Y t-t I,,(X, Y) is symmetric and bilinear. In defining I" we have used a formula for the index I" valid for fields which vanish at the endpoints and with 'Y a nonnull geodesic. Thus when 'Y is a nonnull geodesic, the restriction of I" to variation vector fields which vanish at endpoint is the index I". Thus I" is a sort of extended index form.
620
13. Riemannian and Semi-Riemannian Geometry
Corollary 13.148. Let 'Y : [0, b] -+ M be a cospacelike geodesic of sign E with no points conjugate to 'Y( 0) along 'Y. Suppose that Y is a piecewise smooth vector field along'Y and that J is a Jacobi field along 'Y such that
Y(O) = J(O), Y(b) = J(b), and (Y - J) 1-i'. Then ELy(J, J) :::; EI'Y(Y' Y). Proof of the corollary. From Theorem 13.136, we have 0 :::; cI¢(Y - J, Y - J) = ELy (Y - J, Y - J) and so
0:::; ELy(Y, Y) - 2EI'Y(J, Y)
+ ELy(J, J).
Integrating by parts, we have
ELy(J, Y) = E (\1 atJ, Y)lg
-
fob (\1~tJ, Y) -
= E (\1 aJ, Y)lg = E (\1 aJ, J)lg = EI'Y(J, J) (since J is a Jacobi Thus 0 :::; EI'Y(Y' Y) - 2EI'Y(J, Y)
+ ELy(J, J)
(Rry,Ji', Y) dt
field).
= ELy(Y, Y) - ELy(J, J).
0
Recall that for a Riemannian manifold M, the sectional curvature KM (P) of a 2-plane P c TpM is
(v:t (el/\ e2), el/\ e2) for any orthonormal pair el, e2 that spans P. Definition 13.149. Let M, g and N, h be Riemannian manifolds and let 'YM : [a, b] -+ M and 'YN : [a, b] -+ N be unit speed geodesics defined on the same interval [a, b]. We say that KM :2: KN along the pair ("(M, 'YN) if KM (Q'YM(t)) :2: KN (P'YN(t)) for all t E [a, b] and every pair of 2-planes Qt E T'YM(t)M, Pt E T'YN(t)N. We develop some notation to be used in the proof of Rauch's theorem. Let M be a given Riemannian manifold. If Y is a piecewise smooth vector field along a unit speed geodesic 'YM such that Y (a) = 0, then let
If4 (Y, Y) := =
is is -(\1~tY(t),
(\1atY(t), \1atY(t)) Y(t))
+ (RyM,Yi'M, Y)(t)dt
+ (RyM,yi'M, Y)(t)dt + (\1atY, Y)(s).
If Y is an orthogonal Jacobi field, then
621
13.11. Rauch's Comparison Theorem
Theorem 13.150 (Rauch). Let M, 9 and N, h be Riemannian manifolds of the same dimension and let "fM : [a, b] ---+ M and "fN : [a, b] ---+ N be
unit speed geodesics defined on the same interval [a, b]. Let JM and IN be Jacobi fields along "fM and "fN respectively and orthogonal to their respective curves. Suppose that the following four conditions hold:
(i) JM (a) = IN (a) = 0 and neither of JM (t) or IN (t) is zero for t E (a,b].
IIV'oJM(a)11 = lIV'oJN(a)lI.
(ii)
(iii) L("(M) = dist("(M (a), "fM (b)). (iv) KM 2: KN along the pair ("(M, "fN).
Then IIJM(t)1I ~ IIJN(t)1I for allt E [a,b]. Proof. Let fM be defined by fM(S) := IIJ M (s)1I2 and hM by hM(S) '-
I~(JM, JM)/ IIJ M(s)1I2 for s
E
(a,b]. Define fN and hN analogously. We
have
fk(s) = 2I~(JM, JM) and fk/fM = 2hM and the analogous equalities for fN and hN. If c E (a, b), then In(IIJ M (s)1I2) = In(IIJ M (c)1I2)
+
21
8
hM(s')ds'
with the analogous equation for N. Thus
2 M M In ("J (s)1 ) =In("J (c)12) +21 s [h (s')-hN(s')]ds'. IIJN (s)112 IIJN (c)112 c M From the assumptions (i) and (ii) and L'Hopital's rule, we have
. IIJ M (c)1I2 11m = 1, c-+a+ IIJN (c)112 and so
1
M 8 (s)II:) = 2 lim [hM(S') - hN(s')]ds'. N liJ (s) II c--+a+ c If we can show that hM(S) - hN(S) ~ 0 for s E (a,b], then the result will follow. So fix r E (a,b] and let ZM(s):= JM(s)/IIJM(r)1I and ZN(s):= J N (s ) / J N (r ) Notice the r in the denominator; Z N (s) is not necessarily of unit length for s oF r. We now define a parametrized families of subtangent spaces along "fM by WM(S) := 'YM(s)~ c T"IM(s)M and similarly for WN(S). We can choose a linear isometry Lr : WN(r) ---+ WM(r) such that Lr(ZN (r)) = ZM (r). We now want to extend Lr to a family of linear isometries Ls : WN(S) ---+ WM(S). We do this using parallel transport by In ("J
I
II.
Ls
:=
p("(M):
0
Lr
0
p("(N)~.
13. Riemannian and Semi-Riemannian Geometry
622
Define a vector field Y along
,M
by Y(s):= Ls(ZN(s)). Check that
Y(a) = ZM (a) = 0, Y(r)
= ZM (r),
11Y112 = IIzNI1 2, IIV'atYI12 = IIV'at ZN I1 2. The last equality is a result of Exercise 12.43 where in the notation of that exercise f3(t) := P(rM)i 0 Y(t). Since (iii) holds, there can be no conjugates along up to r. Now Y - ZM is orthogonal to the geodesic and so by Corollary 13.148 we have I~(ZM,ZM):s I~(Y,Y) and in fact, using (iv) and the list of equations above, we have
,M
I;f (ZM, ZM)
,M
:s I~ (Y, Y) =
iT IIV'atYI1 2iT IIV'atYI1 2:s iT IIV'atZNI12 iT IIV'atZN I1 2
iT IIV'at YI1 2+
Y,-rM, Y) b MA YI1 2
=
Kb M
=
K(-rM, Y,-rM, Y)
=
,
RM b M, Y, -rM, Y)
b MI1211Y112
Kb M , ZN,-rM, ZN)
bMI1211zNI12
(by (iv))
+RNbN,zN,-rN,ZN) =I;'(ZN,ZN).
Recalling the definition of ZM and ZN we obtain
and so hM(r) - hN(r) :S O. But r was arbitrary, and so we are done.
0
Corollary 13.151. Let M, g and N, h be Riemannian manifolds of the same dimension and let [a, bj -+ M and ,N : [a, bj -+ N be unit speed geodesics defined on the same interval [a, bj. Assume that KM ~ KN along the pair (rM, ,N). Then if (a) has no conjugate point along ,M, then ,N (a) has no conjugate point along ,N.
,M :
,M
The above corollary is easily deduced from the Rauch comparison theorem above and we invite the reader to prove it. The following famous theorem is also proved using the Rauch comparison theorem but the proof is quite difficult and also uses Morse theory. The proof may be found in
[C-Ej.
13.12. Weitzenbock Formulas
623
Theorem 13.152 (The sphere theorem). Let M, 9 be a complete simply connected Riemannian manifold with sectional curvature satisfying the condition 1 1 1
4: R2 < K::;
for some R
R2
> O. Then M is homeomorphic to the sphere
sn.
13.12. Weitzenbock Formulas The divergence of a vector field in terms of Levi-Civita connection is given by
= trace(V' X). (0 1, ... , on) are dual bases for TpM div(X)
Thus if (el' ...
,en)
and
and T; M, then
n
(div X)p =
L oj (V' ejX) . j=l
Exercise 13.153. Show that this definition is compatible with our previous definition by showing that, with the above definition, we have Lx vol = (div X) vol, where vol is the metric volume element for M. Hint: Use Lx = dix +ixd. Definition 13.154. Let V' be a torsion free covariant derivative on M. The divergence of a (k, £) tensor field A is a (k - 1, £) tensor field is defined by n
(div A)p
(al ... , ak-l, VI,···,
ve) =
L (V' ejA) (oj,
al ... , ak-l, VI,""
ve),
j=l
where (el' ...
,en)
and (0 1, ...
,on)
are dual as above.
Notice that the above definition depends on the choice of covariant derivative but if M is semi-Riemannian, then we will use the Levi-Civita connection. In index notation the definition is quite simple in appearance. For example, if Ai{[ are the components of a (2,2) tensor field, then we have (div A)jk[ = V'rArkl' where V' aAb{[ are the components of V' A. Also, the definition given is really for the divergence with respect to the first contravariant slot, but we could use other slots (the last slot being popular). Thus V' rAj~[ is also a divergence. If we are to define a divergence with respect to a covariant slot (Le., a lower index), then we must use a metric to raise the index. This leads to the following definition appropriate in the presence of a metric.
13. Riemannian and Semi-Riemannian Geometry
624
Definition 13.155. Let \7 be the Levi-Civita covariant derivative on a manifold M. The (metric) divergence of a function is defined to be zero and the divergence of a (0, £) tensor field A is a (0, £ - 1) tensor field defined at any p E M by n
(div A)p (VI, ... , vt'-d
=L
(ej, ej) (\7ej A) (ej, VI,···, VC-l),
j=l
where (el, ... , en) is an orthonormal basis for TpM. Note that the factors (ej, ej) are equal to 1 in the case of a definite metric. On may check that if A il ... il are the components of A in some chart, then the components of div A are given by
Recall that the formal adjoint 0 : O(M) -+ O(M) of the exterior derivative on a Riemannian manifold M is given on Ok(M) by 0:= (_1)n(k+l)+1 *d*. In Problem 8 we ask the reader to show that the restriction of div to Ok (M) is -0: div = -0
(13.11)
on Ok(M) for each k.
In the same problem the reader is asked to show that if /1 E Ok (M) is parallel (\7/1 = 0), then it is harmonic (recall Definition 9.46). In what follows we simplify calculations by the use of a special kind of orthonormal frame field. If (M,g) is a Riemannian manifold and p E M, choose an orthonormal basis (el' ... , en) in the tangent space TpM and parallel translate each ei along the radial geodesics t H expp (tv) for v E TpM. This results in an orthonormal frame field (E l , ... , En) on some normal neighborhood centered at p. Smoothness is easy to prove. The resulting fields are radially parallel and satisfy Ei (p) = ei and (\7 Ed (p) = 0 for every i. Furthermore we have [Ei,Ej] (p)
= \7EiEj(P) -
\7EjEi(p)
=0
for all
~,J.
We refer to an orthonormal frame field with these properties as an adapted orthonormal frame field centered at p. Before proceeding we need an exercise to set things up. Exercise 13.156. Describe how a connection \7 (say the Levi-Civita connection) on M extends to a connection on the bundle I\T* M in such a way that nv X (1 a A··· A a
k)
k
, IA n =, L..t,a"A··· v xa i i=l
A··· A a
k
13.12. Weitzenbock Formulas
625
for Q/ E nl(M). Show that the curvature of the extended connection is given by
R(X, Y)/-l
= \7 x\7Y/-l- \7y\7 x/-l- \7[X,Y]/-l·
Relate this curvature operator to the curvature operator on X(M) and show that the extended connection is flat if the original connection on M is flat. Let M be Riemannian and /-l E nk(M). Define RJ.L by n
RJ.L(VI, ... , Vk)
k
= L L (R(ei' Vj)/-l) (VI' ... ' Vj-l, ei, Vj+1, ... , Vk). i=l j=l
Theorem 13.157 (Weitzenbock formulas). Let M be Riemannian and /-l E nk (M) . Then we have
1
(~/-ll/-l) = 2~ 11/-l112 + 11\7/-l11 2+ (RJ.LI/-l) , ~/-l
where 11\7/-l11 2(p) JorTpM.
:=
= - div \7/-l + RJ.L'
L:i (\7
ei
/-ll\7 ei /-l) Jor any orthonormal basis (el, ... ,en)
Proof. Using the formula of Theorem 12.56, we see that for U, VI, ... , Vk E X(M), we have
(\7/-l - d/-l) (U, VI, ... , Vk ) = \7/-l (U, VI, ... , Vk) - d/-l (U, VI, ... , Vk) k
= L (\7Vj/-l) (VI' ... ' YJ-I' U, YJ+I, ... , Vk). j=l
With this in mind, we fix p E M and VI, . .. ,Vk E TpM. We may choose VI, ... , Vk E X(M) such that Vi(p) = Vi and may assume that (\7Vi) (p) = o. N ow choose an adapted orthonormal frame field E I , ... , En centered at p so that (\7Ed (p) = 0 for all i and [Ei,Ej] (p) = 0 for all i,j. In the following calculation, several steps may appear at first to be wrong. However, if one begins to write out the missing terms, one sees that they vanish because of how the fields were chosen to behave at p. Using equation (13.11) we have
n
i=l
n
= Lei [(\7/-l- d/-l) (Ei' VI, .. . , Vk)] i=l
13. Riemannian and Semi-Riemannian Geometry
626
and using what we know about the fields at p, this is equal to
t
5..N X,y Z and RfJ!,yZ?
Appendix A
The Language of Category Theory
Category theory provides a powerful means of organizing our thinking in mathematics. Some readers may be put off by the abstract nature of category theory. To such readers, I can only say that it is not really difficult to catch on to the spirit of category theory and the payoff in terms of organizing mathematical thinking is considerable. I encourage these readers to give it a chance. In any case, it is not strictly necessary for the reader to be completely at home with category theory before going further into the book. In particular, physics and engineering students may not be used to this kind of abstraction and should simply try to gradually become accustomed to the language. Feel free to defer reading this appendix on category theory until it seems necessary. Roughly speaking, category theory is an attempt at clarifying structural similarities that tie together different parts of mathematics. A category has "objects" and "morphisms". The prototypical category is just the category Set which has for its objects ordinary sets and for its morphisms maps between sets. The most important category for differential geometry is what is sometimes called the "smooth category" consisting of smooth manifolds and smooth maps. (The definition of these terms is given in the text proper, but roughly speaking, smooth means differentiable.) Now on to the formal definition of a category. Definition A.I. A category
-
637
A. The Language of Category Theory
638
denoted Mor(~). In addition, a category is required to have a composition law which is defined as a map 0: HomdX, Y) x HomdY, Z) ~ HomdX, Z) such that for every three objects X, Y, Z E Ob(~) the following axioms hold:
(1) HomdX, Y) and HomdZ, W) are disjoint unless X = Z and Y = W, in which case HomdX, Y) = HomdZ, W). (2) The composition law is associative: f 0 (g 0 h) = (f 0 g) 0 h. (3) Each set of morphisms of the form HomdX, X) must contain a necessarily unique element id x , the identity element, such that f 0 idx = f for any f E HomdX, Y) (and any Y), and idx of = f for any f E HomdY, X). Notation A.2. A morphism is sometimes written using an arrow. For example, if f E HomdX, Y) we would indicate this by writing f : X ~ Y or by X
-4 Y.
The notion of category is typified by the case where the objects are sets and the morphisms are maps between the sets. In fact, subject to putting extra structure on the sets and the maps, this will be almost the only type of category we shall need to talk about. On the other hand there are plenty of interesting categories of this type. Examples include the following. (1) Grp: The objects are groups and the morphisms are group homomorphisms. (2) Rng : The objects are rings and the morphisms are ring homomorphisms. (3) LinIF' : The objects are vector spaces over the field IF and the morphisms are linear maps. This category is referred to as the linear category or the vector space category (over the field IF). (4) Top: The objects are topological spaces and the morphisms are continuous maps. (5) ManT : The category of CT differentiable manifolds and CT maps: One of the main categories discussed in this book. This is also called the smooth or differentiable category, especially when r = 00.
Notation A.3. If for some morphisms Ii : Xi ~ Ii , (i = 1,2), gx : Xl ~ X 2 and gy : Yl ~ Y2 we have gy 0 h = h 0 gx, then we express this by saying that the following diagram "commutes": h
Xl~Yl gx
1
1
gy
X2~Y2
h
A. The Language of Category Theory
Similarly, if h 0
f
639
= g, we say that the diagram
X~Y
~lh Z
commutes. More generally, tracing out a path of arrows in a diagram corresponds to composition of morphisms, and to say that such a diagram commutes is to say that the compositions arising from two paths of arrows that begin and end at the same objects are equal. Definition A.4. Suppose that f : X ---+ Y is a morphism from some category Q:. If f has the property that for any two (parallel) morphisms gl, g2 : Z ---+ X we always have that fogl = fog2 implies gl = g2, i.e. if f is "left cancellable" , then we call f a monomorphism. Similarly, if f : X ---+ Y is "right cancellable", we call f an epimorphism. A morphism that is both a monomorphism and an epimorphism is called an isomorphism. If the category needs to be specified, then we talk about a Q:-monomorphism, Q:-epimorphism and so on).
In some cases we will use other terminology. For example, an isomorphism in the smooth category is called a diffeomorphism. In the linear category, we speak of linear maps and linear isomorphisms. Morphisms which comprise HomdX, X) are also called endomorphisms and so we also write EnddX) := HomdX, X). The set of all isomorphisms in HomdX, X) is sometimes denoted by AutdX), and these morphisms are called automorphisms. We single out the following: In many categories such as the above, we can form a new category that uses the notion of pointed space and pointed map. For example, we have the "pointed topological category". A pointed topological space is a topological space X together with a distinguished point p. Thus a typical object in the pointed topological category would be written as (X,p). A morphism f : (X,p) ---+ (W, q) is a continuous map such that f(p) = q. A functor F is a pair of maps, both denoted by the same letter F, that map objects and morphisms from one category to those of another,
F: Ob(Q:d ---+ Ob(Q:2),
F: Mor(Q:l) ---+ Mor(Q:2), so that composition and identity morphisms are respected. This means that for a morphism f : X ---+ Y, the morphism
F(f) : F(X) ---+ F(Y)
A. The Language of Category Theory
640
is a morphism in the second category and we must have (1) F(id\!:l)
= id\!:2·
(2) If f : X ~ Y and 9 : Y F(g) : F(Y) ~ F(Z) and
~
Z, then F(f)
F(g 0 f) = F(g)
0
F(X)
~
F(Y),
F(f).
Example A.S. Let LinJR be the category whose objects are real vector spaces and whose morphisms are real linear maps. Similarly, let Line be the category of complex vector spaces with complex linear maps. To each real vector space V, we can associate the complex vector space C Q9JR V, called the complexification of V, and to each linear map of real vector spaces f : V ~ W we associate the complex extension fe : C Q9JR V ~ C Q9JR W. Here, C Q9JR V is easily thought of as the vector space V where now complex scalars are allowed. Elements of C Q9IR V are generated by elements of the form c Q9 v, where c E C, v E V and we have i(c Q9 v) = ic Q9 v, where i = J=I. The map fe : C Q9IR V ~ C Q9IR W is defined by the requirement fd c Q9 v) = c Q9 fv. Now the assignments f
H
fe,
V
H
CQ9IR V
define a functor from LinIR to Line. In practice, complexification amounts to simply allowing complex scalars. For instance, we might just write cv instead of c Q9 v. Actually, what we have defined here is a covariant functor. A contravariant functor is defined similarly except that the order of composition is reversed so that instead of (2) above we would have F(g 0 f) = F(f) 0 F(g). An example of a contravariant functor is the dual vector space functor, which is a functor from the category of vector spaces LinIR to itself that sends each space V to its dual V* and each linear map to its dual (or transpose). Under this functor a morphism V ~ W is sent to the morphism V*
..£-
W*.
Notice the arrow reversal. One of the most important functors for our purposes is the tangent functor defined in Chapter 2. Roughly speaking this functor replaces differentiable maps and spaces by their linear parts. Example A.6. Consider the category of real vector spaces and linear maps. To every vector space V, we can associate the dual of the dual V**. This is
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a covariant functor which is the composition of the dual functor with itself: V**
A**l V*
W**
Now suppose we have two functors, Fl : Ob(Q:l) -+ Ob(Q:2), Fl : Mor(Q:l) -+ Mor(Q:2) and
F2 : Ob(Q:l) -+ Ob(Q:2), F2 : Mor(Q:l) -+ Mor(Q:2)' A natural transformation 7 from Fl to F2 is given by assigning to each object X of Q:l, a morphism T(X) : F1(X) -+ F2(X) such that for every morphism f : X -+ Y of Q:l, the following diagram commutes:
Fl(X)!l2 F2(X)
1
Fl(f)
1
F2(f)
Fl (Y) T(Y) F2 (Y) A common first example is the natural transformation i between the identity functor I : LinIR -+ LinIR and the double dual functor ** : LinIR -+ LinIR: ~(V)
V - - V** f
1
W
~
1r* W **
v:
The map V -+ V** sends a vector to a linear function V* -+ lR defined by v(a) := a(v) (the hunted becomes the hunter, so to speak). If there is an inverse natural transformation 7- 1 in the obvious sense, then we say that 7 is a natural isomorphism, and for any object X E Q:l we say that Fl (X) is naturally isomorphic to F2(X), The natural transformation just defined is easily checked to have an inverse, so it is a natural isomorphism. The point here is not just that V is isomorphic to V** in the category LinIR, but that the isomorphism exhibited is natural. It works for all the spaces V in a uniform way that involves no special choices. This is to be contrasted with the fact that V is isomorphic to V*, where the construction of such an isomorphism involves an arbitrary choice of a basis.
Appendix B
Topology
B.1. The Shrinking Lemma We first state and prove a simple special case of the shrinking lemma since it makes clear the main idea at the root of the fancier versions.
Lemma B.1. Let X be a normal topological space and {Ul' U2} an open cover of X. There exists an open set V with V c U1 such that {V, U2} is still a cover of X. Proof. Since U1 UU2 = X, we have (X\Udn(X\U2) = 0. Using normality, we find disjoint open sets 0 and V such that X\U1 c 0 and X\U2 C V. Then it follows that X\O c Ul and X\ V c U2 and so X = U2 U V. But On V = 0 so V c X\O. Thus V c X\O CUI. 0 Proposition B.2. Let X be a normal topological space and {U1 , U2, ... , Un} a finite open cover of X. Then there exists an open cover {VI, V2,···, Vn } such that Vi C Ui for i = 1,2, ... ,n. Proof. Simple induction using Lemma B.1.
o
The proof we give of the shrinking lemma uses transfinite induction. A different proof may be found in the online supplement [Lee, Jeff]. It is a fact that any set A can be well ordered, which means that we may impose a partial order --< on the set so that each nonempty subset SeA has a least element. Every well-ordered set is in an order preserving isomorphism with an ordinal w (which, by definition, is itself the set of ordinals strictly less than w). For purposes of transfinite induction, we may as well assume that the given indexing set is such an ordinal. Let A be the ordinal which is the indexing set. Each Q E A has a unique successor, which is written as Q + 1.
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B. Topology
The successor of ex + 1 is written as ex + 2, and so on. If (3 E A has the form (3 = {ex, ex + 1, ex + 2, ... }, then we say that (3 is a limit ordinal and of course a -< (3 for all a E (3. (This may seem confusing if one is not familiar with ordinals.) Suppose we have a statement P(ex) for all ex E A. Let 0 denote the first element of A. The principle of transfinite induction on A says that if P(O) is true and if the truth of P(ex) for all ex -< (3 can be shown to imply the truth of P((3) for arbitrary (3, then P( ex) is in fact true for all ex E A. In most cases, a transfinite induction proof has three steps: (1) Zero case: Prove that P(O) is true. (2) Successor case: Prove that for any successor ordinal ex + 1, the assumption that P( 0) is true for all 0 -< ex + 1 implies that P( ex + 1) is true. (3) Limit case: Prove that for any limit ordinal w, P(w) follows from the assumption that P(ex) is true for all ex -< w. Definition B.3. A cover {Ua}aEA of a topological space X is called point finite if for every p E X the family A(p) = {ex : p E Ua} is finite. Clearly, a locally finite cover is point finite. Theorem B.4. Let X be a normal topological space and {Ua}aEA a point finite open cover of X. Then there exists an open cover {Va}aEA of X such that Va C Ua for all ex EA. Proof. Assume that A is an ordinal. The goal is to construct a cover {Va}aEA of X such that Va C Ua for all ex E A. We do transfinite induction on A. Let P( ex) be the statement
(*) For all 0 -< ex there exists Vo with Vo C Uo such that {Vo }O- and the module itself T = Tv 1,... ,V k is also referred to as a tensor product of the modules VI, ... , V k. The tensor product is again unique up to isomorphism: Proposition D.1S. If (Tl' ®d and (T2' ®2) are both universal for k-multilinear maps on VI X •.. x V k, then there is a unique isomorphism : Tl -+ T2 such that 0 ®1 = ®2: V 1 X··· XVk
Tl
/~~ ~
T2
Proof. By the assumption of universality, there are maps ®1 and ®2 such that 0 ®1 = ®2 and
