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1. Let (U, p) be a local chart on Mn and P E U. If {zt} are the coordinates of Q E U and {e } the coordinates of p(Q) E Rn, then z' _ i' for 1 < i < n. This is the equality of two real numbers. If we consider the equality of two functions, we must write a' = ` o V. We have V
( W. 0 )P
Indeed, (f being a differentiable function in ((a neighbourhood of Sp(P) in R', I
la
) P1
1P
(f ° gyp) =
\Lf ).{P) =
(')
(P)
(f).
Thus ,p* is a bijection of T(U) onto V(U) x R" C R. is be an atlas for Mn. The set of ]T(UB), Let (U0, an atlas for T(V). Let us show that this atlas is of class Cr1. Sup
pose U. n Us # 0. On Sp.(U. n U8) set 0 = vo o cp;'; then we have 8=(Q, X) = [e(Q), (dO)Q(X)], where Q E p0(Ua n Us) and X E TQ(RR). Thus (dO)Q is of class C''' 1.
2.11. Definition. A differentiable manifold E is a vector fiber bundle of fiber the vector space F if there exist a differentiable manifold M (called the basis) and a differentiable map II of E an M such that, for all P E M, II'1(P) = Ep is isomorphic to F and there exist a neighbourhood U of P in M and a diffeomorphism p of U x F onto RI(U) whose restriction to each Ep is linear, p satisfying 11 o p(P, z) = P for all z E F.
49
The Bracket [X, 11
2.12. Proposition. The tangent bundle T(M) is a vector bundle of fiber IIn.
M is the basis. If X E T(M) and X E Tp(M) for a unique point P E M,
then the map n is x + II(X) = P. Thus II1(P) = Tp(M), which is a vector space of dimension n: F = R". If (U, rp) is a local chart at P, we Itn. So we saw that ip is a diffeommorphism of T(U) = II1(U) onto rp(U) x can choose p = V. 1 o (V, Id), and we know that (rp 1) p is linear. Moreover, p(P, z) E Tp(M), and thus ]a o p(P, z) = P, for all z E Rn.
2.13. Definition. Likewise we can consider the fiber bundles T*(M),
A" r(M),77(M):
T*(M) = U 17(M), PEM P
A7M= PEM U A(M), where A'77(M) is the space of skewsymmetric pforms on Tp(M), and
U 07 p(M)
Tp(M),
PEM
where
0 7P(M) 4 Tp(M) is the space of tensors of type (r, s), r times
covariant, s times contravarlant, on Tp(M).
2.14. Definition. A section of a vector fiber bundle (E, II, M) is a differentiable map { of M into E such that II o e = identity. A vector field is a section of T(M). An (r, s)tensor field is a section of 7, (T(M)). An exterior differential pform is a section of AP 7' (M). In a local chart an exterior differential pform
L
17 =
aj,...jdx" A dxh A ... A dTJP,
1j,<j2 0 are defined in 2.14. Given q E Aq(M) and E AP(M), we define qAf E AP''(M), the exterior product of in and t;, by (n A
C)(X1,...
, Xp+q)
I E e(a)n(Xv(1), ...
, X0(q)X(Xc(1+q), ... , XQ(P+q)),
p'q' VE'P
, X,+q are p + q vector fields and the sum is over the set P of permutations a, e(a) being the signature of a. The exterior product is where X1,
associative and anticommutative: l; A q = (1)Pgq A l;. We also define the inner product i(X)77 of a differential form q E Aq(M) (1 < q< n) by a vector field X. i(X)v7 is a differential (q  1)form defined as follows: If X; (i = 1, 2, ... , q 1) are q 1 vector fields , then [i(X)17](X1,X2,... ,Xq1) = FAX,
X1,X2,... ,Xq1)
We verify that if t E AP(M), then
i(X)(q At) _ [i(X)q] A1: + (1)qq A [i(X).J
and
i(X)[i(X)q] = 0.
2.23. Proposition. Let 0 be a differentiable map of M,a into WP. To any differential qform q on W, we can associate a differential qform 0*q E Aq(M), the inverse image of q by 4b, defined by
(4*q)p(X1, X2, ... , Xq) = fi(n) ('4X 1, ... ,44Xq),
Exterior Differential X1,

53
, Xq being q vectors of Tp(M). For a function f on W (f E A°(W ),
we sets*f =fo4. Here there is no difficulty such as in Definition 2.17. Indeed if P E M, then Q = $(P) is unique and (4*ri)p = (,DP)r7i. We verify that
2.24. Definition. Exterior differential. To q E Aq(M), we associate the exterior differential form dri E Aq+l(M) defined by q+1
A
r dr1(X1, ... , Xq+1) = E(1)i'Xi[r1(X1, .. , Xi,
...
, Xq+l)]
i=1
+,(_1)i+j]7([Xi,Xj],X1,...
,X;,... ,Xj,... ,X4+1)
i<j . where Xi, , Xq+i are vector fields. A caret over a term means that this term is omitted. According to this definition,
f EA°(M)df(X)=X(f), gEA'(M)
di7(X,Y)=X[n(Y))Y[0(X)]n([X,Y])
Let us show that dq is indeed a differential form. It is obvious that dr7(X, Y) = dq(Y, X) and that dri(Xi + X2, Y) = dq(Xj, Y) + dn(X2, Y).
But we have to prove the C°(M)linearity: for any f E C°°(M), dri(fX,Y) = fd>7(X,Y) According to the expression of dry,
di (fX,Y) = fX[r!(Y)] Y[r!(fX)]  rl([fX,Y]) = fX[t(Y)] Y[fri(X)]  rl(f[X,Yj Y(f)X), since V X, X, YJ = dY.(fX)  d(f X ).Y = f [X, Y]  (cf.Y)X.
As Yffr1(X)] = fY[rf(X)] +rl(X)Y(f), we have proved the announced result for a differential 1form. For q > 1 the proof is similar, but we have a simpler expression of the exterior differential, from which it is obvious that dq is an exterior differential (q + 1)form.
2.25. Local expression of the exterior differential. Let (i2, gyp) be a local chart, x1, ... , x" the corresponding coordinates, {8/8x`} (i = 1,  , n) the n vector fields of the natural basis, and (dO} the dual basis. The differential qform n is written
37 _ E aj,...jq(x)d2J1A...Adxj, jl <ja 0.
2.28. Theorem. A differentiable manifold M is orientable if and only if there exists a differential n form that is everywhere nonvanishing.
Suppose M. is orientable. Let be a locally finite atlas, all of whose changes of charts have positive Jacobian, and {a,} a partition of unity subordinate to the covering {R}. Let x,, xf, , xi be the coordinates on i2{, and consider the differential nform w, = a,(x)dz' A dx, A .. A dx?. Let us verify that w = L_,Cl wi is nowhere zero. A given point P belongs only to a finite number of f2,; let f21, 522, ... , ft,,, be these f4. Write all wi(P) in the same coordinate system {xi}: ( m 1 w(P)=1a1(P)+ a,(P)Jdx;Adx A A dx;.
l
I
f=2
l
iI
P
w(P) does not vanish, since each term in the bracket is nonnegative and some of them are strictly positive. Recall that a, (P) = 1, a, (P) > 0
and 84/8I>0.
Conversely, let w be a nonvauishing differential nform, and A = (t2;, Vi)jer an atlas such that all the i2, are connected. From A we will construct an atlas all of whose changes of charts are positive. On ft, there exists a nonvanishing function fi such that w(x) = f; (x)dxi A dxi A .. A dx;. Since w does not vanish and since 52, is connected, f, has a fixed sign. If f{ is positive, we keep the chart (52,, gyp;). In that case we set c3; = Bpi. Otherwise, whenever fj is negative, we consider yaj, the composition of fpj with
56
2. Tangent Space
the transformation (x1, x2, x") i we construct an atlas A = (f4, c )iEI
(xl, x2,
'X") of R". So from A
Now if f, is equivalent to f, but in the chart (f4, Bpi), then f, = fi if
Vi = Vii, fj = fJ otherwise. So fj and f, are positive. At X E fki n f2j, denoting by Al Cthe determinant of the Jacobian of pj o Bpi 1, we have
fjJAI = fi. Thus JAI > 0 and all changes of charts of A have positive Jacobian.
2.29. Definition. Let M be a connected orientable manifold. On the set of nonvanishing differential nforms, consider the following equivalence re
lation: wl  w2 if there exists f > 0 such that wl = fw2. There are two equivalence classes. Choosing one of them defines an orientation of M; then M is called oriented. There are two possible orientations of an orientable connected manifold.
Some examples of orientable manifolds are the sphere, the cylinder, the torus, real projective spaces of odd dimension, the tangent bundle of any manifold, and complex manifolds.
Some examples of nonorientable manifolds are the Mobius band, the Klein bottle, and real projective spaces of even dimension (see 1.9). We can see the Mobius band in R3. a rectangle ABCD in R2: [1, 1) X ]  E, E[, c > 0, A = (1, E), B (1, = (1, E), C = (1, E), D = (1, e), and identify the segment AB with CD. If we do this by identifying A with D and B with C, we have a cylinder. If instead we identify A with C and B with D, we get a Mobius band.
Let us consider the atlas A with two charts, (fZl,col), P2, IM, where
f1i =]4, 4[ x
cR2,
Vl = identity, f12=]1,1( X ]E,E[U [1,2[ X P2 = identity on ]
2, 1
[x
,p2(x, b) _/ (x + 2, y) on [1, ' [ x  E, E[,
Orientable Manifolds
57
(x, y are coordinates on R2). Then
i21f1S22]41[ x
]E,E[U]2,d[ x
]E,E[CR2.
Let (xl, yl) and (x2r y2) be the considered coordinate systems on f1l and respectively. E, e [ the change of coordinates is x2 = X1, y2 = yl, and On I", I [ x
on ] 
[ x ]  E, E[ it is x2 = xl + 2, y2 = yl. On the first open set the change of coordinate chart has positive Jacobian, but on the second it has negative Jacobian. In the proof of Theorem 2.28 we saw that, if a manifold is orientable, from an atlas A we can construct an atlas A all of whose changes of charts are positive only by the eventual change of coordinate x2 p x2. Here if we do that, we always have one positive Jacobian and one negative Jacobian on the two open sets of fl, fl 51,. Thus the Mobius band is not orientable. The Klein bottle is the compact version of the Mobius band. It is in R°. Consider a cylinder with end circles C1 and C2. Let ABCD be four points (in that order) on C1, and A'B'C'D' four points (again, in order) on C2. We identify C1 and G. We get a torus if we identify A with A', B with B', C with C', and D with D. On the other hand, we get a Klein bottle if we identify A with A', B with D', C with C', and D with B. The cylinder is a twosheeted covering manifold of the Mobius band. The torus is a twosheeted covering manifold of the Klein bottle. 4,
2.30. Theorem. If M is nonorientable, M has a twosheeted ori.entable covering manrold M. If M is simply connected, then M is orientable. For the proof see Narasimhan [10]. M is said to be simply connected if any closed curve in M may be reduced to one point by a continuous deformation. More precisely,
2.31. Definition. Let C be a circle, f and g two continuous (respectively differentiable) maps of C into a manifold M. Let f (C) and g(C) be closed (respectively differentiable) curves of M. f (C) is said to be homotopic to g(C) if there exists a continuous map F(s, t) of C x [0,1] into M such that
F(x, 0) = f (x) and F(x,1) = g(x). If f and g are C', we require that F(x, t) be C' in x on C x [0,1].
2.32. Definition. The manifold M is said to be simply connected if any closed curve f (C) in M is homotopic to one point; that is, there exists F(x, t) as above with g(C) reduced to one point. f (C) homotopic to g(C) is an equivalence relation. The equivalence classes are called homotopy classes.
2.33. Definition. Let M be a differentiable orientable manifold. We define the integral of w, a differential nform with compact support, as follows:
2. Tangent Space
58
Let (fZ, A)iEI be an atlas compatible with the orientation chosen, and {ai}iEi a partition of unity subordinate to the covering {Sli};EI. On Sti, w is equal to fi (x)dx, A dx? A dx; . By definition, JM
_ iEI
J0'(a) (ai(x)fi(x)] o Bpi
1dx' n dx2 n ... n dxn.
One may verify that the definition makes sense. The integral does not depend on the partition of unity, and the sum is finite.
Manifolds with Boundary 2.34. Let E be the halfspace xl > 0 of R", XI being the first coordinate of Rn. Consider E C R" with the induced topology. We identify the hyperplane n of R", x1 = 0, with R"1. Letting fl and 8 be two open sets of E,
and ' : 0
0 a homeomorphism, it is possible to prove that the restriction of tp to St n II is a homeomorphism of fl n II onto 0 n II. So the boundary II of the manifold with boundary E is preserved by homeomorphism. E is the standard manifold with boundary, as R" is the standard manifold. LI
IP
2.35. Definition. A separated topological space M" is a manifold with boundary if each point of M" has a neighbourhood homeomorphic to an open set of E. The points of Mn which have a neighbourhood homeomorphic to R" are
called interior points. They form the interior of Mn. The other points are called boundary points. We denote the set of boundary points by W. and call it the boundary of M. As in 1.6, we define a Ckdifferentiable manifold with boundary. By definition, a function is Ckdifferentiable on E if it is the restriction to E of a Ckdifferentiable function on R". 2.36. Theorem. Let M. be a (Ckdifferentiable) manifold with boundary. If 8M is not empty, then aM is a (C"differentiable) manifold of dimension n  1, without boundary: O(OM) _ 0.
Manifolds with Boundary
59
Proof. If Q E 8M, there exists a neighbourhood n of Q homeomorphic by W to an open set 9 C . The restriction Yf of Sp to ft = ii fl 8M is a homieomorphism of the neighbourhood 12 of Q E OM onto an open set 8 = e n n of Rn1. Thus OM is a manifold (without boundary) of dimension n  1 (Definition 1.1). If M is Ckdifferentiable, let (S2i, {p,);EI be a Ckatlas. Clearly, (57,, cp;);El turns out to be a Ckatlas for OM. 2.37. Theorem. If Mn is a Ckdifferentiable oriented manifold with boundary, then 8M is orientable. An orientation of Mn induces a natutl omentation of OM. Proof. Let (12 j, SO3 )jE! be an admissible atlas with the orientation of Mn, M and (S2 j, oj) jE I the corresponding atlas of OM, as above. Let i : OM be the canonical imbedding of 8M into M. We identify Q with i(Q), and
X E TQ(OM) with 1.(X) E TQ(M). Given Q E OM, pick el E TQ(M), el V TQ(&M), el being oriented to the outside, namely, ei(f) > 0 for all differentiable functioos on a neighbourhood of Q which satisfy f < 0 in Mn and f (Q) = 0. We choose a basis {e2, e3, , en} of TQ(8M) such that the basis {el, e2, eg, , en} of TQ(N) belongs to the positive orientation given on Mn. Then {e2, e3, , en} is a positive basis for Tq(8M). This procedure defines a canonical orientation on OM, as one can see.
2.38. Stokes' Formula. Let M be a Ckdifferentiable oriented compact manifold with boundary, and w a differential (n 1)form on Mn; then
I
dw
= Jam aM
1,u
where OM is oriented according to
rthe previow theorem. For convenience
we have written Jam w instead of J and 2.25).
8M
W,
i*w (for the definition of i* see 2.7
8A/
Proof. Let (12{, cp,);EJ be a finite atlas compatible with the orientation of Mn; such an atlas exists, because Mn is compact. Set e, = w(Sl;). Consider {a;}, a Ckpartition of unity subordinate to {1'4}. By definition,
f dw = E iEj
41
j
d(a,w). ;
Thus we have only to prove that d(a;w) =
J
j(o.w).
2. Tangent Space
60
we recall that f4 = (Zi fl am and we have set 6i = w,(SZj) = O1 fl fI. In ((Z;, gyp;) we have n a,w=Efj(x)dx1A...AddjA...Ade,
j=1
where the fj(x) are CAdifferentiable functions with compact support indxj means this term is missing. Now, cluded in
d(aiw) =
dfi(x)Adx'Adx2A...AdxjA...ndx"
j=1 [(_1p_10)1ds1A2A...hdx
by 2.25. According to Fubini's theorem,
J6
d(aaw) =
jf1(x)dx2 A dx3 A ... A Can
where j is the inclusion R E. Indeed, a a a j*(aiw) (5121... i a;w (j.
j*(C
ax
w),
fi(x),
and we identify j*a/axk with a/axk. Observe that
An1(lI)
W1)*w So, we have (gyp 1)* o i1' = j* o 60T.
Exercises and Problems 2.39. Exercise. Let M and W be two differentiable manifolds, and f a diffeomorphism of M onto W. Let P E M, and set Q = f (P). Consider Y E TQ(W) and ,'y E AQ(W), q > 1. Express f*i(Y)%r in terms of ry = f*j.
2.40. Exercise. Let (Z be a bounded connected open set of R3 such that St is a differentiable manifold with boundary. On R3, endowed with an orthonormal coordinate system (x1, x2, x3), we consider a vector field X of components Xi(x): 3
(x)axi.
(x) = i=1
Exercises and Problems
61
Is the boundary 80 orientable? At P E 852, let v (P) be the normal unit vector of 80 oriented to the outside of Q. Prove the wellknown formula
8X
1
ax
dE = fan v . X do, Jest
where dE is the volume element on R3, do, the area element on 8f2, and . X the scalar product of v and Y. Hint. Proceed as follows:
v
a) Find a differential 2form w on R3 such that s
;
dw =
;
dx' A dx2 A dx3.
c=1
b) Let (u, v) be a coordinate system on a neighbourhood 0 C OIL of P,
orthonormal at P and such that (v ,, a) is a positive basis in Tp(R3). We set 3 49
U
a
and
a av =
3
v ax;.
Compute w(&, 8) in terms of the components of c) Use Stokes' formula.
and
.
2.41. Exercise. Consider on R3, endowed with a coordinate system (x, y, z), the following three vector fields:
X = 1(1x2y2zz)ax+(Elz) +(xz+y)
z
Y = (xy+z)
+2(1x2+y2z2)ay+(yzx)1,
Z = (xz  y)
+ (yz+x)
19
+2(1x2y2+z2)z.
a) Verify that the three vectors X(m), Y(m), Z(m) form an orthogonal basis of R3 at each point m E R3 (the components of the Euclidean metric are in the coordinate system ej = 6,J). b) Compute the three Lie brackets [X, Y], [Y, Z], [Z, X] and express them in the basis (X, Y, Z).
c) Let 0 = R3  {0} and consider the map cp of t2 into fl, (x, y, z) (u, v, w), defined by U
x (x2 + y2 + z2)'
2. Tangent Space
62
V = (x2
V z2),
+ y2 + z
w=
(x2 + y2 + x2).
The restrictions of X, Y, Z to fl are still denoted X, Y, Z. Verify that V is a diffeomorphism, and compute 9#X, cp*Y and p.Z. d) Using an atlas with two charts on S3, deduce from c) the existence of three vector fields on S3 forming a basis of the tangent space at each point of S3. Notice that p is the diffeomorphism of change of charts for an atlas with two charts on S3 constructed by stereographic projection.
2.42. Exercise. Let W be the set of real 3 x 3 matrices whose determinant is equal to 1.
a) Exhibit a differentiable manifold structure on W. Show that W is a hypersurface of R", and specify n. b) Identify to a set of matrices the tangent space Tg(W) of W at 1, the identity matrix. 2.43. Problem. In R3, we consider a compact and connected differentiable submanifold M of dimension 2. a) Show that 11 = R3\M has at most two connected components. b) We admit that $I has at least two connected components. Prove that one of them is bounded. We call it W. c) Construct on M a continuous field v(P) of unit vectors in R3 such that v(P) is orthogonal to Tp(M) at each point P E M. The norm of the vectors comes from the scalar product (.,.) defined by the Euclidean metric on R3 endowed with a coordinate system {xi}, i = 1, 2, 3. d) Deduce that M is orientable. e) On R3, consider the differential 3form w = dx' A dx2 A dx3 and a vector field X. Verify that 3
d[i(X )w] = (divX)w,
where divX = k=1
8Xk 8xk
f) Prove that
1
(divX)w = JM
,)j*[i(v)wJ
for an orientation on M. Here j is the inclusion M C W. g) Let f be a C2 function defined on R3. We suppose that f satisfies
i8;;f =O on W. If f I M = 0, or if 8 f = 0 on M, prove that f is constant on W.
Exercises and Problems
63
h) When M is the get of zeros of a C1 function on R3 such that R3 9 x f (x) is of rank 1 at any point x E M, show that Cl has at least two connected components.
2.44. Problem. In this problem R" is endowed with the Euclidean metric E, {x1} is a coordinate system denoted (x, y, z) on R3 , and S1 is the set of the points of R" satisfying a1(x')Z = 1. T is the inclusion Sn_1 c R". a) On R3  {O}, consider the differential 2form
w=(xdyAdzydxAdz+zdxAdy)(x2+y2+z2). Verify thatrw is closed.
b) Compute
J
**&.,. For lack of a better method, one can use spherical
coordinates. c) Is w homologous to zero on R3  {0}? (See 5.18 for the definition.) d) On R"  {0}, consider the differential 1form n
n
I
Compute *a (the adjoint of a, see 5.16 for the definition; here t712..n = 1) and prove that *a is closed.
e) What is the value of fs
+Il*(*a)?
Is *a homologous to zero on
R"{0}? 2.45. Exercise. Let 0(n) be the set of the n x n matrices M such that tMM = I, the identity matrix. a) Show that 0(n) is a manifold. What is its dimension? Hint. If you want, first consider the problem in a neighbourhood of I. b) Identify Tj(O(n)) with a set S of n x n matrices.
c) Show that if A E S and M E 0(n), then MA E TM(O(n)). Verify that the map Is: (M, A) + (M, MA) is a diffeomorphism of 0(n) x S onto T(O(n)). d) Deduce from this result that 0(n) is parallelisable.
2.46. Problem. In this problem w is a C°° differential 1form on Rn+1 which is not zero at 0 E Rn+1
a) Show that if there exist two C°° functions f and g, defined on a 0, such that w = f dg neighbourhood V of 0 E Rn+1 with f (0) on V, then there exists a differential 1form 9 in a neighbourhood of 0 E R"+1 such that dw = 0 A w.
2. Tangent Space
64
b) Exhibit an expression of 0 if w = yzdx + xzdy + dz, (x, y, z) being a coordinate system on R3. Show that we can choose f = e11'. Compute g.
c) Ifw=dz  ydx  dy,do f andgexist? d) Next, put w in the form w = fdg. In the general case, show that the problem may be reduced to the case where, in a neighbourhood of 0,
w=dzA,(x,z)dx'. i=1
Here z E R, x = (xl, X2'.. , x") E R", and the Ai are Coo functions. e) a = (a', a2, , a") E R" and c E R being given, consider the differential equation dz
"
=
Ai(at,z)a',
z(0) = c.
dt
Show that there exists a C°° map F : (t, a, v) ' F(t, a, v) of I x W x J into R with I, W, J neighbourhoods respectively of 0 E R, 0 E R" and c E R, such that OF
_
"
Ai (at, F)ai,
F(0, a, c) = c.
i=1
Verify that F(t, a, v) = F(1, at, v) if one of these terms exists.
f) Prove that u and v defined by u = x and F(1, u, v) = z form a coordinate system in a neighbourhood of (0, c) E R" x R. Hint. Show that (8F/8v) (t, a, v) does not vanish in a neighbourhood of (0, 0, c). 1 Pi (u, v)dui + B(u, v)dv. Show that In this chart w =
Pi(at, v)a' = 0. i=1
g) LetWbethemapofRxR"xRinto R"xRdefined by 1@ (t, a, v) _ (ta, v) = (u, v).
Compute 41*w. Denote by Ri(t, a, v) the coefficient of dai in 4' w. h) We suppose that dw = 0 Aw, 0 being a differential 1form in a neighbourhood of 0 E R"}1. Show that 8Ri/8t = HRi, with H(t, a, v) the coefficient of dt in IIY*B.
Deduce from this result that there exists a neighbourhood of 0 E Rn+1 where w = Bdv.
2.47. Exercise. Let M be a C°° differentiable manifold of dimension n. Consider its cotangent bundle T*(M), and denote by H the canonical pro
jection T*(M) i M.
Exercises and Problems
65
a) (9, gyp) being a local chart on M and {C'} the corresponding coordinate
system, on II1(9) consider the coordinate system (x', x2, ... , xn, y1, y2, . , yn) with x' _ V o II and y' the components of the 1form in the basis {dl;'} (i = 1, 2, , n). Let a E 7= (M), and define a linear form on TQ (7" (M)) by
T.(T*(M)) D u p (II*u, a) = (u, W or).
Show that a  I o defines a differential 1form w on M. What is its expression? n times
b) Compute n = dw A dw A  . A dw. Deduce from the result that T* (M) is orientable.
2.48. Exercise. Prove that for any C°° differentiable manifold M the tangent space T(M) is orientable.
2.49. Problem. a) Describe an atlas of the projective space P2(R) with three local charts (f2i,Oi), i = 1,2,3. b) Compute the changes of charts, and deduce that P2 is not orientable. c) Consider the cylinder H = [1,11 x C, where C is the circle quotient
of R by the equivalence relation in R: 9  9 + 2kx (k E Z). Verify that the Mobius band M may be identified with the quotient of H by the equivalence relation in H: (t, 9)  (t, 9 + x) . Prove that the boundary of M is diffeomorphic to a circle C : C 3 0  ' 0(9) E 8M. d) Let (r, w) be a polar coordinate system on the disk
D={xER2IIxI< 1}. Show that P2(R) may be identified with the quotient of D U M by the equivalence relation: (1, 0) E D is equivalent to ¢(9) E M. 2.50. Problem. On a C°° differentiable manifold M of dimension 2n, we suppose that there exists a closed 2form fZ of rank 2n. That is to say that n times
f2 A A .. A
76 0 everywhere on M.
a) To a C°° vector field X E r(m) we associate the 1form wX = i(X)S2. Verify that wX is closed if and only if CXSl = 0.
b) Show that the map h : X  wx is an isomorphism of r(m) on /l' (M) c) a and Q being two 1forms, we set (a, Q) = h([X,,, X,6]) with X,, = h1(a) and X0 = h1(13). If a is closed, prove that Cx013 = (a,#). Deduce that (a,,S) is homologous to zero if a and j3 are closed.
2. Tangent Space
66
d) Let f, g be two CO° functions on M, and set (f, 9) = fl(Xdq, XdI).
Show that (d f, dg) = d(f, g). Deduce that f is constant along the integral curves of Xd, if and only if g is constant along the integral
curves of X. e) Assume that locally there exists a coordinate system (x1 , ... , x", y1, ... , yn) such that 12 = dx' A V. Compute the local ex1 pression of (f, g) in this coordinate system.
2.51. Exercise. a) Let f be a differentiable map of Rn into R of maximal rank everywhere. Show that f 1(0) is an orientable manifold. b) Prove that the manifold that is the product of two C°° differentiable manifolds M and W is orientable if and only if M and W are orientable.
2.52. Exercise. Let w be a differential 1form on a C°° differentiable mann times
ifold M of dimension 2n + 1. When wA dw A dw A . . A
is never zero on
M, we say that w is a C form. a) Verify that wo = dx2i+1 + E1 x2dldz&2i is a Cform on with coordinates {x?} (j = 1, 2, ... , 2n + 1). b) Let (6k, ^)LEK be an atlas and y a differential 1form such that on each Sk we have y = fk'pkwp, with fk a differentiable function which does not vanish. Show that 7 is a Cform.
c) Conversely, 7 being a Cform, find an atlas (0k,''k)kEK such that on each Bk. For simplicity, do the proof when n = 3. 7=
Solutions to Exercises and Problems Solution to Exercise 2.39. f*i(Y)7 is a (q  I)form on M. Let Xi (1 < i < q 1) be q 1 vectors in Tp(M). We have
[f*i(y)i](Xl, X2, ... , Xq1) = i(Y)7(Yi, ... , Yqi) if Y = f.Xi, i(Y)?'(Y1,... ,Yq1) =
i(Y,Yi,Y2,... ,Yq1) =
Thus f*i(Y). = i(X)7 with X = f,,'Y.
Solution to Exercise 2.40. 0 C W is orientable; thus 811 is orieltabie.
f"7(X,Xl,X2,... ,Xqi)
67
Solutions to Exercises and Problems a) Set w = X1 dx& A dx3 + X2dx3 A dx1 + X3dx1 A dx2, and 3
dw =1: ;X'dxl A dx2 A dx3. i=1
b) w(J, &) = Xl(u2v3  u3v2) + X2(u3v1  v3u1) + X3(t'v2  u2v1). c) Stokes' formula gives a =1
where i is the inclusion a52
(
i*w, on Ion
Q. Let us compute i*w at P:
8 a)duAdv=X. v do.
Indeed, du A dv is the area element and
v = (u2v3  u3v2)
8 ax
+ (u3v1
 v3u1) ay + (4t1v2  u2v1) az8
We can verify that v is orthogonal to
and to
,
.
and that
Il v ll =1 Solution to Exercise 2.41. y + z p x, we obtain X i Y + Z . X. a) fransposing x Thus if we prove that X does not vanish and that X is orthogonal to Y, the three vectors form an orthogonal basis of R3 at each point m E R3. X does not vanish. Indeed, in that case z = xy, y = xz and 1 + x2 = y2 + z2; thus z = x2z and y = x2y. We would have y = z = 0, which is impossible since y2 + z2 > 1. We verify that the scalar product (X, Y) = 0. b) A straighforward computation yields [X, YJ = 2Z Thus [Y, Z) = 2X and [Z, X] = 2Y. c) On 0, , o jP = Identity; thus W is invertible. It is a diffeomorphism since V and io1 are differentiable. Using the fact that cp* = Dip, a computation leads to
rp,X=Z(1+u2v2w2) a (w+uv)
(uw  v)
.
We obtain tp*Y and o*Z by transposing. d) We saw in 1.8 that V is the diffeomorphism of change of charts for an atlas with two charts on S3 constructed by stereographic projection from the north and south poles P and P, respectively. rp*X extends into a vector field U on R3, and V,X at zero is equal to Likewise for p*Y and V,Z in V and W.
2. Tangent Space
68
Thus the vector field on S3 equal to X in one chart (on 53  {P} for instance) extends by U in the other chart.
Solution to Exercise 2.42. a) The set T(3,3) of real 3 x 3 matrices M is in bijection with R9. If M = ((aii) ), 1 < i, j < 3, and {xk } is a coordinate system on R9, the bijection +Il : M _ x = {xk} may be defined by xk = a j with k = 3(i  1) + j. W is the subset of T(3,3) such that det IMO =1. The map $: T(3,3) i detIMI of R9 in R is of rank 1 on W. Indeed, OW /Bail = Mi3, the minor of ai f in det IM 1. Now all minors
cannot vanish on W, since det I.MI = Ls1 a1jMlj = 1 on W. Thus according to Theorem 1.19, W is a submanifold of dimension
8inR9. b) Let t  W(t) be a map of a neighbourhood of zero in R into W such that V(O) = I. If W(t) _ ((aj,(t))), then ail(0) = 6, 3, the Kronecker symbol. Writing det 0 leads to t a'a(0) = 0, and the tangent vector to the curve t + v(t) at t = 0 is the matrix ((a'1(0))). Thus the tangent space Tr (W) may be identified with the set of real 3 x 3 matrices of zero trace.
Solution to Problem 2.43. a) At P E M, there is a local chart (V, V) of R3 (cp(V) a ball in R3) such that cp(V n M) is a disk D of R2. V n Sl has two connected components 01 and 02. 01 is included in a connected component W of i2. Then 8W n v = v1(D) is an open set of M.
Thus 8W n m is an open set in M, and it is nonempty. But it is also compact (OW is closed and included in M, which is compact). Hence 8W = M, since M is connected. At the most there are two connected components: one contains Bl and the other contains 92.
Solutions to Exercises and Problems
69
b) M is compact, so M is included in a ball B. R3\B is connected; thus it is included in a connected component of f), while W is included in B. c) At each point P, we denote by 92 the open set which belongs to the connected component of it which is not bounded. We choose the unit vector v(P) orthogonal to Tp(M) oriented to 92. It is well defined and continuous. d) On each local chart on M (whose coordinates are X1,1;2) we arrange that (v, 8/81, 8/82) is a positive basis in R3 (this means replace 11 by Cl, if necessary). Thus we exhibit an atlas for M, all of whose changes of charts have positive Jacobian. e) We have i(X )w = X ldx2 A dx3  X2dx1 A dx3 + X 3dx1 A dx2,
= d[i(X)w]
3
E
49X k
w.
k=1
f) According to Stokes' formula, for the canonical orientation on M (see 2.37) we have fW div Xw =
fMIi(X)w],
[i(X)w](e1, e2) = w(X, el, e2) = w((X, v)v, el, e2) (X, v)[i(v)w](el, e2),
where el = 8/8t1 and e2 = 8/82. Thus
fdivX1J
= JMN)1g)
We choose X = fV f (on R", the gradient V f of f is the vector V f = X18, f 8/8x' ), and get 3
3
3
f WEai(fb4f)w= fW(Of)2W+ff B fW. W f=1
i=1
Thus 3
JM
f [i(v)w] = f (af)2w  0 Wi=1
according to the result above and the hypothesis
18;; f = 0 in W.
If f6 f = 0 on M, we find that I V f I = 0 in W. Thus f is constant.
2. hngent Space
70
h) The proof is by contradiction. Suppose Il is connected, eo that the exists an are y starting from 01 which goes to 02 without cutting across M (M n y = 0). Now as IVfI(P) # 0 at each point P of M, we have for instance f (x) < f (P) = 0 for x E 01 and f (z) > f (P) = 0 for x E 02. Thus f I.y has the value f (P) somewhere, and y cuts across
M.
Solution to Problem 2.44. a) We have
dw=3(z2+y2+z2)hdxAdyAdz 
3(x2 + y2 + z2)z(x2 + y2 + z2)dx A 4 Adz = 0,
and w is closed.
b) Let (0,,p) be spherical coordinates,
x = coo V Cos 0, y = coo W sin 8,
z = sin W. Thus
dx = (sin
aijaik = O
for all jandailk>j.
There are n n+1 equations. If we prove that the map
r : M  (B1, ... , B., B12, ... , B1n, B23, ... , B(n1)n) is of rank n n+l on 0(n), Theorem 1.19 will imply that 0(n) is a submanifold of Rn' of dimension n n1 We have 13Bj
= 2a,
j,
aBjk
 aik,
OBjk
=
a,j.
At I we obtain aBjk aB;k aB = 1. Oakj aajk  1' aajj 2' The other derivatives vanish. If we suitably arrange the Bj and Bjk, there is in Dr a diagonal matrix with 2 and 1 on the diagonal. Thus 1' is of rank 24R at I. Since I' is CO°, r is of rank n n+1 in a neighbourhood of I in 0(n).


So this neighbourhood is a submanifold of Rn' of dimension 21V11
Now consider the question in a neighbourhood V of A E 0(n). The map of V into T(n, n) defined by M  p tAM is a homeomorphism of V onto a neighbourhood of I. Thus V fl O(n) is a subAs A is arbitrary, 0(n) is a manifold of T(n, n) of dimension !!n1n1 submanifold of T(n, n) of dimension
Solutions to Exercises and Problems
73
We can give an alternative proof. Let W be the map of GL(R')
into the set E of symmetric matrices, defined by co : M 'MM. GL(Rn) is an open set of T(n, n), thus a submanifold, and it is easy to see that E is also a submanifold of T(n, n). We have (DSp)M(H) =
'MH + tHM. The rank of cp at M E O(n) is equal to 24a, the dimension of E, since (Dcp)M is surjective. Indeed, let B E E; then
JMB satisfies (D'p)M('MB) = 2tMMB + LBtMM = B. Thus O(n) = 91(1) is a submanifold of GL(R") of dimension n n1 = n2  14U, since dimGL(Rn) = n2. b) Let t  M(t) be a differentiable curve in O(n) such that M(O) = I. Set M(t) = ((aij(t))). aij(0) = S;, the Kronecker symbol. We have
(dB(O) ao
2a(0) and
akj(O) +a,k(0)
Since Bj(t) and Bjk(t) are constants, it follows that ajj(0) = 0
and akj(0) + ajk(0) = 0 for all j and k 96 j. Thus dM(t)/dt is a tangent vector of O(n) at I. This vector may be identified with the antisynunetric matrix ((a;,(0))). Thus Tj(O(n)) may be identified with the set S of antisymmetric n x n matrices. c) Let u  M(u) be a differentiable curve in O(n) such that M(O) =
M. Then u ' M(u) = tMM(u) is a differentiable curve in O(n) through I, M(u) = MM(u), and dM/du = MdM/du. The result follows.
We saw just above that * is bijective. Moreover 1Y is differentiable. %1 is the map of T(O(n)) onto O(n) x S : (M, B) + (M, tMB), and this map is differentiable.
d) T(O(n)) is trivial since it is diffeomorphic to O(n) x Rn s ' . Indeed, S may be identified to R"('21), since S is defined by n n+i linear equations in T(n,n) identified to Rn2. Moreover, W1 is linear on each fiber. Thus T(O(n)) is parallelisable.
Solution to Problem 2.46. a) We have dw =4f A dg = d log if I A w = dO n w, with O= log l f l in a neighbourhood where f does not vanish. b) We have dw = dz A (ydx + xdy). With f = e'Y, d log f
A w = (xdy + ydx) A [(ydx + xdy)z + dz] = dz A (ydz + xdy).
We have indeed dw = d log an A w. So dg = exd(yzdx + xzdy + dz) = d(ze`y),
74
Z. Tangent Space
and we can choose g = zen. c) If f and g exist, we can write dw = B A w. Now here dw = dx A dy does not have a term with dz like 0 A w with B not zero. So f and g do not exist. d) Since w(O) 96 0, there is a coefficient of the 1form w which does not vanish on a neighbourhood V of 0 E Rs+1: w = B(x, z)dz + with B(0, 0) 36 0. Set w = w/B(x, z). If we can write w = f dg, then w = fdz with f = B(x, z) f, and w has a simpler expression. e) According to Cauchy's theorem, F(t, a, v) exists, and depends differentially on t and on the parameter (a, c). Consider b = Aa with A E R, and let F(u, b, v) be a solution of the equation OF
= E A,(aAu, F)Aa',
F(0, b, c) = c.
i=1
Setting t = Au yields
OF cat
=
°
A,(at, F)a',
F(0, aA, c) = c.
i=1
Thus, since the solution of the equation is unique, F(Au, a, v) = F(u, Aa, v).
Pick u = 1, A = t. This gives us the result. f) Since F(0, a, c) = c in J, OF (0, a, c) = 1. Thus (t, a, v) 36 0 in a neighbourhood of (0, a, c). Applying the inverse function theorem, we can express v in terms of u and z. (u, v) form a coordinate system in a neighbourhood of (0, c). If dv = 0 (c is given) and u = at with a = constant, then w = 0 according to the definition of F. Thus Pi(at, v)a' = 0. g) *w = L1 P,(at,v)tda' + B(at,v)dv according to the result above. Thus R, (t, a, v) = tP;(at, v). h) We have d(l!*w) = *dw = TV A W*w. Thus dR, Adai + dB A dv = Vk*0 A
Rda'+Bdv).
i=1
Since the coefficient of dt A da' must be the same on both sides, OR,
8t
=HR, and R,(0,a,v)=0.
The solution of this equation is unique according to Cauchy's theorem. Thus R, 0 and w = B(u, v)dv.
Solutions to Exercises and Problems
75
Solution to Exercise 2.47.
a) Letu=
1(u'O+v'8). Then 14u
Let a = EL1 a;dx`. Then (II*u, a) _ Thus w = E! 1 y'dC, since y' = a; at a.
u4 Z'FO. 1
a;u'.
b) We have dw = E7 1 dy n de and Q = n1 dy1 A dCI A dya n ... A dC', which is nonzero everywhere. Hence 7'"(M) is orientable.
Chapter 3
Integration of Vector Fields and Differential Forms
The first part of this chapter concerns the integration of vector fields. As a vector field is, by definition, a section of T(M), its components in a local chart are differentiable functions. So, by the Cauchy theorem, we prove that a vector field X is integrable. It defines a oneparameter local group of local diffeotnorphisms. That allows us to define Ex, the Lie derivative with respect to the vector field X, of a vector field Y or a differential form w. Instead of considering a vector field X (x) (1direction field), we can con
sider pdirection fields H. (1 < p < n) and ask the question: Do there exist integral manifolds W of dimension p? That is, do there exist submanifolds
Wsuch that Tx(W)=H., for all xE W? Frobenius' theorem states that, if a certain necessary and sufficient condition is satisfied, there exists, through a given point xo, a unique integral manifold of dimension p.
Integration of Vector Fields 3.1. Definition. A differentiable map (t, P) + cp(t, P) = Wt(P) of R x M into a differentiable manifold M is called a one parameter group of diffeomorphisms on M if a) W(0, P) = P, Wo = identity, and b) for all s and t belonging to R, Va+t = P. o Ot 77
3. Integration of Vector Fields and Differential Forts
78
cpt is indeed a diffeomorphism, since cpt o cpt = cPt o 9t = cpo A one parameter group of diffeomorphisms on M defines a vector field
on M by Xp = [dpt(P)/dt)t=o. It is the tangent vector at t = 0 to the trajectory ,y of P : t +,y(t) = cpt (P). The tangent vector at cp,(P) to ry is
_
= [dy.+u(P)] dt 141,P)l t,

{d 0 such that)  e, e[ x R C Q. We verify easily that B and a exist, since C is a compact set. Let q E N be such that Iti < qe. We can write
&(P) _ (: ore 0...o£4)(P), 4
9
9
the product of q factors £ j, which is invertible. Thus C1 is everywhere of 9 9 rank n on C. Consequently, 6 is of rank n at P, it is locally invertible at
P, and 4_f o t;_f o ... o ct 4
4
(q times).
A one parameter local group of local diffeomorphisms £t defines a vector field Xp = [det(P)/dt)t o. The converse is the goal of
3.3. Theorem. A C'''1 vector field X on M, r > 1, defines a one parameter local group of local difeomorphisms obtained by integrating the following differential equation on M. d q.
Thus 0EI. The necessary and sufficient condition of the Frobenius theorem is
for 1 0 such that jaijI < e for all (i, j) implies P(S2ij + aij) 54 0 on V. According to the continuity of Q. there is a neighbourhood W ofxn such that for x E W we have S2(xo) = dx1 A dx2 + dx3 A dx4 +
//
1152(x)  f2(x0))ijI < e
for all (i, j). Then the rank of Q1 is 2m on W fort E 10, 11. Moreover,
since d(f) !5) = 0, there exists y E A1(V) such that y(ap) = 0 and dy = SZ  fl (see 3.23).
d) We saw in a) that the kernel of g : X  i(X)SZ is reduced to {0}. Thus g is an isomorphism of Tx(V) onto T(V). We can solve the system Xjftij = yj, j = 1, 2, , n (yj being the comporents of the given differential 1form y), at each point x E V. The solution is a C°O function of the yj and S2ij, since the determinant MR AI is nowhere zero. So toy E A' (V) there corresponds X E F(V ). We do the same on the local charts of an atlas. The map y X is linear. e) The existence and uniqueness of fi(x) are given by the Cauchy theorem. Moreover, N(Put)
= atfjf2 +t5ift 0 fl)+fi*0 Cl).
According to 3.23c we find that a (ft* Q0 = f 11xrI1t + Cl  f2]. at
Solutions to Exercises and Problems
95
Since d(It = 0, it follows that Cx,f4t = d[i(Xt)flt] = dry. Thus
(f;nt)=0.
f) We have f152 = fOM = ft according to the previous equality. In the coordinate system corresponding to the local chart (f1(W),* o (fi ') ), n is the constant 2form fl.
Solution to Problem 3.25. a) We have dw = 2xdx n dy  2ydz A dx + 2xdy A dz, and
w A dw = [(1 yz)2x  x(z + x)(2y) + (1 + xy)(2x)]dx A dy A dz = 0.
Thus the equation w = 0 is completely integrable. Its rank is equal to 1 except on M. Indeed, its rank is zero when yz = 1, x(z+x) = 0,
and xy = 1. The second equation gives z = x, since x = 0 is impossible according to the third equation. So the rank is zero on the differential curve M defined by the equations z = x and yz = 1. If the rank of the map W : (x, y, z) (fi, f2) with fi = z + x
and f2 = yz  1 is two on M, then M is a submanifold of Rs. Now
the rank ofD4c=(os
is two on M, since z=Dandy=0donot
happen simultaneously on M.
b) w(Y) = a(1yz)+c(1+xy). We can choose a = 1+xy and c = yz1. Now we have to consider the system
1 = 1 + xy,
= 0,
d
zy  1.
Integrating yields y = yo, log 11 + xyo yot + Constant and log jzlp 11 = yot+Constant. So if yb 0, the integral curves of the vector field Y are given by
x=
(Ae"  1),
z=
y = yo,
(aew + 1),
A = (xoyo + 1) and A = (yozo  1). They are straight lines in the planes y = yo. If yo = 0, the integral curves are straight lines
I x+z y where A,,u, v are constants.
c) We have
dx
1+xyo
_
dz
zyo1
Thus the second first integral may be
zy  1 uxy+f
3. Integration of Vector Fields and Differential Fbrms
96
Putting z = ux + (u + 1)/y in the expression of w yields w
=
u (1+ xy)dx  x
[
x+ux +u+I
dy
+(1+xy) [udx+xdu+ u (u+1)!]
w=
,
(1 +xy)2y2[(u+ 1)dy+ydu].
We could expect this result. As u and y are first integrals, we have dy(Y) = 0 and du(Y) = 0. So w(Y) = 0 and dx(Y) 9k 0 imply that dx cannot appear in the expression of w in the coordinate system (x, y, u). d) We can write w = (1 + xy)(z + x)[du/(1 + u)  dy/y]. If (1 + xy) (z + x) 0 0, then w = 0 is equivalent to w
1
du +u
dy
y
= 0.
Cv = 0 leads to 1 + xy = k(z + x), with k E R. They are the integral manifolds of the equation w = 0. Indeed, we verify that 1 + xy = 0
(k = 0) is an integral manifold (when x = y ,w = 0). The plane z + x = 0 is also an integral manifold. Fork E R  {0}, we have to consider the map cp : (x, y, z) i 1 + xy  k(z + x),
Dip= (yk,x,k),which is of rank 1, except for k = 0 if x = y = 0. Hence the integral manifolds are submanifolds of R. Through a point (xo, bb, zo) of R3  M, there is a unique integral manifold, the one with k = (1 + xoyo)/(zo + xo). Through a point of M, there are two integral manifolds: the plane
x+z=0and the surface 1+xy=0. e) If X is of the form X = bO + c with b = 1 + xy and c = x(z + x), then w(x) = 0. Integrating the equation xdy
1+xy
_
dz
z+x
with x = xo gives at once the first integral V
z+X
97
Solutions to Exercises and Problems
Solution to Problem 3.26. a) A hyperplane H is defined by a vector orthogonal to H. Let {af} (1 s i : n + 1) be its components. If the hyperplane H is not parallel 96 0. Then we choose, as vector orthogonal to H, to Oz, then a bijection between the the vector having hyperplanes which are not parallel to Oz and the set {al, a2, . , which may be considered as the coordinates of a point of E*.
b) The c; = (8f,/8xj)dx' +(Of;/8z)dz (1 < i < q) are independent. A regular contact element will be tangent to W if 8 = a;dx' + dz is a linear combination of the df;. Suppose that the determinant IDI = 18f1 /Oxj I < q) does not vanish. Then, for every k > q,
of
D
8xk
Dk = at
...
aq
(1 : i < q, 1 < j
D =0
and b =
ak
=0. al
W belongs to the space F x E*, which has dimension 2n + 1. But among the n + 1 first coordinates only p = n + 1 q are independent: (x, z) E W. For the n last coordinates, we have p independent conditions. Indeed, ODk/Oak = IDI # 0, and at (k > q) appears only once in the determinants Dj (when j = k). Moreover, b = [DI + via' = 0. 0, at least one of the A' (1 < i < q) is not zero, say Since IDI Aj 36 0. In the Jacobian matrix of (Dq+t,...
(at,a2,... ,a,.) *
,D.,D),
we can exhibit the determinant 0
IDI
ODk Oaj
IDI
,
IDI
0
which does not vanish. The dimension of W is (p + n)  p = n.
c) When p = n, at a point (x, z) E W there is only one contact hyperplane, defined by aj = Oz/8xj. Thus dz = ajdza. When p < n, W is defined by z = ff(x) (1 < i < q) at least when the contact element is regular. And aj = AzOfjl8zi for some A with E a' = 1. Thus we verify that w = 0 on W.
98
3. Integration of Vector Fields and Differential Forms
d) The answer is no. Otherwise the integral manifold depends on d variables. For instance we would have z = f (xl,
 ,
z", ail,   , aaf ),
and dz would be expressed as a function of the dad but also as a function of some daj, which is not the case.
Chapter 4
Linear Connections
In R", there is a natural notion: the parallel transport of a geometric picture. This allows one to compare two vectors in R' which do not have the same origin. On a manifold, that notion does not existit is impossible to compare two vectors which do not belong to the same tangent space. Closely related to this is the fact that if we consider a vector field Y, we do not know
what it means to differentiate Y at a point in a direction. Henceforth, we assume that the manifold is C°° differentiable.
First Definitions 4.1. In the preceding chapter, thanks to the linear tangent map (V_t)*, we transported the vector Y,,e(p) into the tangent space Tp(M), and we defined CXY. But, as we saw, CXY = [X, Y] depends not only on Xp E Tp(M), but also on the vector field X. This is the reason for introducing a connection on a manifold. We define below the derivative of a vector field (then, more generally, of a tensor field)
at a point in a direction. 4.2. Defnftion. A connection on a differentiable manifold M is a mapping D (called the covariant derivative) of T(M) x r(M) into T(M) which has the following properties:
a) If X E Tp(M), then D(X, Y) (denoted by DXY) is in Tp(M). b) For any P E M the restriction of D to Tp(M) x F(M) is bilinear. c) If f is a differentiable function, then
Dx(fY) = X(f)Y + fDxY. 99
4. Linear Connections
100
d) If X and Y belong to f(M), X of class C' and Y of class C''+1, then DxY is in r(M) and is of class C". Recall that l'(M) denotes the space of differential vector fields (2.14). A natural question arises: On a given C°O differentiable manifold, does there exist a connection? The answer is yes, there does. We will prove in Chapter 5 that a particular connection, the Riemannian connection, exists.
So we will be able to differentiate a vector field with respect to a given vector. Then, applying Proposition 4.5, we have all connections. Let us write the covariant derivative DxY of a vector field Y with respect to a vector X E Tp(M) in a local coordinate system {x'} corresponding to a local chart (fl, gyp) with P E Q:
X = X' axi
and Y = Yl 19
axi
are n vector fields on f2 which form at each point {8/ax'} (i = 1, 2, Q E f2 a basis of TQ(M), as we saw in Chapter 2. {Xt} (1 < i < n) and {Yi} (1 < j < n) are the components of X E Tp(M) and Y E F(0). According to (c) and the bilinearity of D, . . , n)
DxY=XID;Y=X'(OYj)
+X'YJD;
where we denote Dal&: by Di and 9f /8xt by &J, , to simplify the notation. According to (d), Di(8/8xi) is a vector field on fl. Writing in the basis {a/axk}, D.
(ate;) = I';`; Oxk.
Christoffel Symbols 4.3. Definition. T are called Christofel symbols of the connection D with respect to the local coordinate system x1, x2, , x". They are COO functions in 11, according to d) (the manifold is assumed to be C°°). They define the local expression of the connection in the local chart (f), g'). Conversely, if for all pairs (i, j) we are given Di(8/axi) = I ;FHB/8xk, then a unique connection D is defined in A. 4.4. Definition. VY is the differential (1,1)tensor which in the local chart (Il, cp) has (D;Y)j as components. (D;YY is the jrn component of the vector field D;Y.
To simplify we write V;YJ instead of (VY);. According to the definition above, (D;Y)' is equal to V;Y1:
V;Yj = 4Y' + l' Yk
Torsion and Curvature
101
of a connection D are not the components of a tensor field. If I' are the Christofel symbols of another connection b, then Ckj = r s  q. are the components of a (2,1)tensor, twice covariant and once contravariant.
4.5. Proposition. T h e C h r i s t o ff e l symbols r
Conversely, if r k are the Christoffel symbols of a connection D and Cikj the components of a (2,1)tensor field. then ]: = Ck, +r are the Christoffei symbols of a connection.
Proof. Let (9, 0) be another local chart at P, {y°} the associated coordi
nate system. Let Ai' = 8}I°/8xi and B336 = 8z'/8V. If X = Xi8/8x' = Xa8/8y°', Y = Y.8/8xi = Y08/O*J, and w = wkdxk = wadyA is a differential 1form, then
Xi(V,Yj)wj = Xa(V"Y3)w'3 X`Aa[8Q(YkAk) +F AkYk[B'@wj.
Xd(ViYj)wj is a real function, so it is the same in any chart. Moreover, we can write X° = ArV, YO = AkYk and w'9 = Biwj. Since the equality above is true for any X and w, 8jYi + rJikYk = 4. [(BQYk)AL$ + Yka,,Ak + ro.AAkYkIBB.
As 0,YJ = As BaYJ, A'O
'' =
,9
Vk
(= 0 if k
j; = 1 if k = j) and since
the equality is true for any Y, we get rjk = BJOAA" + A°AkB'5r0a. k
Christaffel symbols I`iik are not the components of a tensor; in a change of local chart they do not transform like the components of a tensor, whereas 01, are the components of a tensor. Indeed, since 1 = B/S Ak + Ai° Ak B10,I4.a, we get G°ik = Aj
conversely, in SZ, the family of coo functions c9 + r k defines a connection D. Indeed, the mapping
D : (X, Y) + DXY + C,'kX'Yk Sri satisfies conditions (a), (b), (c) and (d) of Definition 4.2.
Torsion and Curvature 4.6. Definition. The torsion of a connection D is the map of r x r into r defined by
(X, Y) ' T (X, Y) = DXY  DyX  [X, Y].
102
4. Linear Connections
Let us verify that the value of T(X, Y) at P depends only on the values of X and Y at P, and does not depend on the first derivatives of the components of X and Y. We have
[T(X,Y)]' = XiVjY' YjVjX'  (X'83Y' Yj8jX') = (I'jk  r=kj)X;Yk = T9kXjyk. The operator of Tp(M) x Tp(M) into Tp(M) defined by T:
p is represented by 7?k(P), which are the components of a (2,1)tensor since contraction with two vectors yields a vector. In fl, the 7lk are C°° functions. 11
It is possible to give an alternative proof. Since 8 Ak = 8ikyO = 06e _ according to the formula in 4.5 with r and 1' we get 7;k = A,°AkB30TQa.
So the T;k are the components of a (2,1)tensor.
4.7. Definition. The curvature of a connection D is a 2form with values in Hom(r, r) defined by
(X, Y) + R(X, Y) = DxDy  DyDx  D[x,y]. For the definition we suppose that the vector fields are at least C2. One
verifies that R(X,Y)Z at P depends only on the values of X,Y and Z at P. In a local chart.
[R(X,Y)Z]k = X'DI[Y)(DjZ)k]  Y'DI[X1(DDZ)k]
 (X'BiYj  Y'BoXj)(D,,Z)k
= ()0OYj  Y'B Xj  [X,Y]j)(D,Z)k + X*Yj[8i(BjZk + rj1Z1) +rki(DjZ)']
 X'Yj [8j (8 Zk + I'k,j Z') + I4j1(Di Z )t]
.
The terms in (D3Z)k, aiZ' and BjZ' vanish. Thus we get
[R(X, Y)Z]k = X'YjZ(air
 ajr ,n + r l jm  rjtl,,1im).
Denote by Ri j the k'h component of R(8/8x', 8/8x) 0&. We have Rj,,jj = &r
 ajr + r,r  rk,,,r
.
R,ij are the components of a (3,1)tensor on 1, since after contraction with three vectors we get the components of a vector R 4ZLX'YI = [R(X, Y)Z]k.
Parallel Transport. Geodesics
103
According to the definition, R(X, Y) _ R(Y, X); thus R , = R jkj,. Since [8/84 8/8z1] = 0, if we choose X = 8/8x' and Y = 8/exi in il, we get
[(DjDjZ)k

a
(a al z = ,
(DjDoZ)k]&,k = R axi,
azi
a jz'&
Parallel Transport. Geodesics 4.8. Definition. A vector field X is said to be parallel along a differentiable curve C if its covariant derivative in the direction of the tangent vector to C is zero. Letting X(t) __ X(C(t)), we get
D fX(t) =
(t)vjx(t) =
*(t) [,%..j(t) + r k(C(t))xk(t)las = o.
Thus X(t) is a parallel vector field along C if, in a local chart,
dj(t) fi
+Nik(C(t))X (t) Wi(t) = 0.
One verifies that if two vector fields are equal to each other at each point
of C, then one is parallel along C if and only if the other is. Thus, for a vector field X to be parallel along C depends only on the values of X along C. This is why we will talk about parallel vector fields along C, even if they are defined only along C.
4.9. Definition. Let P and Q be two points of M, C(t) a differentiable curve from P to Q (C(a) = P, C(b) = Q), and Xo a vector of Tp(M). According to Cauchy's theorem, the initial value problem X(a) = Xo of the equation above has a unique solution X (t) defined for all t E [a, bJ, since the equation is linear. The vector X (b) of this parallel vector field along C (with X (a) = Xo) is called the parallel translated vector of Xo from P to Q along C. The parallel translate along a piecewise differentiable curve C = ( 1 Ci is defined in a natural way. X1 is the parallel translated vector of X0 along
the differentiable curve C1, and by induction X, is the parallel translated vector of Xi_1 along the differentiable curve Cg. X. is called the parallel translated vector of X0 along C. The definition of the parallel translate of a tensor along C is similar, once we define the covariant derivative of a tensor field.
The parallel translate at Q of Xo along C is unique according to Cauchy's
theorem, but it depends on C. That is not the case in R. For instance, on S,,, consider a unit tangent vector Xo at the north pole P, and let C the following piecewise differentiable curve: half meridian tangent to Xo, then
4. Linear Connections
104
a piece of equator AB, and finally another half meridian back to P. Any given unit tangent vector at P may be the parallel translated vector of X0 along C; it depends on B. It is possible to verify this result after Chapter 5 (Exercises 5.36 and 5.51).
The parallel translated vector of Xn along C (resp. PA) is X3 (resp. X1); X2 is the parallel translated vector of Xn along AB. 4.10. Definition. A differentiable curve C(t) of class C2 is a geodesic if its field of tangent vectors is parallel along C(t). That is to say, Ddeldtw = 0. Writing in a local chart, we find that C(t) is a geodesic if and only if d2C?(t) + rj. (C(t)) dC'(t) dCk(t) _ 0 ik dt2 dt dt
This is the geodesic equation. It is obtained from the parallel translation
equation (see 4.8) with X(t) = dC(t)/dt. But this equation is not linearit is much more complicated. We will study it in detail in Chapter 5. It is an equation of the second order. According to Cauchy's theorem,
4.11. Proposition. Given P E M and X E Tp(M), there exists a unique geodesic, starting at P, such that X is its tangent vector at P. This geodesic depends smoothly on the initial conditions at P and X. If X = 0, the geodesic is reduced to the point P.
4.12. Example. Connections on an open set 9 C R' (recall t:liat fl is a manifold). There are atlases with one chart, for instance (St, l.d). If on (fl, Id) we choose arbitrary Christoffel symbols, they define a connection. When there is more than one chart in the atlas, this procedure cannot be applied according to Proposition 4.5. If on (St, Id) we choose r = 0, the curvature vanishes. The parallel transport is the usual one. Indeed, the equation (see 4..3) is then
Covariant Derivative
105
dXJ(t)/dt = 0, so the components of X(t) are constant. The geodesic equation is then d2Ci(t)/dt2 = 0, and so geodesics are straight lines (each Ci(t) is a linear function oft).
Covariant Derivative 4.13. Definition. The definition of covariant derivative extends to differentiable tensor fields as follows:
a) For functions, Dx f = X (f ). b) Dx preserves the type of the tensor. c) Dx (u ®v) = (Dxu) ®v + u ®(Dx v ), where u and v are tensor fields. d) Dx commutes with the contraction. Let us show how we compute the covariant derivative of a differential 1form w in a local chart. Let X = X48/0x', Y = Yj8/8x' and w = wkdxk. According to (a) DX(wkYk) = X 8,(WkYk),
according to (c)
Dx(w ®Y) _ (Dxw) ®Y + w 0 (DxY), and according to (d) Dx(wkYk) = Dx[W(Y)] = (Dxw) (Y) +w(DxY).
Thus
(DXw)kYk = r(8iwk)Yk + X`WkO Yk  wk`V,Y't. Setting X = 8/8x' leads to (Diw)kYk = (aiWk)Yk  wkrijY9. Interchanging the dummy indices k and j in the last term yields, since the equality is true for any Y,
(Diw)k = 8iwk 
By definition, Vw is the twice covariant tensor having components (Vw)ik written Viwk equal to (DiW)k: ViWk = aiwk  1NkWj.
In a similar way we compute the covariant derivative Vi of an (r, s)tensor field u.
The rule is the following: Viu is the sum of the partial derivative 8iu, of a terms with 1;k where j is equal successively to the values of the s contravariant indices and k is a dummy index, and r terms with N?
4. Linear Connections
106
where j is a dummy index and k is equal successively to the values of the r covariant indices.
4.14. Examples. Let g be a (2, 0)tensor field. We have
Vi9jk = a9jk  rij9lk  IlkYjl Let Rlkij be the components of the curvature tensor. Its covariance derivative is
VmRtkij = $mRlkij  W&kij + IkR4°  I'°Rtkaj  I'OmjRtk;Q. 4.15. Exercise. Prove the following formulas (commuting of the covariant derivatives):
V,VjZk
 VjVZk =1 {'Zi  T,jV,Zk
and
ViVjwk  VjViwk = R1 jw1 lijVtwk. The first formula looks strange when we notice that, by Definition 4.7,
(D;DJZ)k  (DjD{Z)k = 1 ijZ'. The difference comes from the fact that DjZ is a vector field whereas VZ is a (1,1)tensor field,
(DiD?Z)k = &(DDZ)k + I'(DfZ)t, whereas
ViVjZk = 8c(V,Zk)+IkVjZl  IjV,Zk. Thus V{VjZk
 VVViZk = (DiDjZ)k  (DjDiZ)k  TijV1Zk,
and the first result follows. For the second result, we can compute A = V*Vj(Zkwk) VjVi(Zkwk) in two different ways:
A = &; (Zkwk)  (f81(Zkwk)  ;i(Zkwk) + I'ji8t(ZkuJk)
= 7 jVt(Zkwk), since for functions Oi f = V 1 f . And if we develop A as
A = wk(V;VjZk VjVjZk) + Zk(ViVjwk  VjViwk), the second formula follows.
4.16. Proposition.
identities are
(VAN  7 '1 k)
nijk = a(ij,k)
o(ij,k)
and
E Vj "ki = E 7 o(iJ,k)
a(i,i,k)
j,
Exercises and Problems
107
where Ea(i,j,k) means the sum on the circular permutations of i,j and k.
When the connection is torsion free, the Bianchi identities are simple. The first identity is E0(i,j,k) R;jk = 0 and the second Ea(i,J,k) OjR1% = 0. We will prove them in Chapter 5.
Exercises and Problems 4.17. Exercise. Consider on R2 the connection defined by 1'11 = xli r12 = 1, r222 = 2x2, the other Christoffel symbols vanishing. Let c be the arc on R2 defined by C'(t) = t, C2(t) = 0, t E (0, 1].
a) Compute the parallel translate along C of the tangent vector 8/8x2 at 0. b) Write the geodesic equation of this connection.
4.18. Exercise. Consider a differentiable manifold endowed with a connection.
a) Give, in local coordinates, the expression of D010 (dx') and of DXT, where T is a (2,1)tensor field, by means of the Christoffel symbols. b) Write the differential equation satisfied by a (1, 1)tensor field parallel along a differentiable curve C.
4.19. Problem. Let Mn (n > 1) be a C°° differentiable manifold endowed with a linear connection (1' are the Christoffel symbols in a coordinate system). a) We say that two vector fields that are never zero have the same direction, if at each point of M the vectors of these vector fields have the same direction. So we define an equivalence relation in the set of vector fields that do not vanish. A differentiable curve C(s) being given, we say that the vector field X preserves a parallel direction along C if there exists in the equivalence class of X a parallel vector field along C. Prove that this property is equivalent to the condition that [4:1k(s)
+l k
dCi(s)1 8
a;] 8xk
and
X
have the same direction. Write the differential system satisfied by the curve C whose tangent vectors preserve a parallel direction along C. Show that these curves C are geodesics up to a change of parameters. b) We consider on M a second linear connection whose Christoffel symbols are and we set T j = ft 1' . Determine the Tk, so that the two connections define the same parallelism: for any differentiable
4. Linear Connections
108
curve, if a vector field preserves a parallel direction for one connection,
it does the same for the other. Deduce from the above result that the necessary and sufficient condition we are looking for is 83 _ pjb.7, where the µ are the cony ponents of a differentiable 1form. c) Given a linear connection on M,,, is it possible to exhibit a connection without torsion which defines the same parallelism?
4.20. Exercise. Let M be a differentiable manifold and let D, D be two connections on M. For two vector fields X and Y we set
A(X, Y) = DXY  DXY and B(X, Y) = A(X, Y) + A(Y, X ). a) Show that a necessary and sufficient condition for D and b to have the same geodesics is that B(X, Y) = 0. b) Verify that D and D are identical if the two connections have the same geodesics and torsion.
c) Prove that for D and b to have the same geodesics up to a change of parameterization, it is necessary and sufficient that that A(X, X) be proportional to X. d) In this case exhibit a Iform w such that B(X, Y) = w(X)Y + w(Y)X.
Solutions to Exercises Solution to Exercise 4.17. a) Let X`(t) = X(C(t)) be the components of the parallel vector field X along C which is equal to C7/8x2 at 0. Then 9
1
+ r,k(C(t))
1
xk =
+X IXI + X2 = 0, dX2 dt
0.
Thus X2 = 1 and (X 1), + x1X 1 + 1 = 0 with XI (0) = 0. The general solution is
X1(t) = ec
r=
J0
z
e$du+ ke4.
Also, X1(0) implies k = 0. So r1
z
57X1
9X2
Solutions to Exercises
109
b) We have 1
1
1
2
1
dC2
+C(dCd2
t=0,
d
(a)2
= o.
2C2 dt2 + 2C2
Solution to Exercise 4.18. a) Delft: (w), with w = dx', is a differential 1form. Also, w = wkdxk,
with w; = i and wk = 0 if k
j. Thus Da18:.(&) = _r? dxk
Finally,
DxT = Xi[8,T  rmTm; 
rj,R
]dx= 0 dx' ®
C)Xk
b) The equation is DdcldtZ = 0 with Z = ZJ dx` ®818xj. That is, for
any (i,j),
Chapter 5
Riemannian Manifolds
In this chapter we will apply what we learned in the previous chapters to the study of Riemannian manifolds. When we have to study a manifold, it is convenient to consider a Riemannian metric on it. We can do this without loss of generality, since there exists a C°O Riemannian metric g on a C°O paracompact differentiable manifold M. We define a distance on (M,9), so (M, 9) is a metric space. Then we introduce normal coordinate systems at a given point. In these systems the computations are easier. We continue with the exponential mapping. Then we define some operators on the differential forms, and we conclude with the proofs of some theorems, using global notation. There are a lot of exercises and problems. Most of them extend the course itself, which is confined to the main topics.
Some Definitions S.I. Definition. A CO° Riemannian manifold of dimension n is a pair (M,,, g), where M. is a CO0 differentiable manifold and g a Riemannian metric. A Riemannian metric is a twicecovariant tensor field g (that is to say, a section of T*(M) ®T*(M)) such that, at each point P E M, gp is a positive definite bilinear symmetric form: gp(X, Y) = gp(Y, X)
for all X, Y,
gp(X,X)>0
if Xt0.
In the sequel, (Ms, g) will always be a connected C°° Riemannian manifold endowed with the Riemannian connection (Definition 5.5). If the manifold is not connected, we study each of its connected components separately. If the manifold is CI, we consider a C' atlas C'equivalent to the C' atlas.
ill
5. Riernannian Manifolds
112
5.2. Theorem. On a paraonnpact C°° differentiable manifold M, there exists a C°° Riemannian metric g.
Proof : Let (Vi,'pi),EJ be an atlas of M and {ai} a partition of unity subordinate to the covering {Y }. This partition exists, since the manifold is paracompact (Theorem 1.12). E = ((4k)) being the Euclidean metric on R" (in an orthonormal basis EJk = bJ ), we consider the tensor field
9EEla+4(E) Let us verify that g is a Riemannian metric on M. Indeed,
g(x, Y) = E aie(Spi.x, Vi*Y) = g(Y, X), iEl
according to the definition of the linear cotangent mapping and since C is symmetric. Moreover, g(X, X) = iE1 0 if X 34 0.
Indeed, at a point P, at least one a, does not vanish. Let aio(P) # 0. Then X # 0 implies (+pia.) pX 96 0. since Vi,),, is invertible. Thus
gp(x,x) >_ ai'(P)0Piofx,
.X) > 0.
By using the Withney Theorem 1.22, we can also define a Riemannian metric on Mn: the imbedding metric. If h is an imbedding M  R2n+', the imbedding metric g is g = W. In a local chart (0, gyp) at a point P E M let us compute the components of g. If {xi} is the coordinate system on R corresponding to (12, cp), let {y°} (a = 1, 2, , 2n + 1) be coordinates in (R2n+1 e)
The imbedding h is defined on 11 by 2n + 1 Cfunctions
a= 1,2,... ,2n+1.
e(xi x2,... ,xn),
The components of the imbedding metric tensor are given on Il by 2n+'
ft w _ E fta Slta $xi 8xj
or I
8xi 8rj
S.S. Definition. The length of a differentiable curve C : R D [a, b] + Mn of lass C' is defined to be
jb L(C) _
gc(t)
(
dt.
If a curve is piecewise differentiable, its length is the sum of the lengths of the pieces.
113
Some Definitions
We verify that the definition makes sense. It does not depend on the atlas or on the parameterization. If t = t(s) with dt/ds > 0, then y(s) _ C(t(s)) and C(t) have the same length. A connected manifold is path connected (Proposition 1.3). For given P, Q in a C°O differentiable manifold M, there is a COO differentiable arc from P to Q. Indeed, the continuous path ry from P to Q is compact, and is coed by a finite set of local charts (St;, Bpi), i = 1, 2, .. , m, such that 9i(Sli) is a ball of R. Let Pe be in Sli (1 R+ I. Instead of the are 7 from P to Pl in il1i we consider a C°O differentiable are Cl in Sl1 from P to A, for means the segment in R". And so where instance w on, until the are C. from P,n_1 to Q. We get a piecewise C°O differentiable arc from P to Q, the union of the arcs Ck (1 < k < m). Then it is possible to smooth the corners at the points P, to obtain a COO differentiable arc from P to Q.
5.4. Theorem. Set d(P, Q) = inf ET1 L(Cj), when the infimwn is overall piecewise C1diferentiable paths C from P to Q, C = LrQ1 Ci. Then (P, Q) + d(P, Q) defines a distance on Mn, and the topology determined by this distance is equivalent to the initial topology on M.
Obviously d(P, Q) = d(Q, P) and d(P, Q) < d(P, R) + d(R, Q), since the inf of the length of the piecewise differentiable arcs from P to Q is less than or equal to the length of the piecewise differentiable arcs from P to Q
through R. It remains to prove that d(P, Q) = 0 implies P = Q; we do it by contradiction. Let (S2, gyp) be a local chart at P with q(P)_= 0. If Q 54 P, there is a ball Br C R", of radius r with center 0, such that Br C V(fl) and Q f tp1(Br). Let us consider the map rY defined by
E1 x
3 (t,x)
E R,
where F"_1 is the unit sphere of dimension n  1 and radius 1 in R", and W;' (e). Since
34 0, it follows that g.(1;, 1:) > 0.
x is a Moreover, 0 is continuous (in fact C) and K = compact set; thus 4' achieves its minimum A2 and its maximum µ2 on K which satisfy 0 < A < p < oo and, for any f E R" and x E A2 If MI2 Ar.
5. Riemannian Manifolds
114
Indeed, dC dt
_ d(V oC) dt
and the length of V(1:) in &k" is at least r. Thus d(P, Q) > 0 if P # Q. Setting Sr = {Q E M I d(P, Q) E r}, we have, according to the proof above, SA,. C p1(Br). Likewise we prove that. 1(90 C SNr Indeed, we have
d(P, R) < L(1') < µ b I) d( dt C) II dt
for any arc r from P to R. Picking V o C to be the straight lint from 0 to V(R), we find that d(P, R) < µr. Hence the topology defined by the metric is the initial topology.
Riemannian Connection 5.5. Definition. The Riemannian connection is the unique connection with vanishing torsion tensor, for which the covariant derivative of the metric
tensor is zero (Vg = 0). Let us compute the expression of the Christoffel symbols in a local coordinate system. The computation gives a proof of the existence aad uniqueness of the Riemannian connection. The connection has no torsion; thus r 1 = r i. Moreover, let us write that Vk9i; = Vi9;k = V;9ki = 0:
ak9i3  rki91;  Iik;9, = 0, t
a;9ki  rjk91,  r,9k, = 0.
Riemannian Connection
115
Taking the sum of the last two equalities minus the first one, we find that Ili = 2 [8igkj + 8j9ki
 8k9ijI9M,
where the gke are, by definition, the components of the inverse matrix of the matrix ((gii)) : giigkj = 6 (the Kronecker symbol).
5.6. With the metric tensor of components gij, the indices go down by con
traction. For example, if 4j (1 < j < n) are the components of a vector field, then ai = giji are the components of a 1form. If Rj are the components of the curvature tensor (see below), then Rklij = gk,,,R7ji are the components of the 4covariant curvature tensor. For the tensor of components gii, the indices go up by contraction. For
example, if ai are the components of a 1form, then i = giiai are the components of a vector, and R1'i = gk.n Rkii 'j.
5.7. Definition. A normal coordinate system at P E M,, is a local coordinate system for which the components of the metric tensor at P satisfy gi (P) = 6,1 and 8kgij(P) = 0 for all i, j, k. (4gii(P) = 0 is equivalent to 1'h(P) = 0; see the proof above in 5.5.)
Let us prove that at each point P, there exists a normal coordinate system. Let (0, W) be a local chart at P with cp(P) = 0, and {x} the corresponding coordinate system. First, by a linear transformation, we may
choose a frame in R" so that 9,j(P) = 4. Then consider the change of coordinates defined by
xkyk=1F 2
%J
(P)yiy1
{yi} is a coordinate system in a neighbourhood of P according to the inverse function theorem. Since the Jacobian matrix ((8xk/8yi)) is invertible at P, it is the identity matrix. In the coordinate system {yi}, the components of the metric tensor are
iii(y) = 9ki(x)[ak 
since 8xk/8yi = &k  q93 p)%'. x and y represent the same point Q in a neighbourhood of P. The first order term in y°L of 9ij(y)  9ii(x) is [1`'im(P)+1m(P)lym=(a2M)pym
(
)Pym.
Hence (8gii /8y'") p = 0 and all the Christoffel symbols t(P) are zero. The value of a normal coordinate system will be obvious when we consider, for instance, the proof of the Bianchi identities.
5. Riemannian Manifolds
116
5.8. The curvature tensor. Consider the 4covariant tensor R(X, Y, Z, T) = g[R(X, Y)T, Z]; its components are Rijw = gi,,,Rm . It has the properties Rijjd = Rijlk (by definition), Rijkl = Rklij, and Rijkl + Riljk + &U., = 0, VmRijkl + VkRijtm + VIRij,,,k = 0. The last two equalities are the Bianchi identities. Let us prove them. Consider at P a normal coordinate system. According to the expression of the components of the curvature tensor,
I4 (P) _ (8ejk)P  (8 L'tik)P, Rijk(P) = (Sjl1ki)P  (8kfji)P,
Rjki(P) _ (8kl ij)P  (8irkj)P. Taking the sum of these three equalities, we get zero. This is the first Bianchi
identity (a tensor equality is proved when it is proved in some coordinate system). Differentiating the components of the curvature tensor at P, we find that (VmRjkl)P = ((8kws j)P  (BImI"kj)P,
(VkRjjlm)P = (8tk mj)P  (8mk1' )P, (VIRmj,,,k)P = (8lmAkj)P  (8lk1"m,)P
Taking the sum of these three equalities, we find that
VmRjkl + Vkim + VI1r,mk = 0, which is the second Bianchi identity. We put it into the form written above by multiplying it by gih. Since Vg = 0, 9ihVmRjkl = Vm(gihRfk!) = VmRhjkl
So we find that VmRijkl + VkRijlm + VIRij,,,k = 0.
The equality remaining to be proved is R{jW = Rjjj. In a normal coordinate system at P we have
Rklij(P) = 2 [Sjkga + 8it9jk  8ki9jl  8jl9ik]P.
In this expression we see that Rijkl = R. Let {tk} be the components of a vector field, and recall the following equality (see 4.15 with zero torsion):
Vivitk  V,Vi£k = Ri{jS.
Exponential Mapping
117
5.9. Sectional curvature. Ricci curvature. Scalar curvature. By definition, o ,(X, Y) = R(X, Y, X, Y) is the sectional curvature of the
2dimensional subspace of T(M) defined by the vectors X and Y, X and Y being orthonormal (i.e., g(X, X) = 1, g(Y, Y) = 1, and g(X, Y) = 0). Otherwise o(X,Y) = R(X, Y, X, Y)/[g(X, X)g(Y, Y)  (g(X, y))2]. From the curvature tensor is it possible to obtain, by contraction, other tensors? We obtain only one nonzero tensor (or its negative); it is called the Ricci tensor. Its components are Rij = Rkk_,. The Ricci tensor is symmetric:
Rtj = gkI Rtikj = g"Rkjti = gtkRkjl, = Rj,
according to the properties of the metric and curvature tensors. The Ricci curvature in the direction of the unit tangent vector X = {e} is Rij The contraction of the Ricci tensor R = Rijgij is called the scalar curvature.
Exponential Mapping 5.10. Let (U, gyp) be a local chart related to a normal coordinate system {x'}
at P; we suppose W(P) = 0. Also, let X = {Ci} 54 0 be a tangent vector of Tp(M). Let C'(t) be the coordinates of the point C(t) belonging to the geodesic defined by the initial conditions C(O) = P and (dC/dt)t=o = X. C(t) is defined for the values oft satisfying 0 < t < 0 (i3 given by the Cauchy theorem).
Then the quantity gij(C(t))
WidC (t) dC (t)
is constant along C. Indeed, the covariant derivative along C of each of the
three terms is zero: Dg = 0 and Dd l 4 = 0. Hence s, the parameter of are length, is proportional to t (s = IIXllt):
gij(C(t))
(t) Wi(t)
=gij(C(0))
(dC*(t))
(dC i=o
dt t)Jt
11X112.
The C'(t) are C°° functions not only of t, but also of the initial conditions. We may consider C'(t, Q, X) = C`(t, xl, x2, ... , xn, £1, ... , V), where the {x'} are the coordinates of Q E U and {C'} the components of X in the basis {8/8x'}. According to the Cauchy Theorem 0.37, /3 may be chosen valid for initial conditions in an open set, for instance for Q E V I (B,.) and IIXII < a (B,. C v(U) being a ball centered at 0 of radius r > 0 and a > 0). As we will consider only geodesics C(t, P, X) starting from P, we will write C(t, X) for C(t, P, X).
5. Riemannian Manifolds
118
Let us verify that C(t, AX) = C(At, X) for all A, when one of the two numbers exists. Set y(u) = C(Au, X). Then dry
_ A and
A2 d2C
d2ry
du dt du2 dt2 Thus, since C is a geodesic, ^y satisfies the geodesic equation (4.10). ry is a geodesic starting from P such that (dry/du)u, o = AX; it is C(u, AX). Hence
in all cases, if a is small enough, we may assume /3 > 1 without loss of generality.
5.11. Theorem. The exponential mapping expp(X), defined by
R"DO X  C(1,P,X)EMn, is a difeomorphism of 0 (a neighbourhood of zero, where the mapping is defined) onto a neighbourhood fl of P, f = expp 0. By definition, expp(O) _ P, and the identification of R" with Tp(M) is made by means of gyp,: X = (W.1)pX.
Proof. We saw in 5.10 that expp(X) is a COO map from a neighbourhoo of 0 E R" into M" (Q may chosen greater than 1). At 0 the Jacobian of this map is 1; then, according to the inverse function theorem, the exponential mapping is locally a diffeomorphism (on 0): f1, f2'... , £n can be expressed as functions of C', C2, . , C". To compute the Jacobian matrix at 0, we compute the derivative in a given direction X:
(fexPpAx) A=O
_
(c(1.xt))
a=o
_
(c(..t)) A=O
=X=(rp;l)pX. Let Bo(r(P)) be the greatest ball with center 0 and radius r(P) in 0. r(P) is called the injectivity radius at P, and r; = infpEM r(P) is called the igjectivity radius of the manifold. 5.12. Corollary. There exists a neighbourhood f2 of P such that every point
Q E fl can be joined to P by a unique geodesic entirely included in f2. (ft, expel) is a local chart, and the corresponding coordinate system is called a normal geodesic coordinate system.
It remains to be proved that this coordinate system is normal at P, since by 5.11 if we have Q (that is to say C', C2, , C"), then we have t 1, e, , r the components of k such that C(1, X) = Q. In this new coordinate system, {!;'} are the coordinates of Q E Q. Let C(t) _ {C`(t)} be the geodesic from P to Q lying in Q. We verify
that C'(t) = t k' for t E [0,1]this comes from the equality C(t, X) _ C(1, tX ).
Exponential Mapping
119
Q_C(1,k)a,prx(ti)
MC(t,k)C(1,tk)=.xpPix (t')
k
P
The arc length s = II X IIt = II X 11 t, so n s
9ij(Q)tto = F(S )2 = IIX112, i=1
and the length of the geodesic from P to Q is IIXII. Since C(t) satisfies the geodesic equation (4.10), d2Ck/dt2 = 0 implies
1'tJ[C(t)]C'41 = 0. Letting t  0 gives 1' (P)s:'t:3 = 0 for all (el. Then I,=I%f implies 1'k, (P)=0 foralli,j,k. 5.13. Proposition. Every geodesic starting from P is perpendicular to Ep(r), the image by expp of the sphere of center 0 and radius r in Rn when r is small enough. Ep(r) Is the subset of the points Q E f2 satisfying 1(f`)2 = r2, where t"' are the geodesic coordinates of Q. Choose an orthonormal frame of RI such that the geodesic coordinates of Q E Ep(r) are CI = r and t;2 = t;3 = _ = 0. The desired result will
be established if we prove that glt(Q) = bl for all i, because a vector Y in TQ(M) tangent to Ep(r) has a zero first component: n j ( L C' if gl,(Q) = 0 for i > 1. 12
Indeed, if y(u) is a differentiable curve in Ep(r) through Q, then we have ELl[, {(u))2 = r3 and E7_1 y'(u)d7'(u)/du = 0 by differentiation. Thus, since y(u) = 0 at Q for i > 1, that implies dyl(u)/du = 0 at Q. We
saw that g j(t;)t;'t;j _ 3 (i)2 t; = {t;'}. At Q that gives gii(Q)
= 1, and differentiation with respect to
yields
8kgi., (Q)eCj + 29gk(Q)C' = 24k.
Hence, at Q, if k 11, r8k911(r) + 291k(r) = 0,
5. Riemannian Manifolcks
120
where g,j(r) are the components of g at the point with coordinates £1 = r, e = 0 for i > 1. Moreover, I jj(r)E = 0 leads to 28rglk(r) = Okgll(r).
Thus we get glk(r) +r8,.glk(r) = 0; that is, N[rglk(r)] = 0 and rglk(r) is constant along the geodesic from P to Q. It is zero at P; hence glk(Q) = 0 fork96 1.
5.14. Theorem. A C2 differentiable curve which minimizes the distance is a geodesic. If HX N is small enough, the distance from P to Q = expp X is IIXII, which is the length of the unique geodesic from P to Q given by Corollary 5.12.
Proof. Let C be a minimizing C2 differentiable curve from P to Q parame
trized by the arc length t ([0, r] 3 t ' C(t) E C), which lies in a chart (R, cp), and let {x'} be the associated coordinate system. We will prove that it is a geodesic. Consider a family {Ca} of C2 differentiable curves from
P to Q (Ca C 11 for A E ]  e, c[), defined by Ca(t) = &(t) + V(t) with C'(0) = Ci(r) = 0 for all i, t fi(t) being Coo functions on [0,r]. Since the quantity
('
r 0
) dt
attains its minimum at A = 0, we have
(dLdAa))
xo
f [+2g,1(C(t))
2

= 0.
As we have
t d
[9<j(C(t))
dCJ
&} =
11
dCk dCj dt dt
&2
integration by parts of the second term gives 28k9t4(C(t))]
2
dC' dt
Pd2C az
 2gct(C(t))
} dt =0
for all t vanishing at 0 and T. Hence we obtain
dt
+F k(C(t))
i dC* =0. dt dt
This is the Euler equation of the problem: "Minimize L(r) for all C2 differentiable curves I' from P to Q." It is also the geodesic equation (4.10). Thus C is a geodesic.
Some Operators on Differential Forms
121
We choose V = expel. If the geodesic coordinates of Q are {l;' }, we saw that the equation of the geodesic from P to Q is {C' (t) = t(.' for 1 < i < n} (t E [0,1]), and its length is r = i=1(Z;') y(t) E M, a differentiable curve from P to Q Consider ry : [0,1] 3 t lying in 12 (y(0) = P, ry(1) = Q). Let us prove that the length of y is greater the than or equal to r. If (p,0) are geodesic polar coordinates (0 E sphere of dimension n  1 and radius 1), Proposition 5.13 implies that
with p(I(t)) =
L(7) =
i=1(ry=(t))
.
Thus
d'y' d7' dt > V9., f7(t)] dt dt 10'r
j
1 dp(7(t))
dt = p(7(1)) = r.
Hence d(P,Q), the distance from P to Q, is equal to r, the length of the geodesic from P to Q.
Some Operators on Differential Forms 5.15. Definition. Let (M,,, g) be an oriented Riemannian manifold and A an atlas compatible with the orientation (all changes of charts have positive Jacobian). In the coordinate system {x'} corresponding to (S2, gyp) E A, the differential nform q is by definition
q = 19(x)Idxl A dxz A ... A dxn, where Ig(x)I is the determinant of the metric matrix ((9,J)). q is a global differentiable nform, called an oriented volume n farm, and it is nowhere zero.
Indeed, in another chart (0, E A such that snit # 0, with coordinate system {y°}, consider the differentiable nform it = I9(y)Idyl A dy2 A . A dyn.
On f2 n 0, let us consider A? = 8y°/8x', Bp = 8x)/8ya. The Jacobian matrix A = ((A°)) E GL(Rn)+, the subgroup of GL(Rn) consisting of those matrices A for which det A = I AI > 0. We set B = ((Bk)). On f2 n 0 we have g.0(y) = and hence I9(y)I = IBI2I9(x)I. Moreover, That yields ij = I BI (9(x)IIAIdx1 A dxz A ... A dxn. Since I B I > 0 and IAIIBI= 1, we have ,' on fin 0. On an oriented Riemannian manifold Mn, we define the following operators on the differentiable forms.
5. Riermannian Maoibide
122
5.16. Adjolnt operator *. The operator * associates to a pform a an (np)form *a, called the acyoint of a, defined as Mows. In a chart (52, p), the components of *a are
(*a)v+=
1
ga u.t 'Pp
..
g p"
1,A2,... ,ay
e
See in 5.6 how the indices go up and down. Let us verify that *1=17,
*,7=1,
**a =
(1)a("P)a
and
a A (*0) = (C"MV,
where 3 is a pform; here (a, $9) denotes the scalar product of a and $9:
(a,3) We will prove the formula above at a point P. We choose a system of normal coordinates at P. Thus t7(P) = dxl A dx2 A ... A dx".
Since * is linear, we can suppose without loss of generality that a = Q1.2, ,p&' A dx2 A .. A dxp. *a has only one component which is Also,
not zero, namely (*a(P))(a+i),(a+2),...,n =
(* * a(P))1,2,...,p = rl(p+1),(F}2),..' ",1,2,... y(*o(P))(p+l),(P+2),...,n
= (1"p)(a(P))1,2,...,p. Since the exterior product is bilinear, a = a1,2,... ,pdx1 A dx2 A .. A dxP. Thus
we can suppose that
(a A (*$))p = (a(P))1,2; ..,pd 1 A dx2 A ... A dxe A[($(P))1.2,...,pdxx+1 A dx"+2 A ... A ds"l
_
(a,fA)prl
Note that the adjoint operator is an isomorphism between the spaces AP(M
and AP(M). 5.17. Codifferential J. Let a E AP(M). We define Sa by its components in a local chart as follows:
On functions, S vanishes: b f = 0. For the definition of 6, the manifold need not be orientable.
Some Operators on Differential Fbrms
123
The differential (p 1)form is called the codj femntial of a and has the following properties: on the pforms 6 = (_1)P *1 d*, 66 = 0.
The last result is a consequence of the first: bd =  *1 dd* = 0. Th prove the first result at P, we choose a system of normal coordinates
at P. Then 1d(*a)]
In
,a,.
9'._ `` µt,lrsr~ ,I+.T
1
1
V1,...
,,...,lAP
fpJ+P+1: ,v,.
8
, v9 isa permutation of pi, I3, ... N; according to the
sip of the permutation f"'P9 = ±1. We saw that * 1 = { It1.... JAq
1)
(P1)(np+i)*
on an (n  p + 1)form; hence (1)P+(p1)(np+l) (&t)n,...,Vj1
 p!(n J" p)r(n  p + 1)! 1'%AP
XfPPPrl,...,An
Since
714.
,).ft
:.. dtn 
y,,... ,VP1
and
I
Mti..,I8P
µP+1,' Jf
Jy. 1i...,l.n I

,W,

(np)(P1)fl+lr...,l'y
1
PVl *,,,& 1,
it follows that eov .~
5.18. The Laplacian operator 0 is defined by 0 = d5 + 8d.
If a is a differential pform, then Aa is a differential pform and the Laplacian commutes with the adjoint operator: A* = *A. Indeed, on differential pforms
*dd = (1)P+'d*d= (1)l+p(np+l}d[{1}(p+1)('aPi) * ' d*] * _ (1)P+i+P(nP)d*d**
_
db *
.
For a function f ,
of = sdf = vvv,,f, a is said to be closed if da = 0, codosed if Sa = 0, and harmonic if Lea = 0. a is said to be enact or homologous to zero if there exists a differential form
f such that a = dfi. a is said to be coexact or cohomologous to zero if
124
5. Riemannian Manifolds
there exists a differential form y such that a = by. Two differential pforms are homologous if their difference is exact.
5.19. Global scalar product. On a compact orientable Riemannian manifold, the global scalar product (a, /3) of two differential pforms a and /3 is defined as follows: (a,13) = fM(a,A)n
Recall that (a, 8) = v
'4. The name of the operator
b comes from the formula (da,y) = (a, 6y)
for all differential (p + 1)forms y. 6 is the adjoint operator of d for the global scalar product. Let us prove this formula. According to 5.16
f(aM)i= f aA*by=(1)x''1 f aAd*y. But d(a A *y) = da A *y + (1)Pa A d * y, and so
(a, by) = fda A *y  f d(a A *i) = (da, y) according to Stokes' formula. Using this result yields (Aa, /3) = (ba, b$) + (da, d,3).
A is an elliptic selfadjoint differential operator. If f E C2(M), then
r (Af,f) = (4,df) = (MVLfVf'7. 5.20. Proposition. On a compact oriented Riemannian manifold M any harmonic form is closed and coclosed. A harmonic function is constant.
This follows from the equality above with 0 = a: (Da, a) = (Sa, be) + (da, da).
Da=0implies be=0andda=0. Ifba=0andda=0, then &a=0 by definition. A harmonic function satisfies df = 0; thus the function f is constant.
5.21. Remark. If M is not orientable, the statement remains true. We consider a twosheeted covering manifold !tit of M such that M is orientable Or : V . V), and g = x*g (Theorem 2.30). We work on a = ,r*a, which is
harmonic. Wehave da=0andba=0;thus da=0andba=0.
Spectrum of a Manifold
125
5.22. Hodge Decomposition Theorem. Let M,, be a compact and orientable Riemannian manifold. A differential pform a may be decomposed into the scam of three differential p forms:
da = dA + Sµ + Ha. where Ha is a harmonic differential pform, dA is exact and 8p coexact. The decom;x9sition is unique. For the proof, see de R.harn [111.
Uniqueness comes from the orthogonality of the three spaces for the global scalar product:
(a, a) = (dA, dA) + (tp, bp) + (Ha, Ha). The dimension by of the space Hp(M,,) of harmonic pforms is called the
ih Betti number of M. It is finite. Since A* = *0 (see 5.18), * defines an isomorphism between the spaces H,(M.) and HRp(Mn). Hence bp(MM) = bnp(Mn). The number
XWO = E(1)pbp(Mn) P=O
bn(Mn) = 1 is called the EulerPoincarl characteristic. Clearly, (Proposition 5.20). H0(M,,) is the space of constant functions and Hn(Mn) is the space of differential nforms proportional to the oriented volume nfiorm rl. Indeed, r! is harmonic since q = *1; or we can see directly that drl = 0 and bq = 0. There is no nonzero (n + 1)form, and, since Vg = 0,

I E1,2,. ,n
V" V 191 = o.
For a connected manifold, bn = 1 or bn = 0 according to whether it is orientable or not. In general, by is the number of connected components.
Spectrum of a Manifold 5.23. The Lebesgue Integral. Let (Mn, g) be a Riemannian manifold and (11, cp) a local chart, with {xi} the associated coordinate system. For a compact support lying in It, we set continuous function f on M with cco; JM
f
dV=
191f) o V1dx1dx2 ... dx"..
For a continuous function f on M with compact support, we set
f Mfdv =
J
iEI M
aifdV,
5. Rienannian Manifolds
126
where {aj}iE1 is a partition of unity subordinate to the covering {Ri}m, (IZj,w,)iE/ being an atlas. In the sum, only a finite number of terms are nonzero, since K fl supp arj is empty except for a finite set of indices i if K is compact. We have to prove that this definition makes sense. It depends neither on the local chart nor on the partition of unity. Indeed, let (0, tai) be another chart, {ya} the associated coordinate system. Suppose supp f c SA fl 8; set
as usual i/8x' = A? and 8xj/8yJ = B'jq (see 5.15). Then (
4
Ie(x)If) o wldxldx' ... dzn
PrA
(
Ig(y)1f)o0'dyldy2...
^,
IIAiIdx'dx2...dx". since Ig(v)I = IB1alg(x)I and dy'dy2...dy"' = Consider another atlas {8j, Oj } jEJ and a partition of unity {,3j }jEJ subordinate to the covering (8j}jEJ. We have
iE1 M
ail
= JM(at[Nj)dV iEt JEJ
OjfdV,
jEJ M
since the sums are finite. Hence f  fm fdV defines a positive Radon measure, and the theory of the Lebesgue integral can be applied. We call dV = (x) I dx'dx2 ... dx" the Riemannian volume element.
5.24. Proposition. Let M,, be a compact Riemannian manifold, and w a diferential 1 form. Then fjW bwdV = 0. In particular, if f E C2(M), then
fmAfdV =0. If M is orientable, by choosing the correct orientation we have
IM
dwdV = f &ori = (6w, 1) = (w, dl) = 0.
If M is nonorientable, we consider an orientable twosheeted Riemannian covering manifold M of M (Theorem 2.30). Let 7r be the covering map
M+ M, and letg=7r*gandw=7r*w. Since M is orientable, fM &JdV = 0. !Moreover,
I &"adV = 2 J & .'dV. M
Spectrum of a Manifold
127
Indeed, let {ai}1 0, then Al > nk/(n  1).
5. Riemannian Manifolds
128
Let f be an eigenfunction: A f = A f with A > 0. Multiplying formula (*) used in 5.25, with a = c(f,, by Vj f , and integrating over Mn lead to
J f RijV`fVJfdV=1M VifV`(VjVif)dV+A JM VJfVjfdV. According to the hypothesis, the 2tensor Rij  kgij is nonnegative. Thus Rij V' f Vi f > kV' f Vi f . Integrating the second integral by parts gives
jf(if)= f
JM
V`[V'fV3Vif]dV 
=  /M VVJJ .Vjf
ju V'VJ fvjVifdV
r
thanks to Proposition 5.24 and the equality Vi V i f = V j V i f . At once we get A > k. But we have more. Since
(ViVif + Af94i/n)(V4VJf +,Afg'J/n) > 0, it follows that fV1VjfV'Vjf > (.f)'/n, becauseg;ig#j =n. Hence
al V,ffVifdV Af
n 1M
fAfdV >_ k f VfVafdV. M
Since
IM ffdV=1M V'(fVjf)dV+IArV'fVifdV _ ( V'fVifdV 96 0, M
the result follows.
5.28. Remark. The inequality of Theorem 5.27 is the best possible. Inof dimension n and radius 1, the sectional curvadeed, for the sphere ture is 1. Thus the Ricci curvature is n 1. If k = n 1, we get Aj > n, and we will see that n is an eigenvalue for the Laplacian on Sn(l) (see Exercise 5.36 or Problem 5.38).
Exercises and Problems 5.29. Exercise. An Einstein metric is a metric for which the Ricci tensor and the metric tensor are proportional at each point. If the metric is Einstein, prove that the scalar curvature R of the Riemannian manifold is constant when the dimension n > 2. 5.30. Exercise. Let (M,,, g) be a compact, oriented Riemannian manifold of dimension n > 1. When X is a C°° vector field on Mn, we associate to X the 1form a such that a(Y) = g(X, Y) for all vectors Y E T(M). a) For an oriented volume nform rt, express Lxq in terms of ba and g.
Exercises and Problems
129
b) In a local coordinate system, express the components of Aa by means of the components a; of a, VjVja,, and the components of the Ricci tensor.
c) Show that a necessary and sufficient condition for LXg = 0 is that
Viaj+Vjaj=0 for all (i,j). d) Prove that Gxg = 0 if and only if dda = 0 and Act = 2RjXjdx', where the X? are the components of X. e) Let G be the one parameter local group of local diffeamorphisms Vt(X) associated to X. Verify that Deduce from this equality that pig = g if LXg = 0. f) Let w be a harmonic pform. Show that Exw = 0 if GXg = 0. Hint. If Gxg = 0, then £x6 = 6LX.
5.31. Problem. Let (M,,, g) be a C°° Riemannian manifold with zero curvature (Ri;M = 0). Prove that there exists an atlas whose local charts ((I, cp) are such that W*E = g, where E is the Euclidean metric on R'2. Hint. Let (9,10) be a local chart at P and {x'} the corresponding coordinate system. We have to exhibit a local chart (Il, w) at P with coordinate system {ya} such that V*E = g. On 9 set 41 = I' ikjdx', and in a neighbour
hood of P set A? = 8y°/8x'.
a) Express d4 in terms of 4. b) Which system of differential equations must the 4 satisfy? c) Represent this system as a Pfaff system of order n 2 on a space of dimension p (p to be specified) and use the Frobenius theorem. d) Solve the problem.
5.32. Problem. Let M be a C°° submanifold of RP endowed with the Euclidean metric 6, and let b be the Riemannian connection on (RP, E). Denote by 4' the inclusion M C RP, and consider on M. the imbedding metric g = 4*E. Let {x'} be the coordinate system in a neighbourhood of Q E M,, associated with the chart (11, gyp), and {ya} the coordinates on RP.
a) Recall the expression of the components gij (x) of g as functions of the coordinates ya(x) of the point 4'(x), x E Q. By 4'* we identify a tangent vector of M to a tangent vector of RP (T(M,) C T(RP)). Let Y C T(RP) be a vector field defined on M,,, and let X E Show that DXY is well defined.
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b) Let irQ be the orthogonal projection of vectors of TQ(RP) onto is a vector field on M,,, prove that DxY = TQ(M,,). If Y C 1rQDXY defines a connection on M which is the Riemannian connection of (M,,, g). c) For two vector fields X and Y on M,,, we set H(X, Y) = DXYDXY. Verify that H(X,Y) = H(Y,X). d) If p = n + 1 and if M,t is orientable, show that there exists on 4b(M" ) a differentiable vector field N C T(R"+i) of unit Euclidean norm, orthogonal to TQ(M,,) at each point Q E M,,. In this case verify that
we can write H(X, Y) = h(X, Y)N, h being a symmetric bilinear form satisfying h(X, Y) = E(Y, DXN). 5.33. Problem. Let (M", g) be a connected and complete COD Riemannian manifold of sectional curvature greater than or equal to k2 > 0. We consider a closed geodesic y; that is, y is a differentiable map of the circle into M. We assume the following result: any pair (P, Q) of points of M can be joined by a geodesic arc whose length is equal to d(P, Q).
a) Let x f y be a point of M. and set d(x, y) = infy d(x, y). Show that there is a point yo of y with yo = y(uo), such that d(x, 4Jo) = d(x, y). We set d(x,7) = a. b) Let us consider a geodesic C from x to yo of length L(C) = a. Let
[0, a] 3 t , C(t) E M with C(O) = x and C(a) = yo. Do we have d(z, yo) = d(z, y) when z E C?
Ebwill be thesetofpoints QEMoftheform Q=C(1,z,X) (see 5.10 for the notations) with g,(X,X) = b2. If b is small enough, prove that d(z, Q) = b when Q E Eb. is orthogonal to (dC/dt)ta. c) Show that Y = Hint. Consider a local chart (St, exp T') at z E C close enough to
yo so that yoE0. d) Let e(t) be the parallel vector field along C such that e(a) = Y. Verify that 9C(c)(e(t), ) = 0 and that gC(e)(e(t), e(t)) = g ,,(Y, Y). e) We consider the family of curves Ca defined for A E ]  e, e[ (e > 0 small) by
[0, a] 3t'C,,(t)=C (Asin
C(t)ie(t) J.
Verify that Ca is a differentiable curve with endpoints x and a point y(ul) of y. What is the value of ul? f) We will admit the existence of a neighbourhood 9 of C and of a local chart (9, cp) such that the corresponding coordinate system is normal
Exercises and Problems
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at any point of C. We set f (A) = L(CA). Prove that
f (0) = T. gciii(v(t),
)dt
for some vector field v. Verify that f'(0) = 0. g) From the fact that f (A) is minimum at A = 0, deduce that d(x, y) < x/2k.
5.34. Exercise. Let (M, 9) be a complete C°° Riemannian manifold and C a geodesic through P: [0, a[ 9 t r C(t) with C(0) = P. a) For an increasing sequence (t, } c R converging to a, show that the sequence C(tj) converges to a point Q E M. b) Consider the map expQ. It is a diffeomorphism of a ball 0 = Bo(r) in R' onto a neighbourhood of Q in M. Verify that there exists an io such that C(t,) E 11 = expQ 0 for i > io, and prove that the geodesic C from P to Q can be extended beyond Q (for t E [a, a + e) with f > 0). c) Deduce from b) that on a complete Riemannian manifold all geodesics are infinitely extendable.
6.35. Exercise. Let (M,,, g) be a compact and oriented C°° Riemannian manifold.
a) Let ry E AP(M). Show that as = y has a solution Cr E AP(M) if and only if y E Ay, the set of the differential pforms which are orthogonal to ff. for the global scalar product. Here H. is the set of harmonic differential pforms.
b) Exhibit a map G : Ap , A. such that AG = GA = Identity on A,,. G is defined for each p (0 p < n). c) Verify that dG = Gd, OG = Gb and G* = *G. d) For two closed differential forms y E AP(M) and E A P(M), we denote by y (respectively ) the set of differential pforms homologous
to y (respectively 0). When y' E y and 0' E 0, show that f y' A 0' depends only on y' and ¢. Let (y, O) = f y A 0. e) Which harmonic forms p satisfy (cp, *cp) = 0?
5.36. Exercise. (x, y, z) is a coordinate system on R3 endowed with the Euclidean metric. S2 is the unit sphere centered at 0 in R3 and 4' is the inclusion of S2 in R3.
Let P be the point of S2 for which z = 1. When M = (x, y, z) E S2. 0 will denote the angle (Ox,OM z ), and r the length of the are PM of meridian.
5. Riemannian Manifolds
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a) What is the greatest open set 11 C S2 where (0, r) is a polar coordinate system? Express the coordinates (x, y, z) of 4'(M) in terms of the coordi
nates (O,r) of ME Il. From that, deduce the components of the metric g = 4*E on fl. b) (S2, g) being endowed with the R.iemannian connection, compute the components of the curvature tensor and of the Ricci tensor, and the value of the scalar curvature. Is the Ricci tensor proportional to the metric tensor?
c) For which values of 0 are the curves 0 = Constant geodesics? Fbr which values of r are the curves r = Constant geodesics? What are the geodesics of the sphere?
d) Prove that the function f = cos r is an eigenfunction of the Laplacian. What is the corresponding eigenvalue? Is it the first (the smallest) nonzero eigenvalue of the Laplacian?
5.37. Exercise.
a) Let (f2, 4') be a local chart of a C0° Riemannian manifold (M,g). On Sl, compute r, in terms of the determinant I9I of the matrix ((gu)). b) Consider a C°° function V on fZ such that cp(Q) = f (r) for all points Q E fl with r = d(P, Q), P being a point of Q. Find a simple expression of App. Hint. Consider a polar geodesic coordinate system at P, and
compute Vi(ghivjw)
5.38. Problem. Let S be the unit sphere in Rn+1 endowed with the Euclidean metric S, 0 its center. We consider on S, the imbedded metric Let P be the point with ong = 4'*E, 4' being the inclusion S C Rr+1.
ordinates xJ = 0 (1 < i < n), xi+1 = 1 (the x' are the coordinates on Rn+' ), and let Q be the point opposite to P on the sphere. To a point we associate the point y, the intersection of the straight x E fl = line Px with the plane it with equation xn+1 = 0: y = 4'(x). We will de. note by {x'} (1 < j c n + 1) the coordinates of x E Ri+1, and by {yk} (1 < k < n) the coordinates of y in 7r. a) Is {yk} a coordinate system on 52 corresponding to a local chart C?
b) Express the coordinates {r} of x E S) in terms of the coordinates {yk}ofyE7r. Hint. Consider the arc or =Qx and compute in polar coordinates on ?r,
c) What are the components of the metric g in the local chart C?
Exercises and Problems
133
d) If £,r is the Euclidean metric on 7r, verify that g = f (p)£,, f being a positive C100 function of p = [En_, (yk)2]
.
e) The straight lines of x through 0 are geodesics for £,r. Are they geodesics for the metric g? (Take care of the parametrization.) f) What are the geodesics of the sphere Sn? g) What is the distance r from Q to x on (Sn,g)? h) What are the components on D of the metric g in a polar geodesic coordinate system (r,9) with 0 E Sn_1? i) Let +p be the trace on Sn of the coordinate function xi±1 : V(x) = x"'1 Express rp as a function of r. j) Show that i+p = aV for a real number A to be determined. Thus 'p is an eigenfunction for the Laplacian A on (Sn, g). k) Does there exist an eigenfunction for A on (Sn, g) equal to cost r + k for some constant k? For j) and k), the result of Exercise 5.37 is useful.
5.39. Problem. Let M and W be two connected, compact and oriented differentiable manifolds of dimension n, and f a differentiable map of M into W.
a) Prove that there is a real number k such that, for all differential nforms w E An (W),
four=k
w.
Hint. Use the Hodge decomposition theorem. b) If f is not onto, show that there exists an open set 0 C W such that
f1(e) = 0. Deduce that k = 0 under this hypothesis. c) For the rest of this problem we suppose f is onto. Let Q E W be a regular value of f (the rank of (f,,) p is n if Q = f (P)). Prove that f1(Q) is a finite set and that there exists an open neighbourhood 0 of Q in W such that f 1(8) is the disjoint union of M. open sets 1 4d) Prove that k is an integer. Hint. Consider a differential nform w with suppw C 0, and compare fw w and fll, f"w for each i. e) Let Sn be the set of unit vectors in R"t1. We endow Sn C IRn+1 with the imbedding metric. Suppose there is on Sn a differentiable field X of unit tangent vectors (g(X, X) = £(X, X) = 1). Let X}(x) be the components of X in R"+1. We consider the map Ft of Sn into IZn+1 : x  y of coordinates yt = xi cos in+Xl (x) sin zrt,
5. Riemannian Manifolds
134
Verify that Ft is where the xt are the coordinates of x E a differentiable map of S" into S. We denote by k(t) the number defined in a) for the map Ft. Show that t  k(t) is continuous. f) What are the values of k(0) and k(1)? When n is even, prove that R"+1.
there does not exist a vector field on S" which is nowhere zero.
5.40. Exercise. Let (M,,, g) be a C°° Riemannian manifold and P E M. For two vectors X, Y of R", show that for any e > 0, there exists it > 0 such
that if0 1, and such that this coordinate system is normal at each point C(t) of C. Prove that the vector fields (8/&) are parallel along C. d) Let W, and Vq be two compact totally geodesic submanifolds of dimension respectively p and q such that W. n Vq = 0. Prove that there exist Po E W and Qo E V such that d(Fb, Qo) = d(W, V). By definition r = d(W, V) = inf d(x, y) for all x E W and y E V. e) We admit that there exist a geodesic C from PO to Qo (Po, Qo as in
d)) of length L(C) = r, and a coordinate system {x'} as in c). Set Fb = C(O) and Qo = C(r). Let H be the set of the vectors of TQo (M) obtained from the vectors of Tpo (W) by parallel transport along C.
For the rest of this problem we suppose p + q > n. Prove that H n TQ° (V) is not reduced to the zero vector. Let Y E H n TQ, (V) be of norm 1 (gQ(,(Y, Y) = 1).
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139
f) Consider the vector field X(t) parallel along C such that X(r) = 1'. For a chosen A E R. we define the are )a by t  Y,\ (t) = eXJ)((1) AX (t).
[0, T]
Verify that y,\ is a C" differentiable are whose endpoints are a point of W and a point, of Al. What can we say about f (A) = L(ya)?
g) Compute first f'(0). then f"(0). in terms of the sectional curvature a(X(t). ). We recall that the curvature tensor is given by 1
RAtij (C(t)) = 2 (a,,g k  c ,xg t  8j,g,k + akg )Cct)
in a coordinate system which is normal at. C(t). h) Deduce from g) that the intersection of WVp and V. is not empty if the sectional curvature is everywhere strictly positive.
5.46. Problem. Let. (Al, 9) be a compact Riemannian manifold. and consider a Riemannian cover (IV, g) of (M, g) such that TV is simply connected. If a : TV , Al is the projection of W on M, then g = lr*g. We suppose that IV has more than one sheet. that is to say, (Al, g) is not simply connected.
a) Let C be a differentiable arc of W. C = ir(C) its projection. Show
that L(C) > L(C) and that the projection of a geodesic are is a geodesic are (L denotes the length). b) Verify that. (W, g) is complete. Al be a differentiable closed curve of Al which is c) Let C : [a, b]
not homotopic to zero. Set P = C(a) = C(b) and choose a point P E 7r'(P). Show that we can construct by continuity a differentiable curve e C TV, locally diffeomorphic to C by 7r/C. such that.
C(a) = P. Why is C(b) = P not the point P? (1) If CO is a curve from P to P, homotopic to C, prove that we find the same endpoint P = Co(b). C0 being constructed from Cc, as C from
C with Co(a) = P. We set f (P) = d(P, P'). So f (P) depends on P and on the chosen homotopy class. e) For a point R of Al, we consider a differentiable curve y : [a, a]  Al homotopic to C and such that y(a) = y(i3) = R. For instance. if r is a differentiable curve from R to P. y may be a regularization of the
following curve: r, then C. then r from P to R.
From a point R E it1(R). we construct (as in c) and d)) a curve y C TV locally diffeomorphic to y by it/y` with y(a) = R. Set R' = ry(j3) and f (R) = d(R, R'). Prove that Al D R  f (R) E IlP is a strictly positive continuous map which achieves its minimum at least at one point. Q E Al.
5. Riemannian Manifolds
140
f) As above, we construct the point Q' from a point Q E 7r1(Q). Then we consider a minimizing geodesic ry" from Q to Q'. Show that 7r(5) is a geodesic from Q to Q, then a closed geodesic
(the curve is also a geodesic at Q). What is its length? We assume that any pair of points (P, Q) of a complete Riemannian manifold (M, g) may be joined by a minimizing geodesic y (that is to say, L(y) = d(P,Q)).
5.47. Exercise. Let (M, g) be a compact Riemannian manifold. Consider the metric g' on M defined by g' = of g, where f is a C°O function on M. a) Show that (M, g') is a Riemannian manifold. b) In a normal coordinate system at P E M, compute the Christoffel symbols I'; (P) of (M, g`) in terms of f. c) In a neighbourhood of P, deduce the expression of c ilk = I ; ,  I'';k where V. are the Christoffel symbols of (M, g). d) In terms of f , compute Rikii  Rt i where RI 'k4 Rik,, are respectively the components of the curvature tensors of (M, g') and of (M, g).
e) Deduce, from the result above, the expression of R;t  RJi, the difference of the components of the Ricci tensors of (M, g') and (M, g).
f) Express R' as a function of R and f, R' and R being the scalar curvatures of (M, g') and (M, g).
5.48. Problem. Let (M, g) be a Riemannian manifold with nonpositive sectional curvature (v(X, Y) < 0). Given a geodesic C : R D [a, b] 3 t , C(t) E M, w e consider an orthonormal f r a m e el, ea, , e,, of Tp(M)
(P = C(a)) such that el = (Vi)a. Then at Q = C(t) we consider the vectors ei(t) obtained from e; by parallel translation along C. We suppose that t is the are length.
a) Do the vectors ei(t) (i = 1, 2, . , n) form a basis of TQ(M)? b) Let x1, x2, , x" be a coordinate system in a neighbourhood of C such that (8/8x')Q = ei(t). What are the values of the components of the metric tensor at Q, and what can we say about the Christoffel symbols at Q? c) We suppose that there exists a coordinate system in a neighbourhood
Sl of C such that at any point Q of C we have gij (Q) = S` and q (Q) = 0. Along C, we consider a vector field X whose components
X'(t) written for X'(C(t)) satisfy
c2 '(t) dt2
_ R;,1(C(t))X'(t).
141
Exercises and Problems
Shaw that such vector fields J exist, and that they form a vector space E. What is the dimension of E? d) Verify that there exist vector fields J which are orthogonal at each point to the geodesic, and that they form a vector subspace of E. e) Prove that a nonzero vector field J may be zero only once. f) If X and Y are two vector fields J along C vanishing at P, show that
g(Y, X') = g(Y', X), where X' _. g) Prove that there exists one and only one vector field J such that X(a) = 0 and X(b) = Xot a given vector of Tc(b)(M). h) Consider n1 vector fields J which satisfy Ye(a) = 0 and Y (b) = ej (b)
(i = 2, 3,
, n).
Verify that n  1 differentiable functions fi(t) on
]a, b] are associated to a differentiable vector field Z along C vanishing
at P and orthogonal to C such that Z(t) = E a f;(t)Y(t). i) We define
I(Z) =J
{(z' Z')  g [R(dt'Z) Z, d] I dt.
Set X(t) = E,n2 fs(b)Y(t), and show that I(X) j) Prove that I(Z) > I(X). Hint_ Consider W (t) =
=g(X'(b),X(b)).
En f; (t)Y=(t), and establish the equality i=2
I(Z) _ f g(W(t),W(t))dt+g(X'(b),X(b)). 0
5.49. Problem. Let (Mn, g) be a connected complete C°O Riemannian manifold of dimension n, and f a map of M into M which preserves the distance: d(P, Q) = d(f (P), f (Q)) for all pairs (P.,Q) E M x M. We assume that any pair of points (P. Q) can be joined by a geodesic. ,
a) Show that the image by f of a geodesic arc is a geodesic arc. Recall that if d(P, Q) < .1(P), the injectivity radius at P, then there exists a unique geodesic arc from P to Q whose length is d(P, Q). b) If X and Y belong to W', prove that
li m[d(expp tX, expp tY)t1] = IIX  9.
c) Prove that any geodesic arc through P can be extended infinitely (that is to say, expp X exists for any X E R"). d) The image by f of a geodesic through P (X is its tangent vector at P) is a geodesic through P' = f(P). We denote by X' = +p(X) its tangent vector at P'. Show that V is a linear map of Tp(M) onto Tp,(M).
5. Riemannian Manifolds
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e) Establish that f is differentiable and that f*(g o f) = g. f is called an isometry. f) Prove that two isometries f and h are identical if there exists a point
P E M such that f (P) = h(P) and (ff)p = (h*) p. g) Define a map of I(M), the set of the isometries of M, into M x 0,,. What can we say when M is the sphere S,,? 5.50. Problem. Let (M., g) be a compact Riemannian manifold of dimension n, and X, Y two C°° vector fields on M. Consider the differential 1forms and a associated to X and Y; gijXj and ai = gijYj are their components in a local chart. We set h = lgI g, where I I is the determinant of ((gij)).
a) Show that, in a neighbourhood of a point where Y is nonzero, there exists a coordinate system {x'} such that [Y, 8/8x'] = 0 for all i. b) Deduce from a) that Gyh = jgj_n [Gyg 2D;Y'g].
c) Set t(a) _ IginGYh. Show that the components of t(a) are t(a)ij = Viaj + Vjai + abagii.
d) Verify that (Aa)j = O'Viaj+Rija', where Rij are the components of the Ricci tensor.
e) Prove that O't(a)ij = 2R;ja'  (da)j  (1  2)(d6a)j. f) Consider the differential 1form u(a) defined by u(a)i = ajt(a)ij. Compute 8u(a) in terms of a1V't(a)ij and (t(a), t(a)). g) Show that Lxh = 0 if and only if
(n1)+ ` 1 n) 64 = where S(k)i = nRij£j. h) Prove that Lxg = 0 if and only if
V2Vitj+RijX° =0 and Deduce f om the result that if GXh = 0 and if the i) Compute scalar curvature R is constant, then
(n1)&5 =R5C. 5.51. Exercise. Let us consider the unit sphere S C R"+1 endowed with the canonical metric g = i'E (i the inclusion), and let C be the geodesic of S through x E S with initial condition (x, v), v belonging to TT(S ). Consider the orthogonal symmetry u with respect to the 2plane P defined
by x and v in R'. a) Prove that C is included in P.
Exercises and Problems
143
b) Deduce that the geodesics of the sphere are the great circles. c) What are the geodesics of the quotient P,,(R) of S. by the antipo
dal map? Show that they are periodic with period r if they are parametrized by are length.
5.52. Exercise. Let us consider the unit sphere Sea+1 C Cn+1 endowed with the canonical metric g. The complex projective space P,, (C) is the quotient of by the equivalence relation R (see 1.41); let q: Pri(C) be the corresponding projection. Let z E S2n+1 and v E T.(S2,,+1). We consider the carve C : (0, x] 3 t + z cos t + v sin t.
a) Verify that C C S27 +1 and that q(C) is a smooth closed curve in
P.M. b) Prove that there is a unique Riemannian metric g on Pn(C) such that
q9=9. c) What is the length of q(C) in (Pn(C), g)?
5.53. Problem. Let M be a C°° Riemannian manifold and let C be a geodesic from P to Q ((0, r] 3 t  C(t) E M). According to Problem 5.41, there exists a coordinate system {x'} on a neighbourhood 0 of C which is normal at each point of C (the coordinates of C(t) are (t, 0, 0, ..., 0)). For IAI < E, let us consider a family {Ca} of differentiable curves in 6 defined by the C' maps ]0, r] x (E, E) 9 (t, A)  xf (t, A) E 0 with xi (0, A) = 0,
z'(t,0)=Iandx'(t,0)=0fori> 1.
a) Verify that the first variation of the length integral of Ca is zero at
A=0. b) Prove that the second variation of the length integral of CA at A = 0 is
a)1
=
f{fl()2 + Ri1 (C(t))v`(t)i(t), ,
= 0,
Jwhere
7/(t) = (&x'(t, A)/8A)a=o
5.54. Problem. Prove that a compact orientable Riemannian manifold of even dimension with positive sectional curvature is simply connected. Hint. Use the result of Problem 5.46 and the expression of the second variation of the length integral found in the problem above. 5.55. Exercise. Let (MM,g) be a Riemannian manifold, endowed with the
Riemannian connection D, and {x'} (i = 1,2,... , n) a normal coordinate system on P E M corresponding to the local chart (1Z, tp). In Rn, with coordinates {w'}, we consider the ball B, of center 0 E Rn and ra
dius e > 0, and a C°O map ¢ of BE into i2 (with 0(0) = P) such that t + ¢(t, 0, 0, .
, 0) is a geodesic on M. Define z = (ws, w+, ... , W"), and
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let Z(t, u, z) = 8,0(t, u, z)/8u be the tangent vector at ¢(t, u, z) to the curve u 0(t, u, z) and T(t, u, z) = 80(t, u, z)/8t the tangent vector to the curve
t+O(t,u,z). a) Compute the bracket [Z(t, u, z), T(t, u, z)].
b) Set X (t) = Z(t, 0, 0) and Y(t) = T(t, 0, 0). Show that DyDyX = R(Y, X )Y.
c) Prove that g(X (t), Y(t)) = at + b, a and b being two real numbers.
5.56. Exercise. Let M" be a compact Cl* orientable manifold of dimension to > 1. We consider a Riemannian metric g on M" and g E A"(M) defined in a coordinate system by
q=
dxlAdx2A...Adx",
where IgI = det((g;,)).
a) Verify that q is a differential reform on M. b) We endow M" with a linear connection whose Christoffel symbols are Ft,. Show that a necessary and sufficient condition for Vq = 0 is that
Ni
= 1A 109 191.
c) Does the Riemannian connection satisfy this condition? d) We suppose (here and for e)) that the manifold is endowed with the Riemannian connection. Let X be a vector field. Show that
Dix' =
18, (X'
I9I).
e) Let a = i(X)q. Compute da and the value of fm VZX`q. 5.57. Exercise. Let M,, be a complete Riemannian manifold of dimension n > 1. For X E S,,1 we define
p(X) = sup{r E V I d[expp(rX), P] = r}, where
{X E R"1 I IXI = 1} and P is a given point.
a) Show that X + u(X) is a continuous map of S,,_1 into I = R+ u {+oo}, endowed with the usual topology. b) Prove that for M" to be compact, it is necessary and sufficient that
p(X) be finite for all X E S1. c:) If M,, is compact and has nonpositive sectional curvature, show that,
given a pair (P, Q) of points of M, there exist only a finite number of minimizing geodesics from P to Q. Then show that the set Cp = {expp[p(X)X] I X E S1} is the union of a finite number of manifolds of dimension n  1. We will assume (or prove) that, when Q E Cp, if there is only one minimizing geodesic from P to
Solutions to Exercises and Problems
145
Q = expp Y then X . expp X is singular at X = Y. A geodesic from P to Q is minimizing if its length is equal to d(P, Q).
Solutions to Exercises and Problems Solution to Exercise 5.29. By hypothesis ;j(P) = f(P)gij(P). Contracting this equality gives R(P) = n f (P) (9ij9ki = dk; if we do i = k = 1, 2,..., n, we find that gijg'j = n). On the other hand, contracting twice the second Bianchi identity (5.8) yields (we multiply it by gjm, then by gi n. Moreover, by definition g(X, Y) = E(X, Y). For a vector field Z on M, applying Dz to this equality at Q gives (Dzg)(X, Y) + g(DzX, Y) + g(X, DZY)
= E(DzX, k) + E(X, DzY),
since Dz = Dz on functions. But E(DzX,Y) = E(DzX,Y), because Y is orthogonal to DzX  DzX. Hence Dz9 = 0. c) We have
H(X, Y)  H(Y, X) = DXY  DyX  (DxY  DyX) = [X, Y]Q  [X, Y]Q = 0 (we saw this result in b)).
Solutions to Exercises and Problems
149
d) Let (Ilj, cp;);E j be an atlas for M compatible with the orientation, and {x; } the associated coordinate systems. At each point Q E M. we choose N such that the basis (8/8x; , , 8/8x°, N) is positive in R"+1 This definition of N makes sense; it does not depend on i. As E(Y, N) = 0, we have £(DXY, N) + E(Y, DXN) = 0. The result follows since £(DXY, N) = 0.
Solution to Problem 5.33. a) y  d(x, y) is a strictly positive and continuous function on a compact set y. Thus there exists yo E y such that d(x, yo) = d(x, y) = a > 0.
b) We have d(z, y) > d(z, yo) for ally E y, as otherwise the are xz U zy would be shorter than the arc ayo. Thus d(z, yo) = d(z, y).
If b is small enough, Eb C f2, the open set of the local chart (11,exps 1). We know that there is a geodesic from z to Q of length d(z, Q), and furthermore there is the geodesic of the exponential map which is included in f2. Its length is b = g=(Jf',X). If the first geodesic is not the second, the first geodesic must go outside f2 and its length would be greater than b (see 5.10). c) Define Sb = {P E M I d(z, P) < b = d(z, yo)}. Then y f1 Sb = 0; thus is tangent to Eb. Moreover, according to Proposition 5.13, (4 )t=a is perpendicular to Eb. The result follows.
Y = ()
d) We have Ddc/dtEgc(t)(e(t),
)] = 0 and DdC/dt[gc(t)(e(t),e(t))] = 0,
since Dg = 0, Da/dt4 = 0 and Ddc/dte(t) = 0. Thus both expressions are constant along C, so they are equal to their values at yo.
e) The maps t  Asin
,
t  C(t) and t y e(t) are differentiable.
Moreover, (u, Q, X)  C(u, Q, X) is differentiable, so the result follows. Also, Ca(0) = C(0, x, e(0)) = x and CC(a) = C(A, yo, Y) _ y(ul), a point of y. Since
dC(Adyo,l')),\_o =1,,
A
both geodesics are the same. Thus ul = no + A. f) We have
= J0 iloc
a (:)
dt
(dC A dCA)f()
dt
On C (A = 0) we know that the square root is constant, so it is equal to 1 since f (0) = a. We can differentiate under the integral sign with
5. Riemannian Manifolds
150
respect to A. Moreover, the first derivative of gU on C is zero. Hence
f'(0) = with v(t)
N
9c{:)
J
(30
)
(v(t),
dt
since we can permute the derivatives
(the function is C2). Now
aA COQ\W) a=o
sin
ste(t) 2a
(see 5.10). Hence
v(t) =
cos
be(t) + sin DFe(t) = ' cos2a't e(t).
The integral vanishes since e(t) is orthogonal to dc. It is obvious that f'(0) = 0, according to the definition of yo. g)
Since f (A) is minimum at A = 0, we have
2 dal
> 0.
,
i
a=o
Now
f"(0) =
dt
2J
iii' sine
2a7rt
+ ( 2a)2 /oaco62 gq(C(t))e`(t)&(t)dt.
But R;k't = 2gij on C, and, moreover, ra Coe2
27rt
a dt = To am2 2a
2.
Thus
0 2. Indeed, the Ricci curvature of the sphere S2 is 1 (9ijC1 i = 1 implies
1).
Solution to Exercise 5.37. a) We have
rilir = 9ij(8i9jk + 8k9ij  8j9ik) = 2049ij. Moreover, 8kI9I _ 19194ek2ij, and hence
I
rsik = 28k log 191 =
"k 10919 V(1_
b) We have
A = 9ijViOj+P = Vi(9"w) = 8i(g"O,p) + r{ikgkj8j,p, since Vg = 0If cp(Q) = f (r), then, in a geodesic coordinate system, 8jW = 0 when j # r. Since g' = 0 and g''" = 1,
aV = f" + f'Or log Solution to Problem 5.38.
191.
a) IF : x  y is a bicontinuous bijection of 0 on ir. (S2, *) is a local chart C, and the corresponding coordinates are {yk}.
b) We have e+z =
cosa, a.=, (x')2 = sin2a and (y')2 = p2.
Since xi = kyi (1
i < n) for some real number k > 0, it follows that
5. Riemannian Manifolds
154
sin2a = k2p2. Thus k = :15. Considering the two triangles POy and PxQ yields
_ P
2 sin i 2sin Ee
_ tan2a
sin a i _
2
1+p2y
P
for1 1. Thus at u = 0 the equation above yields r (C(t))t it j = 0, for all E E Rn1 and
all 1 < k < n. This implies q(C(t)) = 0 when 2 < i, j < n and 1 < k < n. So the Christoffel symbols vanish on the geodesic C.
Solution to Problem 5.42. a) The rank of 4, is 2. Indeed,
(a+bcosv)sinu bsinvcosu D4,=
(a+bcosv)cosu
bsinvsinu
0
bcosv
The 2 x 2 determinants of DO are D1 = (a + b cos v)b sin v
0
if v 3khir (h E Z);
when v = hr 1)2= b(a + b cos v) sin u cos v 0 0
if u 54kir (k E Z);
and when u = k7r and v = h7r D3 = (a+bcosv)bcosvcosu 9k 0.
b) If (u, v) ti (fi., v), then 4'(u, v) = tb(u, v). Thus we can define from V,. Also, 4, is differentiable of rank 2 like 4,. Moreover, .W is injective. Indeed, suppose 4,(u, v) = 4,(u, v). Then the equality
x2 +y2 = (a+bcosv)2 = (a+bcosb)2 implies a + b cos v = a + b cos v, since a > b. Then cos v = cos v and sin v = sin v imply v = v + 2hir (h E Z). Thus u = u + 2kir (k E Z). Finally, 4 is an imbedding since i/, is proper, C x C being a compact set. c) We have x2 + y2 = a2 + tab cos v + b2(1  sine v). On M = 4,(III2) P(x, y, z)
_ (x2 + y2  a2  b2 + z2)2  4a
2(b2
 z2) = 0.
We can write P in the form P(x,y,z) = (x2 + y2 + z2 + a2  b2)2  4a2(x2 + y2).
If P(x, y, z) = 0, then r2 + z2 + a2  b2 = tar with x = r cos u and y = r sin u, since a > b. But (r  a)2 = b2  z2 implies z = b sin v and r= a+ b cos v. Thus M= P1(0).
5. Riemannian Manifolds
160
d) DP is of rank 1 on M. Indeed, if Q = x2 + y2 + z2  a2  b2, then DP = (4xQ, 4yQ, 4zQ + 8a2z). Since Q = tab cos v, Q = 0 implies z2
= 0, and then V= 8a2 z
If Q 96 0 and z j4 0, then
9& 0.
# 0. As (0, 0, 0) 0 M since P(0, 0, 0) = (a2  b2)2 > 0, it folknvs
that DP is of rank 1 on M, and M is a submanifold of R3. t' is differentiable and injective. To r/' there corresponds 0: C x C + M, which is differentiable and bijective. Since D4; is of rank 2, 4i is locally
invertible and its inverse D4' is differentiable. M is diffeomorphic to C x C. e) The components of the metric g in the coordinate system (u, v) are
(,4)2
(,)2+ Ox 8x
= (a+bcosv)2,
8y Oy

49z $z
(,,Z)2 =
(OX)2 +
b;.
The Christoffel symbols that may be nonzero are those in which Bog, appears (those with one v and two u's):
(a+bcosv) b
sin v
and
r:.
bsinv a + b cos v
f) We have R
_
2g Rte
_
2 cos v
b(a + boos v)'
because (see 4.7)
bcosv
_
b2sin2v
a+bcosv + (a+bcosv)2 Since
62sin2v
bcasv
(a+boosv)2 = a+bcosv'
igj = b(a +bcosv), 2x
JM
RdV=4zr
1cosvdv=0.
g) The differential equations satisfied by the arcs of geodesics are
d2u = dt2
du dv (I""e+ I"`°")dt dt
2
b sin v du dv a + b cos v dt dt
and
;2  ru
(du )2 _ (a+bbcoev) sine Idu)
Solutions to Exercises and Problems
161
The equation of a plane through Oz is u = uo. This implies du = 0, and the equations have a solution u = uo, v = vo + At. The intersections of planes through Oz with M are geodesics. The equation of a plane orthogonal to Oz is z = zo; thus v = vo. Then u = uo + At is a solution of the first equation, but the second then implies sinv = 0. Hence v = kir (k E Z). The geodesics which are in a plane orthogonal to Oz are those which are in the plane z = 0. h) If d" = 0 at (uo,ve), then u(t) = uo and v(t) = vo + At is a solution of the equations. When initial conditions are given, the solution is unique according to Cauchy's theorem. We know the solution with initial data (uo, vo, 0, A). For any A, whatever t, du = 0. Thus if du 96 0 at some point, then du does not vanish.
i) The geodesics whose equations are is = uo, v = vo +,ut are those which are in a plane through Oz. For the other geodesics, du never vanishes and we can choose is as parameter. Let v' and v" denote 9Z
and d respectively. We have dv dt
d du dt'
d2u dtz
2
b sin v
a + b cos v
v'
(du) 2 dt J
'
=v"(dt)z+v a2.
d2 Hence we find that (E)
bsinv (v)z
a+bcosv
b
v" always has the same sign as  sin v. j) 0(0, 0) = P is the point of coordinates (a+b, 0, 0) in R3. The geodesic
for which (d')p = 0 has for equation v = 0, is = At; it is the circle of radius a + b centered at O in the plane z = 0. If we substitute v for v in (E), (E) remains satisfied. Then (u/u, v/  v) is a symmetry with respect to the plane z = 0; it permutes the geodesic for which (d') p = a with the one for which (dv) p = a.
k) When u'0we have va(u)Nauand
a+b
2b
z
Thus va(u) decreases from a until zero if a is small enough. On )0, ua[, va is increasing, v,, (t&,,) > 0 and va(u) < au. Then va, V'a, v." are of the order of a. Define w = lima.o(va/a); w satisfies the equation w" = °w and the initial conditions w(0) = 0, w'(0) = 1.
5. Riemannian Manifolds
162
Thus w( u) _
+b1 uJ
b
Va+bsin (V bb
w' vanishes for the first time at uo = z
n+b
If a is small enough,
then ua is close to uo and smaller than 7r/2. va(ua) is close to
n+b
1) Since we have va(0) = 0 and va(ua) = 0, by symmetry that implies va(2ua) = 0. There are three geodesics from P to Q: this geodesic,
the one with (du) p = a and the arc of circle in the plane z = 0. They have the same length 1a. We saw that, when a tends to zero, la to = 2(a + b)uo = b(a + b). We saw also that if a is small enough, then ua exists and la < (a + b)ir.
Solution to Problem 5.43. Questions (a) and (b) are solved in Problem 5.32. c) We have E(vi, Y) = 0. Since DE = 0,
ei(X, Y) = E(DXvi, Y) = E(vi, DXY). Thus k
DXY = DXY
 2 ti (X, Y)vi
As the Euclidean connection is without torsion, E(v., DXY) _ £(v1,DyX) = 4(Y, X) since E(vi, [X, Y]) = 0. Thus f (Y, X) is a symmetric bilinear form. d) We have
DXDyZ = bx[DyZ 
k
ei(Y Z)viJ i=1 k
= DXDyZ k
k
£ (X, DyZ)vi k
 E fi(Y, Z)DXVi  E[DXti(Y, Z)]vii=1
i=1
The result follows since E(vi,T) = 0 for any 1 < i < k. e) We have
R(X,Y, Z, T) = g[R(X,Y)Z,T]
Solutions to Exercises and Problems
163
with R(X, Y)Z = DX Dy Z  Dy DX Z  DpX,y) Z. Moreover,
e(DXDyZ  DybxZ  D1X,y1Z,T) = R(X,Y, Z, T) = 0. The result follows from (d).
f) Ifk=landn=2, then R(X, Y, X, y) = t(X, X)e(Y, y)  [e(X, y)12.
Choose two orthonormal vectors X, Y such that t is diagonal and R(X, Y, X, Y) = A1A2. Thus R = 2A1A2. If k = 1 and n > 2, choose a basis Xt of TQ(M) such that a is diagonal. Then R(X,, X j, X,, Xj) = A{A f. All these values cannot be negative, since two A; have the same sign or some A; are zero.
Solution to Problem 5.44. a) We have
h  h)]. Passing to the limit leads
h) = Bpi [
(cp
a = O Cxh. to d(sh)/dt
u
Necessity. If h is invariant under G, the left hand side above is zero by hypothesis; thus Cxh = 0. Sufficiency. Lx h = 0 implies Ot h = c h = h. b) We have jO*9I = enf I9I ; thus
O*h = IG*9I1/"*9 = ef9/ef I9I'In = h. Necessity. If 10 is a conformal transformation, h is invariant under G. SO 1Cxh = 0.
Sufficiency. Cxh = 0 implies 'Ge h = h. Consequently, tp*g = I'f'*9I1rn919I1/n is of the form efg.
c) We have L = Y. Since Y"
0. So we can express
0,
t as a function of the coordinates of x, in a neighbourhood of P, in a unique way according to the implicit function theorem. Then xe = (x), and x1, x2, , z"`1, t form a coordinate system on this neighbourhood. Indeed, consider the map
r:
(x1,,2,...
,
1, t) . (1pi (xo), 1pt (zo), ... 'on (xo))
defined on a neighbourhood of 0 E R. Then
(Dr)o =
1
0
0
1
... 0 Yl ... 0 y2 1
0 0 ... 0 Yn is invertible. Thus r is a diffeomorphism of a neighbourhood 9 of 0 E Rn onto a neighbourhood of P.
5. Riemannial. Manifolds
164
d) In this coordinate system, 1
£1 h. = I9I1/n 11C y9 
=
n
1911Gy 1919]
1 IgI1/n {c9_ t(9i)GY9ij)9
Let us compute gijCygij. Since Y = and Y' = 1. Then
ViY' _ .qiY' + rijYj = rin =
.
, we have Yi = 1) for i < n
29'j [O pnj + an9ij  8j9, i]
and 91CCygij = 9:3an9ij = 21',n = 2V1Y'
since g'2 = gJ'.
The equality is proved in a coordinate system; since it is a tensor equality, it is valid in any coordinate system.
Solution to Problem 5.45. a) Choose X such that IIXII = 1; in this case t is the arc length. Suppose that expp tX is defined for 0 < t < r, r being the largest real number having this property (r exists according to Theorem 5.11). The proof is by contradiction; we suppose r finite. Let {ti } be an increasing
sequence such that ti  r. Then Qi = expp tiX (i E N) is a Cauchy sequence. Indeed, d(Qi, Qj) < I ti  t j I, and {ti } is a Cauchy sequence. Since M is complete, there exists Q E M such that d(Q, Qi) . 0 when i oc. The geodesic has an end point. Q = exprX. Now we use the theorem on the exponential mapping .'A Q. There
is a neighbourhood It of Q where expel is a diffeomorphism. Since expp tX , Q when t  r, there exists an s such that expp tX E I2 for s < t < r. Now we know that there is a unique geodesic y (u ' y(u)) from expp sX to Q included in 0 (y(0) = Q and y(rs) expp sX). Because of the uniqueness. y(r  t) = expp tX. Let Y = Fu )Q; then y(u) = C(u, Q, Y) exists for any u E ]  e, e[, e > 0 smaL. Thus the geodesic C(s) extends for r < s < r+e. This is in contradiction with the definition of r. b) Let y(u) be the geodesic of W from C(t) such that
Cdulc(c)
d dtt))C(t)
Since W is compact, W is complete and y(u) extends itself infinitely.
According to the hypothesis y(u) is a geodesic of M, but there is only one geodesic with given initial data C(t), namely (d(.'(t)/dt.)C(t) Thus C and y are the same.
Solutions to Exercises and Problems
165
c) We have dC/dt = 8/8x1, so
DA (mss)
=0,
since the Christoffel symbols vanish on C. The vector fields 8/8x' are parallel along C. d) The map W x V 3 ( P , a d(P,Q) > 0 is continuous on a compact set. Thus r is achieved: There exist P,, and Q,, such that
d(P0,Q0)=r>0. e) Let X (t) be a parallel vector field along C. The norm of X (t) is constant along C. Indeed, gr(t)(X(t),X(t)) = Constant since the covariant derivatives D(cfdt of g and X(t) are zero. So the map T, (M) a X(0) + X(T) E TQ®(M) is injective; it is an isomorphism. Thus dim H = p. Since dim TQo (V) = q with p + q > n, it follows that H f) TO (V) does not reduce to zero, because p + q  (n 1) > 0. Indeed, ( )QQ is perpendicular to H and to TQo (V) according to Proposition 5.13. Thus dim(H +TQ0(V)) < n  1. f) t _. X (t) and t _. C(t) are C°° differentiable, and we also know that (H, Z)  expR Z is C°G differentiable. Thus (t, A) + rya(t) is C°° differentiable. Also, A + expQ0 AY is a geodesic of M, included in V since Y E TQa (V). Likewise A exppe AX(0) is a geodesic of M, included in W since X(0) E Tp,(W). Thus f (A) > r. g) The function A
f (A) _
9ai(1'a(t))
0
'a
d2j
ek dt
dt
is C°O differentiable. Since f (A) > f (0), we have f'(0) = 0. Choose A small enough so that the geodesic 7a lies in the coordinate system. Then yo(t) = C(t); thus ryo(t) = t and yo(t) = 0 for i > 1. However,
(&YA(t))
= X(t)
dA
X(t) = > X'
and
;=a
\ a ) c(t)
with each X' = Constant, since 9(X (t), dc) = 0 and according to c). We have
= 0 and 9+i(iu(t))
8A
1 A=O
dt dt =
1
since (8kgtt)c(t) = 0 (the Christoffel symbols vanish on C). Since A i 7,,(t) satisfies the geodesic equation, we have
(
x
_
n(t))
(d Wi(t))
Wi(t)) A=O
Cd
= U, a=o
5. Riemannian Manifolds
166
since the Christoffel symbols vanish on C. So when we compute f"(0), we can get something which is not zero only if we differentiate 9ij twice. But as dryo/dt = 0 when 1 > 1, we obtain f "(O) =
2 f O 911(C(t))Xk(t)X(t)dt. 0
Since 497k(C(t)) = 0, it follows that (191t9jk)C(t) = 0. So if l = j =1 in the expression of Rk1ij, we get
v (X(t), since
1 = Rk111(C(t))Xk(t)X'(t)
_ 28ik911X;(t)Xk(t)
X(t) and dc are unit vectors (gik(C(t))Xi(t)Xk(t) = 1).
h) If o,(X (t), dc) > 0, then
f"(0)= J a(X(t),
)dt0.
This is an elliptic equation (the Laplacian operator is elliptic) which is nonlinear: the exponent in the right hand side is greater than 1. 6.4. The Yamabe problem is a geometrical one: find a metric with constant scalar curvature. We have proven that if we look for a conformal metric, the problem is equivalent to proving that equation (3) with R' = Constant has a strictly positive C00 solution. It is a problem of PDEs (PDEs means partial differential equations).
It is easy to see that, if there are two solutions of (3), the new scalar curvatures of them have the same sign. Indeed, assume that g' has constant scalar curvature R' and g = y4/(a2)g has constant scalar curvature R. Let
us compute R' in the metric g. If we set V = yi, this is possible since 'P and y are strictly positive. Thus g' = 14/(n2)g and (4)
4(n  1)A (n  2)
+ RV, =
jeV,(n+2)/(n2)
9''(d jV,  T 8ku'). Now let us integrate (4)
Here
with respect to the metric (5)
kf
dV R' J 0'67
since f O./'dV = 0. Thus R and R' have the same sign (or are both equal to zero). If R = R' = 0, then 0& = 0; hence u,/ = Constant. The solutions of (4) are proportional. If k = R' < 0 we easily see that (4) has only one solution rG  1. Indeed, at a point P where ip is maximum we have (4 )p > 0; thus R'Vi(P). 2 > Rrp(P), and we find that V,(P)4/(n2) < 1. Now at a point. Q where V, is minimum we have(Da')Q < 0; thus < Rt'(Q), and u'(Q)4/(n2) > 1. Consequently u'(P) = Ii(Q) = 1. we find that We have proven that there are three cases, according to whether R' is positive, negative or zero, and in the negative and zero cases the solutions of (4) are proportional. In the negative case, if k and R' are constant, then 0 = (R/R')(i2)/4 is the unique solution of (4). R'V,(Q)L(^±2i
6.5. The variational method. How can we prove the existence of a solution of (3) with R' = Constant? There are several methods for solving nonlinear equations; one of the most powerful is the variational method. We have to consider, on a set A of functions, a functional I bounded from below on A such that the Euler equation of the variational problem
6. The Yamabe Problem
172
is (3). Then, {u;} (i E N) being a minimizing sequence (i.e. {u$} C A
and limi.,,, I(uj) = p, the inf of 1(u) on A), we try to exhibit a subsequence {u?} c {u;} which converges to a strictly positive solution of (3) with R' = Constant. This is roughly the idea of the method. In practice, it is not so easy, as we will see.
6.6. For some equations, we can imagine several functionals; for of hers there are. none. Here R' appears as a Lagrange multiplier. The constraint which Will give j,(n+2)/(n2) is
K(u) = N
nf
J
I uI NdV with N = n2n2
NIuI4/(ii2)uv. Indeed, Du[(u2)N/2](v) = 2 (u2) i l2uv = The functional which will. give twice the left hand side of (3) is
4(n  1) V'uQ;u + Ru2)dV. (n  2)
I(u) _

Indeed,
DuI(v) = 2
r 4(n  1)
Jv
(n  2)
V`uViv + Ruv dV.
If we formally perform an integration by parts, we obtain Du1(v)
2
j
l
(nn 1) Au + Ru J vdV. 2)
l
We have to use the fact that DI(v) is proportional to DuK(v). Thus the Euler equation is
4n1 (n  2
(6)
Du + Ru =
lrlul4/(n2)u
if the constraint is K(u) = 11N. Indeed, p is the Lagrange multiplier: multiplying (6) by u and integrating lead to I(u) = pNK(u).
6.7. The 'Sobolev imbedding theorem. Now we have to choose the set A. Assume that we choose for A the set of CO° strictly positive functions V such that IIPIIN =
(f
l/N Ic
INdV)
= 1.
It will be very difficult, even impossible, to prove that u, the limit of a subsequence (u,), is a strictly positive C°° function. On the other hand, we can write the Euler equation only if u E A. For this reason, we introduce the Sobolev space H1.
An Introduction to Research
173
Let E be the set of C°O functions on M endowed with the norm IIuIIH; = (IIVu!I2 +
IIu1I2)1/2.
The gradient is ]VuJ = (V{uViu)1/2; its L2 norm is (fv V1uV'udV)'/2 The Sobolev apace H1 is the completion of E with respect to the norm II IIH2. It is a Hilbert space. Moreover, the Sobolev imbedding theorem (see (2]) asserts that H? C LN for compact Riemannian manifolds, and that the inclusion is continuous (i.e. there exists a constant C such that any u E Hi satisfies nuIIN < Coup). To solve the variational problem, we will choose A = {u E H1, u >_ 0 1 IIu1IN = 1}, which makes sense according to the Sobolev imbedding theorem. In general, choosing a constraint like u > 0 implies several difficulties (we cannot write the Euler equation when u is zero). These difficulties are
not present here. Indeed, it is a fact that if u E H2 , then Jul E Hl and IVIull = IVul almost everywhere; we have I(u) = I(Jul), and obviously K(u) = K(Jul). Thus the inf of 1(u) on A is equal to the inf of I(u) on A = {u E H1 I IIUIIN = 1}. Therefore, the Euler equation can be written without any
technical problem, and the limit u will be positive or zero: u = Jul. Hence equation (6) is equation (3) with R' = p. 8.8. We therefore consider the following variational problem: find inf I(u) for U E A. Recall that
I(u) = 4(n  1) fm IVuJ2dV + (rt  2)
JM RuadV
and A = {uEH?,u>0I IIuIIN=1}. Let us prove that p = infA I(u) is finite. Observe that 1(u) > inf(R,0)IIu0I2 >_ inf(R,0)V2/"IIuIIN = inf(R,0)V2/",
according to Holder's inequality. Here V = fu dV; without loss of generality we may suppose that the volume V is equal to 1. Indeed, by a homothetic change of metric we can set the volume equal to 1, and a homothetic change of metric is a conformal one. Let g' = kg with k > 0; then V1
= J dV' = k"/2 J dV = k"/2V.
We have only to choose k = V2/". Henceforth we assume that the volume is equal to 1. Let us consider a minimizing sequence {i4} C A: 1(uj)  p. We can suppose I(ur) < p+1. According to Holder's inequality, IIu D2 < HwlIN =1. Moreover,
4(n
2 < Constant.  2) O Vu p2 3, this problem is equivalent to solving the following equation: (n  1) 4(n)Ocp+R = fcp(" s),
tp>0.
In dimension n = 2, the equation to solve is (see (2) with n = 2)
AV+R= fe`e. This problem is particularly hard on the sphere, where it is the socalled Nirenberg problem.
Bibliography
[11 Aubin, T.: Nonlinear Analysis on Manifolds. MongeAmpere Equations (Grundlehren 252), SpringerVerlag, New York, 1982.
[21 Aubin, T.: Some Nonlinear Problems in Riemannian Geometry, SpringerVerlag, New York, 1998.
[31 Gallot, S., Hulin, D., and Lafontaine, J.: Riemannian Geometry, SpringerVerlag, New York, 1987.
[4J Helgason, S.: Differential Geometry, Lie Groups and Symmetric Spaces. Academic Press, New York, 1978. 151 Kobayashi, S., and Nomizu, K.: Foundations of Differential Geometry. I and II, Interscience, New York, 1963. [61 Lichnerowicz, A.: Ggomktrie des groupes de transformations, Dunod, Paris, 1958. [7) Malliavin, P.: Gdometrie diffenentielle intrinseque, Hermann, Paris, 1972. [81 Milnor, J.: Morse Theory (Annals Studies 51), Princeton Univ. Press, 1963. [91 Milnor, .1.: Topology from the Differentiable Viewpoint, The University Press of Virginia, 1969. [101 Narasimhan, R.: Analysis on Real and Complex Manifolds, Masson, Paris, and NorthHolland, Amsterdam, 1971.
(11] de Rham, G.: Sur la thdorie des formes differentielles harmoniques, Ann. Univ. Grenoble 22 (1946), 135152. 1121 de Rham, G.: Varietes diffdrentiables, Hermann, Paris, 1955. [131 Spivak, M.: Differential Geometry (5 volumes), Publish or Perish, Berkeley, 1979. [141 Sternberg, S.: Lectures on Differential Geometry, PrenticeHall, Englewood Cliffs, NJ, 1965. (151 Warner, F.: Foundations of Differentiable Manifolds and Lie Groups, Academic Press, New York, 1971. [161 Yamabe, H.: On the deformation of Riemannian structures on compact manifolds, Osaka Math J. 12 (1960), 2137.
177
Subject Index
adjoint operator flJfl algebraically equivalent (systems) 311
arc U arc length &3 Aseoli's theorem 0.39 class
d,flnltion L4 equivalent L5 Banach space 0.24 Banach'a theorem 5.9 basis
23j
Betti numbers 5.25 Blanchi's Identities 413, 5.fl boundary 2.35 2.15
bundle 211
conformal metric 03 [Aconnected 03 connection 4.2 Riemannian
contraction (of indices) 022, contravariant QJA coordinates L4 geodesic 512 normal cotangent bundle 23 countable at Infinity Liii covariant (Index., tensor) covenant derivative of a tensor field 413 of a vector field 4.2 covering 0.1 covering manifold LiZ
5.8
thInner Liii aitical
theorem dependance on Initial condition 0.39
point
curvature
chart L4 Chr4.stoffel symbols 4.3. closed
form 3.15
set 03
definitIon 4.1 Ricci &9 scaler 5.2
sectIonal 5.9 tensor 5.8
system 314
set 03
diffeomorphiam ILlS
system 514
differentIable
511 compact set (LB compatible (vector flelde) 2.11 complete 0.24 completely integrable 2.15
013
functIon Lfl manifold 1.8
with boundary 2.35 113 dIfferentIal
2.25
179
I
Notation Basic Notation We use the Einstein summation conv ration. Positive means strictly positive. Nonnegative means positive or zero. Compact manifold means compact manifold without boundary unieas we say otherwiseM is the set of positive integers, n E N.
Rn is the Euclidean nspace, n > 2, with points r = (xi, x2, ..., xn), xi E R, the set of real numbers.
When it is not otherwise stated, a coordinate system (mil I 5,:S. in R" (or (z, V, x, t) in R") is chosen to be orthonormal. We often write 8+ for 8/& and Al for 8i8j. Sometimes we write V 1 for V1V1. ]a, b( or (a, b) means an open interval in R. (a, b) may also be the point of R2 whose coordinates
are a and b.
Notation Index Ck , C, C
f'(x)
CC = nl/(n  P)ipi d exterior differential. 2.24, 2.25
Riemannian metric. 5.1 9.j, g`1 the components of g 5.5 Cx Lie derivative with respect to X. 3.4 M, or Af manifold of dimension n. 1.1 (Mn, 9) Riemannian manifold. 5.1 O(n) 2.45
difermaiable function. 0.23, 0.26 manifold. 1.6
di dtial off. 0.26 dV
Rlemannian volume element. 5.23
DxY or D(X,Y)
P,,(R)
4.2
dxj 0.21, 2.25 d(P, Q) distance from P to Q. 5.4
E={xERn(x1