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ro the set V is disjoint from the boundary, V n OMk = 0. From >i vi = 1 we obtain >i dcpi = 0 and use this to rewrite the exterior derivative of k-1
=
wk-1:
k-1 i=1
-
Pidw k-1 i=1
+
dipi n i=1
wk-1
i=1
80
3. Vector Analysis on Manifolds
This implies the equation r
JM' dwk-1 =
d (hi (Piwk-1))
V is an open subset of Rk for rp + 1 < i < r, and h; is a k-form on V1 whose support is completely contained in V. For each of these indices i we choose a k-chain c in Rk for which supp h, (Wiwk-1) C Int ck C ck C V j.
Applying now Stokes' theorem for chains (Theorem 8, Chapter 2) to ck, we obtain
Jd(h(iwk_1))
wA-1))
=
f
h' (Viwk-1) = 0,
d (ht = Jask since the form vanishes on the boundary of the chain. Now we consider the indices i between 1 and ro. For any of these, V is an open subset of the halfspace Hk, and as before we obtain chains ck in lHlk with the same properties of the supports with respect to h; Applying Stokes' theorem to these as well, we obtain 1.
1v;d (hi Now hi
(,p1wk-1) does
hi (ca1wk-1)
J
.
not necessarily vanish any more on Ock n
(Rk- l x {0} ),
but only at the points of 8c; belonging to the interior of Hk: d (h;
(Vjwk-1))
_
I
VflRk-1
V;
hi
(`Piwk-l).
The pairs (V n Rk-1, h1IRk) with i = 1, ..., ro form a covering of the boundary BAIk. Hence, by the definition of the integral, ro
wk-1
-
t-1 knRlshowing the equation we set out to prove. Jamk
h i (ViwR-1)
,
1
O
In the sections to follow we will be dealing with various applications of Stokes' theorem. As a generalization of the discussion in §2.5, we will first study line integrals and prove an analogue-only for 1-forms, however-of Poincare's lemma. This holds for manifolds in which every closed path can be contracted to a point. Definition 19. A connected manifold Mk is called simply connected if every two C'-curves co, cl : [0, 1] , Mk with coinciding initial and end points are homotopic.
3.7. The Hedgehog Theorem (Hairy Sphere Theorem)
81
By Theorem 9 in §2.5, whose proof immediately carries over to the case of a manifold, on a simply connected manifold the line integral of a closed 1-form w1 depends exclusively on the end points of the curve. Having fixed a point xo E Mk, the line integral along a curve joining the points xo and x, w lox uniquely defines a function on the manifold. Its differential df coincides with w1, and we obtain f(x) =
Theorem 22. Every closed 1-form on a simply connected manifold is exact.
Example 31. The winding form defined on 1R2 - {0} is closed, but not exact. This shows that R2 - {0} is not simply connected.
3.7. The Hedgehog Theorem (Hairy Sphere Theorem) Consider two oriented compact manifolds Mk and Nk without boundary and of equal dimension. Two maps fo, fl : Mk - Nk between them are called homotopic if there exists a smooth map
F : Mk x [0,1] -+ Nk such that F(x, 0) = fo(x) and F(x,1) = fl (x). We prove
Theorem 23. Let wk be a k-form on Nk and let fo, fl : Mk
Nk be
homotopic maps. Then
JMk0 (wk) = fm k fl (wk) . Proof. The oriented manifold Mk x [0, 1] has boundary
8(Mk x [0,1]) = Mk x {1} - Mk x (0}, where the minus sign indicates that the orientation is reversed. Therefore, Stokes' theorem implies
Jf
k
fl (w) -
JMk
f(wk) =
F*(wk) = 8J(Mkx[O,1])
JMk x [OI]
dF(wk) .
But the form dF*(wk) = F'(dwk) = 0 vanishes, since the k-dimensional manifold Nk carries no non-trivial (k + 1)-forms.
0
Theorem 24. The antipodal map from the sphere to itself, A : Sn - Sn, A(x) = -x, is homotopic to the identity Ids. only for odd dimensions n. Proof. Consider on Rn+1 the form n+1
wn =
_
-1x' dx1 A ... A dx' A ... A dx"+1 i=1
3. Vector Analysis on Manifolds
82
whose restriction to the sphere Sn is its volume form dSn (Example 28, equation (17)). If A is homotopic to the identity Ids.., then the previous theorem implies
Jis.
A*(wn) = / n w" = vol(Sn) . S
The induced form A* (,n) = (-1)n+lwn is proportional to the form
n.
Thus we obtain the condition (_1)n+lvol(Sn) = vol(S"), i. e., (n + 1) has to be an odd number.
Theorem 25 (Hedgehog Theorem). A sphere S2k of even dimension has no nowhere vanishing, continuous tangent vector field.
Proof. Suppose that there exists such a vector field on the n-dimensional sphere S". We first approximate this vector field by a smooth vector field V (Stone-Weierstrass theorem), and then normalize it so that the vector V(x) has length one at each point. Next we consider the resulting smooth tangent vector field as a vector-valued function V : Sn - Rn+I satisfying the following two conditions:
(x, V(x)) = 0,
IIV(x)II = 1.
Define the homotopy F : Sn x [0, 1] - Sn from the sphere to itself by the formula
F(x, t) = cos(irt) x + sin(irt) . V(x). The length of F(x, t) is equal to one everywhere, since x and V(x) are perpendicular. Moreover, F(x, 0) = x and F(x, 1) = -x, i. e., F is a homotopy between the identity and the antipodal map of the sphere Sn. But then the dimension n has to be an odd number. In the German mathematical literature, this result is known as the "Hedgehog Theorem" (,,Satz vom Igel"), since its contents can be expressed figuratively by saying that a hedgehog cannot be combed in a continuous way. Because it is so vivid, we prefer this to the name "Hairy Sphere Theorem", which seems to be more common in the Anglo-Saxon world.
3.8. The Classical Integral Formulas Now we will discuss the classical integral formulas already treated in §2.6 for chains. Compared to the preceding case, in this new formulation we benefit
from having notions like divergence, gradient and Laplacian as developed in the differential calculus on manifolds at our disposal. At the same time, the orientation of the manifold and the exterior unit normal vector field on the boundary will play a special role. We will prove these integral formulas
3.8. The Classical Integral Formulas
83
for arbitrary compact and oriented manifolds (in R'). This will be the final formulation of the classical integral formulas as they are needed in many branches of mathematics as well as theoretical electrodynamics and hydrodynamics. We start with the Ostrogradski formula relating the divergence of a vector field to its flow across the surface.
Theorem 26 (Ostrogradski Formula). Let Mk be an oriented, compact manifold, and let N be the exterior unit normal vector field to its boundary. Then, for every vector field V : Mk Tlllk on Mk,
div(V)dMk =
(V, N)
d(OMk).
nik
lL
Proof. We know from Theorem 17 that the divergence and its inner product with the volume form are related by the formula
div(V) dMk = d(V J dMk) . A straightforward application of Stokes' theorem implies
J div(V)dMk = lk
J
d(V i dMk) = f V nl k
k
Let x E 8Mk be a point of the boundary. We decompose the vector V(x) into one part that is proportional to the exterior normal vector, and a vector W(x) that belongs to the tangent space T,ZBMk to the boundary:
V(x) = (V(x), N(x)) N(x) + W(x) . Moreover, note that the restriction of the inner product W J dMk to the boundary 8Mk vanishes identically. This implies that for the inner product of V with the volume form dMk. on the boundary 8Mk
V _j dMk = (V, N) N j dMk + W i dMk = (V, Al) N(x) _j dMk . Hence, we obtainJ
div(V)dMk = ik
f
Mk
(V, N) Ni dll'ik .
By Theorem 19, the inner product N(x) . dMk coincides with the volume form of the boundary, d(8Jik). 0 As a direct application of the Ostrogradski formula we obtain Gauss' theorem.
Theorem 27 (Gauss' Theorem). Let V be a vector field, let f be a function on the oriented, compact manifold Mk, and let N be the exterior unit normal vector field of the boundary. Then
J l (V, grad(f )) dMk + f l
k
f . div(V)dMk = J
alk f
(V, N) d(81bik) .
3. Vector Analysis on Manifolds
84
Proof. This equation immediately follows from the Ostrogradski formula together with the rule from Theorem 11
div(f V) = f div(V) + V(f) = f div(V) + (V, grad(f)) . In a similar way we derive Green's formulas in versions that are not confined to 12. First we generalize Green's first formula.
Theorem 28 (Green's First Formula). Let f, g : Mk
R be smooth func-
tions on the compact, oriented manifold Mk. Then
f
f'O(g) dMk +
g'ad(g)) dMk =
ff
(grad(.9)N) d(8Mk).
aMk
A1k
Proof. By the definition of the Laplacian, we have
f O(g) dMk = J %rk
f
'k f div(g'ad(g)) dAik .
Now apply Gauss' theorem. Applying Green's first formula twice leads to Green's second formula.
Theorem 29 (Green's Second Formula). Let f, g : Mk - III be two smooth functions. Then 1 [g.
(f)-f. (g)] dMk = f
N) ] d(aMk).
aMk
Ark
1-1
Remark. The scalar product (grad(f ), N) defined only on the boundary is
iaN aN
often denoted by the symbol Of ION, since it is the derivative of the function
f in the direction of the exterior normal vector. This leads to a different formulation of Green's second formula:
f [g
(f) - f . %(g)] dMk =
.
fMk
[g.-f.
]
Corollary 1. Let Mk be a compact, oriented manifold without boundary. Then (1)
f
div(V) dMk = 0 for every vector field V;
Mk
(2)
f go(f) dMk = f fo(g) dMk = - f (grad(f ). grad(g)) Alk
Mk
dMk
Mk
for any two functions f, g E CO°(Mk).
0
3.8. The Classical Integral Formulas
85
Hence the Laplacian is symmetric with respect to the L2-scalar product. Moreover, the choice of sign we adopted implies that it is non-positive. Corollary 2 (Hopf's Theorem). Let Mk be a compact, connected, oriented manifold without boundary and assume that the function f : Mk R sat-
isfies at each point the condition 0(f)(x) > 0. Then the function f is constant.
Proof. Integrating the assumption s(f)(x) > 0 over Mk and applying the symmetry of the Laplacian just proved, we first obtain
0< f 1-o(f)-dMk = J f-j(1)-dMk = 0, Mk fk
i.e., the Laplacian of f vanishes identically, A(f) = 0. Inserting f = g into Green's first formula (Theorem 28) and taking into account that the boundary integral vanishes by the assumption concerning Mk, this implies
J
grad(f)I2-dMk
= -Jntkf
0,
and hence grad(f) = 0. Thus f is constant, since Mk is connected.
0
Concluding this section, we formulate Stokes' theorem in its classical form on 1R3. Contrary to the preceding theorems involving the generalizations of divergence and gradient to manifolds as introduced in the second section of this chapter, this only involves the notion of curl on open subsets of 1R3 from §2.3.
Theorem 30 (Stokes' Theorem-Classical Version). Let M2 C R3 be a compact, oriented surface, let V be a vector field defined on an open subset M2 C U C 1R3, let N : M2 S2 be the exterior unit normal field to the surface M2, and let T : aM2 - T(aM2) be the unit tangent vector field on the boundary curve aM2 with the induced orientation. Then
f (curl(V), N) dM2 = f MZ (V, T) d(0M2). Proof. Consider the 1-form 4 := V1dx1 + V2dx2 + V3dx3
on U associated with the vector field V = V la/axl + V2a/axe + Via/ax3 as explained in §2.3 and its derivative
d`4 =
av2
aV1
axl - axe
,
I dx ndx2 +
, aV3 aV2 J dx2 ndx 3. ax, - avl ] dx ndx3 + [ ax3 ax2 - ax3
aV3
3. Vector Analysis on Manifolds
86
Recall that the curl of V corresponds to the 1-form *dwv. If, on the other hand. It : W -+ M2 is a parametrization of the surface with components h', h2, h3 and coordinates yl, y2 from W, then for the exterior normal vector N to the surface we have the relation 09
h x 09h
yl
/ II A
y2
09y1
y2 x ah
ll
Two arbitrary vectors v, w E R3 satisfy the identity 1Iv
x wII2 = det
(V, V)
(v, w)
(v, w) 1 (w, w) J
and hence the preceding equation implies
f.
x
ayl aye II For the first component of the normal vector N written in the coordinates II
y1, y2 this reads, e.g., as
N'dM2 = N' f dy' n dye =
ahe Oh3 _ah2 ah3 , dy' A [ay' aye aye ayl
1
dye .
On the other hand, it is easy to compute the pullback by h of the forms dx2 and dx3:
h*(dx2) =
OhY12
+ A2dy2, h`(dx3) =
ldy' + Oh3dy2.
A direct comparison implies the following formula, which is independent of the coordinates y', y2:
N'dM2 = dx2 A dx3 . Similarly one proves
N2dM2 =
- dx' A dx3,
N3dM2 = dx' A dx2 .
The scalar product of the curl of V with the unit normal field N multiplied by the volume form dM2 is thus simply the differential of wv: (curl(V), N) dM2
aV' 2 = [ aV3 ] N dM + 109x3 axe 09x3 = L aV2
1
dwv.
aV2
9V3
09x1, N2 dM2 + 09x1 - axe, N 3dltil 2 L aV'
Therefore, Stokes' theorem can be applied in the format
J (curl(V),N) dM2 = %f2
J Af2
&4 =
y=J[V1dx1 8M2
8M2
+ V2dx2 + V3dx3].
3.9. The Lie Derivative and the Interpretation of the Divergence
87
If, however, T is the unit tangent vector field to the curve 9M2, then d(8M2)(T) = 1, and hence
T'd(aM2) = dxl, T2d(8M2) = dx2, T3d(8M2) = dx3. Now we can rewrite the line integral above as
J [V ldxl + V2dx2 +V 3dX3]
= J1L12
(V,
T) d(8M2),
and, summarizing, we arrive at Stokes' classical integral formula.
0
3.9. The Lie Derivative and the Interpretation of the Divergence The aim of this section is to interpret the divergence of a vector field geometrically as the infinitesimal volume distortion of its flow. First we recall some results from the local theory of ordinary differential equations and introduce the flow on a manifold as well as the Lie derivative of forms. Then we compute this Lie derivative by means of the exterior derivative, which, in a special case, leads to the interpretation of the divergence mentioned in the title.
Let V be a vector field on the manifold Mk. An integral curve of V is a Mk whose tangent vector -y(t) = ry,(8/8t) at each point curve ry : (a, b) coincides with the value of the vector field there:
?(t) = V('Y(t)) The well-known existence theorem for autonomous differential equations states that for every initial point x E Mk there exists a maximal integral
3. Vector Analysis on Manifolds
88
curve ryx
:
(ax, bx) - Mk defined on an interval, containing the number
0 E R, satisfying
'Y.,(0) = X. Moreover, this maximal integral curve is uniquely determined by the initial condition. Denote by EV the set
EV = {(t,x)E]RxMk:ax) = 4).,m[Vl, V2]m(9)
,
i.e., the vector fields [WI, W2] and [VI, V2] are, in fact, 4)-related.
The commutator of two vector fields measures the extent to which their flows do or do not commute. This explains the name for the vector field [V, W].
Theorem 36. Let V and W be two complete vector fields on the manifold Mk and denote by 4't and X8, respectively, their flows. Then the commutator [V, W] vanishes if and only if 4it o 4' = %P, o tt for all -oo < s, t < oo.
Proof. Because d
= tim (4'-h-h).W - (4'-t1) / = (4'-t1).([V, W]),
the commutator [V, W] vanishes if and only if the vector field W is invariant under the flow 4it, (4)1).W = W. This condition is in turn equivalent to the commutativity 4't o T, = T. o 4it of the diffeomorphism 4't with the flow ID, of W.
3.10. Harmonic Functions A function f : Mk - R is called harmonic if it is a solution of the homogeneous Laplace equation 0(f) = 0. As a special case of Hopf's theorem (Corollary 2), we have the following:
Theorem 37. Every harmonic function on a compact, connected, and oriented manifold without boundary is constant. If the boundary of the manifold Mk is not empty, there exist two particularly important boundary value problems for harmonic functions.
3.10. Harmonic Functions
05
The Dirichlet Problem: Assume that a function V : OMk - IR is given. We IR whose values on the boundary look for a harmonic function f : Mk aMk coincide with those of cp:
0(f) = 0 in Mk and
f IaaMk = cp .
The Neumann Problem: Assume that a function cp : OMk -+ IR is given. We R whose normal derivative on the boundary coincides with yp:
look for a harmonic function f : Aik
0(f) = 0 in Mk and eiv
= cp on aMk.
A solution of the Neumann problem is never unique. For each solution f the sum f + C is, for an arbitrary constant C, a solution of this problem, too. This is the only degree of freedom, since we have
Theorem 38. Let Mk be a compact, connected, and oriented manifold in R", and let cp : aMk -+ IR be a smooth function.
(1) The Dirichlet problem has at most one solution.
(2) If f,, f2 both are solutions of the Neumann problem, then f, - f2 is constant.
(3) The vanishing of the mean value of cp is a necessary condition for the solvability of the Neumann problem: JOMk
p d(aMk) = 0 .
Proof. If fl, f2 both are solutions of the Dirichlet problem, then the difference u := f, - f2 satisfies the equations
0(u) = 0
and u IBMk = 0.
From the first Green formula we obtain
0 = J fk
-Lk
U. 85Jk
a" aN Nd(aMk),
and hence the gradient of u vanishes. Since we have u IBMk= 0, the function u has to vanish identically. In the case of the Neumann problem the argument runs along the same lines. i.e., starting from the equations
0(u) = 0
and
au
= 0 on aMk
and the Green formula, we again conclude that grad(u) = 0. Thus the difference u = fl - f2 is constant. If the Neumann problem has at least one
3. Vector Analysis on Manifolds
96
solution for any given function cp :8Mk integral formulas, we obtain 0
R, then, applying the classical
r L . d(Wk) = J
f O(f) dMA =
d(OMk).
Mk
160
.
The Dirichlet and the Neumann problems have, for a given boundary condition cp : 8Mk R (satisfying f cp = 0 in the case of the Neumann problem) a solution. We will not prove this existence result here for general mani-
folds, but confine ourselves to the case of the ball D"(R) C R' of radius R > 0. For these spaces, there is an explicit classical solution formula which will be derived here. For the sake of simplicity we only discuss the case of
dimensions n > 3 and leave it to the reader to complete the discussion in dimension n = 2, which differs only slightly from the one below. We start with a few preparations. By
r=
(x1)2 +
... + (xn)2
we denote the distance from the point (x', ... , x") E Rn to the point 0 E Rn.
Lemma 3. Let u(x) be a harmonic function defined on the set Rn - {0}, and assume that u depends only on the radius r. Then there exist constants C1 and C2 such that C2
u(x) = C1 + -n-2 Proof. By assumption there exists a smooth function h : (0, oo) -' R satisfying u(x) = h(r(x)). Differentiating this, we obtain the formulas n( )(' i( )r2 - (xi)2 x' Ou OZU
h(r)r, ax = h
112
r \r/
+h
r r3 8x and the latter implies the following differential equation for the function h(r): _
n
0
O(u)
52x1
lh(r).
=
For n > 3 the function h(r) = C1 + C2 r2-n is the solution of this ordinary differential equation.
Let y E Rn be a fixed point. The function u(x) = JJx - y112-n defined on the set Rn - {y} is a translate of r2-n, and hence a harmonic function. On the other hand, the partial derivatives of a harmonic function are harmonic functions as well. Thus
u(x) - 2 .=1
au y ax'
I Ix112
Ilx
I lyl l2
yl In
3.10. Harmonic Functions
97
is a harmonic function defined on iRn - {y}. We will use this family of harmonic functions in the observations to follow.
Lemma 4. Let IIxii < 1 be any point in the interior of the unit ball D". Then
I
IIXII" n
1
IIx
dSn-1(y) = vol(Sn-').
Proof. First we prove that the function
f n-' IIx_ IIyIIn 2
h(x)
.
dSni-1(J)
depends only on the radius r = r(x). In fact, for a linear orthogonal map T : 1R" -+ R" we obtain, from I det(DT)I = I det(T) I = 1 and the corresponding transformation formula for the volume form of the sphere,
T*(dSn-1) = T*(NJ
d1Rn)
= NJ (T'(d1Rn)) = NJ dRn =
dSn-1,
the property to be proved:
h(Tx) = J _ I
dSn-1(y) n-1 IITxIITyIIn
1- IIXII2
= J n-s lixl
TII'(Iy)Iin
dSn-1(y)
. dSn-1(z) = h(x).
Jsn-' IIx - zlln If, in addition, h(x) is a harmonic function, then
z (h)(x)
IS, f
A. (1_ilx2
dSn-1(y))
= 0.
= yIIn The first lemma implies that there are constants C1, C2 with h(x) = C1 + C2 r2-". At the point x = 0 the function h(x) is a regular function and satisfies h(O) = vol(Sn-1). Thus the constant C2 vanishes, and C1 is equal to vol(Sn-'). 1
The solution of the Dirichlet problem for the unit ball D' C ]R" together with an explicit formula are the subjects of the next theorem.
12 is.,
Theorem 39. For every continuous function cp : S"-1 - 1R, the formula _ 1 yII" cp(y) dS"-'(y) AX) = vol(Sn-1) II-
IIx
defines a harmonic function in the interior of the unit ball, and at a boundary point z E Sn-1 lim f (t z) = cp(z).
t-1-
3. Vector Analysis on Manifolds
98
is harmonic with respect to the Proof. The function (1 - IIxII2)IIx variable z, and hence the function f (x) is harmonic, too. We prove that f(z) coincides with ;p(z) on the boundary Sn-1 in the way stated above. y E Sn-1} the maximum of the modulus of p Denote by in := and fix a positive number e > 0. The continuous function p is uniformly continuous on the compact set Sn-'. Hence there exists a number b > 0 such that for any two points y, z E Sn-1 in the sphere, i i - z < 6 implies yll-,,
the estimate I :p(y) - cp(z) I < e. We decompose the sphere Sn-1 = D1 u D2 into the parts
- zII > a} . D1 = {y E S" : IIy - ziI < b}, D2 = {y E Sn-1 For y E D2 and 0 < t < 1 we estimate the distance from y to the line :
II
segment between 0 and z E S"-': fly - tz1I2
=
1 - 2t(y,z) + t2 = (t - (y,z))2 + 1 - (y,z)2 > 1 - (y.
>
1 - (y, z) = 2-(2- 2(y, z)) = -Ily - zII2 >
1
1
Z)2
b2 .
2
We then split the difference _
P ' z) - Y (z) = vol(Sn-1) J$^
2
dSr
Y
Iltz
(?/)
into the integrals over D1 and D2. The modulus of the first integral can be increased to
1-IItzII2
dS"
n
1
1
S^-' Iltz - ylln Di We treat the second integral using the inequality obtained before:
D2
< 2m -
(1-t2)(
a2)n
vol(Sn-').
In summary, for every positive number E > 0 there exists a number 6 > 0 such that for all 0 < t < 1 the following inequality holds:
- t2).
E+2mb Hence the upper limit is bounded by E,
lim suplf(t' z) - V(z)I < E The last inequality holds for all positive numbers E > 0, and this in turn implies lim f (t tt-
z) = p(z)
.
0
99
3.10. Harmonic Functions
Next we will discuss several consequences of the solution formula for the IR defined on the Dirichlet problem. For a harmonic function f : D"(R) closed ball of radius R > 0, the function j (z) := f (R z) is also harmonic on the unit ball. Applying the previous theorem and returning to the vari-
able r = R z E D"(R) finally leads to the Poisson formula for harmonic functions: 2
2
-
2
f(x) = vo1Sn-1) J "-' IIR- RI yl l1"f(R y) dS"-1(y)
Evaluating the Poisson formula at the point x = 0 leads to Gauss' mea value theorem for harmonic functions.
Theorem 40. The value of a harmonic function f at the center of the ball coincides with the mean value of the harmonic function on the boundary of the ball:
We will use Gauss' mean value theorem in the proof of the maximum principle for harmonic functions.
Theorem 41. Every harmonic function f on the connected manifold Mn C R" attaining its maximal value in the interior of Mn is constant.
Proof. Denote by m the maximal value of f and by Q = {x E Al"\8AI" : f (x) = m} the set of all inner points of M" at which f attains this maximal value. By assumption 1 is a non-empty closed set in M"\811". Hence it suffices to prove that 1 is an open subset of M"\8M". Choose a point xo E S2 and a radius Ro such that the ball D"(xo, R0) with center x0 and radius Ra is completely contained in Mn\8M". By Gauss' mean value theorem
f (x0) =
VOl(S1
n-1)
Jsn_l
f (x0 + Ro y) dSn-1(y)
m. = .f (xo)
This implies that f is constant and equal to m on the sphere with center x0 and radius Ra. This observation can also be applied to each radius RR < Ro below R0. Together this implies that f - m is constant on the ball D"(x0i R0), i. e., D" (x0, Ra) is contained in Q. Thus Il is an open subset. 0
Eventually we will prove one more application of the Poisson formula, Liouville's theorem for harmonic functions.
Theorem 42. Every harmonic function f : (above) is constant.
1[P"
R bounded from below
3. Vector Analysis on Manifolds
100
Proof. Changing f, if necessary, by adding a constant, we can suppose without loss of generality that the function f is non-negative. Fix a point xo E Rn and choose the radius R such that xo lies in the ball D"(0, R). By the Poisson formula, Gauss' mean value theorem, and the assumption f > 0 we have Rn 2
f (xo) =
vol(Sn-1) Jsn
Rn-2
R' - II
2
IIo-R
yll In
R2 - I IxOI I2
< vol(Sn-1)
Jn-i IIIxoII - IIR
_
2 - II oIIRIn 2
2(
vo Sn-RIIlxoll
-2(R2 _
RSn
)IIIxoII
Isn
- iRI"
y) dS"-' (y)
. f (R
ylll
n
.f (R y)
dSn- ' (y)
f(R y) dSi-1(y)
f(0)
Taking the limit for R -* oo yields for all points xoi E R" the estimate
f(xo) 0.
19. Prove the orientability of the sphere S", using stereographic projection and Theorem 14.
20. Let AI' be a manifold, h : U _ M' a chart, and -y [a, b] --. h(U) C 111' a curve in Afk completely contained in the image set of h. Represent the curve 7 in the coordinates (h, U) as h-1 o ry(t) = (y1(t), ... , y'(t)). Prove that the length of the curve can be computed by means of the coefficients of the Riemannian metric g1, in the chart (h, U) via the formula :
b
l(7) = f a
1/2
9ti(7(t))d dtt) d dtt)
dl.
21. Let f : [a, b] , R+ be a positive function and let Ale = {(x. y, z) E R3 : y2 + z2 = f2(x)} be the corresponding surface of revolution in R3. Prove the volume formula /b
vol(M2) = 2n / f (x) 1 + (f'(x))2 dx. a
22. Compute the following surface integrals: for M2 = (x, y, z E R3) : x2+y2+:2 = a2. z > 01:
a) ty
-
Exercises
b)
109
/ (x2 + y2)
012.
hr
where M2 is the boundary of the subset of 1R3 x2 + y2 < z < 1.
described by the inequality
23. Compute the following surface integrals:
a) f2(R) (xdyAdz+ydzAdx+zdxAdy); [(y - z)dy A dz + (z - x)dz A dx + (x - y)dx A dy], where M2 is the
b) nJt2
boundary of the subset of R3 described by the inequalities x2 + y2
Oji A wj(V, W) = 0. j=1
and hence all 1-forms w1, ... , w,,,_k vanish on the commutator [V, WJ. There-
fore, this vector field takes values in £k, i.e.. the distribution £k is involutive.
Proof of the implication (2) ; (4). Let £k be a k-dimensional involutive distribution. The form dwi A (wl n ... Aw,,,_k) has degree (m - k + 2). Inserting (m - k + 2) vector fields into this form, we can assume, without loss of generality, that two of these vector fields have values in £k. Denote them by V and W. Because of the involutivity of £k, we have
w;(V) = w;(W) = 0 and dwi(V,W) = -wi([V,W]) = 0. Therefore, the exterior product dwi A (wl A ... A wm-k) vanishes on every (m - k + 2)-tuple of vector fields.
Proof of the implication (1) : (4). The proof of this implication proceeds like the one before. Let £k be an integrable distribution, xo E All a fixed point, and h : W -+ All a parametrization of an integral manifold through this point. Then we have h*(wi) = 0 (1 < i < m - k), and this implies h`(dwi) = 0. Thus the 2-form dwi vanishes on the k-dimensional subspace £k(xo), dwi IEk(zo)XEk(zo)
= 0.
Inserting again (m - k + 2) vectors into the form dwi A (wi A ... Awm_k), we
conclude that at least two of these vectors-call them V and W-lie in the subspace Ek(xo). Hence wj (V) = wj (W) = 0 and &&,i (V, W) = 0. As above, we conclude that the form dwi A (wl A ... A Wm_k) = 0 vanishes.
4. Pfaffian Systems
116
4.2. The Proof of Frobenius' Theorem The implication (3) = (1), which is the core of Fobenius' theorem, is a local statement. Hence, without loss of generality, we can suppose that the manifold Mm is an open subset of R. We will prove a slightly more general result from which the proof of this implication immediately follows.
Theorem 2. Let w1i ... , w,,,_k be linearly independent 1 forms on an open subset M"' C Rm such that m-k
dwi = E B.ij A wj j=1
for certain 1 forms Oij. Then there exist at each point x E Mm a neighborhood U C Mm of x and functions hj and fj defined on the set U satisfying m-k
w= = j=1
Proof of the implication (3)
(1). Let £k be a distribution with the
property stated in condition (3). By Theorem 2, we can represent the forms w; in a neighborhood U of an arbitrary point zo E M' as m-k j=1
for certain functions. By assumption, the 1-forms w1, . . . , w,,,_k are linearly independent. Thus the differentials dfli ... , df,,,_k are linearly independent as well, and the set
Nk = (X E U : f1(x) = fi(xo), ..., fm-k(X) = fm-k(XO) } is a submanifold containing the point xo E M. At an arbitrary point x E Nk, we determine the tangent space:
TxNk = {v E TM' : df1(v) dfm_k(v) = 0 } C {v E TM' : w1(v) _ ... = w,,,-k(v) = 0) = Ek(x). For dimensional reasons, the vector spaces coincide, i.e., Nk is an integral manifold of the distribution £k through the point xo E Mm, and thus the integrability of the distribution £k is proved. 0
Proof of Theorem 2. We proceed in two steps. First we reduce the proof to the case of a system of 1-forms w1, ... , wm_k in a special normal form. To this end, we represent the euclidean space R n as a product RI = Rk X am-k and denote its points accordingly by y = (y', ... , yk) E Rk and z =
4.2. The Proof of Frobenius' Theorem
117
... , zm-k) E Rm-k. It is sufficient to prove the claim in Theorem 2 for 1-forms of the following special type:
(z1,
k
wi = dz`-1: Aii(y,z)'dy'
(1 0 if and only if it satisfies
_I + K2
,i
2
= R2
rc2T
in the natural parametrization. Proof. Differentiating the equation II-'(S)II2 ° R2, we obtain (t(s), -Y(s)) _ 0, and hence y(s) is a linear combination of the vectors h(s) and 6(s),
y(s) = a(s) h(s) +,13(s) b(s) . Furthermore, IIy(S)112 - R2 implies a2(s) + f32(s) (t(s), y(s)) - 0 again leads to K(s) (h(s), y(s)) + 1
R2. Differentiating
0, and thus
a(s) = We differentiate the equation c(s) (h(s), y(s)) + 1 - 0 and obtain the following relation by a simple transformation:
2s) ic'(
a(3)
I£ (S)T(S)
The asserted necessary condition for a spherical curve then immediately follows: R2
=
1
a2(s) + $2(s)
=
a2(s) +
2
r
( K2(3)T(S)
/
5. Curves and Surfaces in Euclidean 3-Space
136
If, conversely, this relation between curvature and torsion holds for a C4curve, then we first differentiate it and obtain the equation
r(s)
d ds
/G(S)
_
KG(s)
!GZ(s)r(s)
0.
Now consider the vector
a(s) := 7(s) +
.6
h(s)
,z() (s)
;R-S)
(s).
Using the Frenet formulas and the preceding relation between curvature and torsion, we compute the derivative of the latter, and find that
id(s) = 0. Hence d:= a(s) is constant, and 11-y(S)
-
aI1z
=
az(s) + \az(s)(s)/z = Rz'
i. e., the curve -y(s) lies on the sphere of radius R with center d.
0
Next we turn to plane curves. Note first that these can be described as the curves with vanishing torsion. Theorem 6. A curve of class C3 with nowhere vanishing curvature, k(s) 0, lies in a plane in R3 if and only if its torsion r(s) - 0 vanishes identically.
Proof. Let a' be a vector perpendicular to the plane in iR3 containing the curve. All the tangent vectors t(s) lie in this plane; hence (t'(s), a) - 0. Since ac(s) # 0, we immediately obtain (h'(s), a") - 0 by differentiating this equation. Thus d coincides with the binormal vector 6(s). In other words, the binormal vector 6(s) is constant. Then 0 = 6'(s) = -r(s) h(s) implies that the torsion vanishes, r(s) - 0. The converse is proved analogously. 0 The curvature of a plane curve can be ascribed a sign. In fact, the principal normal vector is proportional to the vector obtained by rotating the tangent vector through the angle 7r/2 in the positive sense. The curvature of a plane curve is ascribed the positive sign if the corresponding factor is positive. Identifying R2 with the complex numbers, the rotation through 7r/2 in the positive sense corresponds to multiplication by the number i E C. Using the multiplication of complex numbers, this leads to
Definition 7. Let -y : [0, L] - C = R2 be a plane curve in its natural parametrization. The plane curvature k(s) is the function k : [0, L) -' R defined by the equation ast(s) = k(s)
i
t(s) .
5.1. Curves in Euclidean 3-Space
137
The absolute value jk(s)j of the plane curvature coincides with the curvature K(s) of the curve viewed as a space curve.
Example 2. -y(t) = (t, ±t2)
.
A closed curve i' : [0, L] - C = R2 is one that starts and ends at the same point and whose tangent vectors at this point coincide as well, i'(0) = -Y(L) and t(0) = F(L).
Theorem 7. Let
C be a closed curve. Then the integral
[0, L]
r
2 J7
k(s)ds =
2
ILL
is an integer, called the winding number of the closed curve.
Proof. Consider the map t' : [0, L]
t* (s) = exp
C defined by
[k(u)du]
Then dt'(s)/ds = i k(s) - t*(s). and hence (t(s)/t*(s))' = 0. The tangent vectors t(s) are thus described by the formula
F(s) = C exp
Lk(u)duj fL
for a certain constant C. Since t(0) = t(L), the number J k(s)ds is an integral multiple of 27r.
o
0
We conclude the section on curves with the discussion of the Fenchel inequality, which claims that the total curvature of a closed space curve is
5. Curves and Surfaces in Euclidean 3-Space
138
bounded from below by 2;r. We start by considering plane curves, and then generalize the result to space curves. First we need an auxiliary observation. Lemma 1. Let cp : [a, b] lR be a real function of class C', and suppose that the function f (t) := ew 0.
Hint: The function f : M2 -+ R, f (x) = (x, x), has to have a critical point on Ate.
23. Let the group G = SL(2,R) act on the hyperbolic plane ?{2 by _ a b .z _ az+b
cd
6F
cz+d'
Verify that the image point g z actually does belong to ?{2, and that superposition of two of these transformations corresponds to matrix multiplication in G. Show. moreover, that each g E G leaves the metric invariant; hence G is an 3-dimensional group of isometries from N2 onto itself.
24. Let AI'; = 1R3 and denote by DyX the directional derivative of the R3 in the direction of the vector X. Prove that VXY := DXY+2X xY vector-valued function X : R3
defines a covariant derivative having all the properties (1)-(4) from Theorem 42. but violating property (5).
25. Consider on the set {(x,y) E R2 : -7r/2 < y < 7r/2} the pseudoRiemannian metric
1 _ dx2-dye x'2
- y'2
COS2
x'
a) Prove that E _2(y) and P =2(y) are first integrals of the cos
geodesic flow, satisfying, in addition, the inequality p2 - E > 0. b) Discuss the geodesic lines on M2. To do so, assume that y is a function of r and integrate the resulting ordinary differential equation.
c) On Al" there exist points that cannot be joined by a geodesic line.
203
Exercises
26. Prove that every three-dimensional Einstein space is a space of constant curvature.
27. Let M' be a non-flat Einstein space (for example, S"'). Show that MI x M"' with the product metric is an Einstein space, but not a space of constant curvature.
28. It is well-known that a symmetric bilinear form h.(x, y) on a vector space is completely determined by its quadratic form q(x) := h(x, x) (via the polarization formula: 2h(x, y) = q(x + y) - q(x) - q(y)). Prove, in a similar way, that the Riemannian curvature tensor 1 (U. V, W1, W2) is completely determined by the quadratic form K(U, V) := R(U, V, V, U) corresponding to the sectional curvature. 29. Prove that a four-dimensional Riemannian manifold is an Einstein space if and only if for every 2-plane E2 C TM4 and its orthogonal complement (E2)1 the corresponding sectional curvatures coincide, K(E2) = K((E2)1).
30. Because of its symmetry properties, the curvature tensor of a pseudoRiemannian manifold can be interpreted as a transformation R : A2(M'') A2 (M7°) on 2-forms, n 1
R(ai Aaj) = 2 > Ro3i . a, Aa,3. Q,:3=1
In dimension m = 4, this gives rise to an endomorphism R : A2(A14) A2(M4) A2(M4). On the other hand, the Hodge operator * : A2(A14) acts on A2(M4), and its square depends only on the index k of the metric (see
Theorem 5, Chapter 1): ** = (-1)k. In the cases k = 0 (positive definite metric) and k = 2 (neutral metric), the Hodge operator decomposes the real bundle A2(M4) into the corresponding eigenspaces A2 (M4) (see Exercise 8, Chapter 1). Prove that the block representation of the curvature tensor.
R _
R++ R-+
R+- R--
with respect to this decomposition of A2(M4) has the following properties:
a) 7Z is symmetric, i.e., R++ = R++, R__ = R__ and R+_ = R_+.
b) The traces of R++ and R__ coincide, tr(R++) = tr(R_-) = -r/12. c) M4 is an Einstein space if and only if R+_ vanishes.
5. Curves and Surfaces in Euclidean 3-Space
204
Literature: Th. Friedrich, Self-duality of Riemannian manifolds and connections, in: Riemannian geometry and instantons, Teubner-Verlag, Leipzig, 1981, 56-104.
The Einstein equation in the general theory of relativity combines the geometric curvature quantities of a four-dimensional pseudo-Riemannian manifold of signature (1, 3) with its physical properties encoded in the energymomentum tensor T,
Ric-
rcT.
Here Ric. g. r are the Ricci tensor, the metric, and the scalar curvature; K is a constant depending on the chosen system of units. Already for the vacuum, T = 0, there are non-trivial, physically very interesting solutions of this equation, among others the Schwarzschild metric to be discussed now. Obviously, a vacuum solution of the Einstein equation has to be an Einstein space with vanishing Ricci tensor in the sense of the definition given before. 31 (Schwarzschild metric). On a spherically symmetric and static space-time manifold M4 (this is a pseudo-Riemannian manifold of signature (1, 3) with
isometry group SO(3, R) and a distinguished time direction), it is always possible to introduce coordinates from R x R+ x S2 with respect to which the metric can be written as g = e2a(r) dt2 - [e 2b(r) dr2 + r2 (d92 + sin2 O dV2) J .
Here, the functions a(r) and b(r) asymptotically tend to zero for r - co (the metric is "asymptotically flat"). We introduce the following basis of 1-forms:
ao = ea dt, Cl = eb dr,
a2 = rdO, 03 = r sin 9 dcp .
a) Check that the metric satisfies g = 0,02 - 0i - 02 - a3 . b) Compute the forms dai and show, using the first structural equation, that the connection forms are given by wo, = -a'e-bcO,
e-b
WO2 = W03 = 0,
a-b
W12 = -r 0`2, w13 = r- a3, w23 =
cot 9
r
a3-
c) We introduce the notation lid := 2 E.4 Rtasia0 A0,3. (f is the so-called curvature form). Show, using the second structural equation, that
110 = e-2b(a'b'-a"-a2)aoAal, f20 = -a'er"aoAa2, -a'er6
03O =
ao A 03,
5231 = b'erbal A a3,
1121 = bier
6a1
A02,
X32 = '-'-'012 A a3,
Exercises
205
and compute from these formulas the components R kl of the curvature tensor in this basis. d) Compute the components of the Einstein tensor G := Ric - 1g ,r: ) , Gi l = ,1-s _e-2b(77 Goo= I e-2b (13 _ 2b F 7-
-
+),
G22 = G33 = -e-2b (a'2 - a'N + a-+
=d) r
e) Solve the vacuum equation by means of this Ansatz. The result is (M is a constant of integration) r
g = 11-
1
2M dt2 -
r
J
r
11I
2M
r
1_1
dr2 J
- r2 (d92 + sin2 0 d 2)
32. Restrict the Schwarzschild metric g to the two-dimensional submanifold
defined by 0 = x/2 and t = const. Prove that this yields the metric of a paraboloid of revolution (a "lying" parabola!) with the equation z2 = 8M(r - 2M) (see the picture). 33. Light moves along geodesic lines whose tangent vectors have length zero. Making use of the fact that the equations defining a geodesic are precisely
the Euler-Lagrange equations of the length function G = ta gjjx'ia, prove the following assertions:
a) A particle moving at t = 0 in the equatorial plane 0 = x/2 stays there forever.
b) The quantities L := r2cp and E := t(1-2M/r) are first integrals. Moreover, the second Kepler law holds: The orbit ray covers equal areas in equal times.
5. Curves and Surfaces in Euclidean 3-Space
206
c) Set r = r(W) and derive the equation describing the motion of light rays.
Result: It is reasonable to set u := 1/r: u" + u = 3Mu2. d) Solve this differential equation approximately up to second order in V. Result:
uo =
1
sin V +
[1
+
cos 2W
.
Interpret the constant of integration 2 0 ro as 3the scattering length. Which asymptotic value arises for cp if r tends to oo and sp is supposed to stay small? Twice this value, denoted by 5 in the picture, is the relativistic deviation of light in the gravitational field of a very large miss, which can be described by the Schwarzschild metric.
Chapter 6
Lie Groups and Homogeneous Spaces
6.1. Lie Groups and Lie Algebras In the preceding chapter Noether's theorem showed that symmetry considerations simplify the study of geometric problems, and sometimes it is only by symmetry considerations that a solution is possible at all. In fact, beginning in the 1870s, the conviction grew that the basic principle organizing geometry ought to be the study of its symmetry groups. In his inaugural lecture at the University of Erlangen, which later became known as the "Erlanger Programm", Felix Klein said, in 1872, "Es ist eine Mannigfaltigkeit and in derselben eine Transformationsgruppe gegeben; man soil die der Mannigfaltigkeit angehorigen Gebilde hinsichtlich solcher Eigenschaften un-
tersuchen, die durch die Transformationen der Gruppe nicht geandert werdenl.
One has to distinguish whether the groups under consideration are discrete (for example permutation groups) or continuous (for example, oneparameter groups of isometries). The latter were systematically introduced by the Norwegian mathematician Sophus Lie (1842-1899). which is why they bear his name today.
1 "Let a manifold and a transformation group in it be given; the objects belonging to the manifold ought to be studied with respect to those properties which are not changed by the transformations of the group."-quoted from F. Klein, Des Erlanger Programm, Ostwalds Klassiker der exakten Wissenschaften, Band 253. Verlag H. Deutsch, Frankfurt a. M., 1995, p. 34. 207
6. Lie Groups and Homogeneous Spaces
208
The fundamental idea of a Lie group is a very simple one. It ought to be a group which is at the same time a manifold, and hence allows a differential calculus. Moreover, the manifold structure has to be compatible with the group structure, i. e., the product is a differentiable map.
Definition 1. A Lie group is a group G which, at the same time, is a differentiable manifold such that the map is (infinitely often) differentiable.
Remark. Obviously, the last condition is equivalent to requiring that the product map and the inversion (9, h) -. g . h,
g'-' g-1 ,
be differentiable maps.
Example 1. Every finite-dimensional vector space V is an abelian Lie group whose group product is exactly the addition of vectors (v, w) H v + w.
Example 2. The unit circle S' = {z E C : IzI = 1} is an abelian Lie group with the usual multiplication of complex numbers as product.
Example 3. Most Lie groups can be realized as matrix groups. The set of all invertible matrices with entries in K = R or C is an open subset of IK"z and hence a manifold. Endowed with matrix multiplication as product, it forms a Lie group, the general linear group
GL(n,K) := {A E .M"(K) : det A 540} . More generally, GL(V) is meant to denote the group of invertible endomorphisms of the vector space V.
Example 4. The vector space K" and the general linear group GL(n, K) combine to form a new Lie group, the affine group Aff (K") := GL(n, K) x IIS"
with multiplication
(A, v) (B, w) := (AB, Aw + v). This product rule arises in a natural way by defining an action of the affine group on the vector space K" through
Aff(K") x K' V. (A, v)x := Ax + v. and then applying the transformations determined by (B, w) and (A, v) on a vector x one after the other. Each Lie group G acts on itself by means of the left and the right translation with a fixed element g E G,
L9, R9 : G -. G, L9(h) = g h and R9(h) = h g.
6.1. Lie Groups and Lie Algebras
209
The corresponding differentials are the following maps in the tangent bundle of G: Th9G. (dLg)h: ThG - TghG, (dR9)h : ThG To avoid double indices, in this chapter we will use the notation (df)h instead of f.,h for the differential of a map f.
Definition 2. A vector field X on G is called left-invariant (or rightinvariant, respectively) if it is transformed into itself by dL9 (or dR9, i.e., dL9X = X. At a point h E G, this means
(dLg)hX(h) = X(g - h). Since left translation is obviously a diffeomorphism of G, Theorem 35 in §3.9 can be applied to yield, for the commutator of two left-invariant vector fields, the formula
dLg[X,Y] = [dL9X,dLgy] = [X,Y1. This property, together with the fact that vector fields satisfy the Jacobi identity (Theorem 34, §3.9), endows the vector space of left-invariant vector fields with the structure of a Lie algebra, the Lie algebra g of the Lie group G.
Theorem 1. The vector space of left-invariant vector fields on a Lie group G is canonically isomorphic to the tangent space at the neutral element,
g='TeG. Proof. With each left-invariant vector field X, we associate its evaluation at the neutral element e, X - X(e) =: X E TOG. Conversely, every element X E TeG determines a vector field Xx on G by setting
Mx(g) := (dL9)e(X) This satisfies the relation
Xx(gh) = (dLgh)e(X) = (dLg)h(dLh)e(X) = (dLg)hXX(h)
0
hence Xx is left-invariant.
Remark. Because of this fact, we will no longer distinguish between the Lie algebra of left-invariant vector fields and the tangent space to the group
at the neutral element. Its elements will be denoted by upper case Latin letters X, Y, ... E 9. Choosing a basis X1, ..., Xr of 9, we can again write their commutators as linear combinations of the basis elements, r
[Xi,Xj] = k=1
&Xk-
6. Lie Groups and Homogeneous Spaces
210
The antisymmetry of the commutator and the Jacobi identity imply that the constants C have to satisfy the relations
Cij - - Cii '
CijCk,n + Cj' "ki + CrniCkj = 0.
Following E. Cartan, the numbers C are called the structure constants of the Lie group G, since Cartan's structural equations are simple to formulate in their terms. To see this, we agree to call a differential form w on G left-invariant if it satisfies the condition L*9w = w.
Following the argument in the case of vector fields, it is easy to see that the r-dimensional vector space g* of left-invariant 1-forms is canonically isomorphic to Te G. Now let al, ...,o-, be the basis of g* dual to X1. ..., Xr. Theorem 2 (Maurer-Cartan Equations). Let C be the structure constants of a Lie group G with respect to the basis X1, ... , Xr of its Lie algebra g. Then the exterior derivatives of the forms in the dual basis al, ... , or of g* are given by
do, _ -r k of Aak. j 0, and there would exist integers k and I such that
I=
q = Va.
But this would imply q = Ilk E Q. Hence we arrive at a contradiction. As the closed subgroups of IR are precisely the cyclic groups and R itself, only this last possibility remains for Z + qZ. In Example 4, §3.1. we encountered a parametrization of a torus of revolution
by means of S' x S'. The following pictures show the trace of a curve yy in this parametrization for two close rational and irrational values of q, respectively. In Example 8, §7.4, we show that the motion of a spherical pendulum can be parametrized precisely by such a curve on a torus.
The main objective of this section is to prove that a closed subgroup H of a Lie group G is a submanifold of it. Consequently, it is itself a Lie group, and the quotient space C/H carries the structure of a manifold with a smooth G-action. As a preparation, we need a few technical lemmas. Let II - II be any norm on the vector space g. Lemma 3. Let H be a closed subgroup of G. and let Xn # 0 form a sequence converging to zero in 9 such that exp(X,,) belongs to H and Xn/IIXnII tends to an element X E g. Then
exp(tX) E H for all t E R. Proof. For a fixed t > 0 we define a sequence of natural numbers m by
mn := max{kEN: Then the following estimate holds:
mnIIXnII < t < (mn + 1)IIXnII = mnIIXnII + IIXnII But on the right-hand side, the sequence IIXnII tends to zero, and hence
lim mnIIXnII = t. n-or,
6. Lie Groups and Homogeneous Spaces
218
This implies
limn rnXn =
hn-oo m m HIX,,II
IlXnll
= tX,
and, since the exponential map is continuous,
exp(t X).
Iim
n-oo Each term in the sequence exp(m,,Xn) = [exp(X,,)]"'" is an element of H. As H is closed by assumption, the limit of this sequence, exp(t X), has to lie in H, too. The proof for the case t < 0 proceeds along the same lines.
Lemma 4. For every closed subgroup H C G, the set
b :_ {XEg: exp(tX)EHforalltElR} is a linear subspace of g.
Proof. For any vector X E h, its scalar multiples a X also belong to h. It thus suffices to show that h is closed under addition. To this end, let X, Y be elements of h, and suppose that X + Y 36 0. In any case, the product exp(tX) exp(tY) belongs to the subgroup H, and for sufficiently small t we have, by Lemma 1,
exp(tX) exp(tY) = exp(t(X +Y) + 0(t2)). Then Z(t) := O(t2)/t apparently converges to zero for t - 0, and we can rewrite the preceding equation as
exp(tX) exp(tY) = exp (t(X +Y + Z(t))) E H. Choose a sequence of positive numbers that converges to zero, t - 0, and define Xn := tn(X + Y + Z(tn)). Each term exp(XX) lies in H, and --:in X+Y = lim X+Y+Z(t,) lim n-- IIX + Y + Z(t,a)[I = Fix -+Y11 11X-11 Obviously, we have Xn # 0 and Xn -' 0. Thus, Lemma 3 applies, and we can conclude that for every t E R
exp(t
IIX+YII
is an element of H. Hence X + Y belongs to Fj. Suppose that H is, in addition, a closed subgroup of G, and let h be defined as in the preceding lemma. We then choose any linear complement h' of h in 9,
9 = h+h'. Lemma 5. There exists a neighborhood V' C 4' of 0 such that, for every X' 34 0 in V', the element exp(X') does not belong to H.
6.2. Closed Subgroups and Homogeneous Spaces
219
Proof. If the assertion were false, there would exist a sequence X;, E h' converging to zero and satisfying exp(X,,) E H. Now consider the compact set K :_ {X' E h': 1 < JJX'II < 2} and choose natural numbers p,, such that p ,,X,, E K. Since K is compact, we may assume that the sequence p,X;, converges to some 0 9& X' E K. Again, [exp(X;,)]P is an element of H and lim
pnX' 11p-Xnii
_
X' 11X'6
Then Lemma 3 implies that X'/IIX'II E h, contradicting 0 0 X' E '.
0
Now we can turn to the main theorem of this section.
Theorem 7. Let G be a Lie group, and let H be a closed subgroup. Then: (1) H is a submanifold of G and thus itself a Lie group. (2) There exists precisely one differential structure on G/H such that (a) the projection 7r : G - G/H is smooth, (b) for every p E G/H there exist a neighborhood Wp C G/H of p and a smooth map w : Wp G such that 7r o p = Idw,,, (c) the action of G on G/H defined by (g. kH) gkH is smooth. Proof. It obviously suffices to show that there exists a neighborhood W C G of e for which H n W is a submanifold (left translation is a diffeomorphism of G). As before, ddecompose the Lie algebra into g = 1) + h' and consider the map corresponding to this decomposition, 4D :
g = h + 4' ---i G, 4(X + X') = exp(X) exp(X').
In h' we choose a neighborhood V' C h' as in Lemma 5. as well as a subset V C h so small that the exponential map still is a diffeomorphism on V +V'. The image W of V + V' under -ID is an open neighborhood of e E G, and
HnW= by the definition of b and Lemma 5. The set H fl W is thus parametrized by the chart (V, -D Iv+(o}), and hence a submanifold of G.
Now we turn to the proof of the second assertion. Let 7r : G - G/H denote the projection. We define a topology on G/H by the condition
A C G/H is open :a 7r-' (A) C G is open. It is called the quotient topology on G/H, and it is designed to render the map 7r continuous. Endowed with this topology, G/H is a Hausdorff space (see Duistermaat/Koik, Lemma 1.11.3). To verify the properties a manifold
6. Lie Groups and Homogeneous Spaces
220
has to satisfy, consider the distinguished point xo := e H E G/H together with the sets V, V' introduced in the first part of the proof. The map
' : V' - G/H, X'
a(exp X')
,
is continuous and maps V' onto an open neighborhood U of xo. Moreover, V, is injective, since v(X') = O(Y') implies the existence of an element h E H
such that exp(X') = exp(Y') h. Hence
h = exp(X') exp(-Y') = b(0 + (X' - Y')) . Thus h also belongs to the set W, which was defined as the image of V + V
under C Since we already proved H fl W = 4(V + {0}), this implies X' = Y'. In summary, the map 1/i : V' U is continuous and bijective. For an arbitrary point gH E G/H, consider the left translation by g E G on
G/H. L9 : G/H
G/H, kH
gkH
and introduce a chart around gH E G/H by L9(U),
U9H
09H: V' - UgH,
Z'gH := Lg o TV .
For two points gH and kH, the chart transition can be rewritten as follows, kH o y9H =
=
v-1
o Lk-1 o L9 0 V1 = exp-1 o(ir-I o 4-1 0 L9 0 a) o exp
exp-1 oLk-1h
oexp .
Therefore. as a superposition of smooth maps, the chart transition is also smooth. Hence we have proved that G/H is a differentiable manifold, the projection ;r : G - G/H is smooth, and C acts smoothly from the left on G/H. It remains to show (b). For the distinguished coset p = .ro = e H, define y; for each x in U =: Wp by
cp(x) = exp(Vi-1(x)) = 7r-1(x). For an arbitrary point p = gH one again uses the left translation L9.
0
Definition 5. The action of a Lie group G on a manifold Al is called transitive if, for two arbitrarily given points x and y in Al. the one can always be written as the image of the other under the action of G. i. e., there exists a g E G such that y = g x. An equivalent formulation of this requirement is to say that M consists of a single G-orbit, G x = Al. A manifold together with a transitive group action is also called a homogeneous space.
Obviously, the left translation on the quotient Al = G/H is a transitive group action, and thus G/H is a homogeneous space. Theorem 7 can be applied to show that some well-known matrix groups are Lie groups: The following groups apparently are closed subgroups of GL(n, K).
6.3. The Adjoint Representation
221
Example 8. The subgroup of GL(n, K) consisting of all matrices with determinant 1 is a Lie group, the special linear group,
SL(n,K) :_ {A E Mn(K) : det A = 1} .
Example 9. Let H
I =: h u, v E c } be the vector space of {[. Hamilton's quaternions with standard basis V
111
1
110
11'
oil' J Lof OJ, K = Lo i 01 and norm N(h) := uu + vv. The group of all quaternions with norm 1 is
E=
1=
LO
isomorphic to the Lie group
SU(2) := JA E GL(n, IC) : AAt = 12 and det(A) = 1). Example 10. The preceding example can be generalized as follows. The unitary group is embedded into the space of complex matrices as
U(n) :_ {A E
AAt = 1n}
.
The condition AAt = 1n immediately implies I det Al = 1, hence det A E Sl; the special unitary group is defined as the group of all unitary matrices A satisfying det A = 1:
SU(n) := {A E U(n) : det A = 1} . Example 11. The orthogonal group O(n, K) consists of the matrices A E M ,,(K) leaving the euclidean standard scalar product of Kn invariant,
(Ax, Ay) _ (x, y) Realizing the scalar product as (x, y) = xty, we see that this condition is equivalent to AAt = ln. Hence we obtain
O(n, K) = JA E Mn(K) : AA' = 1n} . Obviously, an orthogonal matrix has determinant + 1 or -1. The subgroup of all orthogonal matrices with determinant +1 is called the special orthogonal group SO(n, K),
SO(n,K) = {AEMn(K): AAt=lnand detA=1}.
6.3. The Adjoint Representation Definition 6. Let G be a Lie group with Lie algebra g, and let V be a finite-dimensional vector space.
6. Lie Groups and Homogeneous Spaces
222
(1) A representation of the Lie group C on V is a smooth group homomorphism e : G GL(V), i.e., a smooth map compatible with the group structure,
e(g h) = e(g) e(h) (2) A representation of the Lie algebm g on V is a homomorphism of Lie algebras, p : g - gl(V), i. e., a linear map compatible with the commutator,
e([X, Yl) = [e(X ), e(Y)] = e(X) e(Y) - e(Y), e(X) Sometimes, V is then also called a G-module or a g-module, respectively.
Example 12. The trivial representation of a Lie group G is the group homomorphism that maps every element g E G to the neutral element in GL(V): p(g) = 1V; the trivial representation of g associates the zero map with every element X, o(X) = 0v. Example 13. Matrix groups are defined by means of one of their representations, often called the defining representation. In fact, we introduced the groups GL(n, R), SL(n, R) and SO(n, R) in a way endowing them naturally with a representation on R". A simple example illustrates that these matrix groups and their Lie algebras have many more representations. The Lie algebra sl(2, R), for example, has representations in all dimensions: For every natural number n, define e : s((2, R) - gl(n + 1, R) by
g(H) = diag(n, n - 2, ..., -(n - 2), -n), 0
0 n
1
0
2
e(F) =
e(E) 0
0
n-1
0
nL
1
0
These matrices satisfy the commutator relations of sl(2, R), [e(H), e(E)1 = 2,o (E),
[e (H), e(F)l = -2e(F), [e(E), e(F)] = e(H), and hence form an (n + 1)-dimensional representation of sl(2, R). Properties
that cannot be expressed by the Lie bracket do not have to be preserved under a representation: For example, we have E2 = 0, but e(E)2 0 0. Nevertheless, the property that g(E) is a nilpotent matrix is preserved. There is also a a representation of the Lie group SL(2, R) corresponding to this representation of the Lie algebra; this will be the subject of Exercise 4.
Apart from left and right translation, there is a third remarkable action of a given Lie group G on itself, the so-called conjugation action,
ag : G - G, a9(h) := ghg-' = L9R9-, h.
6.3. The Adjoint Representation
223
It is smooth and satisfies ag(e) = e, and in contrast to left and right translation, it is far from being transitive. In the case G = GL(V), it decomposes the invertible matrices precisely into their similarity classes. In addition, the relation ag(e) = e implies that its differential at e is a map from g to g, d(ag)e: TG 5--- g ----+ TG 2--- 9,
which is obviously invertible, since d(Lg)e and d(Rg-i)e are invertible. We define the adjoint representation of G on g by Ad : G ---+ GL(g),
Ad(g) = d(ag)e E GL(g).
Before we verify that this actually is a representation, recall the definition of the center of a group. It consists of those elements which commute with all the others:
ZG = {gEG: gh=hgdhEG}. Theorem 8. The map Ad : G - GL(g) is a representation of G on the vector space g. The center ZG of G is contained in its kernel, ZG C ker Ad,
and equality holds if and only if G is connected.
Proof. First we check the homomorphism property:
Ad(gh) = d(LghRgh )e = d(LgLhRh-, Rg-i)e = d(agah)e = d(ag)ed(ah)e = Ad(g)Ad(h). Let z belong to the center Z. Then aZ = IdG, and hence Ad(z) = IdGL(9), i.e., z is in the kernel of Ad. Now suppose that G is connected and that
Ad(g) = Ide. Since ag : G - G is a group homomorphism, the map t ' -+ ag (exp tX) is a one-parameter subgroup for every X E g, and, by Theorem 5, there exists an element Y E g such that
exp(tY) = ag(exptX). Differentiating this equation with respect to t, we obtain Y(e) = dt (ag(exp tX )) I e=o = Ad(g) (X (e)) = X (e) , which proves X = Y. For the one-parameter group defined above this means
that exp(tX) = ag(exptX) for all t E R and X E 9. The exponential map is a local diffeomorphism g - G; hence ag = IdG on an open neighborhood W of e. For a connected
224
6. Lie Groups and Homogeneous Spaces
Lie group G this implies ay = IdG, since G can be represented as the union of all powers of W (with respect to the group products), 00
G = UW'. As a9 = Idc is equivalent to g E ZZ, everything is proved.
0
The differential of the adjoint representation of G (in the sense of Definition 4) is a representation of the Lie algebra g which can now be expressed by the commutator.
Theorem 9. The differential ad := Ad. : g -. gl(g) of the adjoint representation is a homomorphism of Lie algebras determined by the formula
ad(X)(Y) = [X, YJ. Proof. By the definition of the differential, we have
Ad(exptX) = exp(tAd.(X)) = 1 + tAd.(X) + ... ; hence
Ad.(X)(Y) =
Xd
Ad(exptX)(Y) - Y
li.o The flow corresponding to the vector field -X is (bt = R P(_tX), since d4ie(e)
dexp(-tX) I
dt LO t=o Applied to a left-invariant vector field Y, however, its differential coincides with Ad(exptX)(Y),
Ad(exptX)(Y) = dL P(tx)dR P(-ex)(Y) = dR p(_tx)(Y) = d4it(Y). Thus the original identity can be rewritten as
Ad.(X)(Y) = li o
`ht(Y) - Y
The right-hand side is precisely the definition of the commutator [Y, -X] _ O
[X, Y].
This representation will also be called the adjoint representation (this time of the Lie algebra g). In case of doubt, the context has to decide whether the representation of the Lie group or that of its Lie algebra is meant. Remark. The definition of the differential immediately implies the identity
Ad(expX) = exp(adX).
6.3. The Adjoint Representation
225
This has to be understood as an identity of operators. Applied to an element Y, it means ad(X)3i3(Y)
exp(X) Y exp(-X) = 1 + ad(X)(Y) +
+ ...
= 1 + [X, Y1 + [X, [2I Y11 + [X, [X 3'XI Y111 + ... .
Example 14. Let g be the three-dimensional Lie algebra which is abstractly defined by the following commutator relations for a basis el, e2, e3: [el, e21 = e3,
[e3, ei1 = e2.
[e2, e31 = el,
The representing matrices of the adjoint representation with respect to this basis can be computed from them. For the operator ad(ei ), we obtain 0
ad(ei)
e2 e3
e3
i
l
=
rO
-e2
0
1
-1
0
e2 e3
Ll
e2 e3
and similarly for the other two operators,
0-1 L2 := ad(e2)
0 rol
0
0 0
10
0 L3 := ad(e3) _
,
1
0
0 0
00
Let us look, on the other hand, more closely at the orthogonal group O(n, R).
It was defined as the set of matrices satisfying f (A) = AAt - 1, = 0. The differential of this map at the point X is
df(A)x = AXt+XAt, and hence, according to Theorem 5 in §3.2, the Lie algebra of O(n, R) is
o(n, R) = Te O(n, R) = {A E M,(R) : A + At = 01. The Lie algebra of the orthogonal group consists precisely of the skewsymmetric matrices, which for n = 3 is apparently spanned by Li, L2 and L3. This proves that the three-dimensional defining representation of o(3, R) is isomorphic to the adjoint representation. In higher dimensions, this fact no longer holds as a simple dimensional consideration shows: A skew-symmetric matrix has exactly as many degrees of freedom as entries above the diagonal. Therefore, 22
dim o(n, R) = and this is equal to n only for n = 3.
- n = n(n2 1)
6. Lie Groups and Homogeneous Spaces
226
Exercises SL(2, R) is not surjective. Hint: 1. The exponential map exp : si(2, R) What are the values that tr exp(A) can attain for A E al(2, R)?
2. Hamilton's quaternions H (Example 9) form not only a vector space, but also a (non-abelian) division algebra, i. e., an associative algebra in which each non-trivial element is invertible. Prove that the standard basis E, I, J, K of Hamilton's quaternions obeys the following algebra relations:
I.K=-J, and compute the inverse of the quaternion h =
0.
3. Identify the quaternions of trace 0,
Ho = {xi . I + x2 J + x3 K j xi, x2, x3 E R}, with the 3-dimensional euclidean space R3. Prove that, for every U E SU(2), the map UxU-1, Ho - Ho, x
defines a special orthogonal transformation of R3. The resulting map e SU(2) - SO(3,R) is a representation which is not injective ("faithful"). 4. The defining representation of SL(2, R) on R2 is the usual matrix action on vectors,
gV
[cx + dy] Ic d] [y] Let V,, be the (n + 1)-dimensional vector space of homogeneous polynomials of degree n in the variables x and y. Define an action of g E SL(2, R) on the polynomial p E V,, by
P(g) ' P ([X])
= P\g_1
[;]).
Prove that B is a representation of SL(2, R).
5. Let G be a Lie group, and let H be a discrete, normal subgroup of G. Prove that H is necessarily contained in the center of G.
Exercises
227
6. Let (p, V) be a representation of the group G. A subspace W of the representation space V is called invariant if. for every g E G. the relation p(g)W C W holds. For trivial reasons, the subspaces W = V and W = {0} are invariant: if the representation has no further invariant subspaces, it is called irreducible. Consider the following two-dimensional representation of the additive group R:
tl
rrI LOW = L0
1
11
Prove that this representation is not irreducible. Does the invariant subspace have an invariant complement?
7. By Theorem 6. the differential of a representation (p. V) of the Lie group G is a representation (Lo., V) of its Lie algebra g. The tensor product of two representations (p, V) and (µ, W) of G is defined by (p ®Fr)(9)(v $ w) := p(9)v (9 p(g)w, g E G. V E V. W E W . Prove that this determines a representation of G on V O W with differential
(e® i),(X)(v ®w) := p,(X )v ®w + v ®p.(X)w, X E g. 8. In order to describe the hyperbolic plane as a homogeneous space. it is useful to introduce a new model for it, the open unit disc.
a) Let D = {z E C I lzj < 1} be the open unit disc with metric
9 = (1_1212)2 I0 Show that the Cayley transform.
I.
`
-i`+i=:x+iy,
c(z) =
c:
i
z-i
is an isometry between D with the metric above and the upper half-plane ?{2 with the fourfold of the hyperbolic metric. b) Let the Lie group
SU(1.1) :_ { act on D via the formula a
b
lb
b
z
:
1a12
- 1b12 = 1}
_ az+b az+b
a Prove that this action is transitive, and that the isotropy group of zero, b
Go := {g E SU(1,1) : g 0 = 0}, is isomorphic to SO(2,IR). Hence D 5 SU(1,1)/SO(2,1R).
Chapter 7
Symplectic Geometry and Mechanics
7.1. Symplectic Manifolds Riemannian geometry is the geometry of a symmetric, bilinear form depending on the point of a manifold. The curvature is a measure of how far two symmetric bilinear forms differ locally. Contrary to this, symplectic geom-
etry is that of an antisymmetric bilinear form depending on the point of a manifold-hence the geometry of a 2-form w. It turns out that all symplectic manifolds are locally equivalent: there cannot be any concept similar to curvature in the sense of Riemannian geometry. Symplectic structures differ, if at all, only globally. Historically, the formulation of mechanics in the sense of Hamilton led to symplectic geometry, hence its essential role in modern mathematical physics.
Definition 1. A symplectic manifold is a pair (1112., w) consisting of a manifold 1112ni of even dimension together with a closed non-degenerate 2form w,
d w = 0 and
A
called the symplectic form or symplectic structure. By Theorem 16 in §3.4, every symplectic manifold is orientable. The volume form is understood to be the 2m-form dM2m
= (-1)
-(--1)/2 '
win
.
m! 229
7. Symplectic Geometry and Mechanics
230
Example 1. In 1R2"' with coordinates {q1, ..., q,", pi, ..., p,,,), the formula m
E dpi A dqi i=1
defines a symplectic form with highest power
w"' = m! -
(-1)-(--1)/2
- dpl A ... A dpm A dq1 A ... A dq,,, .
The volume form in the sense of symplectic geometry is the ordinary volume form of ]R2i'. This symplectic structure is called the canonical symplectic structure.
Example 2. Define a 1-form 0-the so-called Liouville form--in the cotangent bundle T'X"' of an arbitrary m-dimensional manifold as follows: Let V E T,, (T' X ') be a tangent vector at q E T' X' and represent it by a curve
V : (-e, e) - T* X' such that
V(0) = q,
V(0) = V.
Project this curve first by means of the projection 1f : T'X"' - X"' to the manifold, and, after that, apply the 1-form q to the tangent vector of the projected curve:
9(V) := n
d d (T ° V (t)) I=o
The 2-form w := dO is a symplectic structure on T'X'. Any system, {q1, ... , qm }, of coordinates in X' determines-representing a 1-form q as q = >2 pi dqi-coordinates {qt, ... , 9m, pi, ... , p,n } in TX". By the definition of the Liouville form 0, we have m
0= and the 2-form
pi dqi
m
w=d9=EdpiAdgi i=1
is non-degenerate. In particular, the (co-)tangent bundle of every manifold is an orientable manifold (see Exercise 9 in Chapter 3). Further examples of symplectic manifolds arise as the orbits of the coadjoint
representation of a Lie group G. Starting from the adjoint representation, GL(g), of the group G and passing to the dual of the linear Ad : G operator, Ad'(g) := (Ad(g-1))* : g' - g', we obtain a representation
Ad` : G -b GL(g')
231
7.1. Symplectic Manifolds
of the group G in the dual space g' of the vector space g. Through each functional F E g' passes an orbit 01(F) := {Ad*(g)F : g E G}, on which the group G acts transitively. The isotropy group
GF := {g E G : Ad'(g)F = F) is a closed subgroup of G, and 0 *(F) is diffeomorphic to the homogeneous space G/GF. Its Lie algebra can be characterized by a similar condition:
Theorem 1. The Lie algebra OF C g of the isotropy group GF C G is equal to
OF = {XEg :F([X,Y])=0 for all YEg}. Proof. Suppose that F([X,YJ) = 0 holds for all elements Y E g. Then from
(Ad*(exp(t X))F)(Y) = F(Ad(exp(- t X))Y) = F(exp(- t ad(X))Y) =
F(Y - t
z
we immediately obtain Ad*(exp(t X))F = F. The one-parameter group exp(t X) is a subgroup of GF, and hence its tangent X belongs to the Lie algebra OF. This proves one inclusion, the converse is proved analogously.
0 If a Lie group G acts smoothly on a manifold Mm, we can associate with each element X E g of the Lie algebra the unique vector field k on Mm whose integral curves coincide with the trajectories of the one-parameter transformation group exp(t X):
X(x) _
d
The vector field X is called the fundamental vector field corresponding to the element X E g of the Lie algebra. If G acts transitively on Mm, every tangent vector V E TXM'" at a fixed point x E Mm can be realized by a fundamental vector field. This general construction will now be applied to an orbit O' of the coadjoint representation. First, realize a given vector V E TFG' as the value of a
fundamental vector field, f(F) = V. For a further element Y E g of the Lie algebra such that k(F) = V, the equality (X-- Y)(F) = 0 immediately implies
0. e=0
7. Symplectic Geometry and Mechanics
232
Theorem 1 implies X - Y E OF, and the resulting map is injective,
9/9F3X- X(F)ETFO'. For dimensional reasons, it is bijective: the tangent space TFO* to an orbit 0* at F E O' can be identified with the vector space 9/9F. We now define
a symplectic structure wo on each orbit O' C g'.
Definition 2. Let V, W E TFO' be two tangent vectors to the orbit at F E O', and choose elements X, Y E g with f ((F) = V, Y(F) = W. The value of the Kirillov form wo on the vectors V, W is determined by the formula
wo (V, W) := F([X, Y])
Theorem 2. The pair (O',
is a symplectic manifold, and the 2 form
wo. is G-invariant.
Proof. Note first that the 2-form wo is uniquely defined. If the elements X, X1 E g realize the vector V at F, then the difference X - X1 belongs to the Lie algebra OF, and Theorem 1 implies
F([X,Y1) = F([X -X1,Y])+F([X1,Y]) = F([X1,Y]). Moreover, wo is a non-degenerate 2-form. If, in fact, wo (V, W) = 0 for
every tangent vector W E TFO', then we obtain F([X,Y]) = 0 for all elements Y E g. By Theorem 1, X lies in the Lie algebra OF, and hence V = f(F) = 0. It remains to show that wo is a closed form. For two elements X, Y E 9, the function wo (X, k) : O' R is determined by the formula
F([X,YI) Differentiate this relation in the direction of a third fundamental vector field:
2(wo.(X,Y))(F) = d [Ad'(exp(-t.Z)F)[X,Y]]It_o = dtF(Ad(exp(tZ))([X,Y]))It=o = F([Z, [X,YII) Then the expression for the exterior derivative dwo of the 2-form vanishes identically:
dwo. (X, Y, Z) = X (wo (Y, Z)) - Y(wo (X, Z)) + Z(wo (X, f))
-wo.([Y,Z1,X), since it reduces to the Jacobi identity of the Lie algebra g.
0
Corollary 1. Each orbit O' C g' of the coadjoint representation of a Lie group is a manifold of even dimension.
7.1. Sylnplectic Manifolds
233
Example 3. The affine group of R has the matrix representation
G=
{ [0
1 1
a>0,bER}
with Lie algebra 9
=
1[Y l
0
0
:x ,y ER}
.
J11
The computation r[0 0]'[100a -b/a 1 _ 10 ay0bx1 Ad [0 1] [0 1] [0 0] implies that g has one-dimensional orbits. To determine the orbits of the
coadjoint representation, write any element of g' as a pair (a, 0) of real numbers, whose evaluation on the element (x, y) E g is ((a, /3), (x, y)) = ax + /3y
By definition, the group element g =
10
.
1] acts as follows:
(Ad* (g-1)(a, 3), (x, y)) = ((a, R), Ad(g)(x, y)) = ((a, Q), (x, ay - bx)) = ax + /3(ay - bx) = ((a - fib, /3a), (x, y)) . Summarizing, we have Ad*(g-1)(a,0) = (a - f3b,,Oa), and hence for 3 96 0 the coadjoint orbit through (a,,3) is two-dimensional. This example shows that the adjoint and the coadjoint representation of a group G are, in general, not equivalent.
After having discussed examples of symplectic manifolds, we now want to introduce the symplectic gradient, which is the analogue of the gradient of a function on a Riemannian manifold. In this situation, we will make use of the fact that the non-degenerate 2-form w also provides a linear bijection between the tangent bundle TA12m and the cotangent bundle T* M2",.
Definition 3. Let H : M2,
IR be a smooth function on a symplectic manifold. The symplectic gradient s-grad(H) is the vector field on M2in defined by
w(V,s-grad(H)) := dH(V). Example 4. Let {q1, ..., q,,,, p1i ... , pm } be coordinates on M2'" such that the symplectic form w can be written as w = E dpi A dq;. Then
s-grad(H) =
(aH 0
aH
a
This formula immediately follows from the equation defining the symplectic gradient. A curve -y(t) in the symplectic manifold M2rn, represented in the
7. Symplectic Geometry and Mechanics
234
fixed coordinates y(t) = {q1(t), .... q,n(t), pl (t), ... ,p,n(t)}, is thus an integral curve of the vector field s-grad(H) if and only if the so-called Hamilton equations hold:
aH
9i = api
aH and Pi=-aqi
Theorem 3 (Liouville's Theorem). Let H be a function on a symplectic manifold (M2m, w), and suppose that s-grad(H) is a complete vector field with flow 4Dt : M2m , M2m. Then: (1) The Lie derivative of w vanishes,
Gs-g,ad(H)(w) = 0. (2) The flow 't preserves the symplectic volume,
J
dM2m =
dm2m.
J
Proof. From dw = 0 and Theorem 32 in §3.9, we conclude that
Gggrad(H)(w) = d(s-grad(H) J w) = - dd H = 0. ,11m
do not alter the symplectic structure, i.e., 4 (w) = w, and so both assertions are proved. 0 Hence the diffeomorphisms 4)t : A f2m
The existence of this invariant measure has consequences for the dynamics of symplectic gradient fields.
Theorem 4 (Poincare's Return Theorem). Let (M2,, w) be a symplectic manifold of finite volume, and let 4bt : M2m _ hf2m be the flow of the symplectic gradient of a smooth function H. For any set A C M2m of positive measure, the set
B = {x E A : tn(x) 0 A for all n = 1, 2, ...} has measure zero.
Proof. Note first that the intersections 4>_n(B) fl B are empty. Any point
xE
fl B would be a point x E B such that 4n(x) E B C A,
contradicting the definition of the set B. This immediately also implies that
the intersections ' _n(B) fl 4'_m(B) are empty for n
m, and, from the
invariance of the measure, we obtain
x
oc
Evol(B) = Evol((Dn(B)) < Vol(M2m) < x, n=1
n=1
i. e., the measure of the set B has to vanish.
0
This result has several famous generalizations, as the only example of which we quote Birkhoff's ergodicity theorem (without proof).
?.1. Symplectic Manifolds
235
Theorem 5 (Birkhoff's Ergodicity Theorem). Let f be an integrable function on the symplectic manifold (M2,, w), and let Ot be the flow of a symplectic gradient. Then the following limit exists almost everywhere: lim 1 t
Jo
t
f o .0,(x) =: f` (x) .
Furthermore, the function f' is also integrable, and its integral coincides with that of f. Lastly, f* is invariant under the flow fit. Definition 4. The Poisson bracket of two functions f and g on a symplectic manifold is the function
{f,9} = w(s-grad(f),s-grad(9)) = dg(s-grad(f)) = -df(s-grad(g)). Example 5. In the {q, p}-coordinates, we have
{f, 9} _
m Of a9
_ Of a9
5q1
aq1 Opi
i=1 \ apt
In the next theorem, we summarize the properties of the Poisson bracket.
Theorem 6. The ring C'°(M2,) endowed with the Poisson bracket is a Lie-Poisson algebra: for constants cl, c2 E R;
(1)
(2) { f, g} = -{g, f }; (3) { f, {g, h}} + {g, {h, f }} + {h, { f,g}} = 0
(Jacobi identity);
(4) {f,g-h} =g {f,h}+h- {f - g}; (5) s-grad({f, 9}) = [s-grad(f ), s-grrd(9)J.
Proof. The two first identities result immediately from the definition of the Poisson bracket, and the fourth follows from d(g - h) = g - dh + h dg. We prove (5). We insert the vector fields V := s-grad(f ), W := s-grad(g) together with an additional vector field Y into the equation defining the 3-form dw,
0 = dw(V,W,Y) = V(w(W, Y)) - W(w(V, Y)) + Y(w(V, W)) - w([V, W], Y) + w([V, YJ, W) - w([W, Y), V) .
Applying the definition of the symplectic gradient as well as that of the Poisson bracket, we can rewrite this equation as
0 = -V (Y(9)) + W (Y (f )) + Y({ f, 9}) - w([V, W], Y)
+ [V,Y)(9)-[W,Yj(f) = -Y({f,9})+w(Y,[V,WI)
7. Symplectic Geometry and Mechanics
236
Hence s-grad({ f, g}) = [V, W] = [s-grad(f ), s-grad(g)J. The Jacobi identity is a consequence of formula (5). In fact, we obtain
{f. {g,h}}+{g.{h, f}}+ {h.{f,g}} = s-grad(f)(s-grad(g)(h)) - s-grad(g)(s-grad(f)(h)) -s-grad({f,g})(h) = [s-grad(f ), s-grad(g)] (h) - [s-grad(f ), s-grad(g)] (h) = 0. A Hamiltonian system consists of a symplectic manifold (111211, W, H) to-
gether with a function H. The integration of the corresponding Hamilton equation relies on determining the integral curves of the vector field s-grad(H). For this, there exists an analogous notion of first integrals as in the Riemannian case.
Definition 5. A function f : 1112" -. R is called a first integral of the Hamilton function if it is constant on each integral curve of the vector field s-grad(H).
Theorem 7. (1) A function f is a first integral of the Hamilton function H if and only if its Poisson bracket with H vanishes,
{ f, H} = 0. (2) The set of all first integrals of a Hamilton function is a Lie-Poisson algebra.
Proof. We compute the derivative of a function f along an integral curve y(t) of s-grad(H):
dt f o -y(t) = df (j(t)) = df (s-grad(H)) = {H, f } . This implies the first assertion. The second follows from the Jacobi identity for the Poisson bracket.
7.2. The Darboux Theorem The Darboux theorem states that all symplectic manifolds are locally equivalent.
Theorem 8 (Darboux Theorem). Near each point x E 1112n of a symplectic manifold (M2i/.w), there exists a chart h : U C 1112m - R2m in which the symplectic form w is the pullback of the usual symplectic form.
w = h*
dpi n dqi) e=1
7.2. The Darboux Theorem
237
Coordinates with these properties are called symplectic (canonical) coordinates.
Proof. In the cotangent space TTM2," to the manifold at the point x E M2",, we choose a basis al, ... , o , ILl...... m in which the symplectic form at this point is represented in normal form, Iii
w(x) = E ai A pi . i=1
Consider, moreover, a chart (D : V - 1R 2»i around x such that
4t(x) = 0 and w(x) = 4D` (dPAdQi(0)) Denoting the corresponding symplectic form on V by
Wi := fi' I
dpi A dqi I
,
i=1
we see that there exists a neighborhood U C V of x such that for all parameters t E [0,1] the form
wt :_ (1 - t)w + twl is a symplectic structure on U. In fact, dwt = 0, and since wt(x) = w(x) for all t, a compactness argument shows that all the forms wt (t E [0,11) do not degenerate at the same time in a neighborhood of x. Since d(wl - wo) = 0, Poincare's lemma shows the existence of a 1-forma such that the difference
w, -wo = da is the exterior derivative of this 1-form. By subtracting locally, if necessary, a 1-form with constant coefficients from a, we may assume that a vanishes at the point x, a(x) = 0. Dualizing a by means of the symplectic forms wt, we obtain a family Wt of vector fields on U parametrized by t,
wt W, Wt) = a(V) .
Let cp(y, t) E Mgr" be the solution of the (non-autonomous) differential equation Ve(t) = Wt('p(t)), V(0) = y All the vector fields W1(x) 0 vanish at the point x, and the solution
corresponding to the initial condition x is constant, p(x, t) - x. Hence there exists a neighborhood U1 C U of x such that, for every initial condition y E U1, the corresponding solution p(y, t) is at least defined in the interval
7. Symplectic Geometry and Mechanics
238
[0, 1]. Let t : Ul -+ A12°' be the corresponding map. The formula for the Lie derivative of a differential form (Theorem 32, §3.9) implies
(L)
+ 0i ('Ca , ,a (Wt)) = V P1 - w) + Vi (Gww, ((WO) dt 4 (wt) _ 'pi = ipi (wi - w + d(Wt J wt) + W1 J dwt) = tipi (wt - w - da) = 0.
Thus Spi (wl) = pp(w) = w, and 4 o cpl is the chart we were looking for,
(DoVI)" Edpindyi
w.
O
7.3. First Integrals and the Moment Map As in the Riemannian case, some first integrals can be derived from symmetry considerations. The isometries (which are not available on symplectic manifolds) giving rise to these first integrals will be replaced by symplectic diffeomorphisms which, however, satisfy a compatibilty condition with respect to the Hamilton function under consideration. We will describe this in detail in the case of an exact symplectic manifold, i. e. , a symplectic manifold whose symplectic form is the exterior derivative of a 1-form, w = d9.
Suppose that a Lie group G acts from the left on M2" in such a way that each diffeomorphism 1g : M2i' -+ M2ni leaves the form 9 invariant,
l9(0) = 9. These diffeomorphisms 1g are then symplectic, i.e., they preserve the sym-
plectic structure w. Now let X E g be an element of the Lie algebra of the group G, and let k be the fundamental vector field corresponding to X under the G-action on M2in. The evaluation of the 1-form 9 on X is a function,
.6(X) := 0(.k). This construction determines a linear Map '1 : g _ Coo(M2,n) from the Lie algebra g to the space of functions COD(M2in) on the symplectic manifold. Its properties are the subject of the following symplectic variant of Noether's theorem.
Theorem 9 (Noether's Theorem). (1) $ : g - C, (M2,) is a homomorphism of Lie algebras, 'DQX, Y]) = (4'(X ), CY)} . (2) s-grado4' corresponds to the transition to fundamental vector fields,
s-grad(qD(X)) = f(.
239
7.3. First Integrals and the Moment Map
(3) If the Hamilton function H is G-invariant, then
is a first
integral of H, 0.
Proof. Fix an element X E g in the Lie algebra and consider the oneparameter group of diffeomorphisms corresponding to the group elements exp(-t X). Its generating vector field is the fundamental vector field X. The relation l9 (0) = 0 implies that the Lie derivative of 0 with respect to X vanishes,
0 = cX(0) = X-jd0+d(XJ0) = X_jw+d(-t(X)). Thus, for every vector field V, we have the equation
-w(X,V) = V(4(X)) = w(V,s-grad(4'(X))) as well as X = s-grad(4(X)). Using this formula, we compute the difference
{4(X),4(Y)} - .0([X, Y]) = w(X,Y) - 4,([X, Y]) = dO(X,Y) - 4,([X, Y])
= X(4(Y)) -Y(,t(X)) -24([X,Y]) = 2({'F(X), 4'(Y)} - -NX,Y])) , and this yields
{4(X), F(Y)} = C[X, Y]) . If, finally, the Hamilton function is G-invariant, we obtain
-X(H) = 0,
{H,fi(X)} =
0
i.e., 4(X) is a first integral.
The elements of the Lie algebra g provide first integrals for every G-invariant Hamilton function. These first integrals can be combined into a single vector-valued first integral by passing to the dual space g'.
Definition 6. The moment map of a Hamiltonian system with symmetry group G is the map IF : M2m -+ g' from the symplectic manifold to the dual of the Lie algebra defined by
CX)(m)
IF(m)(X)
Theorem 10. (1) The map %P is Ad'-equivariant, i. e., the following diagram commutes: M2m
19
M2.
7. Symplectic Geometry and Mechanics
240
(2) %V is a first integral of H.
Proof. The fundamental vector field X of a G-action has the following invariance property: d X (1,(x)) = dt [exp(-tX) g x] Jt_o _
d lg
d
dt
[exp(-t Ad(g-1)X)
xJ I t-0
= dlg(Ad(g-1)X(x)).
N
N
The 1-form 0 is G-invariant by assumption, and from this we obtain
'P(lgx)X = 8(X(lg.x)) = B(Ad(g-1)X(x)) = '(x)(Ad(g-1)X). 0 In Exercise 11, we discuss the case M'' = T'R3 with symmetry group SO(3, R) and its usual representation on R3. In particular, it is shown that the moment map P : T`R3 --+ so(3, R) = R3 coincides with classical angular momentum, hence justifying its name. Closely related to this situation is the following example:
Example 6. Consider the 2-dimensional representation of G = SL(2, R) on
M2 = V = R2. Its cotangent bundle is T'M = V x V' ^' V x V, since the representation V is self-dual. A group element g E SL(2, R) acts on an element (p, q) E V x V of the cotangent bundle by g - (p, q) = (gR gq)We call two elements (p, q) and (p', q') equivalent if they lie in the same G-orbit, (p, q) - (p', q'). Let p = (pl, p2) and q = (ql, q2) be the components of the vectors p and q, respectively. One easily computes that the moment map is given by V x V - + s1(2, R),
2 (g1P2
(q, p) --
+ g2P1)
q2P2
[
1
-gipi
- 2(g1p2 + g2Pl)
In particular, this map is equivariant with respect to the adjoint action of SL(2, R) on sl(2, R). The moment map is best studied by examining its action on SL(2, R)-orbits. For this, observe that the quantity
det(q, p) := det [qi pi g2
= g1P2 - 92P1
P2
is SL(2, R)-invariant. It thus allows to parametrize the orbit space; we omit the easy proof here'.
1For details on this SL(2,R)-action, we refer to §1.4. of the book by Hanspeter Kraft, Ceometnsche Afethoden in der Invariantentheorie. Vieweg, 1985.
7.4. Completely Integrable Hamiltonian Systems
241
Theorem 11. (1) If det(q,p) _: A 56 0, then
(q, p) -
0 A ( (1)1(0))
(2) det(q,p) = 0 if and only if p and q are linearly dependent. In this case, there exist infinitely many G-orbits, for which one can choose the following representatives: 0 0 0 0 0 0 ( (0)1(0)), ((,U) W) ((1)1(0)),
with AERR.
Thus, the moment map acts on G-orbits as follows:
((01),(A0))
LA02
-A/2J'
\(0)' 0// ~
[0
0
J
((0) ' (1)) ' (CO) ' Co)) [0 0 In particular, the generic orbits with parameter A 54 0 are mapped to semisimple elements of the Lie algebra sl(2, R). Their orbits are 2-dimensional closed submanifolds of sl(2, J). 7.4. Completely Integrable Hamiltonian Systems In this section we will make use of the following fact concerning the structure of discrete subgroups IF of the additive group IRA.
Theorem 12. Let t C ]RI be a discrete subgroup. Then there exist linearly independent vectors vl, ... , vk such that
r=
k
m; vi
:
m; an integer
i. e., F is the lattice generated by the vectors vi, .... vk.
Proof. If I' # {0} is not trivial, we choose a vector ryl E r such that I I7i I I 0 0 and consider the ball D" (0; I I7i I I) The intersection D" (0; I I71 I I) nr
is a compact and discrete subset of IR^, hence finite. Thus, on the straight line generated by 71i there exists a vector 7i E D° (0; 117, 1I) n r realizing the minimum of the distance to 0 E lR'. For this vector we have Ht
7i n r = {m 7j : m an integer),
since any vector x 36 m7i belonging to the intersection (1R 7i) n r would have to lie in one of the segments and then (m+1)7i -x would be a vector on the line through 7t with a smaller distance to 0 than 7l*. If the group r contains only integer multiples of -y,*, the proof is completed.
7. Syrnplectic Geometry and Mechanics
242
Otherwise, there has to exist a vector 1'2 E r\{m - ryi : man integer}. We project ry2 orthogonally to the straight line passing through ry1 and denote by Y2 the resulting vector. It lies in one of the half-closed segments y2 E [m - 'y , (m + 1) ryi ). Let E be the cylinder with axis [m' . (m + 1) - 7i) whose radius is equal to the distance from ry2 to the line through 'Y1. In this cylinder, there are again only finitely many elements of the group r. Let rye be the vector in r n E whose distance to the axis of the cylinder is minimal and which is not a multiple of ryi . Then we have 2
r n {R 7i ED R7;)
mi ry,
:
mi an integer
.
i=1
In fact, if there were a point x 0 miel + m2e2 in r n {Rryi ® R-y }, then x would belong to the interior of a parallelogram in the {'y , 7s }-plane. Taking the difference with a vertex of this parallelogram, we obtain a vector in r lying closer than -y; to the axis of the cylinder. Repeating this construction finitely many times proves the assertion. 0
Corollary 2. Let F C R" be a discrete subgroup. Then R"/r is diffeomorphic to the product of a k-dimensional torus Tk with R"-k,
R"/r
Tk x
Rn-k
Theorem 13 (Arnold-Liouville Theorem). Let (M2m, w, H) be a Hamiltonian system, and let fl = H, f2, ..., fm be m functions with the following properties:
(1) all functions fi are first integrals of H: (2) the functions fi commute, { fi, f;} = 0; (3) the differentials dfl, ..., df,,, are linearly independent at each point; (4) the symplectic gradients s-grad(fi) are complete vector fields.
For a given point c = (cl, ... , c,,) E R'", we consider the level manifold
Al, = {x E M2m : fl(x) = C1, ..., fm(x) = C,n}. Then:
a) The connected components of Al, are diffeomorphic to Tk x
R'-k
b) The vector field s-grad(H) is tangent to Mc. In particular, each integral curve of this vector field is completely contained in one of the level manifolds. c) If Mc is compact and connected, angle coordinates y91, ... , cp,,, can be introduced in Mc -- T' so that the integral curves of s-grad(H)
7.4. Completely Integrable Hamiltonian Systems
243
are described by the system of differential equations
y , = vi,
v; = constant.
Proof. Consider the flows 4i , ... , 4if
:
M2m
M2, of the symplectic
gradients s-grad(fi). Since
0 = s-grad{fi,f3) = [s-grad(fi),s-grad(f3)1, all these flows commute with one another (Theorem 36, §3.9). This determines an action of the additive group R' on the manifold M2'":
(tl, ... , tm) x = .01 0 ... The orbits of this R'-action coincide with the connected components of the level manifolds. In fact, since
0 = {ff,fi} = s-grad(fi)(f3), each function fi is constant on every orbit. Hence, the orbits of the R'"action are contained in the level manifolds. On the other hand, both are m-dimensional submanifolds of M2in, since the differentials dfl, ..., dfm are linearly independent. The isotropy group r(xo) = {t E Rm : t xo = xo} of a point xo E M2rn for the R'-action is discrete. Hence each component of a level manifold is diffeomorphic to the product of a torus and euclidean space:
Rm/r(xo) = T" x ][ Assertions a) and b) are proved, so we turn to the remaining one. Suppose that a level manifold Mc is compact and connected. Choose a point xo E Mf and denote by v1, ... , v,1 E Rm a basis of the isotropy group r(xo). Representing the basis vectors {v;) of the vector space I(tm as linear combinations of the vectors in the standard basis e1, ... , em of R.. M
vi = 1: ajaeo , a=1
we obtain a quadratic matrix A := (aid). Let m
r(m) _
{ni.ei
:
ni an integer
i=1
denote the standard integral lattice in Rm. Then m
0: Rm/r(m) -. Rm/r(xo) _ -
m
o Exi. ei) = E xi . Vi, i=1
defines a diffeomorphism. The inverse of this map, 4D-1 : MM -+ Rm/r(m) = S1 x ... x S',
i=1
7. Symplectic Geometry and Mechanics
244
as well as its components, 4b-1 = (WI, ..., cp"), lead to the angle coordinates
for the level manifold M, In fact, if yl, ... , y' are the coordinates in R"'/r(xo) = MM determined by 1
m
M. VM,
1
then, by the construction of the R'-action on Af,
s-grad(fi) = y; With respect to the
gyom}-coordinates, this yields m a aj,
s-grad(fi) = a=1
awa
In particular. the symplectic gradient s-grad(H) is a vector field with constant coefficients on the torus MM = T', and the third assertion results by taking v. := alb. W e want to discuss more closely h o w the angle coordinates ( 1, ... , m) of a compact and connected level manifold can be determined directly from the commuting first integrals. This will lead to an explicit algorithm for the integration of a Hamiltonian system (M2m, w, H) provided with sufficiently many commuting first integrals. Because of this procedure, these systems are called completely integrable (or integrable by quadrature). Denote by wl (c), ... , ul,,, (c) the frame of 1-forms on Al, dual to the vector fields s-grad(fl ), ..., s-grad(f,,,). The representation of the vector fields s-grad( f;) in terms of the vector fields 8/&pj immediately implies the following formula for the differentials: M
d'pi =
=1
aia wa(c)
Let ik be the closed curve in MM corresponding to the parameter values rpm=0. Then m
aik = f dcpi = F, ai0 f wn(c) 'Y k
a=1
k
Hence, first the coefficients aid and then the angle coordinates can be computed directly from the first integrals. We summarize this in the form of an algorithm comprising five steps.
Step 1. Fix c = (cl, ... , c,,,) E IIt"' and let Al, be compact and connected. Choose a homology basis y', ... , y", for the first homology group H, (Al,: 7L).
7.4. Completely Integrable Hamiltonian Systems
245
Step 2. Compute the symplectic gradients s-grad(fi) of the first integrals
f1=H,f2.....fm Step 3. Determine the frame of 1-forms w1 (c), ..., wm(c) dual to the frame of vector fields s-grad(f l ), ... , s-grad (f.. ) on M.
Step 4. Compute the line integrals frk w0(c), and invert the resulting (m x m) matrix. This yields the matrix A = (aij(c)). Step 5. Compute the angle coordinates (pi(c) on the level set MM from the equations m
dvi = E ai0(c) wa(c), 1 < i < m. Q=1
This procedure computes the angle coordinates Vi(c) on one level manifold Mc. Note that, according to Step 5, these are only determined up to constants. As we vary the parameters c = (cl, ... , c,,,), the Cpl, ..., cp,,, become functions on an open neighborhood of a level manifold Mc C M2m. Since the symplectic gradients are tangent to Mc, the Poisson bracket with the original functions f1, . . . , fm is computed by
IVi,fjI = dpi(s-gradfj) = aij(fl, . . . , fm). Moreover, it is a function exclusively depending on fl, . . . , fm. Similarly, we prove that the functions {Vi, cpj } are also constant on the level sets.
Lemma 1. The Poisson brackets {cpj,Vj} = bij(fl, ...,fm) are functions depending only on fl, ... , fm. In particular, they are constant on each level set 't1c.
Proof. We compute the derivative of {vi, wj } with respect to the vector field s-grad(fk):
{{Vi,'j},fk} _ {Wj,fk},'Pi} - {{fk,'ci},'pj} _ -{aik(fl, ...,fm),'Pi} + {aik /lfl, ...,fm),pj} m 49
m COajk a=1 m
Oya
8ajk
fta
}
a_I
as -
aaEk
aj.)
Thus all the derivatives cpj} (1 < k < m) are constant on every torus T' = Afc. But then the Poisson bracket {Vi, oj} itself is constant on every level manifold Mc. 0
7. Symplectic Geometry and Mechanics
246
Now we alter the angle coordinates, which up to now were considered only on a single level manifold, by a suitable constant on adjacent level manifolds. The aim of choosing these constants of integration for the angle coordinates is to obtain functions cpi commuting on M2n'.
... , B," (yl, ... , y') such
Lemma 2. There exist functions Bl (yl, that the angle coordinates
:= Wi+Bi(fl,...,fm)
Bpi
commute on the symplectic manifold M2m, {(pi , Vj*) = 0.
Proof. Using the notation {cpi, f;} = aij and {Wi,co } = bij, we apply the Jacobi identity to the triples (cpi, ip,, fk) and (ipi, Wj, cpk), and take into account the fact that the Poisson brackets { fi, fl) vanish. Thus we obtain the relations m aaik Oa,j m
E
E
a;a-
Obi;
ab;k
y u 'aka +
ft la
= 0,
aia
1
'
afa +8bki d ajQ J = 0 .
Inserting also the coefficients air of the inverse of the matrix A = (aid), we consider the 2-form m m
Il :_
> bijai°a'p dy° A dyO. i,j=1 a,p=1
The above relations say that fl is a closed 2-form. In fact, the first relation means that the 1-forms m
E a° dy°
of
a=1
are closed, doi = 0. Hence we can choose coordinates z1, ... , z'" such that of = dzi, and 1 becomes M
fl = > bij
dzi Adzj .
i,j=1
Since
Obi; Ozk
_
m
1
Obi,
ay°
Oya
8zk
8bij a= 1
or or aka
the second relation then precisely expresses the vanishing of the exterior derivative dfl = 0. The angle coordinates we set out to find are now taken to have the form m Bpi
aiaBa
Wi + O=1
247
7.4. Completely Integrable Hamiltonian Systems
with functions Bl*, ..., Bm depending only on fl, ... , f,,,. Then m
ai.
aY
a;l_ M
B.
syv
L
aB;
a (aiaaj3 - ajasid)
+ a.8=1
Taking into account the first of the relations above, the condition 0 turns out to be equivalent to
bij =
1:
".
a.3=1
a 33
(aiflaja - ai.aj,3)
This system of differential equations can be reformulated as
aB M 8ii - aya a
r3
m
[: bijaiaajZ, i;=1
and, by Poincare's lemma, it has a solution, since the 2-form f) considered above is closed.
Thus we can determine the constants of integration occurring in the transition from the differentials m
d'pi(c) = E afa(r) wa(c) a=1
to the angle coordinates on the individual level manifolds in such a way that the functions Vi, defined in a neighborhood of a level manifold, commute
on M2'". Now we add so-called action coordinates J1, ..., J,, and thereby bring the symplectic structure near a level manifold into normal form. The Hamilton function H = f, is itself constant on the level manifolds and only depends on the action variables, H = H(J1, ...,J,,,).
Theorem 14 (Action and Angle Coordinates). In a neighborhood of any compact, connected level manifold M,: C At" of a Hamiltonian system determined by m commuting integrals fl, ... , fm, there exist angle coordinates Cpl, ... , v,,, and first integrals J1, .... J,,, such that in
w = dindJi. i=1
In particular, this implies
Vj} = 0 = {Ji, Jj} and {y'i, Jj} = bij.
Proof. First, we determine the angle coordinates near the compact, connected level manifold MM such that
J(pj, ypj} = 0 and
{tpi, fj} = aij(fi,
fm)
248
7. Symplectic Geometry and Mechanics
We look for the functions J1, ..., J,,, using the Ansatz J, = Ai(fl, and compute the Poisson bracket m
8Aj ai. OY.
The condition {,..1j} = 8ij leads to the system of equations OA
ayj
= aij
where a'j is the inverse of the matrix aij. By Poincare's lemma, the integrability condition is
8aij
8aik
8yk = $yjj
On the other hand, we obtain from the Jacobi identity 0 = {`r'k,{iPi,.fj}}+{Vi.{fj, 7k}}+{.f .{Y"'k,Y^'i}}, and, taking into account {j,9i.k} = 0, this immediately yields
8aq
E fta aka
= L 8ukj
8 a aip..
a=1
Q=1
y
A simple computation shows that this relation is equivalent to the integrability condition for the coefficients aij of the inverse matrix. 0
Example 7 (Two-dimensional Hamiltonian System). Consider in R2 with the symplectic structure w = dp A dq a Hamilton function H(q, p) for which the level curves (q, P) E 1R2
H(q. p) = c}
:
are closed. The action variable J = J(H) is a function of H, and, together with the angle coordinate , we have
dpAdq = dV AV = Applying the Hodge operator * of R2, we see that dy: is proportional to the 1-form *dH, * dH .
dV
J'(H) IIdHII2 The integral of d,o over each level curve is an integer, which we take to be equal to -1. This condition is called the classical Bohr-Sommerfeld condition. The equation J'(c)
1
nor IIdHII2
* dH
7.4. Completely Integrable Hamiltonian Systems
249
uniquely determines the action variable J = J(H) in terms of the Hamilton function H. Consider the domain bounded by the level curve A4,,
Q, = {(q, p) E R2 : H(9, p) 5 C1. The vector field W := grad(H)/Ilgrad(H)II2 satisfies W(H) - 1; hence its flow +t maps the set flc onto tl +t. We compute the change of the area of the domain: d
(vol(S2c)) =
r
Ji1c
tli o
t (J$2) -J t k
d(W i dlt2) =
st4
J
W I dR2 =
lo
2)
If
,1 4
=
Js24 Gbb'(dR2)
IIdHII2 *
dH = X(c)
.
Thus the action variable J = J(H) can be interpreted as follows: J(c) is the volume vol(Q,) of the domain bounded by the level curve hf, _ {(q, p) E R2 : H(q,p) = c}.
Example 8 (Spherical Pendulum). Consider spherical coordinates on the sphere S2\{N, S} with the north and south pole deleted, h(4,) = (cos cp cos ti, sin yp cos 0 , sin
g) .
The Riemannian metric of the sphere is then described by the matrix (see Example 14 in §3.2) 9
_
1
L
0
O cost'
Denote the coordinates in the cotangent bundle by (cp, o, pw, p v) and consider the Hamilton function
H = 2v+2 H describes the motion of a pendulum of length one suspended in the center of the sphere. The meaning of the angles tp and t'} can be seen in the picture to follow. For simplicity, the gravitational constant was taken to be one.
250
7. Symplectic Geometry and Mechanics
Z
The variable (p does not explicitly occur in the Hamilton function; hence P:= p,, = Ocos2 iii is a first integral, {H, P} = 0. The Hamiltonian system (T*S2, H) is thus completely integrable. The level manifold
M2(ci,c2) :_ {pw,po) E T* S2 : P = c1, H = C2} is empty for negative values of the parameter c2, and it consists of the south pole remaining at rest in the case c2 = 0. Hence we suppose that the parameter is positive, c2 > 0. The equations describing the manifold A12(cl,c2) are
C1 = pr,,
c2 =
2 + 2 cs2 i + 1 + sin ?P.
The relation cl = 0 implies that cp has to be constant. In this case, the second equation of motion reduces to that of a planar pendulum, so we will henceforth exclude this case. Depending upon the sign of c1, the function is monotone increasing or decreasing, i. e., the pendulum does not change its direction of motion. Rewriting the second equation and setting z = sin 4i, we obtain
p=
c2-1-z- 2(1
C- 2
z2)
U(z)
and see that the function U(z) thus defined cannot be negative. The limiting
case p = 0 corresponds to the pendulum moving in a fixed plane, hence on a meridian. To see where the function U(z) can be strictly positive, we multiply it by the denominator 1- z2 and look for the zeroes of the resulting cubic polynomial,
V(z) = (1 - z2)U(z) =
c2 (c2 - 1 - z)(1 - z2) - 2
.
7.4. Completely Integrable Hamiltonian Systems
251
At the boundaries of the interval, V(±1) = -c/2 < 0 is negative, and at +oe the function V diverges to +oc. Hence one of the three zeroes has to lie above 1, and, since it cannot correspond to any angle 1;i, this zero has no physical relevance. In the generic case, the other zeroes belong to the interval (-1, 1); between them V, and hence also U, is positive. We conclude that V(z) has the qualitative behavior of the graph on the previous page. The mass point can only move between the two meridians corresponding to the zeroes in the interval (-1, 1). The boundary values t ,'4'2 defined this way are actually reached at the end of every up or down swing. Summarizing,
the level manifold can be parametrized by the two parameters 4z = V and
0=
via ( P Cl!
cos
- 2 - 2sini ) .
It consists of 2 two-dimensional tori, where ' only takes values in (>G1, 021. The corresponding coordinate vector fields are a a a a apti; a
a = TT' a = a + o '
ap
,
We express the symplectic gradients of the first integrals in the coordinates of the level manifold: P;, a a d P,2. a s-grad(H)
= cost ;, a + p av - a (2 cost V, + 1 + sin V) aplo Pro
a
a
= cost' ap +Pva +Pv a i a a _ 2 Zi app. + Pv a,. pw
s-grad(P) = a =
ate,
.
"PV
7. Symplectic Geometry and Mechanics
252
The dual forms w1(cl , c2) and w2(cl, c2) are thus 1
wl(Cl,C2) =
de*,
PO
C12
w2(cl,c2) = dw - p o Cos '+G dip
.
We compute the periods with respect to the homology cycle 'rl which is parametrized in M2(c1i C2) by 't()'. The factor 2 is a consequence of the fact that the boundaries, y1,'Y2, of the interval correspond to the minimum and the maximum of the motion, whereas a cycle is meant to be a motion between two extremal points of the same kind:
b1l =
f wl = 2 ti
J
%
c>>
Py
b12 =
f
w2 = -2c1
7i
r 02 diP Jv Pb oos2
Let, similarly, 72 denote the homology cycle determined by 0 < gyp` < 2Tr. Then b21 =
Jwi = 0,
Jw2
b22 =
= 27r .
The matrix occurring in the algorithm for computing the angle coordinates is now easily calculated:
-l [b21
- [a21
b221
a22]
1
-j w2
f7twl
2vfnwl
0
1/27r
The resulting quotient of the basic frequencies is
V1 =all =_1f1-'2= 21r V2
a12
,
C1
7r
112
.//ol
di&
cost tai ' PW
This is nothing but the perihelion precession, i.e. the total variation of the angle V for a complete cycle: O +a
dye = 2
d` d1/b = 2
02
. dpi = 2c1
d
27r L1 .
COS2t' ' PG = .1,1 J 'rI J d it fo , fo, I In general, the motion is quasi-periodic. The trajectory is closed if and 1
only if vl /v2 is rational; otherwise, it is dense on the torus. The transition to action angle coordinates allows us to determine the physically relevant basic frequencies of the system without having to explicitly perform the integration. This accounts for their importance in astronomic perturbation theory.
7.5. Formulations of Mechanics Newton's equations describe the motion of a mechanical system under the impact of a force. The latter is understood as a vector field depending upon position and velocity, and is central for Newton's formulation of mechanics.
7.5. Formulations of Mechanics
253
During the 18th century, this view changed in that Lagrange considered the action integral as the fundamental quantity for the description of dynamics. Newton as well as Lagrange formulated mechanics within the tangent bundle of configuration space. In the 19th century, by transition to the cotangent bundle, Hamilton succeeded in formulating the dynamics of mechanical systems within the framework of symplectic geometry. The aim of this section is to explain these fundamental ideas of mechanics and the related mathematical structures.
Newton (1643-1727)
Newtonian systems
Lagrange (1736-1813)
Hamilton (1805-1865)
Lagrangian systems Hamiltonian systems
i Newtonian systems with potential energy
1
hyper-regular Lagrangian systems
Legendre transformations
Mathematical Contents: Riemannian geometry
Finsler geometry
symplectic geometry
Formulation of Mechanics According to Newton In Newtonian mechanics, the state of a mechanical system is described by finitely many real parameters. This leads to the notion of configuration space. which is a smooth and finite-dimensional manifold Mm. A motion of the mechanical system is a curve 'y : (a, b) - M' in configuration space.
Its tangent-the velocity-is then a curve ' : (a, b) - TM' in the tangent bundle. and this space is called the phase space. According to Newton, the forces acting on the mechanical system are described by vector fields depending only on position and velocity, that is, vector fields X on TM"'. However, not all vector fields are allowed, since a force can act only in space.
7. Symplectic Geometry and Mechanics
254
The force vector field X has to satisfy the condition
drr o X = Id. Here 7r : TM" - Mm denotes the projection of the tangent bundle, and dir : TTM' --+ TM' is its differential. Summarizing, we arrive at the notion of a Newtonian system.
Definition 7. An autonomous Newtonian system is a triple (Mm, g, X ) consisting of a manifold M'", a Riemannian metric g, and a vector field on the space TM'" such that d7r o X = IdTMm .
The function T : TM'" -+ R defined by T(v) := 2 g(v,v)
is called the kinetic energy.
Definition 8. A motion of the Newtonian system (M"', g, X) is understood to be a curve y : (a, b) - M' in configuration space whose curve of tangents ry : (a, b) - TM'" is an integral curve of X,
'1'(t) = X MW This is an invariant formulation of Newton's equation.
Example 9. Consider R with the coordinate x and identify TR = R2 with R2. Here the coordinates are denoted by {x, i}. The vector field 8 1 0 2
X=
wax+m(-k x-P)8i
on TR has the required projection property, and a curve x(t) in R is a motion in this Newtonian system if x(t) is a solution of the oscillator equation
ml(t) = -k2x(t) - pe(t) . Example 10. Let (Mm,g) be a Riemannian manifold. We define a vector field S : TM°1 - TTMm on its tangent bundle-the so-called geodesic spray-as follows: If v E TAM' is a tangent vector, then there exists precisely one geodesic line y,,,,, : (-e, c) -> Mm such that
7'r,.(0) = X, % AO) = v. Consider its tangent curve, %,v : (-e, e) - TM'", and set S(v)
dt (%"(t)) t=o
The relation 7r o ryx,,, (t) = ry v (t) implies dir o S = IdTMm. Hence (M', g, S) is a Newtonian system, and the motions of this system are the geodesic lines
7.5. Formulations of Mechanics
255
of the Riemannian manifold (Mm, g). In the coordinates {x', ii} of the tangent bundle, the geodesic spray is given by the formula
S=
x' i=1
axi
- E r;k x'xA axi i,j,k=1
a straightforward consequence of the system of differential equations describing geodesic lines-see §5.7. Many Newtonian systems have a potential energy. This is a smooth function
V : Mm --+ R defined on configuration space. The gradient grad(V) is a vector field on Mm locally determined by grad(V)
'"
av a
- `i,j=1 g'' (x) axi axj
In the sequel we will need, however, a different vector field, denoted by grad(V). This will be a vector field on phase space. At the point v E TM'", it is defined by the following equation: d
grad
(v + t - grad(V)(ir(v))) e-o
(V) (v) = dt In the manifold Mm, the curve v + t - grad(V)(a(v)) projects to the base point rr(v) E M' of the vector v E TM'. Hence the vector field grad(V) projects to zero under the differential dir, d7r o grad(V) = 0,
and, for any potential energy V, the vector field X := S - grad(V) is an admissible vector field on TM'" in the sense of Newtonian mechanics. In local coordinates, we obtain the formula in
av a
E e(x) axi aij i,j=1 Definition 9. A Newtonian system with potential energy is a triple (M, g, V)
consisting of a Riemannian metric g and a potential energy V : M' -+ R. The corresponding force vector field is
X = S-grad(V). A motion in a Newtonian system with potential energy is defined to be a Mm whose tangential curve (a, b) -+ TM'" is an curve 7 : (a, b) integral curve of X.
Definition 10. The energy of a Newtonian system (Mm, g, V) with potential energy is the sum of the kinetic and the potential energy,
E:TM' R, E=T+Voir.
7. Symplectic Geometry and Mechanics
256
Theorem 15 (Energy Conservation for Newtonian Systems). Let (Mm, g, V) be a Newtonian system with potential energy, and let X = S - grad(V) be the force vector field. Then dE(X) = 0.
In particular, E(y(t)) is constant for every motion-y(t) of the system.
Proof. In local coordinates, the energy E and the force vector field are given by the formulas I M
E = 9 E 9ij(x)xY +V(x), i,j=1
X =
i=1
x' a ii -
k=1
I'v (x) +
8i 9ikW I i=1
82k
.
Using the expression for the Christoffel symbols riki from §5.7, 2
gkQ (x) (Lgii (x) + 8x (x)
-ij axQ
(X))
we obtain dE(X) = X (E) = 0 by an elementary calculation.
,
0
Motions with large energy in a Newtonian system (M'", g, V) with potential energy are-up to a change of parametrization-geodesic lines with respect to a new Riemannian metric. This construction will lead to the MaupertuisJacobi principle. Suppose that the potential energy V : Mm - R is bounded from above by Eo,
sup{V(x): xEMm} < Eo. Then g' = (Eo-V) -g is a Riemannian metric on the manifold Mm. Consider a motion y : (a, b) - MI of the Newtonian system (MI, g, V) with energy Eo,
Eo = 29('Y(t),'Y(t))+V(7(t)) Since Eo > V(y(t)), the tangent vector y(t) vanishes nowhere, and the function
s(t) := f r' (Eo - V(7(µ)))dp a
becomes a strictly monotone function a : (a, b) - (0, b* ). We invert this function and thus view t E (a, b) as a function of the parameter s E (0, b*), t = t(s). Let the curve .y*(s) be the initial curve y in this new parametrization.
7.5. Formulations of Mechanics
257
Theorem 16 (Maupertuis-Jacobi principle). Let -y(t) be a motion of the Newtonian system (Mm, g, V) with energy E0. Then ry' (s) is a geodesic line
in Mm with respect to the Riemannian metric g' = (Eo - V) g.
Proof. The Christoffel symbols1) r and 't V of the metrics g and g' are, in local coordinates, related by the formula k
rtj
=
1
k
_ 8V
1
8V
r'3 + 2 (Eo - V)
ajk
C7xi
8V ak
-&k +
02-a
9 9ij
We write the motion -y(t) _ (x' (t), ... , x'(t)) in local coordinates. Then dxk
_
ds d2xk
dsk dt
_
dt
_
- V) dt
dxk m 8V dxa 2(Eo - V)3 dt Ox- dt '
d2xk
1
1
1
2(Eo - V)2 dt2
ds2
dxk
1
1
72= (Eo
Ws-
and, using the equation of motion, d2xk dt2
--
m
rk Ix' dx'
m
ij dt i,j=1
OV ka
dt - a=1 8xa 9
as well as the energy condition m
dxi dxj
E i,j=i gij dt
dt
2(Eo - V),
we obtain the claimed result: d2xk WS-2 +
k dx' dx'
m
I'i ds ds ij=1 1
+ `1(Ec, __V) 3
m
a=1
=
8V ka 8xa9
8V
1
ka
-2(Eo - V)2 a=1 8xa9 in
dx' dx-I
ij=1 go dt dt
= 0.
Formulation of Mechanics According to Lagrange The transition to Lagrangian mechanics proceeds by considering the Lagrange function of a Newtonian system with potential energy.
Definition 11. Let (M'", g, V) be a Newtonian system with potential energy. The Lagrange function (or Lagrangian) L : TM' R is the difference of kinetic and potential energy,
L = T-Voir.
7. Symplectic Geometry and Mechanics
258
Theorem 17 (d'Alembert-Lagrange). A curve 7 : (a, b) -+ Mm is a motion of the Newtonian system (Mm, g, V) with potential energy if and only if the Euler-Lagrange equations hold: dt
(ex (y(t))) = 8x (7(t))
Proof. We prove this theorem again using local coordinates. The curve 7(t) _ (x1(t), .. . , xm (t)) is a motion of the Newtonian system with potential energy if and only if it solves the system of differential equations
ik
=-
»i
m
I
ji`ij
ij=1
- a=1 E
Va9ak
For brevity, we denoted by V. the partial derivative of the potential function, 4G, := 8V/8x°. The Lagrange function is
L('y) =
2
gijii2j - V(1'), i,7=1
and this leads to the difference d dt
8L
m
8L
(8zi
a,;3=1
d9ia
1 99. Q
ax-*l
2
axi)
axa +
m
9iaya+ V; . a=1
Multiplying the Euler-Lagrange equations by gik and summing over the index i, this system of equations turns out to be equivalent to m m a9ia 109.e M 0 = xk + M V 'k +
E i-l
i9
E ` 8x3 i=1 a.3= 1
2 axi
9
The claim now immediately follows from the formula for the Christoffel symbols r 13.. o Thus the equations of motion of Newtonian mechanics are, in the case of a potential force, equivalent to the Euler-Lagrange equations. For the latter, it does not matter that the Lagrange function arises as the difference of a kinetic and a potential energy. Hence we define:
Definition 12. An autonomous Lagrangian system is a pair (Mm, L) con-
sisting of a manifold Al" and a smooth function L : TMm - R. A Lagrangian motion is a curve -y : (a, b) -- M" which solves the system of Euler-Lagrange equations Wt
(er ('i (t))) = dL ('Y(t))
259
7.5. Formulations of Mechanics
Example 11. Let (M'", g) be a pseudo-Riemannian manifold and A a 1form on it. The Lagrange function
L(v) =
g(v, v) - A(v)
2 generalizes, in a sense to be discussed in Chapter 9, the motion of a charged particle of mass m under the influence of the Lorentz force of the electromagnetic field generated by A. Motions in a Lagrangian system can be understood from the point of view that they are critical points of the action integral L. This measures for a curve -y : [a. b] -p Mm the mean value of the Lagrange function on this curve, b
L(y) :=
Ja
Theorem 18 (Principle of Least Action). A curve -y
:
[a. b]
Mm is a
motion of the Lagrangian system (Mm, L) if and only if the variation of the action integral vanishes for every variation y1, of the curve with fixed initail and end points, y1,(a) = y(a), -y. (b) = y(b): d dµ
(jb)
0. LO =
Proof. We compute the derivative of the action integral with respect to the parameter p in coordinates -t,, (t) = (x1(µ, t), ... , x'"(µ, t)) by partial integration: =Jab
d (,C(-t,.)) 1,0
[aL((t))
- d (a (7(t)))
(0, t)dt.
The functions 8x'(0, t)/8µ are arbitrary functions vanishing at the end points of the interval [a, b], and hence the Euler-Lagrange equations are equivalent to the condition dµ (L(yµ)) Iµ=o = 0.
For general Lagrangian systems, there exists a notion of energy which, on the one hand, generalizes the energy of a Newtonian system with potential energy, and is, on the other hand, a preserved quantity. This Lagrangian energy is obtained by first introducing the Legendre transformation. Let
L : TM'
R be a Lagrange function, and let v E .,Mm be a vector at the point x E Mm. Now restrict L to the tangent space .,Aim and consider the differential D(LIT=SIm)(v) at the point v E TTMm. This is a linear map TTll1"' IR. hence a covector in T *M'.
7. Symplectic Geometry and Mechanics
260
Definition 13. The Legendre transformation C : TM'
T'M' of an
arbitrary Lagrangian system is the map ,C(v) := D(LIT=Mm)(v).
Example 12. If the Lagrange function L = Zg - V is the difference of a
kinetic and a potential energy, the following relation holds for v, w E TM':
G(v)(w) = 2 . D(g)(v)(w) = &,w). Hence the Legendre transformation L : TM"' T* M' is simply the identification of the tangent bundle with the cotangent bundle via the metric.
Definition 14. The energy of a Lagrangian system (MI, L) is the function E : TM' -' R on the tangent bundle defined by
E(v) = L(v)(v) - L(v). In the case of a Newtonian system with potential energy, we have
E(v) = L(v)(v) - L(v)
2g(v, v) + V(r(v)) .
This shows that the energy in the sense of Lagrangian mechanics coincides with the Newtonian energy.
Example 13. We compute the Legendre transformation for the Lagrange function from Example 11. Let (x, y) E TTM" with local coordinates
xl...
, xm
and y', ... , y'. Then, m
...,y"') = 2
m Egjjy'1!r - EAiy', i
1,3
and, as an element of T,M, its differential is m
m
D(LITTMm)(yl,...,ym) = mEgify'dhe -EAidy'.
i,j
i
We evaluate this map at (x, v), v = (v', ... , v'"): G(y)(v) = D(LI T:MO1) (y)(v) = m . g(y, v) - A(v) . By definition, this yields for the energy the value
E(v) = C(v)(v) - L(v) =
2 g(v, v),
which has the remarkable property of not depending on A. A more careful physical analysis shows that the energy of a charged particle indeed only
depends on the electric, not on the magnetic field. This is related to the fact that the magnetic field does no work on the particle (see §9.5).
7.5. Formulations of Mechanics
261
Remark. Denoting the coordinates on TM' by {x', , X-, x', ...'e } and the coordinates on T` Mm by {qj, ... , qm. pi, .... p,"}. we see that the Legendre transformation f- is given by
qi = x'
and
aL pi =5p,
and the expression for the energy E takes the following form:
E(x,
_ "' OL xi - L(x, x) x) - ==Y 8ii
.
Theorem 19 (Conservation of Energy for Lagrangian Systems). The energy E(y' (t)) of each motion ti(t) of a Lagrangian system (Mm, L) is constant.
Proof. The energy of a curve is
E(7(t)) _ t-1
8L
dxi
f72i
dt
- L(ti(t))
and by differentiation we obtain "'
d
dt E(Y(t))
_
ij=1
8L
dx' dx-
8L
dx'
1
C 8ii8xj dt dt + C7x'&i dt dtZ J -
' &L dx' i=Y
dxi dt
Using the Euler-Lagrange equation m
dx' "` c7L OL d2xi EY Ox-. = O±'Oxi dt + axiaxj dt2 i,7=Y ij= i3L
we immediately obtain the assertion.
0
Thus the energy is a first integral for any motion in a Lagrangian system. As in §7.3. further first integrals can be derived from symmetries of the Lagrange function. To this end, consider a one-parameter group of diffeomorphisms ,P6 : Al' M' of the configuration space as well as its generating vector field, d
V(x) := d('ts(x))j8=0 on Al'. The differentials d(4 Q : TM' TM"' are diffeomorphisms of the phase space TM"' into itself.
Theorem 20 (Noether's Theorem). Let the Lagrange function L be invari-
ant under the action of a one-parameter group of diffeomorphisms, L(d(4s)(v)) = L(v). Then the function fv : TM' --+ R defined by
fv(w) = lim
µ is a constant of motion for the Lagrangian system (Mm, L). µ-.O
7. Symplectic Geometry and Mechanics
262
Proof. Let the one-parameter group of diffeomorphisms $t be determined in coordinates by
4tt(xil ...,xm) = ('Ft(x1, ...,X"), .... Dm(x'....,xm The invariance of the Lagrange function implies that m
m aL a(De
ax= as
I
aL O4
dx'
+ E axj . ax=as
and the vector field V has the components m
V=E =1
dt
= 0,
a
s
s=IUC7-
x
a3
'
i
Thus we obtain
fv(7(t)) = 1 8x as Is.o' and from the Euler-Lagrange equation we conclude that d
d
m
m IL 81' ax= WS=0 I8=0
+
aL 02V ax' ax_as o
dxj
dt = o. o
Example 14. To each transformation group'Ft : Mm - Mm preserving the metric g and the potential function V there corresponds a first integral,
fv(w) = g(w,V), which is linear in every fiber of the tangent bundle Tlblt. We made use of this first integral in Theorem 37, Chapter 5, to integrate the geodesic flow on surfaces of revolution (Clairaut's theorem). Hence first integrals of the geodesic flow which are linear in the fibers arise from isometries of the metric.
Example 15. If 't preserves the pseudo-Riemannian metric g as well as the 1-form A from Example 11, then the Lagrangian of a charged particle in an electromagnetic field is invariant, and hence Noether's theorem can be applied to yield the invariant
fv(w) = m g(w, V) - A(V).
Formulation of Mechanics According to Hamilton We will arrive at the formulation of mechanics according to Hamilton by starting from a Lagrangian system (M, L) with bijective Legendre transformation C : TM' T'Mm. These Lagrangian systems are called hyperregular. A regular Lagrangian system is one whose Legendre transformation is locally invertible. For simplicity, we confine ourselves here to the case
7.5. Formulations of Mechanics
263
of a globally invertible Legendre transformation. For example, Newtonian systems with potential energy or the system of a charged particle moving in an electromagnetic field (Example 11) are always hyper-regular.
Definition 15. Let (Mm, L) be a hyper-regular Lagrangian system with Legendre transformation C : TM' --+ T* M' and energy E : TM' --+ R. The function
H := EoL
1
defined on the cotangent bundle is called the Hamilton function (or Hamiltonian) of the system.
Example 16. In the case of a Newtonian system (Mm, g, V) with potential energy, the Legendre transformation is determined by the formulas m
qi =
(9L and pi= axi = E gi.±
xi
a=1
Inverting this transformation leads to m
x' = qi
and
i' = E gtnPa , U=1
and the formulas for the energy E, the Lagrange function L, and the Hamilton function H read as follows:
(1) E = 2
i j=1
gijx'i +V(xl, ...,x'"),
m
(2) L = 2 E gijz'xi - V(xl, ...,xm), ij=1
(3) H = 2 E g''rpipj+V(gl,...,gm). ij=1 Through this change of phase space-i. e., replacing the tangent bundle TM"' by the cotangent bundle T'M'-we enter the realm of symplectic geometry, since T*M'" always carries the symplectic form w = d9.
Theorem 21 (Hamilton's Theorem). Let (M'", L) be a hyper-regular Lagrangian system. A curve -y : (a, b) -' AM"' is a Lagrangian motion if and only if the curve G(ry) : (a, b) -. T' M"' is an integral curve of the symplectic gradient s-grad(H) of the Hamilton function.
Proof. The Legendre transformation C is defined by i
OL
qi = x, Pi = axi . For its inverse map, we introduce the notation
x' = qi
and
i = I (ql, ... , qm, Pi, ... , pm).
7. Symplectic Geometry and Mechanics
264
The Hamilton function is then m
H= a=1
and we compute its partial derivatives:
+
1: (P. -
apt OH
a
M
a=l
E51-aL aV _ m
a
aqt
_ OL axt
OL
axt =
.
(i+)
Thus we obtain a formula for the symplectic gradient:
s-grad(H) = E t=l
A curve
G(?'(t)) = 1 21(t), ... , 2m(t), 'oil (fi(t)),
--
,
OL
is an integral curve of the vector field s-grad(h) if and only if
it = V and
Wt
(5p IL
(ti(t))) = axt ('Y(t))
The first equation is trivialy satisfied by setting it = fit, and this proves the 0 assertion.
Exercises 1. In R4 with the symplectic structure w = dxl Adx3 +dx2 Adx4. we choose the following four diffeomorphisms a, b, c, d:
a(xl,x2,x3,x4) = (21,x2+1,x3,x4), b(xl, x2, x3, x4) = (xl, x2, x3, x4 + 1), C(xl, x2, x3, x4) = (xl + 1,x2, x3, x4) , d(xl, x2, x3, x4) = (x1, 22 + 24,x3 + 1,x4 ),
and denote by r the group of motions of R4 generated by them. Prove that w is a F-invariant 2-form. Hence w induces a symplectic form on the manifold M4 = R4/F. Prove that M4 is compact. Finally, denote by [F, F) the commutator group of F. Then F/[t, rJ is a free abelian group of rank three (Thurston, 1971).
Exercises
265
2. Prove that the symplectic form w of a compact symplectic manifold M2m can never be an exact differential form. Hence the second de Rham cohomology HDR(M2m) is non-trivial. In particular, for m 1, even-dimensional
spheres St' have no symplectic structure.
3. Consider M = R2\{0} with the symplectic form w = dx n dy and the vector field
V_
x
8
y
8
x2+y2 8x+x2+y2 8y
a) Let cp = arctan(y/x) be the polar angle defined in every sufficiently small neighborhood of (x, y)
(0, 0). Compute grad(W) and
b) Conclude from this that V is no Hamiltonian vector field on all of M.
4 (Continuation of Example 3). Prove that the Liouville form on the twodimensional coadjoint orbit through the element (a, Q) E g` (3 54 0) is given by the expression
w=
daAdf
.
Hint: Show first that the fundamental vector fields corresponding to the Lie algebra elements (1, 0) and (0, 1) are
aa, and - Qom. 5. Let V be a vector field on a manifold Mm and 1 = V(x) the associated differential equation. A first integral of V is a smooth function h : Mm - R which is constant along every solution of the differential equation. a) Prove that h : Mm -p JR is a first integral of V if and only if dh(V) = 0.
b) Prove that the set Cv (M'n; R) of all first integrals of V is a subring of
C' (M'; R). c) Show that h(x, y) = y2 - 2x2 + x9 is a first integral of the vector field
V = 2y 8/8x + (4x - 4x3) 8/ay in R2. Describe the geometric shape of the integral curves in the (x, y)-plane by means of h.
d) Find a vector field in the plane which has no non-trivial first integral.
6. Let f = a + b i : U -p C be a holomorphic function on an open subset U of JR2 ^_' C. Consider the vector field V = grad (a).
a) Prove that b : U -. R is a first integral for V.
266
7. Symplectic Geometry and Mechanics
b) Describe the geometric shape of the integral curves of this vector field for the functions f(z) = zk and f(z) = z + 1/z. 7. Consider on ]R3 a vector field B (a magnetic field) and its 2-form
B= as well as the symplectic form on the phase space R3 x JR3 with coordinates (9, ii) = (x, y, z, Vx, Vy, Vz),
wB = m(dxAdvx+dyAdvy+dzAdvz) - - B. C
As Hamilton function, we choose the kinetic energy
Ho = 2 (vZ+vy2+vZ). a) Show that the defining equation for the Hamiltonian vector field associated with Ho is equivalent to the Lorentz equation dv
_
e
c.vxB.
mdt
(*)
b) If B = dA is exact and A denotes the associated vector field, show that the map
f:
1R3 x JR3 -R 3 x 1R3,
(4> ) ' (q, mii + A)
(q, p)
is a canonical transformation, i.e., for the canonical symplectic form wo and the Hamilton function HB,
wo = dxndpx+dyAdpy+dzndpzi HB =
-lip-
-All
the following equations hold:
f*wo = 'B, f*HB = Ho, and equation (*) does not change.
c) If B is constant, any particle moves on a helix. Hint: Interpret torsion and curvature as first integrals. If both these quantities are constant for a curve, then this curve has to be a helix (Exercise 6 in Chapter 5). 8 (Plane Toda Lattice). Consider in JR2 the second order differential equations
i=
-ex-y,
y=
ex-y
which in phase space (x, y, i, y) E JR4 are equivalent to t; = X
a a
X(x,y,i,y) = a TX + y ay -
ex-y a +
ai
The energy E = (i2 + y2)/2 + ex-y is a first integral.
ex-y
ay
with
267
Exercises
a) Show that this system has further first integrals, for example P = i + y or K = (i - 2y)(y - 2i)/9 - e---y. These quantities are related by E + K = 5P2/18.
b) The set M2 (E, P)
(x, y, i, y) E 11 P4 :
(i2 + y2)/2 + eZ-v = E, ±+y=P}
is not empty if and only if 4E - P2 > 0. In this case M2(E, P) is a smooth two-dimensional submanifold of R4 without boundary . This leads to a decomposition of ]R4 into a family of submanifolds, and every integral curve of X lies completely in one of them. c) Show that each of the submanifolds M2(E, P) lies in an affine subspace of dimension three and is diffeomorphic to R2.
d) If {(t) = (x(t), y(t), i(t), y(t)) is an integral curve of X in M2(E, P), then
x y i+y = P and f 4E-P -4ez-V =
t
Show that (A > 0) dz
A -Bez
_=
1
In
v/-A
vA- A -Bee ( I + A -Bee
'
and use this formula to integrate the equations for the integral curves in M2(E, P) completely.
9 (Euler Equations). Let I : ]R3 - R3 be a symmetric positive definite operator. Consider the differential equation I(i) = 1(w) x w, where x denotes the vector product.
a) Prove that this differential equation has two first integrals, the energy 2E = (I(w),w) and the momentum M2 = (I(m), I(w)). b) Conclude from this that the integral curves of the differential equation are intersections of two ellipsoids with center 0 E IR3 (I 0 IdR3). In particular, all integral curves are closed. Let Il be the smallest and 13 the largest eigenvalue of I. Then there exists an integral curve only for 2EIl < M2 < 2E13. . 10 (Motion in a Central Force Field). Let a potential force F act on a point of mass one in R3, dU
-grad(U(r)) F drr) r with a function U(r) depending on the radius r = IIxil only. The energy E = 11±112/2 + U(r) is a first integral.
7. Symplectic Geometry and Mechanics
268
a) Show that M = x x . is a further first integral (M is the angular momentum), and conclude from this that every trajectory of the point lies in a plane in R3. b) Determine for which values of the parameter M E JR3 the level surface
A3(M) = {(x,x)ER6:xx:i=M} is a three-dimensional submanifold of phase space R6.
c) Let r(t) be the distance to the origin of a motion whose an;;ular momentum is M. Then d2r
_ _dr dU
dt2
1IM112
+ _73-
i.e., r(t) describes the motion of a point in R1 under the effective force F2 = -grad(V) with effective potential V(r) = U(r) + IIMII2/2r2. The energy of this motion is
E* = 2 + U(r) +
I I2r2Iz
Prove that this energy E* coincides with the energy E (for fixed angular momentum M).
Hence, if x(t) E R3 is the trajectory of a point moving under a force F and r(t) is its distance to 0 E K3, then
V2E -
IIMII2
- 2U(r) = r',
i.e.
r
J
-dt = t
11 (Classical Moment Map). Consider K3 with the defining representation of SO(3, K).
a) Prove that this representation is equivalent to the adjoint representation of SO(3, K) on its Lie algebra so(3, K), if R3 and so(3, K) are identified via the map vl R3
so(3, f8),
v = v2 u.- v =
0
-V3
v::
v3
0 vi
-V1 0
-V2
V3
Moreover, the adjoint and the coadjoint representation of SO(3, K) are also equivalent.
b) Prove that this identification satisfies the equation
v(w) = v x w, [v, w] = [v, w],
(v, w)
Ztr (vv-) .
269
Exercises
c) By a), the moment map' : T'R3 - so(3,R)* can be interpreted as a map from T*R3 to R3. Show that it can then be written in the form
W(9,P) = 9XP, and hence it coincides with the classical angular momentum.
Chapter 8
Elements of Statistical Mechanics and Thermodynamics
8.1. Statistical States of a Hamiltonian System The Hamiltonian formulation of mechanics starts from a configuration space
X"' and makes use of the phase space T'X' with its canonical symplectic structure. A state of the mechanical system under consideration is a point in the phase space T`X'", and the motions of states are the integral curves of the symplectic gradient s-grad(H) of a Hamilton function H : T* X' --+ R. In this formulation of mechanics, the only essential data to be given are a symplectic manifold M2rn and a function H. The point of view of statistical mechanics is based on the idea that, for instance because of the size of the mechanical system or as a consequence of the imprecision of measurements, the state of the mechanical system cannot be determined precisely by fixing 2m real parameters. Instead, to each open set U C M2"', we can only ascribe the probability p(U) that the state belongs to the set U. This leads to the concept of considering mechanical states no longer as points in the phase space M2, but as probability measures on M2'".
Definition 1. Let a Hamiltonian system (M2"', w, H) consisting of a symplectic manifold and a Hamilton function be given. A statistical state is a probability measure p defined on the a-algebra of all Borel sets on M2'".
271
8. Elements of Statistical Mechanics and Thermodynamics
272
Example 1. For each point x E M2n', consider the measure 62 concentrated at this point,
a:(U)
1 0 ifx¢U, l 1 if a E U.
Therefore, classical mechanical states are particular statistical states.
Let -tt M2" -+ M2in be the flow of the symplectic gradient s-grad(H). The motion of the classical state x E MZ"' is the trajectory 0t(x). For the probability measure b+,(s) corresponding to the point it(x), we have :
t
xE (U) 1 = bx(-tt I(u)), 60&)(U) = 1 0, 1 , x E t 1(U ) and this formula leads to the following definition.
Definition 2. Let a Hamiltonian system (M2-, w, H) and a statistical state p be given. The motion of µ under the impact of the Hamiltonian system is the curve At of measures
At (U) := µ(4 1(U)) Definition 3. An equilibrium state of a Hamiltonian system (M2m,W, H) is a state A which does not change in the course of the motion of s-grad(H), At = P.
Definition 4. A statistical state p of a Hamiltonian system (M2in,W, H) has a stationary terminal distribution if the equation
'U-(U) := lim t 00 MO) defines a Borel measure on M2"'.
Theorem 1. A stationary terminal distribution µ«, is always an equilibrium state. Consider the volume form (-1)m(m-1U2
dM2m _
Wm
m!
of the symplectic manifold and the Borel measure induced from it (see the final remark at the end of §3.5). If the measure p is absolutely continuous with respect to the volume measure,
µ(U) := je(x).dM2m(x), then a is called the density function of the state µ.
8.1. Statistical States of a Hamiltonian System
273
be a state with density Theorem 2 (Liouville's Equation). Let p = function, and let at be the motion of this state in a Hamiltonian system. Then the measures pt = pt dM2" are also states with density function, and
d
dtet =
-{H,e}o$_t.
Proof. The flow ibt consists of symplectic transformations and preserves, in particular, the symplectic volume form dM2in, 4 (dM2i`) = dM2m. Transforming the integral, we see that pt = P o &_t is the density function of the state µt. This implies dt of = dt
P o D_t = do o dt
-t = -
o
-t = -{H, p} o O_t.
0 Corollary 1. A statistical state µ with density function P is an equilibrium state if and only if a is a first integral of the Hamiltonian H, (H, g} = 0. If x E M2m is a classical state in Hamiltonian mechanics, the value H(x) of the Hamilton function H is the energy E of the state,
E(x) = H(x) = JM2rn The definition of the energy of a statistical state generalizes this relation:
Definition 5. The energy of a statistical state p is the integral
E(µ) =
JMom
H(x) dp(x),
if this integral exists.
Theorem 3 (Conservation of Energy in Statistical Mechanics). If u t is the motion of a statistical state in a Hamiltonian system, the energy E(pt) is constant.
Proof. After transformation of/ the integral, we immediately obtain
E(pt) = l
H o I 1(x) dtt(x),
J M2m
and differentiating this with respect to the parameter t yields the formula
d E(µt) = fM2m {H, H}(x) du(x) = 0.
0
Now we want to study how the probability 1At(N2m) that the state µ = B dM2ri is in the compact subset N2ni C M21 at time t changes in time. The following lemma serves as a preparation.
8. Elements of Statistical Mechanics and Thermodynamics
274
Lemma 1. Let (M2i', w) be a symplectic manifold.
Then, for any two
functions f, g : lbl2in --+ R, the following formula holds:
df AdgAw,,,1 =
1 If, M
Proof. To prove this formula, we choose on 1112»' local syinplectic coordinates (ql, ..,P.),
w=
dp0 A dqq ,r= I
Setting Ai := dpi Adgi, we see that the exterior product satisfies the relations
AiAAj = AjAAi and AiAAi = 0. Since the forms Ai commute with one another, we can use the binomial formula to compute the exterior power wk, m
wn, _
(A1)'
>
1
m.
Aa' A
A A"
Because Ai A Ai = 0, only summands with rri < 1 occur in this sum, and hence it reduces to a single term:
writ = in!.AlA...AAna = m!.(dpiAdgi)A...A(dpmAdq»,) We compute WI-1 in a similar way, and obtain ",
(m-1)!
AlA...A,;...A
Thus, the following exterior products vanish:
(1) dpi Adpj Aw"'-1 = 0 = deli Adgj
Awl"-
= 0 if i # j
(2) dpi A dqj A w'
and we obtain fit
df A dg
Awrr-1 =
(.!m- dg dp, A dqi + 2f ag dq, Adpi) api Oqi OCli dpi rn
('m - 1 ) !
Of yg pi
arr.-1 mi
.
Tqi-
Of 09
di
w"'
A I A ... A A,,,
CdR
{f,g} wr
The relations m. d(f dg A w'"-1) = in df A dg A w"'- I = {f, gj w"' together with Stokes' theorem imply, after integration.
8.1. Statistical States of a Hamiltonian System
275
Theorem 4. Let (A12mw) be a compact symplectic manifold with boundary. Then
J12in
f.dg nwm-1
{f,g} w"r i)At 2°
.
In particular, the integral
J r2" {f,9} w"` = 0 vanishes for any compact symplectic manifold without boundary.
Corollary 2. If the Poisson bracket if, g} of two functions defined on a compact symplectic manifold 1112" without boundary does not change sign, then it vanishes identically, If, g} = 0.
We will apply these formulas to a statistical state with density function e.
Definition 6. The (2m. - 1)-form
At)
A'dHAw"'-1
(m-1)!
is called the probability current of the statistical state it = e dM2"' in the symplectic manifold (Al2m,w) with Hamilton function H.
Theorem 5. Let lit be the motion of the statistical state p = e dA12"' in the Hamiltonian system (1112"', W, H), and let N2", C AI2si be a compact submanifold. Then (µr(N2-)) (r=o
Wt
= JNFIJ
Proof. We compute the derivative with respect to time at t = 0 and use Lionville's equation as well as the preceding integral formulas: d N2", {He} dM2"r dt r=e dM2", o-
_J
J
N2m
N2s'
(m - 1)!
J
e dH n w"'-i =
UN2"4
f
j(Ic) .
aN2"
The probability current j(µ) is a hypersurface measure on all (2m - 1)dimensional submanifolds of the symplectic manifold M21. It expresses the infinitesimal change of the probability for the state it to be in the set N2i' C A12"' as an integral over the boundary of N2'".
Example 2. If the subinanifold N2i" C M2" is described by inequalities of the Hamilton function,
N2n = {x E M2" : C1 < H(x) < C21,
8. Elements of Statistical Mechanics and Thermodynamics
276
then the differential dH vanishes on the tangent bundle T(8N2m) of the boundary, and so does the form j(p). We hence obtain, for the motion At of every statistical state p = p dM2,, dtpt({x E M2in :
C1 < H(x) < C2}) = 0.
Thus the probability for a state p to be in N2in at time t is constant.
Example 3. In case m = 1, the formulas for the probability current of a two-dimensional Hamiltonian system (M2, w, H) simplify:
j(p) = e dH and dpt(N2) = JaN2 p dH. t Now we turn to the notion of information entropy for a statistical state. As a motivation, we first recall the notion of information of an event introduced by C. E. Shannon (1948). The heuristics is the following: If, in a series of experiments, an event occurs with probability close to one, then the "information" contents of this event is small. Conversely, if the probability of this event is close to zero, the occurrence of this event contains a large amount of "information". Modeling the probability computation by a triple (i, 21, p) consisting of a set i, a or-algebra 21 of subsets of S2, and a measure p defined on 21 such that A(fl) = 1, we arrive at the following definition for the Shannon information.
Definition 7. Let (S2, 21,,u) be a probability space. The amount of information contained in an event A E 21 is
1(A) := - log(p(A)). For a finite set S2, one forms the mean of these information amounts and gets in this way the so-called information entropy.
Definition 8. Let (12, p) be a finite probability space. The information entropy of this space is S(Q, p)
f
t
1(w)
. dp(w) = - E pi
log(pi) ,
i=1
where the set 11 consists of the elements {w1, ...,wn}, and pi := p({wi}) is the probability for the event wi. The information entropy is a measure for the uncertainty of the probability space. In fact, if St = {wl, ... , w } consists of n points and we denote by p' the measure corresponding to the equidistribution, p'(wi) := 1/n, then the following holds.
8.1. Statistical States of a Hamiltonian System
277
Theorem 6. The information entropy of a finite probability space (S),µ) does not exceed the information entropy of the equidistribution: S(Q, Ft)
0
for all x E M2s` .
Assumption 2. For every positive number B > 0, the following integral exists:
Z(B) :=
r
e-N(x)/B. dM2m(x)
JM
The function Z(B) is called the partition function of the Hamiltonian system.
8.1. Statistical States of a Hamiltonian System
279
Assumption 3. For every positive number B > 0, the following integral exists: H(x) . e-H(=)lB . dM2m(W ) . bl2m
Remark. The parameter B will later be identified with absolute temperature (multiplied by the Boltzmann constant k). Its appearance here indicates the exceptional role that temperature plays among all thermodynamic parameters. The function
E(B) :=
I
H(z)e-H(:)/B , dA12-(z)
Z(B) M2m is called the inner energy of the Hamiltonian system (M2m, w, H); its inner entropy is E(B) S(B) := log(Z(B)) + B Finally, the free energy F(B) is defined by
F(B) := -B log(Z(B)) = E(B) - B - S(B). First we note that the inner energy E(B) is a non-decreasing function on the interval (0, oo). Its derivative is
dE _ TB
1
Z2(B) B
[(I
M2m
e-
dM2m(x)) (ML H2( l2
_
- (ML H(x )e
z)e-dM2m(x))
dM2m(x))
and the Cauchy-Schwarz inequality shows that this derivative is positive
(H j4 const). Denote by En and Em., respectively, the bounds of the range of the inner energy E : (0, oo) - R. We calculate the derivative of the partition function in a similar way: dZ
1
dB = B2
JM2m
BZ Z(B) E(B).
Hence the inner energy as well as the inner entropy can be expressed in terms of the partition function. We summarize these formulas. Theorem 8 (Simple Equation of State for a Hamiltonian System).
(1) E(B) = B2gR (log(Z(B)));
(2) S(B) = W(B log(Z(B)));
8. Elements of Statistical Mechanics and Thermodynamics
280
(3) in the sense of 1 forms on the one-dimensional manifold (0, oo),
dS = B dE. Proof. The third equation is a straightforward consequence of the first two.
a The Gibbs state (canonical ensemble) is distinguished by the property that it realizes the maximum of the information entropy among all stag of fixed energy.
Theorem 9 (Gibbs State, Canonical Ensemble). Let an energy value Eo between the minimum and the maximum of the energy function E(B) for the Hamiltonian system (M2m, w, H) be given. Among all statistical states
p = P dM2' with density function of energy E(p) = Eo, there exists precisely one state, PGibbs = PGibbs dM2m, of maximal information entropy
S(pcibb.). The density function of this state is 1 -H(x)/Bo , PGibbs(x) = Z(Bo)e
where the parameter Bo is determined as a solution of the equation E(Bo) _ Eo. The value of the maximal entropy S(pCibbs) is
S(pGibbs) = S(Bo) = log(Z(Bo)) +
E(BO) o)
Proof. Choose Bo as a solution of the equation E(Bo) = E0, and )et e-H(x)/Bo dM2m 1 Z(Bo) be the corresponding Gibbs state. Its inner energy is pGibbs =
E(pcibb.) = Z(B0)
H(x)e-H(-)/Bo
JM2m
. dM2m = E(Bo) = Eo,
and hence UGibbe is a statistical state of energy Eo. For every oti er state p = P dM2m with energy
E(IL) = fM2m H(x)P(x) dM2'n = E0, we consider the function
f (t) = S((i - t)pGibbs + tp)
,
the information entropy of the statistical state (1 - t)pGibbs + tp. Differentiating f (t) with respect to the parameter t, we obtain
d2f(t) dt2
- _ JU2-
(e - PGib.)2 (1 - t)PGib, + tP
dM2'
f (1), we obtain
S(P)
0 such that Ilnl12
- (n,
)2 = k2
and
11,7112 = k2+(S+ a)2.
A short calculation for the curvature of the particle's curve of motion yields
K = Itrill =
1117 x ells
=
2TmvJ lot I
1-ln,n>2/1117112
__
sin(n,il) 2'rmvl
In11s
=
1 -oos (2, _ 27mvI InI
111711'-117,17)2
__
k
2 ymvllnl l2
2-tmvl ln113 2'rrmll,II3 If k = 0, the curvature vanishes and the curve is a straight line, i. e., there exist vectors v", w' E R3 such that n(s) = s v + v7. The equation of motion (*) then implies v' x u7 = 0, i. e., v' and 0 are linearly dependent, so thet we finally obtain n(s) = (s + A)t7, A E It , which is the equation of a line through the origin. Let us assume from now on that k > 0. We shall derive the expression for the torsion of the curve. By Frenet's formulas, the principal normal vector hh is
h=-=- k7xil. 1
The binormal vector then becomes
b = ,7xh = -i4x(rlx1), with derivative
=
-kn x (n x ii) =
n
2kmv7II,II3 )) Using Grassmann's identity u7 x (u' x ) = u (v, w-) - v' (u, m'), one shows that 6 x (u" x (i x ii)) = - (u, v3) i x *7, so that the former expression can further dsb
323
9.5. The Lorentz Force
be simplified to dg _
ds
(n, n) 2mvry11i1113
__
(-k r) X rl
2mvryIIi11I3h.
The torsion is thus equal to
tail)
r=
(b, h) _ -2mvryll71113
__
s+a 2mvry
k
+(a+a)23
In particular, the quotient of torsion and curvature satisfies the simple identity r/ic = -(s + a)/k. We prove next that the curve lies on a cone, that is, that there exists a constant vector ii whose angle p with 71 is also constant.
Define the functions
s+a
_ f(s)
2km yv
_
k + (a + a)
'
g(s)
1
2m yv
k + -(s+ a)2
They are primitives of K and -T and have the same quotient, d f (s) f (s) _ k dg(s) r ds = K' ds g(s) Frenet's formulas thus imply immediately that the vector iZ :=
is constant. To compute its angle V with rl, we derive an exact expression for q by applying Grassmann's identity to the binormal, b
1
1
= -krl x (71 x n) _ -k [rl -il(s+a)], q = (s+a)t-k6,
and remark that ii can be rewritten 11= rl/2mkryvllr7l l + h. For the opening angle of the cone we obtain
cos() =
(fi, u')
110 141
-
1
1 + (2mkyv)
9. Elements of Electrodynamics
324
By definition, a curve is a geodesic on a given surface if and only if its principal normal vector is orthogonal to the tangent plane of the surface at any of its points. Denoting by 7P the angle between h and i , we get (h, u)
-
1
Iluli
_
2mkvy 1 + (2mkvy)2
and we see that this is equal to 1 - cos2(cp) = sin(e). Hence, W and V are related through V-V = it/2, and we conclude that h is indeed perpendicular to the tangent plane of the curve. The particle moves along a geodesic of the cone, as claimed.
Exercises
325
Exercises 1 (Kirchhoff Formula). Let u(x, t) be a function defined on JR3 x Ht and S C JR3 a compact domain with smooth boundary. Prove, for each point (Xo, to) E SZ x JR, the Kirchhoff formula:
Jsl
u(xo, to) =
Ou(a,to+
cr 8t
u(1l, to - r/c) or
or(a,to)
- u(a'to
eN
c
r f au (or, to -
dy + J
r
IL
r 8N
8r-1(a,to) C
OM
where r := Ilxo - yII is the distance to the spatial point xo E R3, N is the normal vector to the boundary 00, and = Ox - 1/c28it is the wave operator. Deduce from this the solution formulas for the Cauchy problem of the wave equation by choosing for H a 3-dimensional ball. 2. Under the assumptions of Helmholtz' theorem, the electric and the magnetic field E and B, as well as the current density field J can be written as the sum of a divergence-free and a curl-free vector field,
E = Ediv + Ecuri, B = Bdiv + Bcuri and J = Jdiv + Jcurl a) Prove that the Maxwell equations can then be written as follows: curl(Ed;v) = curl(Bd;Y) =
1
c
-
18Bdiv
8t
8Ed;v
at +
47r
Jai,,,
chv(Bcuri) = 0 div(Ecuri) = 47r LO.
The continuity equation then becomes div(Jcuri) +
OLO
= 0.
Hint: In the proof of Theorem 2, we proved that, under the assumptions made here, any vector field which is at the same time divergence- and curl-free, has to vanish identically. b) On the other hand, for the electric field E we already know the decomposition
E_
c
+grad(-O).
9. Elements of Electrodynamics
326
Prove that this is precisely the Helmholtz decomposition of E if the magnetic potential satisfies the condition called Coulomb gauge:
div(A) = 0. 3. Deduce from Theorem 4 that if Eo(x) and Bo(x) are C3-functions on R3 such that divE0 = divB0 = 0, then the Cauchy initial value problem for an electromagnetic wave
curl(B) = c 5 ,
curl(E)
c at , div(B) = 0, div(E) = 0, E(x,O) = Eo(x), B(x, 0) = Bo(x) has the unique solution E(x, ct)
4act
f f
curl Bo(y)dy +
B(x, ct)
4act
f Eo(y)dy
4ac 8t
,
S2(x,d)
S2(x,d)
curl Eo(y)dy + 41
S2(x,d)
f Bo(y)dy
8t
1S2(x,ct)
4. Using Theorems 1 and 2, determine the electric and the magnetic field in 1R3 which is generated in the following situations:
a) a homogeneously charged ball of radius R with constant charge density P;
b) a charged spherical shell of radius R with constant surface charge density a; c) an infinitely extended straight wire of radius R through which a current with constant current density j flows. 5. Describe the solution of the classical Lorentz equation for a homogeneous, static electric field (E = const) and vanishing magnetic field (B = 0). What happens in the non-relativistic limit of small velocities v 0,
atu(x,0) = u1(x)
Part 1. General Shape of the Solution. Prove that there exist two C2functions f and g satisfying
u(x,t) = f(x+ct)+g(x-ct). The solution is hence the superposition of a wave traveling to the left and one traveling to the right. Hint: Introduce the new coordinates xt = x ± ct and show that the wave equation is equivalent to the differential equation a2u
ax+(9x- -
0.
Part 2. Solution of the Cauchy Problem. With the above Ansatz for the solution, the following relations have to hold: uo(x) = f(x) + g(x),
u1 (x) = c(f'(x) - g'(x))
By integration of the second equation over the interval [0, x], prove that the general solution has to be given by
Ed
u(x,t) = 2[uo(x + ct) + ua(x - ct)) + 2cu1(s)ds.
In particular, this argument shows the uniqueness of the solution. How can the result be understood qualitatively by means of the light-cone ?
9. Elements of Electrodynamics
328
9 (One-sided Infinitely Extended Oscillating String). An oscillating string extending infinitely in the positive x-direction is modeled by the one-dimensional wave equation,
attu = c2axxu,
t > 0,
x > 0.
In addition to the initial conditions,
u(x, O) = uo(x), 8tu(x, 0) = u, (x), one has to impose boundary condition which describe the behavior of th "wall" at x = 0 for all times t: u(0, t) = cp(t).
Here uo, 4o E C2(R+) and ul E C'(R+) are required. Prove (using a similar Ansatz as in the preceding exercise) that the solution is x+d
2cjx-d
[uo(x + ct) + uo(x - ct)] + -L
u(x,t) =
2 [uo(x +
ui(s)ds,
d+x
Ct) - up(ct - x)] + p(t - x/C) + ZJ
d-x
ul (s) ds,
where the upper line is to be taken for points (x, t) in the region I, i. e. below
the line x = ct, and the lower line for points (x, t) in the region II, above the line x = ct. What can be said concerning the behavior of the lightcone, in particular in II? Describe, moreover, the regularity properties of the solution on the line x = ct, and explain carefully under which additional conditions it is of class C2 there. x=ct /1
I x
10 (Oscillating String Fixed on Either Side). For the wave equation on bounded domains a separation Ansatz going back to Bernoulli proved to be successful, in that it reduces the problem to one for Fourier series. We are looking for a solution of the one-dimensional wave equation on the interval [0, 1],
attu = c28xxu,
t>0, 0<x 0 necessarily lead to the trivial solution u = 0; in the case A < 0 there has to exist a natural number k such that A = -k27r2 and uk(x, t) = Xk(x)Tk(t) = sin kirx (ak sin klrct + bk cos kirct) . The general solution is then obtained as the series 00
u(x, t) _
uk(x, t) _ E sin kirx (ak sin k7rct + bk cos klrct) . k=1
k=1
c) Find an integral formula for the coefficients ak, bk depending on the initial conditions uo and u1. Solution: ak
lore
1
1
2
ul (x) sin k7rx dx,
bk = 2 fo uo(x) sin k7rx dx .
0
11 (Validity of the Fourier Method). By a theorem of Dirichlet, the Fourier series of a function f E C1([O,1]) satisfying f (0) = f (1) = 0 is uniformly convergent and tends pointwise to f. Prove the following lemma:
For a function f E Ck([O,1]) such that f(0), ..., f(k-1) vanish at 0 and 1, there exists a constant A for which the Fourier coefficients of f satisfy the following inequality:
j f(x)sinn7rxdx
