¬‚ux homomorphism. Let p ∈ Pe Sp(„¦) be chosen, and let γ: S 1 ’ M be a closed curve

representing an element of H1 (M, Z). Then we can de¬ne

(p · γ)— („¦)

F (p, γ) =

[0,1]—S 1

where (p · γ): [0, 1] — S 1 ’ M is de¬ned by (p · γ)(t, θ) = p t, γ(θ) . The number F (p, γ)

is called the ¬‚ux of p through γ.

(i) Show that F (p, γ) = F (p , γ ) if p is homotopic to p and γ is homologous to γ ).

(Hint: Use Stokes™ Theorem several times.)

Thus, F is actually well de¬ned as a map F : Sp0 („¦) — H1 (M, R) ’ R.

(ii) Show that F : Sp0 („¦) — H1 (M, R) ’ R is linear in its second variable and that, under

the obvious multiplication, we have

F (p p , γ) = F (p, γ) + F (p , γ).

(Hint: Use Stokes™ Theorem again.)

Thus, F may be transposed to become a homomorphism

¦: Sp0 („¦) ’ H 1 (M, R).

Show (by direct computation) that if ζ is a closed 1-form on M for which the symplectic

vector ¬eld Z = ζ is complete on M, then the path p in Sp(M) de¬ned by the ¬‚ow

of Z from t = 0 to t = 1 satis¬es ¦(p) = [ζ] ∈ H 1 (M, R). Conclude that the ¬‚ux

homomorphism ¦ is always surjective and that its derivative ¦ (˜): sp(„¦) ’ H 1 (M, R)

e

is just the operation of taking cohomology classes. (Recall that we identify sp(„¦) with

Z 1 (M).)

E.5.4 98

(iii) Show that if M is a compact surface of genus g > 1, then the ¬‚ux homomorphism

is actually well de¬ned as a map from Sp(„¦) to H 1 (M, R). Would the same result

be true if M were of genus 1? How could you modify the map so as to make it well-

de¬ned on Sp(„¦) in the case of the torus? (Hint: Show that if you have two paths

p and p with the same endpoint, then you can express the di¬erence of their ¬‚uxes

across a circle γ as an integral of the form

Ψ— („¦)

S 1 —S 1

where Ψ: S 1 — S 1 ’ M is a certain piecewise smooth map from the torus into M.

Now use the fact that, for any piecewise smooth map Ψ: S 1 — S 1 ’ M, the induced

map Ψ— : H 2 (M, R) ’ H 2 (S 1 — S 1 , R) on cohomology is zero. (Why does this follow

from the assumption that the genus of M is greater than 1?) )

In any case, the subgroup ker ¦ (or its image under the natural projection from Sp0 („¦)

to Sp(„¦)) is known as the group H(„¦) of exact or Hamiltonian symplectomorphisms. Note

that, at least formally, its Lie algebra is h(„¦).

13. In the case of the geodesic ¬‚ow on a surface of revolution (see Lecture 4), show that

™

the energy f 1 = EL and the conserved quantity f 2 = F (r)r + G(r)θ are in involution. Use

™

the algorithm described in Theorem 3 to compute the functions a1 and a2 , thus verifying

that the geodesic equations on a surface of revolution are integrable by quadrature.

E.5.5 99

Lecture 6:

Symplectic Manifolds, II

The Space of Symplectic Structures on M.

I want to turn now to the problem of describing the symplectic structures a manifold M

can have. This is a surprisingly delicate problem and is currently a subject of research.

Of course, one fundamental question is whether a given manifold has any symplec-

tic structures at all. I want to begin this lecture with a discussion of the two known

obstructions for a manifold to have a symplectic structure.

The cohomology ring condition. If „¦ ∈ A2 (M 2n ) is a symplectic structure on a

compact manifold M, then the cohomology class [„¦] ∈ HdR (M, R) is non-zero. In fact,

2

[„¦]n = [„¦n ], but the class [„¦n ] cannot vanish in HdR (M) because the integral of „¦n over

2n

M is clearly non-zero. Thus, we have

Proposition 1: If M 2n is compact and has a symplectic structure, there must exist an

element u ∈ H 2 (M, R) so that un = 0 ∈ HdR (M).

2n

Example. This immediately rules out the existence of a symplectic structure on S 2n for

all n > 1. One consequence of this, as you are asked to show in the Exercises, is that there

cannot be any simple notion of connected sum in the category of symplectic manifolds

(except in dimension 2).

The bundle obstruction. If M admits a symplectic structure „¦, then, in particu-

lar, this de¬nes a symplectic structure on each of the tangent spaces Tm M which varies

continuously with m. In other words, T M must carry the structure of a symplectic vector

bundle. There are topological obstructions to the existence of such a structure on the tan-

gent bundle of a general manifold. As a simple example, if M has a symplectic structure,

then T M must be orientable.

There are more subtle obstructions than orientation. Unfortunately, a description of

these obstructions requires some acquaintance with the theory of characteristic classes.

However, part of the following discussion will be useful even to those who aren™t familiar

with characteristic class theory, so I will give it now, even though the concepts will only

reveal their importance in later Lectures.

De¬nition 1: An almost symplectic structure on a manifold M 2n is a smooth 2-form

„¦ de¬ned on M which is non-degenerate but not necessarily closed. An almost complex

structure on M 2n is a smooth bundle map J : T M ’ T M which satis¬es J 2 v = ’v for all

v in T M.

The reason that I have introduced both of these concepts at the same time is that

they are intimately related. The really deep aspects of this relationship will only become

apparent in the Lecture 9, but we can, at least, give the following result now.

L.6.1 100

Proposition 2: A manifold M 2n has an almost symplectic structure if and only if it has

an almost complex structure.

Proof: First, suppose that M has an almost complex structure J . Let g0 be any Rie-

mannian metric on M. (Thus, g0 : T M ’ R is a smooth function which restricts to each

Tm M to be a positive de¬nite quadratic form.) Now de¬ne a new Riemannian metric by

the formula

g(v) = g0 (v) + g0 (J v).

Then g has the property that g(J v) = g(v) for all v ∈ T M since

g(J v) = g0 (J v) + g0 (J 2 v) = g0 (J v) + g0 (’v) = g(v).

Now let , denote the (symmetric) inner product associated with g. Thus, v, v =

g(v), so we have J x, J y = x, y when x and y are tangent vector with the same base

point. For x, y ∈ Tm M de¬ne „¦(x, y) = J x, y . I claim that „¦ is a non-degenerate 2-form

on M. To see this, ¬rst note that

„¦(x, y) = J x, y = ’ J x, J 2 y = ’ J 2y, J x = ’ J y, x = ’„¦(y, x),

so „¦ is a 2-form. Moreover, if x is a non-zero tangent vector, then „¦(x, J x) = J x, J x =

g(x) > 0, so it follows that x „¦ = 0. Thus „¦ is non-degenerate.

To go the other way is a little more delicate. Suppose that „¦ is given and ¬x a

Riemannian metric g on M with associated inner product , . Then, by linear algebra

there exists a unique bundle mapping A: T M ’ T M so that „¦(x, y) = Ax, y . Since „¦ is

skew-symmetric and non-degenerate, it follows that A must be skew-symmetric relative to

, and must be invertible. It follows that ’A2 must be symmetric and positive de¬nite

relative to , .

Now, standard results from linear algebra imply that there is a unique smooth bundle

map B: T M ’ T M which positive de¬nite and symmetric with respect to , and which

satis¬es B 2 = ’A2 . Moreover, this linear mapping B must commute with A. (See the

Exercises if you are not familiar with this fact). Thus, the mapping J = AB ’1 satis¬es

J 2 = ’I, as desired.

It is not hard to show that the mappings (J, g0 ) ’ „¦ and („¦, g) ’ J constructed in the

proof of Proposition 1 depend continuously (in fact, smoothly) on their arguments. Since

the set of Riemannian metrics on M is contractible, it follows that the set of homotopy

classes of almost complex structures on M is in natural one-to-one correspondence with

the set of homotopy classes of almost symplectic structures.

(The reader who is familiar with the theory of principal bundles knows that at the

heart of Proposition 1 is the fact that Sp(n, R) and GL(n, C) have the same maximal

compact subgroup, namely U(n).)

L.6.2 101

Now I can describe some of the bundle obstructions. Suppose that M has a symplectic

structure „¦ and let J be any one of the almost complex structures on M we constructed

above. Then the tangent bundle of M can be regarded as a complex bundle, which we will

denote by T J and hence has a total Chern class

c(T J ) = 1 + c1 (J ) + c2 (J ) + · · · + cn(J )

where ci (J ) ∈ H 2i (M, Z). Now, by the properties of Chern classes, cn (J ) = e(T M), where

e(T M) is the Euler class of the tangent bundle given the orientation determined by the

volume form „¦n .

These classes are related to the Pontrijagin classes of T M by the Whitney sum formula

(see [MS]):

p(T M) = 1 ’ p1 (T M) + p2 (T M) ’ · · · + (’1)[n/2] p[n/2] (T M)

= c(T J • T ’J )

= 1 + c1 (J ) + c2 (J ) + · · · + cn(J ) 1 ’ c1 (J ) + c2 (J ) ’ · · · + (’1)n cn (J )

Since p(T M) depends only on the di¬eomorphism class of M, this gives quadratic equations

for the ci (J ),

2

’ 2ck’1 (J )ck+1 (J ) + · · · + (’1)k 2c0 (J )c2k (J ),

pk (T ) = ck (J )

to which any manifold with an almost complex structure must have solutions. Since not

every 2n-manifold has cohomology classes ci (J ) satisfying these equations, it follows that

some 2n-manifolds have no almost complex structure and hence, by Proposition 2, no

almost symplectic structure either.

Examples. Here are two examples in dimension 4 to show that the cohomology ring

condition and the bundle obstruction are independent.

• M = S 1 — S 3 does not have a symplectic structure because H 2 (M, R) = 0. However

the bundle obstruction vanishes because M is parallelizable (why?). Thus M does have

an almost symplectic structure.

• M = CP2 # CP2 . The cohomology ring of M in this case is generated over Z by

two generators u1 and u2 in H 2 (M, Z) which are subject to the relations u1 u2 = 0 and

u2 = u2 = v where v generates H 4 (M, Z). For any non-zero class u = n1 u1 + n2 u2 , we

1 2

have u = (n2 + n2 )v = 0. Thus the cohomology ring condition is satis¬ed.

2

1 2

However, M has no almost symplectic structure: If it did, then T = T M would have

a complex structure J , with total Chern class c(J ) and the equations above would give

2

p1 (T ) = c1 (J ) ’ 2c2 (J ). Moreover, we would have e(T ) = c2 (J ). Thus, we would have

to have

2

c1 (J ) = p1 (T ) + 2e(T ).

L.6.3 102

For any compact, simply-connected, oriented 4-manifold M with orientation class

µ ∈ H 4 (M, Z), the Hirzebruch Signature Theorem (see [MS]) implies p1 (T ) = 3(b+ ’

2

’ ±

b2 )µ, where b2 are the number of positive and negative eigenvalues respectively of the

intersection pairing H 2 (M, Z) — H 2 (M, Z) ’ Z. In addition, e(T ) = (2 + b+ + b’ )µ.

2 2

2 ’

Substituting these into the above formula, we would have c1 (J ) = (4 + 5b2 ’ b2 )µ for

+

any complex structure J on the tangent bundle of M.

2

In particular, if M = CP2 # CP2 had an almost complex structure J , then c1 (J )

would be either 14v (if µ = v, since then b+ = 2 and b’ = 0) or ’2v (if µ = ’v, since

2 2

’

then b2 = 0 and b2 = 2). However, by our previous calculations, neither 14v nor ’2v is

+

the square of a cohomology class in H 2 (CP2 # CP2 , Z).

This example shows that, in general, one cannot hope to have a connected sum

operation for symplectic manifolds.

The actual conditions for a manifold to have an almost symplectic structure can be

expressed in terms of characteristic classes, so, in principle, this can always be determined

once the manifold is given explicitly. In Lecture 9 we will describe more fully the following

remarkable result of Gromov:

If M 2n has no compact components and has an almost symplectic structure Υ, then

there exists a symplectic structure „¦ on M which is homotopic to Υ through almost

symplectic structures.

Thus, the problem of determining which manifolds have symplectic structures is now

reduced to the compact case. In this case, no obstruction beyond what I have already

described is known. Thus, I can state the following:

Basic Open Problem: If a compact manifold M 2n satis¬es the cohomology ring condi-

tion and has an almost symplectic structure, does it have a symplectic structure?

Even (perhaps especially) for 4-manifolds, this problem is extremely interesting and

very poorly understood.

Deformations of Symplectic Structures. We will now turn to some of the features

of the space of symplectic structures on a given manifold which does admit symplectic

structures. First, we will examine the “deformation problem”. The following theorem due

to Moser (see [We]) shows that symplectic structures determining a ¬xed cohomology class

in H 2 on a compact manifold are “rigid”.

Theorem 1: If M 2n is a compact manifold and „¦t for t ∈ [0, 1] is a continuous 1-parameter

family of smooth symplectic structures on M which has the property that the cohomology

classes [„¦t ] in HdR (M, R) are independent of t, then for each t ∈ [0, 1], there exists a

2

di¬eomorphism φt so that φ— („¦t ) = „¦0 .

t

L.6.4 103

Proof: We will start by proving a special case and then deduce the general case from it.

Suppose that „¦0 is a symplectic structure on M and that • ∈ A1 (M) is a 1-form so that,

for all s ∈ (’1, 1), the 2-form

„¦s = „¦0 + s d•

is a symplectic form on M as well. (This is true for all su¬ciently “small” 1-forms on M

since M is compact.) Now consider the 2-form on (’1, 1) — M de¬ned by the formula