for some functions gij on M. Let us regard the functions gij as unknowns for a moment.

Let Y1 , . . . , Yn be the vector ¬elds which satisfy

Yi „¦ = βi ,

¯

with Yi denoting the corresponding quantities when the gij are set to zero. Then it is easy

to see that

Yi = Yi ’ gij Xf j .

¯

Now, by construction,

„¦(Xf i , Xf j ) = 0 and „¦(Yi , Xf j ) = δij .

Moreover, as is easy to compute,

¯¯

„¦(Yi , Yj ) = „¦(Yi , Yj ) ’ gji + gij .

L.5.13 92

¯¯

Thus, choosing the functions gij appropriately, say gij = ’ 1 „¦(Yi , Yj ), we may assume that

2

„¦(Yi , Yj ) = 0. It follows that the sequence of 1-forms df 1 , . . . , df n , β1 , . . . , βn is the dual

basis to the sequence of vector ¬elds Y1 , . . . , Yn , Xf 1 , . . . , Xf n . In particular, we see that

„¦ = df 1 § β1 + · · · + df n § βn ,

since the 2-forms on either side of this equation have the same values on all pairs of vector

¬elds drawn from this basis.

Now, since „¦ is closed, we have

d„¦ = df 1 § dβ1 + · · · + df n § dβn = 0.

If, for example, we wedge both sides of this equation with df 2 , . . . , df n , we see that

df 1 § df 2 § . . . n

§ df § dβ1 = 0.

Hence, it follows that dβ1 lies in the ideal generated by the forms df 1 , . . . , df n . Of course,

there was nothing special about the ¬rst term, so we clearly have

dβi ≡ 0 mod df 1 , . . . , df n for all 1 ¤ i ¤ n.

In particular, it follows that, if we pull back the 1-forms βi to any n-dimensional level set

Mc ‚ M de¬ned by equations f i = ci where the ci are constants, then each βi becomes

closed.

Let m ∈ M be ¬xed and choose functions g1 , . . . , gn on a neighborhood U of m in M

so that gi (m) = 0 and so that the functions g1 , . . . , gn , f 1 , . . . , f n form a coordinate chart

on U. By shrinking U if necessary, we may assume that the image of this coordinate chart

in Rn — Rn is an open set of the form B1 — B2 , where B1 and B2 are open balls in Rn

(with B1 centered on 0). In this coordinate chart, the βi can be expressed in the form

j

βi = Bi (g, f) dgj + Cij (g, f) df j .

De¬ne new functions ai on B 1 — B 2 by the rule

1

j

hi (g, f) = Bi (tg, f)gj dt.

0

(This is just the Poincar´ homotopy formula with the f™s held ¬xed. It is also the ¬rst

e

place where we use “quadrature”.) Since setting the f™s equals to constants makes βi a

closed 1-form, it follows easily that

βi = dhi + Aij (g, f) df j

for some functions Aij on B 1 — B 2 . Thus, on U, the form „¦ has the expression

„¦ = df i § dhi + Aij df i § df j .

L.5.14 93

It follows that the 2-form A = Aij df i §df j is closed on (the contractible open set) B 1 — B 2 .

Thus, the functions Aij do not depend on the g-coordinates at all. Hence, by employing

quadrature once more (i.e., the second time) in the Poincar´ homotopy formula, we can

e

write A = ’d(si df ) for some functions si of the f™s alone. Setting ai = hi + si , we have

i

the desired local normal form „¦ = df i §dai .

In many useful situations, one does not need to restrict to a local neighborhood U

to de¬ne the functions ai (at least up to additive constants) and the 1-forms dai can be

de¬ned globally on M (or, at least away from some small subset in M where degeneracies

occur). In this case, the construction above is often called the construction of “action-

angle” coordinates. We will discuss this further in Lecture 6.

L.5.15 94

Exercise Set 5:

Symplectic Manifolds, I

1. Show that the bilinear form on Rn de¬ned in the text by the rule B(x, y) = tx A y (where

A is a skew-symmetric n-by-n matrix) is non-degenerate if and only if A is invertible. Show

directly (i.e., without using Proposition 1) that a skew-symmetric, n-by-n matrix A cannot

be invertible if n is odd. (Hint: For the last part, compute det(A) two ways.)

2. Let (V, B) be a symplectic space and let b = (b1 , b2 , . . . , bm ) be a basis of B. De¬ne

the m-by-m skew-symmetric matrix Ab whose ij-entry is B(bi , bj ). Show that if b = bR

is any other basis of V (where R ∈ GL(m, R) ), then

Ab = tR Ab R.

Use Proposition 1 and Exercise 1 to conclude that, if A is an invertible, skew-symmetric

2n-by-2n matrix, then there exists a matrix R ∈ GL(2n, R) so that

0n In

A = tR R.

’In 0n

In other words, the GL(2n, R)-orbit of the matrix Jn de¬ned in the text (under the “stan-

dard” (right) action of GL(2n, R) on the skew-symmetric 2n-by-2n matrices) is the open

set of all invertible skew-symmetric 2n-by-2n matrices.

3. Show that Sp(n, R), as de¬ned in the text, is indeed a Lie subgroup of GL(2n, R) and

has dimension 2n2 + n. Compute its Lie algebra sp(n, R). Show that Sp(1, R) = SL(2, R).

4. In Lecture 2, we de¬ned the groups GL(n, C) = {R ∈ GL(2n, R) | Jn R = R Jn } and

O(2n) = {R ∈ GL(2n, R) | tR R = I2n }. Show that

GL(n, C) © Sp(n, R) = O(2n) © Sp(n, R) = GL(n, C) © O(2n) = U(n).

5. Let „¦ be a symplectic form on a vector space V of dimension 2n. Let W ‚ V be a

subspace which satis¬es dim W = d and dim (W © W ⊥ ) = ν. Show that there exists an

„¦-standard basis of V so that W is spanned by the vectors

e1 , . . . , eν+m , f1 , . . . , fm

where d ’ ν = 2m. In this basis of V , what is a basis for W ⊥ ?

E.5.1 95

6. The Pfaffian. Let V be a vector space of dimension 2n. Fix a basis b = (b1 , . . . , b2n ).

For any skew-symmetric 2n-by-2n matrix F = (f ij ), de¬ne the 2-vector

1

f ij bi § bj = 1 b § F t

¦F = b.

§

2 2

Then there is a unique polynomial function Pf, homogeneous of degree n, on the space of

skew-symmetric 2n-by-2n matrices for which

(¦F )n = n! Pf(F ) b1 § . . . § b2n .

Show that

Pf(F ) = f 12 when n = 1,

Pf(F ) = f 12 f 34 + f 13 f 42 + f 14 f 23 when n = 2.

Show also that

Pf(A F tA) = det(A) Pf(F )

for all A ∈ GL(2n, R). (Hint: Examine the e¬ect of a change of basis b = b A. Compare

Problem 2.) Use this to conclude that Sp(n, R) is a subgroup of SL(2n, R). Finally, show

2

that (Pf(F )) = det(F ). (Hint: Show that the left and right hand sides are polynomial

functions which agree on a certain open set in the space of skew-symmetric 2n-by-2n

matrices.)

The polynomial function Pf is called the Pfa¬an. It plays an important role in

di¬erential geometry.

7. Verify that, for any B ∈ A2 (V ), the symplectic reduction (V , B) is a well-de¬ned

symplectic space.

8. Show that if there is a G-invariant non-degnerate pairing ( , ): g — g ’ R, then g and

g— are isomorphic as G-representations.

9. Compute the adjoint and coadjoint representations for

ab

a ∈ R+ , b ∈ R

G=

01

Show that g and g— are not isomorphic as G-spaces! (For a general G, the Ad-orbits of G

in g are not even of even dimension in general, so they can™t be symplectic manifolds.)

10. For any Lie group G and any ξ ∈ g— , show that the symplectic structures „¦ξ and „¦a·ξ

on G · ξ are the same for any a ∈ G.

E.5.2 96

11. This exercise concerns the splitting properties of the two Lie algebras sequences

associated to any symplectic structure „¦ on a connected manifold M:

0 ’’ R ’’ C ∞ (M) ’’ h(„¦) ’’ 0

and

0 ’’ h(„¦) ’’ sp(„¦) ’’ HdR (M, R) ’’ 0.

1

De¬ne the “divided powers” of „¦ by the rule „¦[k] = (1/k!) „¦k , for each 0 ¤ k ¤ n.

(i) Show that, for any vector ¬elds X and Y on M,

„¦(X, Y ) „¦[n] = ’(X „¦) § (Y „¦) § „¦[n’1] .

Conclude that the ¬rst of the above two sequences splits if M is compact. (Hint: For

the latter statement, show that the set of functions f on M for which M f „¦[n] = 0

forms a Poisson subalgebra of C ∞ (M).)

(ii) On the other hand, show that for R2 with the symplectic structure „¦ = dx§dy, the

¬rst sequence does not split. (Hint: Show that every smooth function on R2 is of the

form {x, g} for some g ∈ C ∞ (R2 ). Why does this help?)

(iii) Suppose that M is compact. De¬ne a skew-symmetric pairing

β„¦ : HdR (M, R) — HdR (M, R) ’ R

1 1

by the formula

a § ˜ § „¦[n’1] ,

β„¦ (a, b) = ˜b

M

where a and ˜ are closed 1-forms representing the cohomology classes a and b respec-

˜ b

tively. Show that if there is a Lie algebra splitting σ: HdR (M, R) ’ sp(„¦) then

1

β„¦ (a, b)

„¦ σ(a), σ(b) = ’

vol(M, „¦[n] )

for all a, b ∈ HdR (M, R). (Remember that the Lie algebra structure on HdR (M, R) is

1 1

the abelian one.) Use this to conclude that the second sequence does split for a sym-

plectic structure on the standard 1-holed torus, but does not split for any symplectic

structure on the 2-holed torus. (Hint: To show the non-splitting result, use the fact

that any tangent vector ¬eld on the 2-holed torus must have a zero.)

E.5.3 97

12. The Flux Homomorphism. The object of this exercise is to try to identify the

subgroup of Sp(„¦) whose Lie algebra is h(„¦). Thus, let (M, „¦) be a symplectic manifold.

First, I remind you how the construction of the (smooth) universal cover of the identity

component of Sp(„¦) goes. Let p: [0, 1] — M ’ M be a smooth map with the property that

the map pt : M ’ M de¬ned by pt (m) = p(t, m) is a symplectomorphism for all 0 ¤ t ¤ 1.

Such a p is called a (smooth) path in Sp(M). We say that p is based at the identity map

e: M ’ M if p0 = e. The set of smooth paths in Sp(M) which are based at e will be

denoted by Pe Sp(„¦) .

Two paths p and p in Pe Sp(„¦) satisfying p1 = p1 are said to be homotopic if there

is a smooth map P : [0, 1] — [0, 1] — M ’ M which satis¬es the following conditions: First,

P (s, 0, m) = m for all s and m. Second, P (s, 1, m) = p1 (m) = p1 (m) for all s and m.

Third, P (0, t, m) = p(t, m) and P (1, t, m) = p (t, m) for all t and m. The set of homotopy

classes of elements of Pe Sp(„¦) is then denoted by Sp0 („¦). In any reasonable topology

on Sp(„¦), this should to be the universal covering space of the identity component of

Sp(„¦). There is a natural group structure on Sp0 („¦) in which ˜, the homotopy class of

e

the constant path at e, is the identity element (cf., the covering spaces exercise in Exercise

Set 2).