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for some functions gij on M. Let us regard the functions gij as unknowns for a moment.
Let Y1 , . . . , Yn be the vector п¬Ѓelds which satisfy

Yi в„¦ = ОІi ,

ВЇ
with Yi denoting the corresponding quantities when the gij are set to zero. Then it is easy
to see that
Yi = Yi в€’ gij Xf j .
ВЇ

Now, by construction,

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

Moreover, as is easy to compute,

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

L.5.13 92
ВЇВЇ
Thus, choosing the functions gij appropriately, say gij = в€’ 1 в„¦(Yi , Yj ), we may assume that
2
в„¦(Yi , Yj ) = 0. It follows that the sequence of 1-forms df 1 , . . . , df n , ОІ1 , . . . , ОІn is the dual
basis to the sequence of vector п¬Ѓelds Y1 , . . . , Yn , Xf 1 , . . . , Xf n . In particular, we see that

в„¦ = df 1 в€§ ОІ1 + В· В· В· + df n в€§ ОІn ,

since the 2-forms on either side of this equation have the same values on all pairs of vector
п¬Ѓelds drawn from this basis.
Now, since в„¦ is closed, we have

dв„¦ = df 1 в€§ dОІ1 + В· В· В· + df n в€§ dОІn = 0.

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

df 1 в€§ df 2 в€§ . . . n
в€§ df в€§ dОІ1 = 0.

Hence, it follows that dОІ1 lies in the ideal generated by the forms df 1 , . . . , df n . Of course,
there was nothing special about the п¬Ѓrst term, so we clearly have

dОІi в‰Ў 0 mod df 1 , . . . , df n for all 1 в‰¤ i в‰¤ n.

In particular, it follows that, if we pull back the 1-forms ОІi to any n-dimensional level set
Mc вЉ‚ M deп¬Ѓned by equations f i = ci where the ci are constants, then each ОІi becomes
closed.
Let m в€€ M be п¬Ѓxed 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 Г— Rn is an open set of the form B1 Г— B2 , where B1 and B2 are open balls in Rn
(with B1 centered on 0). In this coordinate chart, the ОІi can be expressed in the form
j
ОІi = Bi (g, f) dgj + Cij (g, f) df j .

Deп¬Ѓne new functions ai on B 1 Г— B 2 by the rule
1
j
hi (g, f) = Bi (tg, f)gj dt.
0

(This is just the PoincarВґ homotopy formula with the fвЂ™s held п¬Ѓxed. It is also the п¬Ѓrst
e
place where we use вЂњquadratureвЂќ.) Since setting the fвЂ™s equals to constants makes ОІi a
closed 1-form, it follows easily that

ОІi = dhi + Aij (g, f) df j

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

в„¦ = df i в€§ dhi + Aij df i в€§ df j .

L.5.14 93
It follows that the 2-form A = Aij df i в€§df j is closed on (the contractible open set) B 1 Г— 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 PoincarВґ homotopy formula, we can
e
write A = в€’d(si df ) for some functions si of the fвЂ™s alone. Setting ai = hi + si , we have
i

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

In many useful situations, one does not need to restrict to a local neighborhood U
to deп¬Ѓne the functions ai (at least up to additive constants) and the 1-forms dai can be
deп¬Ѓned 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 вЂњaction-
angleвЂќ 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 deп¬Ѓned 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. Deп¬Ѓne
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 в€€ 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 в€€ GL(2n, R) so that

0n In
A = tR R.
в€’In 0n

In other words, the GL(2n, R)-orbit of the matrix Jn deп¬Ѓned in the text (under the вЂњstan-
dardвЂќ (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 deп¬Ѓned 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 deп¬Ѓned the groups GL(n, C) = {R в€€ GL(2n, R) | Jn R = R Jn } and
O(2n) = {R в€€ GL(2n, R) | tR R = I2n }. Show that

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

5. Let в„¦ be a symplectic form on a vector space V of dimension 2n. Let W вЉ‚ V be a
subspace which satisп¬Ѓes dim W = d and dim (W в€© W вЉҐ ) = ОЅ. Show that there exists an
в„¦-standard basis of V so that W is spanned by the vectors

e1 , . . . , eОЅ+m , f1 , . . . , fm

where d в€’ ОЅ = 2m. In this basis of V , what is a basis for W вЉҐ ?

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 ), deп¬Ѓne the 2-vector

1
f ij bi в€§ bj = 1 b в€§ F t
О¦F = b.
в€§
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

(О¦F )n = n! Pf(F ) b1 в€§ . . . в€§ 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 в€€ GL(2n, R). (Hint: Examine the eп¬Ђect 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 Pfaп¬ѓan. It plays an important role in
diп¬Ђerential geometry.

7. Verify that, for any B в€€ A2 (V ), the symplectic reduction (V , B) is a well-deп¬Ѓned
symplectic space.

8. Show that if there is a G-invariant non-degnerate pairing ( , ): g Г— g в†’ R, then g and
gв€— are isomorphic as G-representations.

9. Compute the adjoint and coadjoint representations for

ab
a в€€ R+ , b в€€ R
G=
01

Show that g and gв€— 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 canвЂ™t be symplectic manifolds.)

10. For any Lie group G and any Оѕ в€€ gв€— , show that the symplectic structures в„¦Оѕ and в„¦aВ·Оѕ
on G В· Оѕ are the same for any a в€€ G.

E.5.2 96
11. This exercise concerns the splitting properties of the two Lie algebras sequences
associated to any symplectic structure в„¦ on a connected manifold M:

0 в€’в†’ R в€’в†’ C в€ћ (M) в€’в†’ h(в„¦) в€’в†’ 0

and
0 в€’в†’ h(в„¦) в€’в†’ sp(в„¦) в€’в†’ HdR (M, R) в€’в†’ 0.
1

Deп¬Ѓne the вЂњdivided powersвЂќ of в„¦ by the rule в„¦[k] = (1/k!) в„¦k , for each 0 в‰¤ k в‰¤ n.

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

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

Conclude that the п¬Ѓrst 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 в„¦[n] = 0
forms a Poisson subalgebra of C в€ћ (M).)
(ii) On the other hand, show that for R2 with the symplectic structure в„¦ = dxв€§dy, the
п¬Ѓrst sequence does not split. (Hint: Show that every smooth function on R2 is of the
form {x, g} for some g в€€ C в€ћ (R2 ). Why does this help?)
(iii) Suppose that M is compact. Deп¬Ѓne a skew-symmetric pairing

ОІв„¦ : HdR (M, R) Г— HdR (M, R) в†’ R
1 1

by the formula

a в€§ Лњ в€§ в„¦[nв€’1] ,
ОІв„¦ (a, b) = Лњb
M

where a and Лњ are closed 1-forms representing the cohomology classes a and b respec-
Лњ b
tively. Show that if there is a Lie algebra splitting Пѓ: HdR (M, R) в†’ sp(в„¦) then
1

ОІв„¦ (a, b)
в„¦ Пѓ(a), Пѓ(b) = в€’
vol(M, в„¦[n] )

for all a, b в€€ 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 п¬Ѓeld 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(в„¦) whose Lie algebra is h(в„¦). Thus, let (M, в„¦) be a symplectic manifold.
First, I remind you how the construction of the (smooth) universal cover of the identity
component of Sp(в„¦) goes. Let p: [0, 1] Г— M в†’ M be a smooth map with the property that
the map pt : M в†’ M deп¬Ѓned by pt (m) = p(t, m) is a symplectomorphism for all 0 в‰¤ t в‰¤ 1.
Such a p is called a (smooth) path in Sp(M). We say that p is based at the identity map
e: M в†’ M if p0 = e. The set of smooth paths in Sp(M) which are based at e will be
denoted by Pe Sp(в„¦) .
Two paths p and p in Pe Sp(в„¦) satisfying p1 = p1 are said to be homotopic if there
is a smooth map P : [0, 1] Г— [0, 1] Г— M в†’ M which satisп¬Ѓes 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(в„¦) is then denoted by Sp0 (в„¦). In any reasonable topology
on Sp(в„¦), this should to be the universal covering space of the identity component of
Sp(в„¦). There is a natural group structure on Sp0 (в„¦) in which Лњ, the homotopy class of
e
the constant path at e, is the identity element (cf., the covering spaces exercise in Exercise
Set 2).
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