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1 ¤ i ¤ n.
B(ei , w) = δi ,

We know that these n equations are linearly independent, so there exists a solution f0 . Of
course, once one particular solution is found, any other solution is of the form f j = f0 +aji ei
for some n2 numbers aji . Thus, we have found the general solutions f j to the equations
B(ei , f j ) = δi .

L.5.2 81
We now show that we can choose the aij so as to satisfy the last remaining set of
conditions, B(f i , f j ) = 0. If we set bij = B(f0 , f0 ) = ’bji , then we can compute

j j
B(f i , f j ) = B(f0 , f0 ) + B(aik ek , f0 ) + B(f0 , ajl el ) + B(aik ek , ajl el )
i i

= bij + aij ’ aji + 0.

Thus, it su¬ces to set aij = ’bij /2. (This is where the hypothesis that the characteristic
of R is not 2 is used.)
Finally, it remains to show that the vectors e1 , . . . , en , f 1 . . . , f n form a basis of V .
Since we already know that dim(V ) = 2n, it is enough to show that these vectors are
linearly independent. However, any linear relation of the form

ai ei + bj f j = 0,

implies bk = B(ek , ai ei + bj f j ) = 0 and ak = ’B(f k , ai ei + bj f j ) = 0.

We often say that a basis of the form found in Proposition 1 is a symplectic or standard
basis of the symplectic space (V, B).

Symplectic Reduction of Vector Spaces. If B: V — V ’ R is a skew-symmetric
bilinear form which is not necessarily non-degenerate, then we de¬ne the null space of B
to be the subspace
NB = {v ∈ V | B(v, w) = 0 for all w ∈ V } .
On the quotient vector space V = V /NB , there is a well-de¬ned skew-symmetric bilinear
form B: V — V ’ R given by
B(x, y) = B(x, y)
where x and y are the cosets in V of x and y in V . It is easy to see that (V , B) is a
symplectic space.
De¬nition 2: If B is a skew-symmetric bilinear form on a vector space V , then the
symplectic space (V , B) is called the symplectic reduction of (V, B).

Here is an application of the symplectic reduction idea: Using the identi¬cation of
A2 (V ) with Λ2 (V — ) mentioned earlier, Proposition 1 allows us to write down a normal
form for any alternating 2-form on any ¬nite dimensional vector space.

Proposition 2: For any non-zero β ∈ Λ2 (V — ), there exist an integer n ¤ 1
dim(V ) and
linearly independent 1-forms ω 1 , ω 2 , . . . , ω 2n ∈ V — for which

β = ω 1 § ω 2 + ω 3 § ω 4 . . . + ω 2n’1 § ω 2n.

Thus, n is the largest integer so that β n = 0.

L.5.3 82
Proof: Regard β as a skew-symmetric bilinear form B on V in the usual way. Let
(V , B) be the symplectic reduction of (V, B). Since B = 0, we known that V = {0}. Let
1 n
dim(V ) = 2n ≥ 2 and let e1 , . . . , en , f 1 . . . , f n be elements of V so that e1 , . . . , en , f . . . , f
forms a symplectic basis of V with respect to B. Let p = dim(V ) ’ 2n, and let b1 , . . . , bp
be a basis of NB .
It is easy to see that

b = e1 f 1 e2 f 2 · · · en f n b1 · · · bp

forms a basis of V . Let
ω 1 · · · ω 2n+p
denote the dual basis of V — . Then, as the reader can easily check, the 2-form

„¦ = ω 1 § ω 2 + ω 3 § ω 4 . . . + ω 2n’1 § ω 2n

has the same values as β does on all pairs of elements of b. Of course this implies that
β = „¦. The rest of the Proposition also follows easily since, for example, we have

β n = n! ω 1 § · · · 2n
§ω = 0,

although β n+1 clearly vanishes.
If we regard β as an element of A2 (V ), then n is one-half the dimension of V . Some
sources call the integer n the half-rank of β and others call n the rank. I use “half-rank”.
Note that, unlike the case of symmetric bilinear forms, there is no notion of signature
type or “positive de¬niteness” for skew-symmetric forms.
It follows from Proposition 2 that for β in A2 (V ), where V is ¬nite dimensional, the
pair (V, β) is a symplectic space if and only if V has dimension 2n for some n and β n = 0.

Subspaces of Symplectic Vector Spaces. Let „¦ be a symplectic form on a vector
space V . For any subspace W ‚ V , we de¬ne the „¦-complement to W to be the subspace

W ⊥ = {v ∈ V | „¦(v, w) = 0 for all w ∈ W }.

The „¦-complement of a subspace W is sometimes called its skew-complement. It is an

exercise for the reader to check that, because „¦ is non-degenerate, W ⊥ = W and that,
when V is ¬nite-dimensional,

dim W + dim W ⊥ = dim V.

However, unlike the case of an orthogonal with respect to a positive de¬nite inner product,
the intersection W © W ⊥ does not have to be the zero subspace. For example, in an
„¦-standard basis for V , the vectors e1 , . . . , en obviously span a subspace L which satis¬es
L⊥ = L.

L.5.4 83
If V is ¬nite dimensional, it turns out (see the Exercises) that, up to symplectic linear
transformations of V , a subspace W ‚ V is characterized by the numbers d = dim W and
ν = dim (W © W ⊥ ) ¤ d. If ν = 0 we say that W is a symplectic subspace of V . This
corresponds to the case that „¦ restricts to W to de¬ne a symplectic structure on W . At
the other extreme is when ν = d, for then we have W © W ⊥ = W . Such a subspace is
called Lagrangian.

Symplectic Manifolds.
We are now ready to return to the study of manifolds.
De¬nition 3: A symplectic structure on a smooth manifold M is a non-degenerate, closed
2-form „¦ ∈ A2 (M). The pair (M, „¦) is called a symplectic manifold. If „¦ is a symplectic
structure on M and Υ is a symplectic structure on N , then a smooth map φ: M ’ N
satisfying φ— (Υ) = „¦ is called a symplectic map. If, in addition, φ is a di¬eomorphism, we
say that φ is a symplectomorphism.
Before developing any of the theory, it is helpful to see a few examples.

Surfaces with Area Forms. If S is an orientable smooth surface, then there exists
a volume form µ on S. By de¬nition, µ is a non-degenerate closed 2-form on S and hence
de¬nes a symplectic structure on S.

Lagrangian Structures on T M. From Lecture 4, any non-degenerate Lagrangian
L: T M ’ R de¬nes the 2-form dωL , which is a symplectic structure on T M.

A “Standard” Structure on R2n . Think of R2n as a smooth manifold and let „¦
be the 2-form with constant coe¬cients

dx = dx1 § dxn+1 + · · · + dxn § dx2n .
„¦= 2 dx Jn

Symplectic Submanifolds. Let (M 2m , „¦) be a symplectic manifold. Suppose that
P 2p ‚ M 2m be any submanifold to which the form „¦ pulls back to be a non-degenerate
2-form „¦P . Then (P, „¦P ) is a symplectic manifold. We say that P is a symplectic sub-
manifold of M.
It is not obvious just how to ¬nd symplectic submanifolds of M. Even though being
a symplectic submanifold is an “open” condition on submanifolds of M, is is not “dense”.
One cannot hope to perturb an arbitrary even dimensional submanifold of M slightly so
as to make it symplectic. There are even restrictions on the topology of the submanifolds
of M on which a symplectic form restricts to be non-degenerate.
For example, no symplectic submanifold of R2n (with any symplectic structure on
R2n ) could be compact for the following simple reason: Since R2n is contractible, its
second deRham cohomology group vanishes. In particular, for any symplectic form „¦
on R2n , there must be a 1-form ω so that „¦ = dω which implies that „¦m = d ω§„¦m’1 .
Thus, for all m > 0, the 2m-form „¦m is exact on R2n (and every submanifold of R2n ).

L.5.5 84
By Proposition 2, if M 2m were a submanifold of R2n on which „¦ restricted to be non-
degenerate, then „¦m would be a volume form on M. However, on a compact manifold the
volume form is never exact (just apply Stokes™ Theorem).

Example. Complex Submanifolds. Nevertheless, there are many symplectic subman-
ifolds of R2n . One way to construct them is to regard R2n as Cn in such a way that
the linear map J : R2n ’ R2n represented by Jn becomes complex multiplication. (For
example, just de¬ne the complex coordinates by z k = xk + ixk+n .) Then, for any non-zero
vector v ∈ R2n , we have „¦(v, J v) = ’|v|2 = 0. In particular, „¦ is non-degenerate on every
complex subspace S ‚ Cn . Thus, if M 2m ‚ Cn is any complex submanifold (i.e., all of
its tangent spaces are m-dimensional complex subspaces of Cm ), then „¦ restricts to be
non-degenerate on M.

The Cotangent Bundle. Let M be any smooth manifold and let T — M be its
cotangent bundle. As we saw in Lecture 4, there is a canonical 2-form on T — M which
can be de¬ned as follows: Let π: T — M ’ M be the basepoint projection. Then, for every
v ∈ T± (T — M), de¬ne
ω(v) = ± π (v) .

I claim that ω is a smooth 1-form on T — M and that „¦ = dω is a symplectic form on T — M.
To see this, let us compute ω in local coordinates. Let x: U ’ Rn be a local coordinate
chart. Since the 1-forms dx1 , . . . , dxn are linearly independent at every point of U, it follows
that there are unique functions ξi on T — U so that, for ± ∈ Ta U, —

± = ξ1 (±) dx1 |a + · · · + ξn (±) dxn |a .

The functions x1 , . . . , xn , ξ1 , . . . , ξn then form a smooth coordinate system on T — U in which
the projection mapping π is given by

π(x, p) = x.

It is then straightforward to compute that, in this coordinate system,

ω = ξi dxi .

Hence, „¦ = dξi §dxi and so is non-degenerate.

Symplectic Products. If (M, „¦) and (N, Υ) are symplectic manifolds, then M — N
carries a natural symplectic structure, called the product symplectic structure „¦ • Υ,
de¬ned by
— —
„¦ • Υ = π1 („¦) + π2 (Υ).

Thus, for example, n-fold products of compact surfaces endowed with area forms give
examples of compact symplectic 2n-manifolds.

L.5.6 85
Coadjoint Orbits. Let Ad— : G ’ GL(g— ) denote the coadjoint representation of G.
This is the so-called “contragredient” representation to the adjoint representation. Thus,
for any a ∈ G and ξ ∈ g— , the element Ad— (a)(ξ) ∈ g— is determined by the rule

Ad— (a)(ξ)(x) = ξ Ad(a’1 )(x) for all x ∈ g.

One must be careful not to confuse Ad— (a) with Ad(a) . Instead, as our de¬nition

shows, Ad— (a) = Ad(a’1 ) .
Note that the induced homomorphism of Lie algebras, ad— : g ’ gl(g— ) is given by

ad— (x)(ξ)(y) = ’ξ [x, y]

The orbits G · ξ in g— are called the coadjoint orbits. Each of them carries a natural
symplectic structure. To see how this is de¬ned, let ξ ∈ g— be ¬xed, and let φ: G ’ G · ξ
be the usual submersion induced by the Ad— -action, φ(a) = Ad— (a)(ξ) = a · ξ. Now let ωξ
be the left-invariant 1-form on G whose value at e is ξ. Thus, ωξ = ξ(ω) where ω is the
canonical g-valued 1-form on G.

Proposition 3: There is a unique symplectic form „¦ξ on the orbit G·ξ = G/Gξ satisfying
φ— („¦ξ ) = dωξ .

Proof: If Proposition 3 is to be true, then „¦ξ must satisfy the rule

„¦ξ φ (v), φ (w) = dωξ (v, w) for all v, w ∈ Ta G.

What we must do is show that this rule actually does de¬ne a symplectic 2-form on G · ξ.
First, note that, for x, y ∈ g = Te G, we may compute via the structure equations that

dωξ (x, y) = ξ dω(x, y) = ξ ’[x, y] = ad— (x)(ξ)(y).

In particular, ad— (x)(ξ) = 0, if and only if x lies in the null space of the 2-form dωξ (e). In
other words, the null space of dωξ (e) is gξ , the Lie algebra of Gξ . Since dωξ is left-invariant,
it follows that the null space of dωξ (a) is La (gξ ) ‚ Ta G. Of course, this is precisely the
tangent space at a to the left coset aGξ . Thus, for each a ∈ G,

Ndωξ (a) = ker φ (a),

It follows that, Ta·ξ (G · ξ) = φ (a)(Ta G) is naturally isomorphic to the symplectic quotient
space (Ta G)/ La (gξ ) for each a ∈ G. Thus, there is a unique, non-degenerate 2-form „¦a

on Ta·ξ (G · ξ) so that φ (a) („¦a ) = dωξ (a).
It remains to show that „¦a = „¦b if a · ξ = b · ξ. However, this latter case occurs only
if a = bh where h ∈ Gξ . Now, for any h ∈ Gξ , we have

R— (ωξ ) = ξ R— (ω) = ξ Ad(h’1 )(ω) = Ad— (h)(ξ)(ω) = ξ(ω) = ωξ .
h h

L.5.7 86
Thus, R— (dωξ ) = dωξ . Since the following square commutes, it follows that „¦a = „¦b .
Ta G Tb G
¦ ¦
¦ ¦
φ (a) φ (b)


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