2

where ω: T G ’ g is, as usual, the canonical left-invariant form on G. Then, using the

basepoint map π: T G ’ G, we compute that

ωL = ω, π — (ω) .

Q

As we saw in Lecture 3, the assumption that Q is non-degenerate implies that dωL is

a symplectic form on T G. Now, since the ¬‚ow of a right-invariant vector ¬eld Yx is

multiplication on the left by etx , it follows that, for this action, we may de¬ne

= ’ ω, Ad(g ’1 )x

ρ(x) = ’ωL (Yx ) = ’ ω, ω(Yx ) Q Q

L.7.5 120

(where g: T G ’ G is merely a more descriptive name for the base point map than π).

Now, there is an isomorphism „Q : g ’ g— , called transpose with respect to Q that

satis¬es „Q (x)(y) = x, y Q for all x, y ∈ g. In terms of „Q , we can express the momentum

mapping as

µ(v) = ’Ad— (g) „Q (ω(v))

for all v ∈ T G. Note that µ is G-equivariant, as promised by the theorem.

According to the Proposition 1, the function µ is a conserved quantity for the solutions

of the Euler-Lagrange equations. In one of the Exercises, you are asked to show how this

information can be used to help solve the Euler-Lagrange equations for the L-critical

curves.

Example: Coadjoint Orbits. Let G be a Lie group and consider ξ ∈ g— with stabilizer

subgroup Gξ ‚ G. The orbit G · ξ ‚ g— is canonically identi¬ed with G/Gξ ( identify a · ξ

with aGξ ) and we have seen that there is a canonical G-invariant symplectic form „¦ξ

—

on G/Gξ that satis¬es πξ („¦ξ ) = dωξ where πξ : G ’ G/Gξ is the coset projection, ω is

the tautological left-invariant 1-form on G, and ωξ = ξ(ω).

Recall also that, for each x ∈ g, the right-invariant vector ¬eld Yx on G is de¬ned so

that Yx (e) = x ∈ g. Then the vector ¬eld »— (x) on G/Gξ is πξ -related to Yx , so

—

πξ »— (x) „¦ξ = Yx dωξ = d ’ωξ (Yx ) .

(This last equality follows because ωξ , being left-invariant, is invariant under the ¬‚ow

of Yx .) Now, the value of the function ωξ (Yx ) at a ∈ G is

ωξ (Yx )(a) = ξ ω(Yx (a)) = ξ Ad(a’1 )(x) = Ad— (a)(ξ)(x) = (a · ξ)(x).

Thus, it follows that the natural left action of G on G·ξ is Poisson, with momentum

mapping µ : G·ξ ’ g— given by

µ(a · ξ) = ’ a · ξ.

(Note: some authors do not have a minus sign here, but that is because their „¦ξ is the

negative of ours.)

Reduction. I now want to discuss a method of taking quotients by group actions in

the symplectic category. Now, when a Lie group G acts symplectically on the left on a

symplectic manifold M, it is not generally true that the space of orbits G\M can be given

a symplectic structure, even when this orbit space can be given the structure of a smooth

manifold (for example, the quotient need not be even dimensional).

However, when the action is Poisson, there is a natural method of breaking the orbit

space G\M into a union of symplectic submanifolds provided that certain regularity criteria

are met. The procedure I will describe is known as symplectic reduction. It is due, in its

modern form, to Marsden and Weinstein (see [GS 2]).

The idea is simple: If µ: M ’ g— is the momentum mapping, then the G-equivariance

of µ implies that there is a well-de¬ned set map

µ: G\M ’ G\g— .

¯

L.7.6 121

The theorem we are about to prove asserts that, provided certain regularity criteria are

met, the subsets Mξ = µ’1 (ξ) ‚ G\M are symplectic manifolds in a natural way.

¯

¯

De¬nition 4: Let f: X ’ Y be a smooth map. A point y ∈ Y is a clean value of f if

the set f ’1 (y) ‚ X is a smooth submanifold of X and, moreover, if Tx f ’1 (y) = ker f (x)

for each x ∈ f ’1 (y).

Note: While every regular value of f is clean, not every clean value of f need be

regular. The concept of cleanliness is very frequently encountered in the reduction theory

we are about to develop.

Theorem 2: Let »: G — M ’ M be a Poisson action on the symplectic manifold M.

Let µ: M ’ g— be a momentum mapping for ». Suppose that, ξ ∈ g— is a clean value of µ.

Then Gξ acts smoothly on µ’1 (ξ). Suppose further that the space of Gξ -orbits in µ’1 (ξ),

say, Mξ = Gξ \ µ’1 (ξ) , can be given the structure of a smooth manifold in such a way

that the quotient mapping πξ : µ’1 (ξ) ’ Mξ is a smooth submersion. Then there exists a

—

symplectic structure „¦ξ on Mξ that is de¬ned by the condition that πξ („¦ξ ) be the pullback

of „¦ to µ’1 (ξ).

Proof: Since ξ is a clean value of µ, we know that µ’1 (ξ) is a smooth submanifold of M.

By the G-equivariance of the momentum mapping, the stabilizer subgroup Gξ ‚ G acts

on M preserving the submanifold µ’1 (ξ). The restricted action of Gξ on µ’1 (ξ) is easily

seen to be smooth.

Now, I claim that, for each m ∈ µ’1 (ξ), the „¦-complementary subspace to Tm µ’1 (ξ)

is the space Tm G · m , i.e., the tangent to the G-orbit through m. To see this, ¬rst

note that the space Tm G · m is spanned by the values at m assumed by the vector

¬elds »— (x) for x ∈ g. Thus, a vector v ∈ Tm M lies in the „¦-complementary space

of Tm G · m if and only if v satis¬es „¦ »— (x)(m), v = 0 for all x ∈ g. Since, by

de¬nition, „¦ »— (x)(m), v = d ρ(x) (v), it follows that this condition on v is equivalent

to the condition that v lie in ker µ (m). However, since ξ is a clean value of µ, we have

ker µ (m) = Tm µ’1 (ξ) , as claimed.

Now, the G-equivariance of µ implies that µ’1 (ξ)© G·m = Gξ ·m for all m ∈ µ’1 (ξ).

In particular, Tm Gξ ·m ⊆ Tm µ’1 (ξ) ©Tm G·m . To demonstrate the reverse inclusion,

suppose that v lies in both Tm µ’1 (ξ) and Tm G·m . Then v = »— (x)(m) for some x ∈ g,

and, by the G-equivariance of the momentum mapping and the assumption that ξ is clean

(so that Tm µ’1 (ξ) = ker µ (m)) we have

0 = µ (m)(v) = µ (m) »— (x)(m) = Ad— — (x)(µ(m)) = Ad— — (x)(ξ)

so that x must lie in gξ . Consequently, v = »— (x)(m) is tangent to the orbit Gξ · m and

thus,

Tm µ’1 (ξ) © Tm G · m = ker µ (m) © Tm G · m = Tm Gξ · m .

As a result, since the „¦-complementary spaces Tm µ’1 (ξ) and Tm G · m intersect in the

tangents to the Gξ -orbits, it follows that if „¦ξ denotes the pullback of „¦ to µ’1 (ξ), then

˜

˜

the null space of „¦ξ at m is precisely Tm Gξ · m .

L.7.7 122

Finally, let us assume, as in the theorem, that there is a smooth manifold structure

on the orbit space Mξ = Gξ \µ’1 (ξ) so that the orbit space projection πξ : µ’1 (ξ) ’ Mξ is

˜

a smooth submersion. Since „¦ξ is clearly Gξ invariant and closed and moreover, since its

null space at each point of M is precisely the tangent space to the ¬bers of πξ , it follows

that there exists a unique “push down” 2-form „¦ξ on Mξ as described in the statement of

the theorem. That „¦ξ is closed and non-degenerate is now immediate.

The point of Theorem 2 is that, even though the quotient of a symplectic manifold

by a symplectic group action is not, in general, a symplectic manifold, there is a way to

produce a family of symplectic quotients parametrized by the elements of the space g— .

The quotients Mξ often turn out to be quite interesting, even when the original symplectic

manifold M is very simple.

Before I pass on to the examples, let me make a few comments about the hypotheses

in Theorem 2.

First, there will always be clean values ξ of µ for which µ’1 (ξ) is not empty (even

when there are no such regular values). This follows because, if we look at the closed

subset Dµ ‚ M consisting of points m where µ (m) does not reach its maximum rank,

then µ Dµ ) can be shown (by a sort of Sard™s Theorem argument) to be a proper subset

of µ(M). Meanwhile, it is not hard to show that any element ξ ∈ µ(M) that does not lie

in µ Dµ ) is clean.

Second, it quite frequently does happen that the Gξ -orbit space Mξ has a manifold

structure for which πξ is a submersion. This can be guaranteed by various hypotheses that

are often met with in practice. For example, if Gξ is compact and acts freely on µ’1 (ξ),

then Mξ will be a manifold. (More generally, if the orbits Gξ · m are compact and all of the

stabilizer subgroups Gm ‚ Gξ are conjugate in Gξ , then Mξ will have a manifold structure

of the required kind.)

Weaker hypotheses also work. Basically, one needs to know that, at every point m

of µ’1 (ξ), there is a smooth slice to the action of Gξ , i.e., a smoothly embedded disk D

in µ’1 (ξ) that passes through m and intersects each Gξ -orbit in Gξ · D transversely and

in exactly one point. (Compare the construction of a smooth structure on each G-orbit in

Theorem 1 of Lecture 3.)

Even when there is not a slice around each point of µ’1 (ξ), there is very often a near-

slice, i.e., a smoothly embedded disk D in µ’1 (ξ) that passes through m and intersects

each Gξ -orbit in Gξ · D transversely and in a ¬nite number of points. In this case, the

quotient space Mξ inherits the structure of a symplectic orbifold, and these ˜generalized

manifolds™ have turned out to be quite useful.

Finally, it is worth computing the dimension of Mξ when it does turn out to be a

manifold. Let Gm ‚ G be the stabilizer of m ∈ µ’1 (ξ). I leave as an exercise for the

reader to check that

dim Mξ = dim M ’ dim G ’ dim Gξ + 2 dim Gm

= dim M ’ 2 dim G/Gm + dim G/Gξ .

L.7.8 123

Since we will see so many examples in the next Lecture, I will content myself with

only mentioning two here:

• Let M = T — G and let G act on T — G on the left in the obvious way. Then the reader

—

can easily check that, for each x ∈ g, we have ρ(x)(±) = ± Yx (a) for all ± ∈ Ta G where,

as usual, Yx denotes the right invariant vector ¬eld on G whose value at e is x ∈ g. Hence,

µ: T — G ’ g— is given by µ(±) = R— (±).

π(±)

Consequently, µ’1 (ξ) ‚ T — G is merely the graph in T — G of the left-invariant 1-form

ωξ (i.e., the left-invariant 1-form whose value at e is ξ ∈ g— ). Thus, we can use ωξ as a

section of T — G to pull back „¦ (the canonical symplectic form on T — G, which is clearly

G-invariant) to get the 2-form dωξ on G. As we already saw in Lecture 5, and is now

borne out by Theorem 2, the null space of dωξ at any point a ∈ G is Ta aGξ , the quotient

by Gξ is merely the coadjoint orbit G/Gξ , and the symplectic structure „¦ξ is just the one

we already constructed.

Note, by the way, that every value of µ is clean in this example (in fact, they are all

regular), even though the dimensions of the quotients G/Gξ vary with ξ.

• Let G = SO(3) act on R6 = T — R3 by the extension of rotation about the origin in

R3 . Then, in standard coordinates (x, y) (where x, y ∈ R3 ), the action is simply g · (x, y) =

(gx, gy), and the symplectic form is „¦ = dx · dy = tdx§dy.

We can identify so(3)— with so(3) itself by interpreting a ∈ so(3) as the linear func-

tional b ’ ’tr(ab). It is easy to see that the co-adjoint action in this case gets identi¬ed

with the adjoint action.

We compute that ρ(a)(x, y) = ’txay, so it follows without too much di¬culty that,

with respect to our identi¬cation of so(3)— with so(3), we have µ(x, y) = xty ’ y tx.

The reader can check that all of the values of µ are clean except for 0 ∈ so(3). Even

this value would be clean if, instead of taking M to be all of R6 , we let M be R6 minus

the origin (x, y) = (0, 0).

I leave it to the reader to check that the G-invariant map P : R6 ’ R3 de¬ned by

P (x, y) = (x · x, x · y, y · y)

maps the set µ’1 (0) onto the “cone” consisting of those points (a, b, c) ∈ R3 with a, c ≥ 0

and b2 = ac and the ¬bers of P are the G0 -orbits of the points in µ’1 (0).

For ξ = 0, the P -image of the set µ’1 (ξ) is one nappe of the hyperboloid of two sheets

described as ac ’ b2 = ’tr(ξ 2 ). The reader should compute the area forms „¦ξ on these

sheets.

L.7.9 124

Exercise Set 7:

Classical Reduction

1. Let M be the torus R2 /Z2 and let dx and dy be the standard 1-forms on M. Let

„¦ = dx§dy. Show that the “translation action” (a, b) · [x, y] = [x + a, y + b] of R2 on M is

symplectic but not Hamiltonian.

2. Let (M, „¦) be a connected symplectic manifold and let »: G—M ’ M be a Hamiltonian

group action.

(i) Prove that, if ρ: g ’ C ∞ (M) is a linear mapping that satis¬es »— (x) „¦ = d ρ(x) ,

then »— [x, y] „¦ = d {ρ(x), ρ(y)} .

(ii) Show that the associated linear mapping cρ: g — g ’ R de¬ned in the text does indeed

satisfy cρ [x, y], z + cρ [y, z], x + cρ [z, x], y = 0 for all x, y, z ∈ g. (Hint: Use the

fact the Poisson bracket satis¬es the Jacobi identity and that the Poisson bracket of

a constant function with any other function is zero.)

3. Lie Algebra Cohomology. The purpose of this exercise is to acquaint the reader

with the rudiments of Lie algebra cohomology.

The Lie bracket of a Lie algebra g can be regarded as a linear map ‚: Λ2 (g) ’ g.

The dual of this linear map is a map ’δ: g— ’ Λ2 (g— ). (Thus, for ξ ∈ g— , we have

δξ(x, y) = ’ξ [x, y] .) This map δ can be extended uniquely to a graded, degree-one

derivation δ: Λ— (g— ) ’ Λ— (g— ).

(i) For any c ∈ Λ2 (g— ), show that δc(x, y, z) = ’c [x, y], z ’ c [y, z], x ’ c [z, x], y .

(Hint: Every c ∈ Λ2 (g— ) is a sum of wedge products ξ§· where ξ, · ∈ g— .) Conclude

that the Jacobi identity in g is equivalent to the condition that δ 2 = 0 on all of Λ— (g— ).

Thus, for any Lie algebra g, we can de¬ne the k™th cohomology group of g, denoted H k (g),

as the kernel of δ in Λk (g— ) modulo the subspace δ Λk’1 (g— ) .

(ii) Let G be a Lie group whose Lie algebra is g. For each ¦ ∈ Λk (g— ), de¬ne ω¦ to be the

left-invariant k-form on G whose value at the identity is ¦. Show that dω¦ = ωδ¦ .

(Hint: the space of left-invariant forms on G is clearly closed under exterior derivative

and is generated over R by the left-invariant 1-forms. Thus, it su¬ces to prove this

formula for ¦ of degree 1. Why?)

Thus, the cohomology groups H k (g) measure “closed-mod-exact” in the space of left-

invariant forms on G. If G is compact, then these cohomology groups are isomorphic to

the corresponding deRham cohomology groups of the manifold G.

(iii) (The Whitehead Lemmas) Show that if the Killing form of g is non-degenerate,

then H 1 (g) = H 2 (g) = 0. (Hint: You should have already shown that if κ is non-

degenerate, then [g, g] = g. Show that this implies that H 1 (g)=0. Next show that for

¦ ∈ Λ2 (g— ), we can write ¦(x, y) = κ(Lx, y) where L: g ’ g is skew-symmetric. Then

show that if δ¦ = 0, then L is a derivation of g. Now see Exercise 3.3, part (iv).)

E.7.1 125

4. Homogeneous Symplectic Manifolds. Suppose that (M, „¦) is a symplectic mani-

fold and suppose that there exists a transitive symplectic action »: G— M ’ M where G is

a group whose Lie algebra satis¬es H 1 (g) = H 2 (g) = 0. Show that there is a G-equivariant

symplectic covering map π: M ’ G/Gξ for some ξ ∈ g— . Thus, up to passing to covers,

the only symplectic homogeneous spaces of a Lie group satisfying H 1 (g) = H 2 (g) = 0 are

the coadjoint orbits. This result is usually associated with the names Kostant, Souriau,

and Symes.

(Hint: Since G acts homogeneously on M, it follows that, as G-spaces, M = G/H for

some closed subgroup H ‚ G that is the stabilizer of a point m of M. Let φ: G ’ M be

φ(g) = g · m. Now consider the left-invariant 2-form φ— („¦) on G in light of the previous

Exercise. Why do we also need the hypothesis that H 1 (g) = 0?)

Remark: This characterization of homogeneous symplectic spaces is sometimes mis-

quoted. Either the covering ambiguity is overlooked or else, instead of hypotheses about

the cohomology groups, sometimes compactness is assumed, either for M or G. The ex-

ample of S 1 — S 1 acting on itself and preserving the bi-invariant area form shows that

compactness is not generally helpful. Here is an example that shows that you must allow

for the covering possibility: Let H ‚ SL(2, R) be the subgroup of diagonal matrices with

positive entries on the diagonal. Then SL(2, R)/H has an SL(2, R)-invariant area form,

but it double covers the associated coadjoint orbit.

5. Verify the claim made in the text that, if there exists a G-invariant 1-form ω on M so

that dω = „¦, then the formula ρ(x) = ’ω »— (x) yields a lifting ρ for which cρ = 0.