that, for some c ∈ Rn , the f-level set Mc = f ’1 (c) is compact. Replacing f by f ’ c, we

may assume that c = 0, which we do from now on.

Show that M0 ‚ M is a closed Lagrangian submanifold of M.

Use Proposition 4 to show that there is an open neighborhood B of 0 ∈ Rn so that

f ’1 (B), „¦ is symplectomorphic to a neighborhood U of the zero section in T — M0 (en-

dowed with its standard symplectic structure) in such a way that, for each b ∈ B, the sub-

manifold Mb = f ’1 (b) is identi¬ed with the graph of a closed 1-form ωb on M0 . Show that

it is possible to choose b1 , . . . , bn in B so that the corresponding closed 1-forms ω1 , . . . , ωn

are linearly independent at every point of M0 .

Conclude that M0 is di¬eomorphic to a torus T = Rn /Λ where Λ ‚ Rn is a lattice,

in such a way that the forms ωi become identi¬ed with dθi where θi are the corresponding

linear coordinates on Rn .

Now prove that for any b ∈ B, the 1-form ωb must be a linear combination of the ωi

with constant coe¬cients. Thus, there are functions ai on B so that ωb = ai (b)ωi . (Hint:

Show that the coe¬cients must be invariant under the ¬‚ows of the vector ¬elds dual to

the ωi .)

Conclude that, under the symplectic map identifying MB with U, the form „¦ gets

identi¬ed with dai §dθi . The functions ai and θi are the so-called “action-angle coordi-

nates”.

Extra Credit: Trace through the methods used to prove Proposition 4 and show that,

in fact, the action-angle coordinates can be constructed using quadrature and “¬nite”

operations.

E.6.4 115

Lecture 7:

Classical Reduction

In this section, we return to the study of group actions. This time, however, we

will concentrate on group actions on symplectic manifolds that preserve the symplectic

structure. Such actions happen to have quite interesting properties and moreover, turn

out to have a wide variety of applications.

Symplectic Group Actions. First, the basic de¬nition.

De¬nition 1: Let (M, „¦) be a symplectic manifold and let G be a Lie group. A left

action »: G — M ’ M of G on M is a symplectic action if »— („¦) = „¦ for all a ∈ G.

a

We have already encountered several examples:

Example: Lagrangian Symmetries. If G acts on a manifold M is such a way that

it preserves a non-degenerate Lagrangian L: T M ’ R, then, by construction, it preserves

the symplectic 2-form dωL .

Example: Cotangent Actions. A left G-action »: G — M ’ M, induces an action

» of G on T — M. Namely, for each a ∈ G, the di¬eomorphism »a : M ’ M induces a

˜

di¬eomorphism »a : T — M ’ T — M. Since the natural symplectic structure on T — M is

˜

˜

invariant under di¬eomorphisms, it follows that » is a symplectic action.

Example: Coadjoint Orbits. As we saw in Lecture 5, for every ξ ∈ g— , the coadjoint

orbit G · ξ carries a natural G-invariant symplectic structure „¦ξ . Thus, the left action of

G on G · ξ is symplectic.

Example: Circle Actions on Cn . Let z 1 , . . . , z n be linear complex coordinates on Cn

and let this vector space be endowed with the symplectic structure

dz 1 § d¯1 + · · · + dz n § d¯n

i

„¦= z z

2

= dx1 § dy 1 + · · · + dxn § dy n

where z k = xk + iy k . Then for any integers (k1 , . . . , kn ), we can de¬ne an action of S 1 on

Cn by the formula « 1 « ik1 θ 1

z e z

. .

eiθ · .

.

=

. .

zn eikn θ z n

The reader can easily check that this de¬nes a symplectic circle action on Cn .

Generally what we will be interested in is the following: Y will be a Hamiltonian

vector ¬eld on a symplectic manifold (M, „¦) and G will act symplectically on M as a

group of symmetries of the ¬‚ow of Y . We want to understand how to use the action of G

to “reduce” the problem of integrating the ¬‚ow of Y .

L.7.1 116

In Lecture 3, we saw that when Y was the Euler-Lagrange vector ¬eld associated to a

non-degenerate Lagrangian L, then the in¬nitesimal generators of symmetries of L could

be used to generate conserved quantities for the ¬‚ow of Y . We want to extend this process

(as far as is reasonable) to the general case.

For the rest of the lecture, I will assume that G is a Lie group with a symplectic

action » on a connected symplectic manifold (M, „¦).

Since » is symplectic, it follows that the mapping »— : g ’ X(M) actually has image

in sp(„¦), the algebra of symplectic vector ¬elds on M. As we saw in Lecture 3, »— is

an anti-homomorphism, i.e., »— [x, y] = ’ »— (x), »— (y) . Since, as we saw in Lecture

5, [sp(„¦), sp(„¦)] ‚ h(„¦), it follows that »— [g, g] ‚ h(„¦). Thus, H» : g ’ HdR (M, R)

1

de¬ned by H» (x) = »— (x) „¦ is a homomorphism of Lie algebras with kernel containing

the commutator subalgebra [g, g].

The map H» is the obstruction to ¬nding a Hamiltonian function associated to each

in¬nitesimal symmetry »— (x) since H» (x) = 0 if and only if »— (x) „¦ = ’df for some

f ∈ C ∞(M).

De¬nition 2: A symplectic action »: G — M ’ M is said to be Hamiltonian if H» = 0,

i.e., if »— (g) ‚ h(„¦).

There are a few particularly interesting cases where the obstruction H» must vanish:

• If HdR (M, R) = 0. In particular, if M is simply connected.

1

• If g is perfect, i.e., [g, g] = g. For example this happens whenever the Killing form

on g is non-degenerate (this is the ¬rst Whitehead Lemma, see Exercise 3). However, this

is not the only case: For example, if G is the group of rigid motions in Rn for n ≥ 3, then

g has this property, even though its Killing form is degenerate.

• If there exists a 1-form ω on M that is invariant under G and satis¬es „¦ = dω.

(This is the case of symmetries of a Lagrangian.) To see this, note that if X is a vector

¬eld on M that preserves ω, then

0 = LX ω = d(X ω) + X „¦,

so X „¦ is exact.

For a Hamiltonian action », every in¬nitesimal symmetry »— (x) has a Hamiltonian

function fx ∈ C ∞. However, the choice of fx is not unique since we can add any constant

to fx without changing its Hamiltonian vector ¬eld. This non-uniqueness causes some

problems in the theory we wish to develop.

To see why, suppose that we choose a (linear) lifting ρ: g ’ C ∞(M) of ’»— : g ’ h(„¦).

(The choice of ’»— instead of »— was made to get rid of the annoying sign in the formula

for the bracket.)

g

’»—

“

ρ

C ∞(M)

’R ’ ’ h(„¦) ’

0 0

L.7.2 117

Thus, for every x ∈ g, we have »— (x) „¦ = d ρ(x) . A short calculation (see the Exercises)

now shows that {ρ(x), ρ(y)} is a Hamiltonian function for ’»— ([x, y]), i.e., that

„¦ = d {ρ(x), ρ(y)} .

»— [x, y]

In particular, it follows (since M is connected) that there must be a skew-symmetric

bilinear map cρ: g — g ’ R so that

{ρ(x), ρ(y)} = ρ [x, y] + cρ (x, y).

An application of the Jacobi identity implies that the map cρ satis¬es the condition

cρ [x, y], z + cρ [y, z], x + cρ [z, x], y = 0 for all x, y, z ∈ g.

This condition is known as the 2-cocycle condition for cρ regarded as an element of A2 (g) =

Λ2 (g— ). (See Exercise 3 for an explanation of this terminology.)

For purposes of simplicity, it would be nice if we could choose ρ so that cρ were

identically zero. In order to see whether this is possible, let us choose another linear map

ρ: g ’ C ∞(M) that satis¬es ρ(x) = ρ(x) + ξ(x) where ξ: g ’ R is any linear map. Every

˜ ˜

possible lifting of ’»— is clearly of this form for some ξ. Now we compute that

ρ(x), ρ(y) = ρ(x), ρ(y) = ρ [x, y] + cρ(x, y)

˜ ˜

= ρ [x, y] + cρ (x, y) ’ ξ [x, y] .

˜

Thus, cρ(x, y) = cρ(x, y) ’ ξ [x, y] . Thus, in order to be able to choose ρ so that cρ = 0,

˜

˜ ˜

—

we see that there must exist a ξ ∈ g so that cρ = ’δξ where δξ is the skew-symmetric

bilinear map on g that satis¬es δξ(x, y) = ’ξ [x, y] (see the Exercises for an explanation

of this notation). This is known as the 2-coboundary condition.

There are several important cases where we can assure that cρ can be written in the

form ’δξ. Among them are:

• If M is compact, then the sequence

0 ’ R ’ C ∞ (M) ’ H(„¦) ’ 0

splits: If we let C0 (M, „¦) ‚ C ∞ (M) denote the space of functions f for which M f „¦n =

∞

0, then these functions are closed under Poisson bracket (see Exercise 5.6 for a hint as to

why this is true) and we have a splitting of Lie algebras C ∞(M) = R • C0 (M, „¦). Now

∞

∞

just choose the unique ρ so that it takes values in C0 (M, „¦). This will clearly have cρ = 0.

• If g has the property that every 2-cocycle for g is actually a 2-coboundary. This hap-

pens, for example, if the Killing form of g is non-degenerate (this is the second Whitehead

Lemma, see Exercise 3), though it can also happen for other Lie algebras. For example,

for the non-abelian Lie algebra of dimension 2, it is easy to see that every 2-cocycle is a

2-coboundary.

L.7.3 118

• If there is a 1-form ω on M that is preserved by the G action and satis¬es dω = „¦.

(This is true in the case of symmetries of a Lagrangian.) In this case, we can merely take

ρ(x) = ’ω »— (x) . I leave as an exercise for the reader to check that this works.

De¬nition 3: A Hamiltonian action »: G — M ’ M is said to be a Poisson action if

there exists a lifting ρ with cρ = 0.

Henceforth in this Lecture, I am only going to consider Poisson actions. By my

previous remarks, this case includes all of the Lagrangians with symmetries, but it also

includes many others.

I will assume that, in addition to having a Poisson action »: G — M ’ M speci¬ed,

we have chosen a lifting ρ: g ’ C ∞ (M) of ’»— that satis¬es ρ(x), ρ(y) = ρ [x, y] for

all x, y ∈ g. Note that such a ρ is unique up to replacement by ρ = ρ + ξ where ξ : g ’ R

˜

satis¬es δξ = 0. Such ξ (if any non-zero ones exist) are ¬xed under the co-adjoint action

of the identity component of G.

The Momentum Mapping. We are now ready to make one of the most important

constructions in the theory.

The momentum mapping associated to » and ρ is the mapping µ: M ’ g—

De¬nition 4:

that satis¬es

µ(m)(y) = ρ(y)(m).

Note that, for ¬xed m ∈ M, the assignment y ’ ρ(y)(m) is a linear map from g to

R, so the de¬nition makes sense.

It is worth pausing to consider why this mapping is called the momentum mapping.

The reader should calculate this mapping in the case of a free particle or a rigid body

moving in space. In either case, the Lagrangian is invariant under the action of the group

G of rigid motions of space. If y ∈ g corresponds to a translation, then ρ(y) gives the

function on T R3 that evaluates at each point (i.e., each position-plus-velocity) to be the

linear momentum in the direction of translation. If y corresponds to rotation about a ¬xed

axis, then ρ(y) turns out to be the angular momentum of the body about that axis.

One important reason for studying the momentum mapping is the following formula-

tion of the classical conservation of momentum theorems:

Proposition 1: If Y is a symplectic vector ¬eld on M that is invariant under the action

of G, then µ is constant on the integral curves of Y .

In particular, µ provides conserved quantities for any G-invariant Hamiltonian.

The main result about the momentum mapping is the following one.

Theorem 1: If G is connected, then the momentum mapping µ: M ’ g— is G-equivariant.

L.7.4 119

Proof: Recall that the coadjoint action of G on g— is de¬ned by Ad— (g)(ξ)(x) =

ξ Ad(g ’1 ) x . The condition that µ be G-equivariant, i.e., that µ(g · m) = Ad— (g) µ(m)

for all m ∈ M and g ∈ G, is thus seen to be equivalent to the condition that

ρ Ad(g ’1 )y (m) = ρ(y)(g · m)

for all m ∈ M, g ∈ G, and y ∈ g. This is the identity I shall prove.

Since G is connected and since each side of the above equation represents a G-action,

if we prove that the above formula holds for g of the form g = etx for any x ∈ g and any

t ∈ R, the formula for general g will follow. Thus, we want to prove that

ρ Ad(e’tx )y (m) = ρ(y)(etx · m)

for all t. Since this latter equation holds at t = 0, it is enough to show that both sides

have the same derivative with respect to t.

Now the derivative of the right hand side of the formula is

d ρ(y) »— (x)(etx · m) = „¦ »— (y)(etx · m), »— (x)(etx · m)

= „¦ »— Ad(e’tx )y (m), »— Ad(e’tx )x (m)

= „¦ »— Ad(e’tx )y (m), »— (x)(m)

where, to verify the second equality we have used the identity

»a »— (y)(m) = »— Ad(a)y (a · m)

and the fact that „¦ is G-invariant.

On the other hand, the derivative of the left hand side of the formula is clearly

ρ [’x, Ad(e’tx )y] (m) = ’ ρ(x), ρ Ad(e’tx )y (m)

= „¦ »— Ad(e’tx )y (m), »— (x)(m)

so we are done. (Note that I have used my assumption that cρ = 0!)

Example: Left-Invariant Metrics on Lie Groups. Let G be a Lie group and let

Q: g ’ R be a non-degenerate quadratic form with associated inner product , Q . Let

L: T G ’ R be the Lagrangian