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L = 1Q ω
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) .

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

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
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— (±).
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

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.


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