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L = 1Q П‰
2
where П‰: T G в†’ g is, as usual, the canonical left-invariant form on G. Then, using the
basepoint map ПЂ: T G в†’ G, we compute that

П‰L = П‰, ПЂ в€— (П‰) .
Q

As we saw in Lecture 3, the assumption that Q is non-degenerate implies that dП‰L is
a symplectic form on T G. Now, since the п¬‚ow of a right-invariant vector п¬Ѓeld Yx is
multiplication on the left by etx , it follows that, for this action, we may deп¬Ѓne

= в€’ П‰, Ad(g в€’1 )x
ПЃ(x) = в€’П‰L (Yx ) = в€’ П‰, П‰(Yx ) Q Q

L.7.5 120
(where g: T G в†’ G is merely a more descriptive name for the base point map than ПЂ).
Now, there is an isomorphism П„Q : g в†’ gв€— , called transpose with respect to Q that
satisп¬Ѓes П„Q (x)(y) = x, y Q for all x, y в€€ g. In terms of П„Q , we can express the momentum
mapping as
Вµ(v) = в€’Adв€— (g) П„Q (П‰(v))
for all v в€€ T G. Note that Вµ is G-equivariant, as promised by the theorem.
According to the Proposition 1, the function Вµ 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 Оѕ в€€ gв€— with stabilizer
subgroup GОѕ вЉ‚ G. The orbit G В· Оѕ вЉ‚ gв€— is canonically identiп¬Ѓed with G/GОѕ ( identify a В· Оѕ
with aGОѕ ) and we have seen that there is a canonical G-invariant symplectic form в„¦Оѕ
в€—
on G/GОѕ that satisп¬Ѓes ПЂОѕ (в„¦Оѕ ) = dП‰Оѕ where ПЂОѕ : G в†’ G/GОѕ is the coset projection, П‰ is
the tautological left-invariant 1-form on G, and П‰Оѕ = Оѕ(П‰).
Recall also that, for each x в€€ g, the right-invariant vector п¬Ѓeld Yx on G is deп¬Ѓned so
that Yx (e) = x в€€ g. Then the vector п¬Ѓeld О»в€— (x) on G/GОѕ is ПЂОѕ -related to Yx , so
в€—
ПЂОѕ О»в€— (x) в„¦Оѕ = Yx dП‰Оѕ = d в€’П‰Оѕ (Yx ) .

(This last equality follows because П‰Оѕ , being left-invariant, is invariant under the п¬‚ow
of Yx .) Now, the value of the function П‰Оѕ (Yx ) at a в€€ G is

П‰Оѕ (Yx )(a) = Оѕ П‰(Yx (a)) = Оѕ Ad(aв€’1 )(x) = Adв€— (a)(Оѕ)(x) = (a В· Оѕ)(x).

Thus, it follows that the natural left action of G on GВ·Оѕ is Poisson, with momentum
mapping Вµ : GВ·Оѕ в†’ gв€— given by
Вµ(a В· Оѕ) = в€’ a В· Оѕ.
(Note: some authors do not have a minus sign here, but that is because their в„¦Оѕ 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 Вµ: M в†’ gв€— is the momentum mapping, then the G-equivariance
of Вµ implies that there is a well-deп¬Ѓned set map

Вµ: G\M в†’ G\gв€— .
ВЇ

L.7.6 121
The theorem we are about to prove asserts that, provided certain regularity criteria are
met, the subsets MОѕ = Вµв€’1 (Оѕ) вЉ‚ G\M are symplectic manifolds in a natural way.
ВЇ
ВЇ

Deп¬Ѓnition 4: Let f: X в†’ Y be a smooth map. A point y в€€ Y is a clean value of f if
the set f в€’1 (y) вЉ‚ X is a smooth submanifold of X and, moreover, if Tx f в€’1 (y) = ker f (x)
for each x в€€ f в€’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 О»: G Г— M в†’ M be a Poisson action on the symplectic manifold M.
Let Вµ: M в†’ gв€— be a momentum mapping for О». Suppose that, Оѕ в€€ gв€— is a clean value of Вµ.
Then GОѕ acts smoothly on Вµв€’1 (Оѕ). Suppose further that the space of GОѕ -orbits in Вµв€’1 (Оѕ),
say, MОѕ = GОѕ \ Вµв€’1 (Оѕ) , can be given the structure of a smooth manifold in such a way
that the quotient mapping ПЂОѕ : Вµв€’1 (Оѕ) в†’ MОѕ is a smooth submersion. Then there exists a
в€—
symplectic structure в„¦Оѕ on MОѕ that is deп¬Ѓned by the condition that ПЂОѕ (в„¦Оѕ ) be the pullback
of в„¦ to Вµв€’1 (Оѕ).

Proof: Since Оѕ is a clean value of Вµ, we know that Вµв€’1 (Оѕ) is a smooth submanifold of M.
By the G-equivariance of the momentum mapping, the stabilizer subgroup GОѕ вЉ‚ G acts
on M preserving the submanifold Вµв€’1 (Оѕ). The restricted action of GОѕ on Вµв€’1 (Оѕ) is easily
seen to be smooth.
Now, I claim that, for each m в€€ Вµв€’1 (Оѕ), the в„¦-complementary subspace to Tm Вµв€’1 (Оѕ)
is the space Tm G В· m , i.e., the tangent to the G-orbit through m. To see this, п¬Ѓrst
note that the space Tm G В· m is spanned by the values at m assumed by the vector
п¬Ѓelds О»в€— (x) for x в€€ g. Thus, a vector v в€€ Tm M lies in the в„¦-complementary space
of Tm G В· m if and only if v satisп¬Ѓes в„¦ О»в€— (x)(m), v = 0 for all x в€€ g. Since, by
deп¬Ѓnition, в„¦ О»в€— (x)(m), v = d ПЃ(x) (v), it follows that this condition on v is equivalent
to the condition that v lie in ker Вµ (m). However, since Оѕ is a clean value of Вµ, we have
ker Вµ (m) = Tm Вµв€’1 (Оѕ) , as claimed.
Now, the G-equivariance of Вµ implies that Вµв€’1 (Оѕ)в€© GВ·m = GОѕ В·m for all m в€€ Вµв€’1 (Оѕ).
In particular, Tm GОѕ В·m вЉ† Tm Вµв€’1 (Оѕ) в€©Tm GВ·m . To demonstrate the reverse inclusion,
suppose that v lies in both Tm Вµв€’1 (Оѕ) and Tm GВ·m . Then v = О»в€— (x)(m) for some x в€€ g,
and, by the G-equivariance of the momentum mapping and the assumption that Оѕ is clean
(so that Tm Вµв€’1 (Оѕ) = ker Вµ (m)) we have

0 = Вµ (m)(v) = Вµ (m) О»в€— (x)(m) = Adв€— в€— (x)(Вµ(m)) = Adв€— в€— (x)(Оѕ)
so that x must lie in gОѕ . Consequently, v = О»в€— (x)(m) is tangent to the orbit GОѕ В· m and
thus,
Tm Вµв€’1 (Оѕ) в€© Tm G В· m = ker Вµ (m) в€© Tm G В· m = Tm GОѕ В· m .
As a result, since the в„¦-complementary spaces Tm Вµв€’1 (Оѕ) and Tm G В· m intersect in the
tangents to the GОѕ -orbits, it follows that if в„¦Оѕ denotes the pullback of в„¦ to Вµв€’1 (Оѕ), then
Лњ
Лњ
the null space of в„¦Оѕ at m is precisely Tm GОѕ В· m .

L.7.7 122
Finally, let us assume, as in the theorem, that there is a smooth manifold structure
on the orbit space MОѕ = GОѕ \Вµв€’1 (Оѕ) so that the orbit space projection ПЂОѕ : Вµв€’1 (Оѕ) в†’ MОѕ is
Лњ
a smooth submersion. Since в„¦Оѕ is clearly GОѕ invariant and closed and moreover, since its
null space at each point of M is precisely the tangent space to the п¬Ѓbers of ПЂОѕ , it follows
that there exists a unique вЂњpush downвЂќ 2-form в„¦Оѕ on MОѕ as described in the statement of
the theorem. That в„¦Оѕ 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 gв€— .
The quotients MОѕ 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 Оѕ of Вµ for which Вµв€’1 (Оѕ) is not empty (even
when there are no such regular values). This follows because, if we look at the closed
subset DВµ вЉ‚ M consisting of points m where Вµ (m) does not reach its maximum rank,
then Вµ DВµ ) can be shown (by a sort of SardвЂ™s Theorem argument) to be a proper subset
of Вµ(M). Meanwhile, it is not hard to show that any element Оѕ в€€ Вµ(M) that does not lie
in Вµ DВµ ) is clean.
Second, it quite frequently does happen that the GОѕ -orbit space MОѕ has a manifold
structure for which ПЂОѕ is a submersion. This can be guaranteed by various hypotheses that
are often met with in practice. For example, if GОѕ is compact and acts freely on Вµв€’1 (Оѕ),
then MОѕ will be a manifold. (More generally, if the orbits GОѕ В· m are compact and all of the
stabilizer subgroups Gm вЉ‚ GОѕ are conjugate in GОѕ , then MОѕ will have a manifold structure
of the required kind.)
Weaker hypotheses also work. Basically, one needs to know that, at every point m
of Вµв€’1 (Оѕ), there is a smooth slice to the action of GОѕ , i.e., a smoothly embedded disk D
in Вµв€’1 (Оѕ) that passes through m and intersects each GОѕ -orbit in GОѕ В· 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 Вµв€’1 (Оѕ), there is very often a near-
slice, i.e., a smoothly embedded disk D in Вµв€’1 (Оѕ) that passes through m and intersects
each GОѕ -orbit in GОѕ В· D transversely and in a п¬Ѓnite number of points. In this case, the
quotient space MОѕ inherits the structure of a symplectic orbifold, and these вЂ˜generalized
manifoldsвЂ™ have turned out to be quite useful.
Finally, it is worth computing the dimension of MОѕ when it does turn out to be a
manifold. Let Gm вЉ‚ G be the stabilizer of m в€€ Вµв€’1 (Оѕ). I leave as an exercise for the
reader to check that
dim MОѕ = dim M в€’ dim G в€’ dim GОѕ + 2 dim Gm
= dim M в€’ 2 dim G/Gm + dim G/GОѕ .

L.7.8 123
Since we will see so many examples in the next Lecture, I will content myself with
only mentioning two here:
вЂў Let M = T в€— G and let G act on T в€— G on the left in the obvious way. Then the reader
в€—
can easily check that, for each x в€€ g, we have ПЃ(x)(О±) = О± Yx (a) for all О± в€€ Ta G where,
as usual, Yx denotes the right invariant vector п¬Ѓeld on G whose value at e is x в€€ g. Hence,
Вµ: T в€— G в†’ gв€— is given by Вµ(О±) = Rв€— (О±).
ПЂ(О±)
Consequently, Вµв€’1 (Оѕ) вЉ‚ T в€— G is merely the graph in T в€— G of the left-invariant 1-form
П‰Оѕ (i.e., the left-invariant 1-form whose value at e is Оѕ в€€ gв€— ). Thus, we can use П‰Оѕ as a
section of T в€— G to pull back в„¦ (the canonical symplectic form on T в€— G, which is clearly
G-invariant) to get the 2-form dП‰Оѕ on G. As we already saw in Lecture 5, and is now
borne out by Theorem 2, the null space of dП‰Оѕ at any point a в€€ G is Ta aGОѕ , the quotient
by GОѕ is merely the coadjoint orbit G/GОѕ , and the symplectic structure в„¦Оѕ is just the one
we already constructed.
Note, by the way, that every value of Вµ is clean in this example (in fact, they are all
regular), even though the dimensions of the quotients G/GОѕ vary with Оѕ.
вЂў Let G = SO(3) act on R6 = T в€— R3 by the extension of rotation about the origin in
R3 . Then, in standard coordinates (x, y) (where x, y в€€ R3 ), the action is simply g В· (x, y) =
(gx, gy), and the symplectic form is в„¦ = dx В· dy = tdxв€§dy.
We can identify so(3)в€— with so(3) itself by interpreting a в€€ so(3) as the linear func-
tional b в†’ в€’tr(ab). It is easy to see that the co-adjoint action in this case gets identiп¬Ѓed
with the adjoint action.
We compute that ПЃ(a)(x, y) = в€’txay, so it follows without too much diп¬ѓculty that,
with respect to our identiп¬Ѓcation of so(3)в€— with so(3), we have Вµ(x, y) = xty в€’ y tx.
The reader can check that all of the values of Вµ are clean except for 0 в€€ 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 в†’ R3 deп¬Ѓned by

P (x, y) = (x В· x, x В· y, y В· y)

maps the set Вµв€’1 (0) onto the вЂњconeвЂќ consisting of those points (a, b, c) в€€ R3 with a, c в‰Ґ 0
and b2 = ac and the п¬Ѓbers of P are the G0 -orbits of the points in Вµв€’1 (0).
For Оѕ = 0, the P -image of the set Вµв€’1 (Оѕ) is one nappe of the hyperboloid of two sheets
described as ac в€’ b2 = в€’tr(Оѕ 2 ). The reader should compute the area forms в„¦Оѕ 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
в„¦ = dxв€§dy. Show that the вЂњtranslation actionвЂќ (a, b) В· [x, y] = [x + a, y + b] of R2 on M is
symplectic but not Hamiltonian.

2. Let (M, в„¦) be a connected symplectic manifold and let О»: GГ—M в†’ M be a Hamiltonian
group action.
(i) Prove that, if ПЃ: g в†’ C в€ћ (M) is a linear mapping that satisп¬Ѓes О»в€— (x) в„¦ = d ПЃ(x) ,
then О»в€— [x, y] в„¦ = d {ПЃ(x), ПЃ(y)} .
(ii) Show that the associated linear mapping cПЃ: g Г— g в†’ R deп¬Ѓned in the text does indeed
satisfy cПЃ [x, y], z + cПЃ [y, z], x + cПЃ [z, x], y = 0 for all x, y, z в€€ g. (Hint: Use the
fact the Poisson bracket satisп¬Ѓes 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 в€‚: О›2 (g) в†’ g.
The dual of this linear map is a map в€’Оґ: gв€— в†’ О›2 (gв€— ). (Thus, for Оѕ в€€ gв€— , we have
ОґОѕ(x, y) = в€’Оѕ [x, y] .) This map Оґ can be extended uniquely to a graded, degree-one
derivation Оґ: О›в€— (gв€— ) в†’ О›в€— (gв€— ).
(i) For any c в€€ О›2 (gв€— ), show that Оґc(x, y, z) = в€’c [x, y], z в€’ c [y, z], x в€’ c [z, x], y .
(Hint: Every c в€€ О›2 (gв€— ) is a sum of wedge products Оѕв€§О· where Оѕ, О· в€€ gв€— .) Conclude
that the Jacobi identity in g is equivalent to the condition that Оґ 2 = 0 on all of О›в€— (gв€— ).
Thus, for any Lie algebra g, we can deп¬Ѓne the kвЂ™th cohomology group of g, denoted H k (g),
as the kernel of Оґ in О›k (gв€— ) modulo the subspace Оґ О›kв€’1 (gв€— ) .
(ii) Let G be a Lie group whose Lie algebra is g. For each О¦ в€€ О›k (gв€— ), deп¬Ѓne П‰О¦ to be the
left-invariant k-form on G whose value at the identity is О¦. Show that dП‰О¦ = П‰ОґО¦ .
(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 suп¬ѓces to prove this
formula for О¦ of degree 1. Why?)
Thus, the cohomology groups H k (g) measure вЂњclosed-mod-exactвЂќ 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 Оє is non-
degenerate, then [g, g] = g. Show that this implies that H 1 (g)=0. Next show that for
О¦ в€€ О›2 (gв€— ), we can write О¦(x, y) = Оє(Lx, y) where L: g в†’ g is skew-symmetric. Then
show that if ОґО¦ = 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, в„¦) is a symplectic mani-
fold and suppose that there exists a transitive symplectic action О»: GГ— M в†’ M where G is
a group whose Lie algebra satisп¬Ѓes H 1 (g) = H 2 (g) = 0. Show that there is a G-equivariant
symplectic covering map ПЂ: M в†’ G/GОѕ for some Оѕ в€€ gв€— . 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 вЉ‚ G that is the stabilizer of a point m of M. Let П†: G в†’ M be
П†(g) = g В· m. Now consider the left-invariant 2-form П†в€— (в„¦) 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 Г— 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 вЉ‚ 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 П‰ on M so
that dП‰ = в„¦, then the formula ПЃ(x) = в€’П‰ О»в€— (x) yields a lifting ПЃ for which cПЃ = 0.

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