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the property that the coordinate functions f i are in involution, i.e., {f i , f j } = 0. Suppose
that, for some c в€€ Rn , the f-level set Mc = f в€’1 (c) is compact. Replacing f by f в€’ c, we
may assume that c = 0, which we do from now on.
Show that M0 вЉ‚ M is a closed Lagrangian submanifold of M.
Use Proposition 4 to show that there is an open neighborhood B of 0 в€€ Rn so that
f в€’1 (B), в„¦ is symplectomorphic to a neighborhood U of the zero section in T в€— M0 (en-
dowed with its standard symplectic structure) in such a way that, for each b в€€ B, the sub-
manifold Mb = f в€’1 (b) is identiп¬Ѓed with the graph of a closed 1-form П‰b on M0 . Show that
it is possible to choose b1 , . . . , bn in B so that the corresponding closed 1-forms П‰1 , . . . , П‰n
are linearly independent at every point of M0 .
Conclude that M0 is diп¬Ђeomorphic to a torus T = Rn /О› where О› вЉ‚ Rn is a lattice,
in such a way that the forms П‰i become identiп¬Ѓed with dОёi where Оёi are the corresponding
linear coordinates on Rn .
Now prove that for any b в€€ B, the 1-form П‰b must be a linear combination of the П‰i
with constant coeп¬ѓcients. Thus, there are functions ai on B so that П‰b = ai (b)П‰i . (Hint:
Show that the coeп¬ѓcients must be invariant under the п¬‚ows of the vector п¬Ѓelds dual to
the П‰i .)
Conclude that, under the symplectic map identifying MB with U, the form в„¦ gets
identiп¬Ѓed with dai в€§dОёi . The functions ai and Оёi are the so-called вЂњaction-angle coordi-
natesвЂќ.
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 вЂњп¬ЃniteвЂќ
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 deп¬Ѓnition.
Deп¬Ѓnition 1: Let (M, в„¦) be a symplectic manifold and let G be a Lie group. A left
action О»: G Г— M в†’ M of G on M is a symplectic action if О»в€— (в„¦) = в„¦ for all a в€€ 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 в†’ R, then, by construction, it preserves
the symplectic 2-form dП‰L .
Example: Cotangent Actions. A left G-action О»: G Г— M в†’ M, induces an action
О» of G on T в€— M. Namely, for each a в€€ G, the diп¬Ђeomorphism О»a : M в†’ M induces a
Лњ
diп¬Ђeomorphism О»a : T в€— M в†’ T в€— M. Since the natural symplectic structure on T в€— M is
Лњ
Лњ
invariant under diп¬Ђeomorphisms, it follows that О» is a symplectic action.
Example: Coadjoint Orbits. As we saw in Lecture 5, for every Оѕ в€€ gв€— , the coadjoint
orbit G В· Оѕ carries a natural G-invariant symplectic structure в„¦Оѕ . Thus, the left action of
G on G В· Оѕ 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 в€§ dВЇ1 + В· В· В· + dz n в€§ dВЇn
i
в„¦= z z
2
= dx1 в€§ dy 1 + В· В· В· + dxn в€§ dy n

where z k = xk + iy k . Then for any integers (k1 , . . . , kn ), we can deп¬Ѓne an action of S 1 on
Cn by the formula пЈ« 1 пЈ¶ пЈ« ik1 Оё 1 пЈ¶
z e z
.пЈё пЈ­ .
eiОё В· пЈ­ . пЈё
.
=
. .
zn eikn Оё z n
The reader can easily check that this deп¬Ѓnes a symplectic circle action on Cn .

Generally what we will be interested in is the following: Y will be a Hamiltonian
vector п¬Ѓeld on a symplectic manifold (M, в„¦) and G will act symplectically on M as a
group of symmetries of the п¬‚ow of Y . We want to understand how to use the action of G
to вЂњreduceвЂќ the problem of integrating the п¬‚ow of Y .

L.7.1 116
In Lecture 3, we saw that when Y was the Euler-Lagrange vector п¬Ѓeld associated to a
non-degenerate Lagrangian L, then the inп¬Ѓnitesimal generators of symmetries of L could
be used to generate conserved quantities for the п¬‚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 О» on a connected symplectic manifold (M, в„¦).
Since О» is symplectic, it follows that the mapping О»в€— : g в†’ X(M) actually has image
in sp(в„¦), the algebra of symplectic vector п¬Ѓelds on M. As we saw in Lecture 3, О»в€— is
an anti-homomorphism, i.e., О»в€— [x, y] = в€’ О»в€— (x), О»в€— (y) . Since, as we saw in Lecture
5, [sp(в„¦), sp(в„¦)] вЉ‚ h(в„¦), it follows that О»в€— [g, g] вЉ‚ h(в„¦). Thus, HО» : g в†’ HdR (M, R)
1

deп¬Ѓned by HО» (x) = О»в€— (x) в„¦ is a homomorphism of Lie algebras with kernel containing
the commutator subalgebra [g, g].
The map HО» is the obstruction to п¬Ѓnding a Hamiltonian function associated to each
inп¬Ѓnitesimal symmetry О»в€— (x) since HО» (x) = 0 if and only if О»в€— (x) в„¦ = в€’df for some
f в€€ C в€ћ(M).
Deп¬Ѓnition 2: A symplectic action О»: G Г— M в†’ M is said to be Hamiltonian if HО» = 0,
i.e., if О»в€— (g) вЉ‚ h(в„¦).
There are a few particularly interesting cases where the obstruction HО» must vanish:
вЂў If HdR (M, R) = 0. In particular, if M is simply connected.
1

вЂў 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 п¬Ѓrst 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 в‰Ґ 3, then
g has this property, even though its Killing form is degenerate.
вЂў If there exists a 1-form П‰ on M that is invariant under G and satisп¬Ѓes в„¦ = dП‰.
(This is the case of symmetries of a Lagrangian.) To see this, note that if X is a vector
п¬Ѓeld on M that preserves П‰, then

0 = LX П‰ = d(X П‰) + X в„¦,

so X в„¦ is exact.

For a Hamiltonian action О», every inп¬Ѓnitesimal symmetry О»в€— (x) has a Hamiltonian
function fx в€€ C в€ћ. However, the choice of fx is not unique since we can add any constant
to fx without changing its Hamiltonian vector п¬Ѓeld. This non-uniqueness causes some
problems in the theory we wish to develop.
To see why, suppose that we choose a (linear) lifting ПЃ: g в†’ C в€ћ(M) of в€’О»в€— : g в†’ h(в„¦).
(The choice of в€’О»в€— instead of О»в€— was made to get rid of the annoying sign in the formula
for the bracket.)
g
в€’О»в€—
в†“
ПЃ

C в€ћ(M)
в†’R в†’ в†’ h(в„¦) в†’
0 0

L.7.2 117
Thus, for every x в€€ g, we have О»в€— (x) в„¦ = d ПЃ(x) . A short calculation (see the Exercises)
now shows that {ПЃ(x), ПЃ(y)} is a Hamiltonian function for в€’О»в€— ([x, y]), i.e., that

в„¦ = d {ПЃ(x), ПЃ(y)} .
О»в€— [x, y]

In particular, it follows (since M is connected) that there must be a skew-symmetric
bilinear map cПЃ: g Г— g в†’ R so that

{ПЃ(x), ПЃ(y)} = ПЃ [x, y] + cПЃ (x, y).

An application of the Jacobi identity implies that the map cПЃ satisп¬Ѓes the condition

cПЃ [x, y], z + cПЃ [y, z], x + cПЃ [z, x], y = 0 for all x, y, z в€€ g.

This condition is known as the 2-cocycle condition for cПЃ regarded as an element of A2 (g) =
О›2 (gв€— ). (See Exercise 3 for an explanation of this terminology.)
For purposes of simplicity, it would be nice if we could choose ПЃ so that cПЃ were
identically zero. In order to see whether this is possible, let us choose another linear map
ПЃ: g в†’ C в€ћ(M) that satisп¬Ѓes ПЃ(x) = ПЃ(x) + Оѕ(x) where Оѕ: g в†’ R is any linear map. Every
Лњ Лњ
possible lifting of в€’О»в€— is clearly of this form for some Оѕ. Now we compute that

ПЃ(x), ПЃ(y) = ПЃ(x), ПЃ(y) = ПЃ [x, y] + cПЃ(x, y)
Лњ Лњ
= ПЃ [x, y] + cПЃ (x, y) в€’ Оѕ [x, y] .
Лњ

Thus, cПЃ(x, y) = cПЃ(x, y) в€’ Оѕ [x, y] . Thus, in order to be able to choose ПЃ so that cПЃ = 0,
Лњ
Лњ Лњ
в€—
we see that there must exist a Оѕ в€€ g so that cПЃ = в€’ОґОѕ where ОґОѕ is the skew-symmetric
bilinear map on g that satisп¬Ѓes ОґОѕ(x, y) = в€’Оѕ [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 cПЃ can be written in the
form в€’ОґОѕ. Among them are:
вЂў If M is compact, then the sequence

0 в†’ R в†’ C в€ћ (M) в†’ H(в„¦) в†’ 0

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

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 в€ћ(M) = R вЉ• C0 (M, в„¦). Now
в€ћ
в€ћ
just choose the unique ПЃ so that it takes values in C0 (M, в„¦). This will clearly have cПЃ = 0.
вЂў 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
вЂў If there is a 1-form П‰ on M that is preserved by the G action and satisп¬Ѓes dП‰ = в„¦.
(This is true in the case of symmetries of a Lagrangian.) In this case, we can merely take
ПЃ(x) = в€’П‰ О»в€— (x) . I leave as an exercise for the reader to check that this works.
Deп¬Ѓnition 3: A Hamiltonian action О»: G Г— M в†’ M is said to be a Poisson action if
there exists a lifting ПЃ with cПЃ = 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 О»: G Г— M в†’ M speciп¬Ѓed,
we have chosen a lifting ПЃ: g в†’ C в€ћ (M) of в€’О»в€— that satisп¬Ѓes ПЃ(x), ПЃ(y) = ПЃ [x, y] for
all x, y в€€ g. Note that such a ПЃ is unique up to replacement by ПЃ = ПЃ + Оѕ where Оѕ : g в†’ R
Лњ
satisп¬Ѓes ОґОѕ = 0. Such Оѕ (if any non-zero ones exist) are п¬Ѓxed 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 О» and ПЃ is the mapping Вµ: M в†’ gв€—
Deп¬Ѓnition 4:
that satisп¬Ѓes
Вµ(m)(y) = ПЃ(y)(m).

Note that, for п¬Ѓxed m в€€ M, the assignment y в†’ ПЃ(y)(m) is a linear map from g to
R, so the deп¬Ѓnition 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 в€€ g corresponds to a translation, then ПЃ(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 п¬Ѓxed
axis, then ПЃ(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 п¬Ѓeld on M that is invariant under the action
of G, then Вµ is constant on the integral curves of Y .

In particular, Вµ 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 Вµ: M в†’ gв€— is G-equivariant.

L.7.4 119
Proof: Recall that the coadjoint action of G on gв€— is deп¬Ѓned by Adв€— (g)(Оѕ)(x) =
Оѕ Ad(g в€’1 ) x . The condition that Вµ be G-equivariant, i.e., that Вµ(g В· m) = Adв€— (g) Вµ(m)
for all m в€€ M and g в€€ G, is thus seen to be equivalent to the condition that

ПЃ Ad(g в€’1 )y (m) = ПЃ(y)(g В· m)

for all m в€€ M, g в€€ G, and y в€€ 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 в€€ g and any
t в€€ R, the formula for general g will follow. Thus, we want to prove that

ПЃ Ad(eв€’tx )y (m) = ПЃ(y)(etx В· 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 ПЃ(y) О»в€— (x)(etx В· m) = в„¦ О»в€— (y)(etx В· m), О»в€— (x)(etx В· m)
= в„¦ О»в€— Ad(eв€’tx )y (m), О»в€— Ad(eв€’tx )x (m)
= в„¦ О»в€— Ad(eв€’tx )y (m), О»в€— (x)(m)

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

О»a О»в€— (y)(m) = О»в€— Ad(a)y (a В· m)

and the fact that в„¦ is G-invariant.
On the other hand, the derivative of the left hand side of the formula is clearly
ПЃ [в€’x, Ad(eв€’tx )y] (m) = в€’ ПЃ(x), ПЃ Ad(eв€’tx )y (m)
= в„¦ О»в€— Ad(eв€’tx )y (m), О»в€— (x)(m)

so we are done. (Note that I have used my assumption that cПЃ = 0!)
Example: Left-Invariant Metrics on Lie Groups. Let G be a Lie group and let
Q: g в†’ R be a non-degenerate quadratic form with associated inner product , Q . Let
L: T G в†’ R be the Lagrangian
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