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More generally, we can take v ∈ g to ’Xv using the identi¬cation g = g—— ⊆
C ∞ (g— ). However, this homomorphism might be trivial. In fact, it seldom provides
the pointwise faithful representation needed to prove the Poincar´-Birkho¬-Witt e
theorem. Instead, we use the following trick.
For a regular point in g— , Theorem 4.1 states that there is a neighborhood U with
‚ ‚
‚qi § ‚pi (cf.
canonical coordinates q1 , . . . , qk , p1 , . . . , pk , c1 , . . . , c such that Π =
Example 2 of Section 3.4). In terms of Π, we have


Π(dqi ) =
‚pi

Π(dpi ) =
‚qi

Π(dci ) = 0.
This implies that the hamiltonian vector ¬eld of any function will be a linear combi-
‚ ‚
nation of the vector ¬elds ‚qi , ‚pi . Unless the structure de¬ned by Π on the regular
part of g— is symplectic (that is l = 0), the representation of g as di¬erential oper-
ators on C ∞ (g— ) will have a kernel, and hence will not be faithful.
To remedy this, we lift the Lie-Poisson structure to a symplectic structure on
a larger manifold. Let U — R have the original coordinates q1 , . . . , qk , p1 , . . . , pk ,
c1 , . . . , c lifted from the coordinates on U , plus the coordinates d1 , . . . , d lifted from
the standard coordinates of R . We de¬ne a Poisson structure {·, ·} on U — R by
‚ ‚ ‚ ‚
§ §
Π= + .
‚qi ‚pi ‚ci ‚di
i i

We now take the original coordinate functions µi on U and lift them to functions,
still denoted µi , on U — R . Because the µi ™s are independent of the dj ™s, we
cijk µk . Thus the homomorphism g ’ C ∞ (U ),
see that {µi , µj } = {µi , µj } =
vi ’ µi , lifts to a map

E C ∞ (U — R ) ’Π —¦ d E χ(U — R )
g

E ’Π (dµi ) = ’Xµ .
vi E µi
i

The composed map is a Lie algebra homomorphism. The di¬erentials dµ1 , . . . , dµn
are pointwise linearly independent on U and thus also on U — R . Since ’Π is
an isomorphism, the hamiltonian vector ¬elds ’Xµ1 , . . . , ’Xµk are also pointwise
linearly independent, and we have the pointwise faithful representation needed to
complete the proof of the Poincar´-Birkho¬-Witt theorem.
e
Remarks.
1. Section 2.4 explains how to go from a pointwise faithful representation to a
local Lie group. In practice, it is not easy to ¬nd the canonical coordinates
in U , nor is it easy to integrate the Xµi ™s.
2. The integer is called the rank of the Lie algebra, and it equals the dimen-
sion of a Cartan subalgebra when g is semisimple. This rank should not be
confused with the rank of the Poisson structure.
™¦
4.3 The Splitting Theorem 19


4.3 The Splitting Theorem
We will prove Theorem 4.1 as a consequence of the following more general result.

Theorem 4.2 (Weinstein [163]) On a Poisson manifold (M, Π), any point O ∈
M has a coordinate neighborhood with coordinates (q1 , . . . , qk , p1 , . . . , pk , y1 , . . . , y )
centered at O, such that
‚ ‚ 1 ‚ ‚
§ §
Π= + •ij (y) and •ij (0) = 0 .
‚qi ‚pi 2 ‚yi ‚yj
i i,j

The rank of Π at O is 2k. Since • depends only on the yi ™s, this theorem gives
a decomposition of the neighborhood of O as a product of two Poisson manifolds:
one with rank 2k, and the other with rank 0 at O.
Proof. We prove the theorem by induction on ρ = rank Π(O).

• If ρ = 0, we are done, as we can label all the coordinates yi .
• If ρ = 0, then there are functions f, g with {f, g}(O) = 0. Let p1 = g and
look at the operator Xp1 . We have Xp1 (f )(O) = {f, g}(O) = 0. By the ¬‚ow
box theorem, there are coordinates for which Xp1 is one of the coordinate

vector ¬elds. Let q1 be the coordinate function such that Xp1 = ‚q1 ; hence,
{q1 , p1 } = Xp1 q1 = 1. (In practice, ¬nding q1 amounts to solving a system
of ordinary di¬erential equations.) Xp1 , Xq1 are linearly independent at O
and hence in a neighborhood of O. By the Frobenius theorem, the equation
[Xq1 , Xp1 ] = ’X{q1 ,p1 } = ’X1 = 0 shows that these vector ¬elds can be
integrated to de¬ne a two dimensional foliation near O. Hence, we can ¬nd
functions y1 , . . . , yn’2 such that
1. dy1 , . . . , dyn’2 are linearly independent;
2. Xp1 (yj ) = Xq1 (yj ) = 0. That is to say, y1 , . . . , yn’2 are transverse to
the foliation. In particular, {yj , q1 } = 0 and {yj , p1 } = 0.


Exercise 10
Show that dp1 , dq1 , dy1 , . . . , dyn’2 are all linearly independent.


‚ ‚
Therefore, we have coordinates such that Xq1 = ’ ‚p1 , Xp1 = ‚q1 , and by
Poisson™s theorem
{{yi , yj }, p1 } = 0
{{yi , yj }, q1 } = 0
We conclude that {yi , yj } must be a function of the yi ™s. Thus, in these
coordinates, the Poisson structure is
‚ ‚ 1 ‚ ‚
§ §
Π= + •ij (y) .
‚q1 ‚p1 2 ‚yi ‚yj
i,j


• If ρ = 2, we are done. Otherwise, we apply the argument above to the
structure 1 ‚ ‚
•ij (y) ‚yi § ‚yj .
2

2
20 4 NORMAL FORMS


4.4 Special Cases of the Splitting Theorem
1. If the rank is locally constant, then •ij ≡ 0 and the splitting theorem recovers
Lie™s theorem (Theorem 4.1). Hence, by the argument in Section 4.2, our
proof of the Poincar´-Birkho¬-Witt theorem is completed.
e
2. At the origin of a Lie-Poisson manifold, we only have yi ™s, and the term
‚ ‚
‚qi § ‚pi does not appear.

3. A symplectic manifold is a Poisson manifold (M, Π) where rank Π = dim M
everywhere. In this case, Lie™s theorem (or the splitting theorem) gives canon-
ical coordinates q1 , . . . , qk , p1 , . . . , pk such that
‚ ‚
§
Π= .
‚qi ‚pi
i

In other words, Π : T — M ’ T M is an isomorphism satisfying
‚ ‚
Π(dqi ) = ’ and Π(dpi ) = .
‚pi ‚qi
Its inverse ω = Π’1 : T M ’ T — M de¬nes a 2-form ω ∈ „¦2 (M ) by ω(u, v) =
ω(u)(v), or equivalently by ω = (Π’1 )— (Π). With respect to the canonical
coordinates, we have
dqi § dpi ,
ω=
which is the content of Darboux™s theorem for symplectic manifolds. This
also gives a quick proof that ω is a closed 2-form. ω is called a symplectic
form.

4.5 Almost Symplectic Structures
Suppose that (M, Π) is an almost symplectic manifold, that is, Π is non-
degenerate but may not satisfy the Jacobi identity. Then Π : T — M ’ T M is an
isomorphism, and its inverse ω = Π’1 : T M ’ T — M de¬nes a 2-form ω ∈ „¦2 (M )
by ω(u, v) = ω(u)(v).
Conversely, any 2-form ω ∈ „¦2 (M ) de¬nes a map
ω : T M ’ T —M by ω(u)(v) = ω(u, v) .
We also use the notation ω(v) = iv (ω) = v ω. Suppose that ω is non-degenerate,
meaning that ω is invertible. Then for any function h ∈ C ∞ (M ), we de¬ne the
hamiltonian vector ¬eld Xh by one of the following equivalent formulations:
• Xh = ω ’1 (dh) ,
• Xh ω = dh , or
• ω(Xh , Y ) = Y · h .
There are also several equivalent de¬nitions for a bracket operation on C ∞ (M ),
including
{f, g} = ω(Xf , Xg ) = Xg (f ) = ’Xf (g) .
It is easy to check the anti-symmetry property and the Leibniz identity for the
bracket. The next section discusses di¬erent tests for the Jacobi identity.
4.6 Incarnations of the Jacobi Identity 21


4.6 Incarnations of the Jacobi Identity

Theorem 4.3 The bracket {·, ·} on an almost symplectic manifold (de¬ned in the
previous section) satis¬es the Jacobi identity if and only if dω = 0.


Exercise 11
Prove this theorem. Hints:
1
• With coordinates, write ω locally as ω = ωij dxi § dxj . The condi-
2
tion for ω to be closed is then
‚ωjk
‚ωij ‚ωki
+ + =0.
‚xk ‚xi ‚xj

Since (ωij )’1 = (’πij ), this equation is equivalent to
‚πj
‚πij ‚π i
πk + πki + πkj =0.
‚xk ‚xk ‚xk
k

Cf. Exercise 8 in Section 3.4.
• Without coordinates, write dω in terms of Lie derivatives and Lie brack-
ets as
LX (ω(Y, Z)) + LY (ω(Z, X)) + LZ (ω(X, Y ))
dω(X, Y, Z) =
’ω([X, Y ], Z) ’ ω([Y, Z], X) ’ ω([Z, X], Y ) .
At each point, choose functions f, g, h whose hamiltonian vector ¬elds at
that point coincide with X, Y, Z. Apply LXf (ω(Xg , Xh )) = {{g, h}, f }
and ’ω([Xf , Xg ], Xh ) = {{f, g}, h}.



Remark. For many geometric structures, an integrability condition allows us
to drop the “almost” from the description of the structure, and ¬nd a standard
expression in canonical coordinates. For example, an almost complex structure is
complex if it is integrable, in which case we can ¬nd complex coordinates where the
almost complex structure becomes multiplication by the complex number i. Simi-
larly, an almost Poisson structure Π is integrable if Π satis¬es the Jacobi identity,
in which case Lie™s theorem provides a normal form near points where the rank
is locally constant. Finally, an almost symplectic structure ω is symplectic if ω is
closed, in which case there exist coordinates where ω has the standard Darboux
™¦
normal form.
We can reformulate the connection between the Jacobi identity and dω = 0 in
terms of Lie derivatives. Cartan™s magic formula states that, for a vector ¬eld X
and a di¬erential form ·,

LX · = d(X ·) + X d· .

Using this, we compute

LXh ω = d(Xh ω) + Xh dω
= d(dh) + Xh dω
= Xh dω .

We conclude that dω = 0 if and only if LXh ω = 0 for each h ∈ C ∞ (M ). (One
implication requires the fact that hamiltonian vector ¬elds span the whole tangent
bundle, by invertibility of ω.) It follows that another characterization for ω being
22 4 NORMAL FORMS


closed is ω being invariant under all hamiltonian ¬‚ows. This is equivalent to saying
that hamiltonian ¬‚ows preserve Poisson brackets, i.e. LXh Π = 0 for all h. Ensuring
that the symplectic structure be invariant under hamiltonian ¬‚ows is one of the
main reasons for requiring that a symplectic form be closed.
While the Leibniz identity states that all hamiltonian vector ¬elds are deriva-
tions of pointwise multiplication of functions, the Jacobi identity states that all
hamiltonian vector ¬elds are derivations of the bracket {·, ·}. We will now check
directly the relation between the Jacobi identity and the invariance of Π under
hamiltonian ¬‚ows, in the language of hamiltonian vector ¬elds. Recall that the
operation of Lie derivative is a derivation on contraction of tensors, and therefore

{{f, g}, h} = Xh {f, g} = Xh (Π(df, dg))
(LXh Π)(df, dg) + Π(LXh df, dg) + Π(df, LXh dg)
=
= (LXh Π)(df, dg) + Π(dLXh f, dg) + Π(df, dLXh g)
= (LXh Π)(df, dg) + Π(d{f, h}, dg) + Π(df, d{g, h})
(LXh Π)(df, dg) + {Xh f, g} + {f, Xh g}
=
(LXh Π)(df, dg) + {{f, h}, g} + {f, {g, h}} .
=

We conclude that the Jacobi identity holds if and only if (LXh Π)(df, dg) = 0 for all
f, g, h ∈ C ∞ (M ).
5 Local Poisson Geometry
Roughly speaking, any Poisson manifold is obtained by gluing together symplectic
manifolds. The study of Poisson structures involves both local and global concerns:
the local structure of symplectic leaves and their transverse structures, and the
global aspects of how symplectic leaves ¬t together into a foliation.

5.1 Symplectic Foliation

At a regular point p of a Poisson manifold M , the subspace of Tp M spanned by the
hamiltonian vector ¬elds of the canonical coordinates at that point depends only
on the Poisson structure. When the Poisson structure is regular (see Section 4.1),
the image of Π (formed by the subspaces above) is an involutive subbundle of T M .
Hence, there is a natural foliation of M by symplectic manifolds whose dimension
is the rank of Π. These are called the symplectic leaves, forming the symplectic
foliation.
It is a remarkable fact that symplectic leaves exist through every point, even
on Poisson manifolds (M, {·, ·}) where the Poisson structure is not regular. (Their
existence was ¬rst proved in this context by Kirillov [95].) In general, the symplectic
foliation is a singular foliation.
The symplectic leaves are determined locally by the splitting theorem (Sec-
tion 4.3). For any point O of the Poisson manifold, if (q, p, y) are the normal
coordinates as in Theorem 4.2, then the symplectic leaf through O is given locally
by the equation y = 0.
The Poisson brackets on M can be calculated by restricting to the symplectic
leaves and then assembling the results.
0
Remark. The 0-th Poisson cohomology, HΠ , (see Section 3.6) can be interpreted
as the set of smooth functions on the space of symplectic leaves. It may be useful
1
™¦
to think of HΠ as the “vector ¬elds on the space of symplectic leaves” [72].

Examples.

1. For the zero Poisson structure on M , HΠ (M ) = C∞ (M ) and HΠ (M ) consists
0 1

of all the vector ¬elds on M .

2. For a symplectic structure, the ¬rst Poisson cohomology coincides with the
¬rst de Rham cohomology via the isomorphisms
ω
Poisson vector ¬elds ’’ closed 1-forms
ω
hamiltonian vector ¬elds ’’ exact 1-forms

1 1
’’
HΠ (M ) HdeRham (M ) .

In the symplectic case, the 0-th Poisson cohomology is the set of locally con-
0
stant functions, HdeRham (M ). This agrees with the geometric interpretation
of Poisson cohomology in terms of the space of symplectic leaves.
1 1
On the other hand, on a symplectic manifold, HΠ HdeRham gives a ¬-
nite dimensional space of “vector ¬elds” over the discrete space of connected
components


23
24 5 LOCAL POISSON GEOMETRY


™¦

Problem. Is there an interesting and natural way to give a “structure” to the
point of the leaf space representing a connected component M of a symplectic
manifold in such a way that the in¬nitesimal automorphisms of this “structure”
1
™¦
correspond to elements of HdeRham (M )?


5.2 Transverse Structure

As we saw in the previous section, on a Poisson manifold (M, Π) there is a natural
singular foliation by symplectic leaves. For each point m ∈ M , we can regard M as
¬bering locally over the symplectic leaf through m. Locally, this leaf has canonical
coordinates q1 , . . . , qk , p1 , . . . , pk , where the bracket is given by canonical symplectic
relations. While the symplectic leaf is well-de¬ned, each choice of coordinates
y1 , . . . , y in Theorem 4.2 can give rise to a di¬erent last term for Π,

1 ‚ ‚
§
•ij (y) ,
2 ‚yi ‚yj
i,j


called the transverse Poisson structure (of dimension ). Although the trans-
verse structures themselves are not uniquely de¬ned, they are all isomorphic [163].
Going from this local isomorphism of the transverse structures to a structure of
“Poisson ¬ber bundle” on a neighborhood of a symplectic leaf seems to be a di¬cult
problem [90].
Example. Suppose that Π is regular. Then the transverse Poisson structure
is trivial and the ¬bration over the leaf is locally trivial. However, the bundle
structure can still have holonomy as the leaves passing through a transverse section
™¦
wind around one another.
Locally, the transverse structure is determined by the structure functions πij (y) =
{yi , yj } which vanish at y = 0. Applying a Taylor expansion centered at the origin,
we can write
cijk yk + O(y 2 )
πij (y) =
k

where O(y 2 ) can be expressed as dijkl (y)yk y , though the dijkl are not unique
outside of y = 0.
Since the πij satisfy the Jacobi identity, it is easy to show using the Taylor
expansion of the jacobiator that the truncation

πij (y) = cijk yk
k

also satis¬es the Jacobi identity. Thus, the functions πij de¬ne a Poisson structure,
called the linearized Poisson structure of πij .
From Section 3.4 we know that a linear Poisson structure can be identi¬ed
with a Poisson structure on the dual of a Lie algebra. In this way, for any point
m ∈ M , there is an associated Lie algebra, called the transverse Lie algebra. We
will now show that this transverse Lie algebra can be identi¬ed intrinsically with
the conormal space to the symplectic leaf Om through m, so that the linearized
5.3 The Linearization Problem 25


transverse Poisson structure lives naturally on the normal space to the leaf. When
the Poisson structure vanishes at the point m, this normal space is just the tangent
space Tm M .
Recall that the normal space to Om is the quotient

N Om = Tm M Tm Om .

The conormal space is the dual space (N Om )— . This dual of this quotient space
of Tm M can be identi¬ed with the subspace (Tm Om )—¦ of cotangent vectors at m
which annihilate Tm Om :

(N Om )— (Tm Om )—¦ ⊆ Tm M .



To de¬ne the bracket on the conormal space, take two elements ±, β ∈ (Tm Om )—¦ .
We can choose functions f, g ∈ C ∞ (M ) such that df (m) = ±, dg(m) = β. In order
to simplify computations, we can even choose such f, g which are zero along the
symplectic leaf, that is, f, g|Om ≡ 0. The bracket of ±, β is

[±, β] = d{f, g}(m) .

This is well-de¬ned because

• f, g|Om ≡ 0 ’ {f, g}|Om ≡ 0 ’ d{f, g}|Om ∈ (Tm Om )—¦ . That the set of func-
tions vanishing on the symplectic leaf is closed under the bracket operation
follows, for instance, from the splitting theorem.

• The Leibniz identity implies that the bracket {·, ·} only depends on ¬rst
derivatives. Hence, the value of [±, β] is independent of the choice of f and g.

There is then a Lie algebra structure on (Tm Om )—¦ and a bundle of duals of Lie
algebras over a symplectic leaf. The next natural question is: does this linearized
structure determine the Poisson structure on a neighborhood?

5.3 The Linearization Problem

Suppose that we have structure functions

cijk yk + O(y 2 ) .
πij (y) =
k

Is there a change of coordinates making the πij linear? More speci¬cally, given πij ,
is there a new coordinate system of the form

zi = yi + O(y 2 )

such that {zi , zj } = cijk zk ?
This question resembles Morse theory where, given a function whose Taylor
expansion only has quadratic terms or higher, we ask whether there exist some
coordinates for which the higher terms vanish. The answer is yes (without further
assumptions on the function) if and only if the quadratic part is non-degenerate.
When the answer to the linearization problem is a¬rmative, we call the structure
πij linearizable. Given ¬xed cijk , if πij is linearizable for all choices of O(y 2 ),
26 5 LOCAL POISSON GEOMETRY


then we say that the transverse Lie algebra g de¬ned by cijk is non-degenerate.
Otherwise, it is called degenerate.
There are several versions of non-degeneracy, depending on the kind of coordi-
nate change allowed: for example, formal, C ∞ or analytic. Here is a brief summary
of some results on the non-degeneracy of Lie algebras.

• It is not hard to see that the zero (or commutative) Lie algebra is degenerate
for dimensions ≥ 2. Two examples of non-linearizable structures in dimension
2 demonstrating this degeneracy are
2 2
1. {y1 , y2 } = y1 + y2 ,
2. {y1 , y2 } = y1 y2 .

• Arnold [6] showed that the two-dimensional Lie algebra de¬ned by {x, y} = x
is non-degenerate in all three versions described above. If one decomposes
this Lie algebra into symplectic leaves, we see that two leaves are given by
the half-planes {(x, y)|x < 0} and {(x, y)|x > 0}. Each of the points (0, y)
comprises another symplectic leaf. See the following ¬gure.


y
T



E
x




• Weinstein [163] showed that, if g is semi-simple, then g is formally non-
degenerate. At the same time he showed that sl(2; R) is C ∞ degenerate.

• Conn [27] ¬rst showed that if g is semi-simple, then g is analytically non-
degenerate. Later [28], he proved that if g is semi-simple of compact type
(i.e. the corresponding Lie group is compact), then g is C ∞ non-degenerate.

• Weinstein [166] showed that if g is semi-simple of non-compact type and has
real rank of at least 2, then g is C ∞ degenerate.

• Cahen, Gutt and Rawnsley [22] studied the non-linearizability of some Poisson
Lie groups.

Remark. When a Lie algebra is degenerate, there is still the question of whether a
change of coordinates can remove higher order terms. Several students of Arnold [6]
looked at the 2-dimensional case (e.g.: {x, y} = (x2 + y 2 )p + . . .) to investigate
which Poisson structures could be reduced in a manner analogous to linearization.
Quadratization (i.e. equivalence to quadratic structures after a coordinate change)
has been established in some situations for structures with su¬ciently nice quadratic
™¦
part by Dufour [49] and Haraki [80].
5.4 The Cases of su(2) and sl(2; R) 27


We can view Poisson structures near points where they vanish as deformations
of their linearizations. If we expand a Poisson structure πij as

{xi , xj } = π1 (x) + π2 (x) + . . . ,

where πk (x) denotes a homogeneous polynomial of degree k in x, then we can de¬ne
a deformation by
{xi , xj }µ = π1 (x) + µπ2 (x) + . . . .
This indeed satis¬es the Jacobi identity for all µ, and {xi , xj }0 = π1 (x) is a linear
Poisson structure. All the {·, ·}µ ™s are isomorphic for µ = 0.

5.4 The Cases of su(2) and sl(2; R)

We shall compare the degeneracies of sl(2; R) and su(2), which are both 3-dimensional
as vector spaces. First, on su(2) with coordinate functions µ1 , µ2 , µ3 , the bracket
operation is de¬ned by
{µ1 , µ2 } = µ3
{µ2 , µ3 } = µ1
{µ3 , µ1 } = µ2 .
The Poisson structure is trivial only at the origin. It is easy to check that the
function µ2 + µ2 + µ2 is a Casimir function, meaning that it is constant along the
1 2 3
symplectic leaves. By rank considerations, we see that the symplectic leaves are
exactly the level sets of this function, i.e. spheres centered at the origin. This
foliation is quite stable. In fact, su(2), which is semi-simple of real rank 1, is C ∞
non-degenerate.
On the other hand, sl(2; R) with coordinate functions µ1 , µ2 , µ3 has bracket
operation de¬ned by
{µ1 , µ2 } = ’µ3
{µ2 , µ3 } = µ1
{µ3 , µ1 } = µ2 .
In this case, µ2 + µ2 ’ µ2 is a Casimir function, and the symplectic foliation consists
1 2 3
of

• the origin,

• two-sheeted hyperboloids µ2 + µ2 ’ µ2 = c < 0,
1 2 3

• the cone µ2 + µ2 ’ µ2 = 0 punctured at the origin, and
1 2 3

• one-sheeted hyperboloids µ2 + µ2 ’ µ2 = c > 0.
1 2 3

There are now non-simply-connected symplectic leaves. Restricting to the hori-
zontal plane µ3 = 0, the leaves form a set of concentric circles. It is possible to
modify the Poisson structure slightly near the origin, so that the tangent plane to
each symplectic leaf is tilted, and on the cross section µ3 = 0, the leaves spiral
toward the origin. This process of “breaking the leaves” [163] requires that there
be non-simply-connected leaves and that we employ a smooth perturbation whose
derivatives all vanish at the origin (in order not to contradict Conn™s results listed
in the previous section, since such a perturbation cannot be analytic).
Part III
Poisson Category
6 Poisson Maps
Any Poisson manifold has an associated Poisson algebra, namely the algebra of
its smooth functions equipped with the Poisson bracket. In this chapter, we will
strengthen the analogy between algebras and spaces.

6.1 Characterization of Poisson Maps

Given two Poisson algebras A, B, an algebra homomorphism ψ : A ’ B is called a
Poisson-algebra homomorphism if ψ preserves Poisson brackets:

ψ ({f, g}A ) = {ψ(f ), ψ(g)}B .

A smooth map • : M ’ N between Poisson manifolds M and N is called a
Poisson map when

•— ({f, g}N ) = {•— (f ), •— (g)}M ,

that is, •— : C ∞ (N ) ’ C ∞ (M ) is a Poisson-algebra homomorphism. (Every
homomorphism C ∞ (N ) ’ C ∞ (M ) of the commutative algebra structures arising
from pointwise multiplication is of the form •— for a smooth map • : M ’ N [1,
16].) A Poisson automorphism of a Poisson manifold (M, Π), is a di¬eomorphism
of M which is a Poisson map.
Remark. The Poisson automorphisms of a Poisson manifold (M, Π) form a
group. For the trivial Poisson structure, this is the group of all di¬eomorphisms.
In general, ¬‚ows of hamiltonian vector ¬elds generate a signi¬cant part of the
automorphism group. In an informal sense, the “Lie algebra” of the (in¬nite di-
mensional) group of Poisson automorphisms consists of the Poisson vector ¬elds
™¦
(see Section 3.6).
Here are some alternative characterizations of Poisson maps:

• Let • : M ’ N be a di¬erentiable map between manifolds. A vector ¬eld
X ∈ χ(M ) is •-related to a vector ¬eld Y on N when

for all x ∈ M .
(Tx •) X(x) = Y (•(x)) ,

If the vector ¬elds X and Y are •-related, then • takes integral curves of X
to integral curves of Y .
We indicate that X is •-related to Y by writing

Y = •— X ,

though, in general, •— is not a map: there may be several vector ¬elds Y on
N that are •-related to a given X ∈ χ(M ), or there may be none. Thus we
understand Y = •— X as a relation and not as a map.


29
30 6 POISSON MAPS


This de¬nition extends to multivector ¬elds via the induced map on higher
wedge powers of the tangent bundle. For X ∈ χk (M ) and Y ∈ χk (N ), we say
that X is •-related to Y , writing Y = •— X, if

§k Tx • X(x) = Y (•(x)) , for all x ∈ M .

Now let ΠM ∈ χ2 (M ), ΠN ∈ χ2 (N ) be bivector ¬elds specifying Poisson
structures in M and N . Then • is a Poisson map if and only if

ΠN = •— ΠM .


Exercise 12
Prove that this is an equivalent description of Poisson maps.



• • being a Poisson map is also equivalent to commutativity of the following
diagram for all x ∈ M :

ΠM (x) E

Tx M Tx M
T

Tx • Tx •

ΠN (•(x)) c

T•(x) N T•(x) N
E

That is, • is a Poisson map if and only if

ΠN (•(x)) = Tx • —¦ ΠM (x) —¦ Tx • , for all x ∈ M .

Since it is enough to check this assertion on di¬erentials of functions, this
characterization of Poisson maps translates into X•— h being •-related to Xh ,
for any h ∈ C ∞ (N ):


Xh (•(x)) = ΠN (•(x)) (dh (•(x))) = (Tx •) ΠM (x) (Tx • (dh (•(x))))
= (Tx •) ΠM (x) (d (•— h(x)))
= (Tx •) (X•— h (x)) ,

where the ¬rst equality is simply the de¬nition of hamiltonian vector ¬eld.

The following example shows that X•— h depends on h itself and not just on the
hamiltonian vector ¬eld Xh .
Example. Take the space R2n with coordinates (q1 , . . . , qn , p1 , . . . , pn ) and Pois-
‚ ‚ n
‚qi § ‚pi . The projection • onto R with coordi-
son structure de¬ned by Π =
nates (q1 , . . . , qn ) and Poisson tensor 0 is trivially a Poisson map. Any h ∈ C ∞ (Rn )
has Xh = 0, but if we pull h back by •, we get

‚h ‚
Xh—¦• = ’ .
‚qi ‚pi
i
6.2 Complete Poisson Maps 31


This is a non-trivial vertical vector ¬eld on R2n (vertical in the sense of being killed
by the projection down to Rn ) . ™¦



p1 , . . . , pn

q1 , . . . , qn





c
Rn
q1 , . . . , qn



6.2 Complete Poisson Maps

Although a Poisson map • : M ’ N preserves brackets, the image is not in general
a union of symplectic leaves. Here is why: For a point x ∈ M , the image •(x) lies
on some symplectic leaf O in N . We can reach any other point y ∈ O from •(x) by
following the trajectory of (possibly more than one) hamiltonian vector ¬eld Xh .
While we can lift Xh to the hamiltonian vector ¬eld X•— h near x, knowing that
Xh is complete does not ensure that X•— h is complete. Consequently, we may not
be able to lift the entire trajectory of Xh , so the point y is not necessarily in the
image of •. Still, the image of • is a union of open subsets of symplectic leaves.
The following example provides a trivial illustration of this fact.
Example. Let • : U ’ R2n be the inclusion of an open strict subset U of the
space R2n with Poisson structure as in the last example of the previous section.
Complete hamiltonian vector ¬elds on R2n will not lift to complete vector ¬elds on
™¦
U.
To exclude examples like this we make the following de¬nition.
A Poisson map • : M ’ N is complete if, for each h ∈ C ∞ (N ), Xh being a
complete vector ¬eld implies that X•— h is also complete.

Proposition 6.1 The image of a complete Poisson map is a union of symplectic
leaves.

Proof. From any image point •(x), we can reach any other point on the same
symplectic leaf of N by a chain of integral curves of complete hamiltonian vector
¬elds, Xh ™s. The de¬nition of completeness was chosen precisely to guarantee that
the X•— h ™s are also complete. Hence, we can integrate them without restriction,
32 6 POISSON MAPS


and their ¬‚ows provide a chain on M . The image of this chain on M has to be the
original chain on N since Xh and X•— h are •-related. We conclude that any point
on the leaf of •(x) is contained in the image of •. 2

Remarks.

1. In the de¬nition of complete map, we can replace completeness of Xh by the
condition that Xh has compact support, or even by the condition that h has
compact support.

2. A Poisson map does not necessarily map symplectic leaves into symplectic
leaves. Even in the simple example (previous section) of projection R2n ’ Rn ,
while R2n has only one leaf, each point of Rn is a symplectic leaf.

™¦
The example of projecting R2n to Rn is important to keep in mind. This projec-
tion is a complete Poisson map, as Xh is always trivial (and thus complete) on Rn
‚h ‚
and the pull-back ’ ‚qi ‚pi is a complete vector ¬eld. However, if we restrict the
projection to a subset of R2n , then the map will in general no longer be complete.
The subsets of R2n for which the projection restricts to a complete map are those
which are open collections of full vertical p-¬bers.
Here is another justi¬cation of our terminology.

Proposition 6.2 Let • : M ’ R be a Poisson map. Then • is complete if and
only if X• is complete.

Proof. First, assume that • is complete. The hamiltonian vector ¬eld Xt for the
identity (or coordinate) function t : R ’ R is trivial, and thus complete. Thus the
vector ¬eld X•— t = X• is complete.
Conversely, assume that X• is complete, let h : R ’ R be any function, and
compute
X•— h = Xh—¦• = ΠM (d(h —¦ •))
= ΠM (h · d•)
= h · ΠM (d•)
= h · X• .
More precisely, X•— h (x) = h (•(x)) · X• (x). At this point, recall that Xf · f =
{f, f } = 0 for any f ∈ C ∞ (M ) (the law of conservation of energy). Therefore, along
any trajectory of X• , h (•(x)) is constant, so X•— h , being a constant multiple of
X• , must be also complete. 2


6.3 Symplectic Realizations

A Poisson map • : M ’ N from a symplectic manifold M is called a symplectic
realization of the Poisson manifold N .
Examples.

1. A basic example of symplectic realization is the inclusion map of a symplectic
leaf into the ambient Poisson manifold.
6.3 Symplectic Realizations 33


2. A more signi¬cant example is provided by our construction in Section 4.2 of a
faithful representation of g. We took an open subset U of g— with coordinates
(q, p, c) and formed the symplectic space U — R with coordinates (q, p, c, d).
The map projecting U — R back to g— is a symplectic realization for g— .
It is certainly not a complete Poisson map. It was constructed to have the
property that functions on g— with linearly independent di¬erentials pull back
to functions on U — R with linearly independent hamiltonian vector ¬elds.
™¦
If a symplectic realization • : M ’ N is a submersion, then locally there is
a faithful representation of the functions on N (modulo the constants) by vector
¬elds on M , in fact, by hamiltonian vector ¬elds. Example 2 above turns out to be
quite general:

Theorem 6.3 (Karasev [89], Weinstein [34]) Every Poisson manifold has a
surjective submersive symplectic realization.

The proof of this theorem (which is omitted here) relies on ¬nding symplectic
realizations of open subsets covering a Poisson manifold and patching them together
using a uniqueness property. It is often di¬cult to ¬nd the realization explicitly.
We do not know whether completeness can be required in this theorem.
Example. Let N = R2 with Poisson bracket de¬ned by {x, y} = x. (This is the
dual of the 2-dimensional nontrivial Lie algebra.)

Exercise 13
Study 2-dimensional symplectic realizations of N . Find a surjective realization
de¬ned on the union of three copies of R2 . Show that the inverse image of any
neighborhood of the origin must have in¬nite area. Can you ¬nd a surjective
submersive realization with a connected domain of dimension 2?

We next look for a symplectic realization R4 ’ N . In terms of symplectic
coordinates (q1 , p1 , q2 , p2 ) on R4 , the two functions

f = q1 and g = p1 q1

satisfy the same bracket relation as the coordinates on N

{f, g} = {q1 , p1 q1 } = q1 {q1 , p1 } = q1 = f .

The map (f, g) : R4 ’ N is a symplectic realization with a singularity at the origin.
To make it a non-singular submersion, simply rede¬ne g to be p1 q1 + q2 . For this
new representation, we compute the hamiltonian vector ¬elds:

’Xf = ‚p1
‚ ‚ ‚
’Xg p1 ‚p1 ’ q1 ‚q1 +
= .
‚p2



Exercise 14
Is this realization complete? If we can integrate the vector ¬elds Xf and Xg ,
we have essentially constructed the Lie group with Lie algebra R2 , [x, y] = x.


™¦
34 6 POISSON MAPS


6.4 Coisotropic Calculus

A submanifold C of a Poisson manifold M is called coisotropic if the ideal

IC = {f ∈ C ∞ (M ) | f |C = 0}

is closed under the bracket {·, ·}. Recalling that (T C)—¦ is the subspace of T — M
which annihilates T C, we can restate the condition above as

Π ((T C)—¦ ) ⊆ T C .

Example. Suppose that (M, ω) is symplectic. Then C is coisotropic whenever

(T C)⊥ ⊆ T C ,

where ⊥ denotes the symplectic orthogonal space. The term coisotropic is linked
to the concept of isotropic submanifolds in symplectic geometry. A submanifold C
is called isotropic if
T C ⊆ (T C)⊥ .
In other words, C is isotropic if i— ω = 0, where i : C ’ M is the inclusion. For
more on isotropic submanifolds, see the lecture notes by Bates and Weinstein [11].
™¦
Coisotropic submanifolds play a special role with regard to Poisson maps:

Proposition 6.4 (Weinstein [168]) A map f : M1 ’ M2 between Poisson man-
ifolds is a Poisson map if and only if its graph is coisotropic in M1 — M 2 , where
M 2 has Poisson structure given by minus the Poisson tensor of M2 .

This suggests de¬ning a Poisson relation from M1 to M2 to be a coisotropic
submanifold R ⊆ M1 — M 2 . For relations R and S from M1 to M2 and M2 to M3 ,
respectively, we can de¬ne the composition S —¦ R by

S —¦ R = {(p1 , p3 ) | ∃ p2 ∈ M2 , (p1 , p2 ) ∈ R, (p2 , p3 ) ∈ S} .

We can then view Poisson relations as generalized Poisson maps using the following:

Proposition 6.5 If R and S are Poisson relations as above with clean compo-
sition [11] in the sense that the composition S —¦ R is a smooth submanifold and
T (S —¦ R) = T S —¦ T R, then S —¦ R is a Poisson relation.

6.5 Poisson Quotients

Suppose that ∼ is an equivalence relation on a Poisson manifold M such that the
quotient M/ ∼ has a C ∞ structure for which the quotient map • : M ’ M/ ∼
is a submersion. Then ∼ is called a regular equivalence relation. We say that
the relation is compatible with the Poisson structure if M/ ∼ has a Poisson
structure for which • is a Poisson map. Equivalently, the relation is compatible
when •— (C ∞ (M/ ∼)) forms a Poisson subalgebra of C ∞ (M ). The manifold M/ ∼
is called a Poisson quotient. Theorem 6.3 implies that all Poisson manifolds can
be realized as Poisson quotients of symplectic manifolds.
6.5 Poisson Quotients 35


A regular equivalence relation de¬nes a foliation on M . For the relation to be
compatible, the set of functions constant along the leaves of this foliation should
be closed under the bracket operation. With this notion of compatibility, it makes
sense to refer to ∼ as compatible even if it is not regular.
Let G be a Lie group acting on a Poisson manifold M by Poisson maps. Then the
set of G-invariant functions on M , C ∞ (M )G , is closed under the bracket operation.
Hence, if the orbit equivalence relation on M is regular, the orbit space M/G
becomes a Poisson manifold, and the quotient map M ’ M/G is a Poisson map.
When M is symplectic, this gives a symplectic realization of the quotient space. In
fact, we have:

Proposition 6.6 Under the assumptions above, the map M ’ M/G is complete.

Given a complete function h ∈ C ∞ (M/G) C ∞ (M )G and a point
Proof.
x ∈ M , we need to show that the vector ¬eld Xh—¦• has a full integral curve through
x. We shall suppose that this is not the case and ¬nd a contradiction.
Assume that there is a maximal interval (t’ , t+ ) of de¬nition for the integral
curve through x for which t+ is ¬nite (the case of t’ ¬nite is essentially the same).
If we project down to •(x), then there is no obstruction to extending the integral
curve σ = σh of Xh through •(x). At time t+ , the curve σ reaches some point
σ(t+ ) ∈ M/G. Because • is a projection, there is some y ∈ •’1 (σ(t+ )). We can
lift the integral curve σ to an integral curve of Xh—¦• through y and follow the curve
back to a lift yt+ ’µ of σ(t+ ’ µ). On the integral curve of Xh—¦• through x, there is
also a lift xt+ ’µ of σ(t+ ’ µ), and so there is some element g of G which maps yt+ ’µ
to xt+ ’µ . Because Xh—¦• is G-invariant, we can translate the integral curve through
yt+ ’µ by g to extend the curve through x past t+ , giving us a contradiction. Thus
t+ must be ∞.




yt+ ’µ ys
s

g M
xt+ ’µ ©
s s
x




c

s s M/G
σ(t+ ’ µ)
•(x)

2

Remark. The proof of Proposition 6.6 shows that any vector ¬eld invariant under
a regular group action is complete if the projected vector ¬eld on the quotient is
™¦
complete.
36 6 POISSON MAPS


For any manifold Q, the cotangent bundle T — Q has a canonical symplectic
structure. One way to construct it is to take local coordinates x1 , . . . , xn on
an open set U ⊆ Q. If π : T — Q ’ Q is the natural projection, then we can
put a corresponding coordinate system (q1 , . . . , qn , p1 , . . . , pn ) on T — Q|U such that

qi = xi —¦ π and pi = ·, ‚xi . We de¬ne the canonical symplectic structure by
‚ ‚
dqi § dpi (or by Π = ‚qi § ‚pi ). This expression for ω is preserved by
ω=
changes of coordinates on U .
Alternatively, there is a canonical 1-form ± on T — Q, de¬ned at any element
v ∈ Tb (T — Q) by ±(v) = b(π— v). The canonical symplectic form is ω = ’d±. One
can check the equivalence of these two constructions by writing ± in coordinates:
±= pi dqi . This shows clearly that ω is independent of the choice of coordinates.
If γ : Q1 ’ Q2 is a di¬eomorphism, the natural lift of γ to a di¬eomorphism
T Q1 ’ T — Q2 is a Poisson map.



Example. Let Q = G be a Lie group. It acts on itself by left translations and
this action lifts to an action of G on T — G by Poisson maps. The orbit space T — G/G

g— . This gives a
is then a Poisson manifold, which can be identi¬ed with Te G
Poisson structure on g— .

Exercise 15
Show that this Poisson structure on g— is the negative of the one constructed
in Section 3.1.


The quotient map T — G ’ g— provides a symplectic realization of g— which is, in
general, larger than the one that we found in Section 4.2 (moreover, the symplectic
™¦
realization here requires the existence of G).


6.6 Poisson Submanifolds

When a Poisson map • is an embedding, we often say that the image of • is a
Poisson submanifold, although sometimes the term is applied only when • is
also proper. If M ⊆ N is a closed submanifold, then M is a Poisson submanifold if
any of the following equivalent conditions holds:

1. The ideal IM ⊆ C ∞ (N ) de¬ned by

IM = {f ∈ C ∞ (N ) | f |M = 0} .

is a Poisson ideal. That is, IM is an ideal under the bracket multiplication as
well as the pointwise multiplication of functions. In this case, the inclusion
M ’ N corresponds to the quotient

C ∞ (M ) C ∞ (N )/IM ←’ C ∞ (N ) .

2. Every hamiltonian vector ¬eld on N is tangent to M .

3. At each point x in M , Π(Tx N ) ⊆ Tx M .

4. At each x ∈ M , Πx ∈ §2 Tx M , where we consider §2 Tx M as a subspace of
§ 2 Tx N .
6.6 Poisson Submanifolds 37


Remark. Symplectic leaves of a Poisson manifold N are minimal Poisson subman-
ifolds, in the sense that they correspond (at least locally) to the maximal Poisson
ideals in C ∞ (N ). They should be thought of as “points,” since each maximal ideal
of smooth functions on a manifold is the set of all functions which vanish at a
™¦
point [16].
Suppose that M and N are symplectic with Poisson structures induced by the
symplectic 2-forms ωM and ωN . For a map • : M ’ N , the symplectic condition
•— ωN = ωM does not make • a Poisson map, unless • is a local di¬eomorphism.
The following two examples illustrate this di¬erence.
Examples.

1. The inclusion R2 ’ R4 of symplectic manifolds de¬ned by mapping the
coordinates (q1 , p1 ) ’ (q1 , p1 , 0, 0) is a symplectic embedding, but it is not a
Poisson map, since {q2 , p2 }R4 = 1, while the bracket in R2 of their pull-backs
is 0.

2. On the other hand, the projection R4 ’ R2 given by mapping

(q1 , p1 , q2 , p2 ) ’’ (q1 , p1 )

is a Poisson map, but is not symplectic, since •— ωN = dq1 § dp1 = ωM .

™¦
In general, the condition •— ωN = ωM requires the map to be an immersion,
while Poisson maps between symplectic manifolds are always submersions.
7 Hamiltonian Actions
A complete Poisson map from a Poisson manifold M to a Lie-Poisson manifold g—
gives rise to a left action of the connected, simply connected Lie group G with Lie
algebra g on M by Poisson automorphisms, as we will now explain and explore.

7.1 Momentum Maps

Each element v of a Lie algebra g corresponds to a linear function hv ∈ C ∞ (g— )
de¬ned by hv (µ) = µ(v). Moreover, this correspondence is a Lie algebra homomor-
phism: {hv , hw } = h[v,w] ; see Section 3.1. Given a Poisson map J : M ’ g— , the
composition
J—
h X
C ∞ (g— ) ’’ C ∞ (M )
· ·
’’ ’’ χ(M )
g
’’ ’’
v hv h Xh

is a Lie algebra anti-homomorphism ρ : g ’ χ(M ) (because the last arrow is
an anti-homomorphism). In other words, J induces a left action of g on M by
hamiltonian vector ¬elds.
Suppose that J is complete. For each v ∈ g, the vector ¬eld Xhv ∈ χ(g— ) is
complete. Hence, each XJ — (hv ) is also complete. In this case, the action ρ can be
integrated to a left action of the connected, simply connected Lie group G with Lie
algebra g on M by Poisson automorphisms [134].
Let JM : M ’ g— , JN : N ’ g— and • : M ’ N be Poisson maps such that
the diagram

M EN

d  
d  
d  
JM d   JN

d ©
 
g—
commutes. Then • will necessarily be compatible with the group actions induced
by JN and JM .
Let M = g— and let J be the identity map. The induced action
Example.
of G on g— is just the dual of the adjoint representation, called the coadjoint
action. In this case, G can be any connected (not necessarily simply connected)
Lie group whose Lie algebra is g. This action of G restricts to a transitive action
on each symplectic leaf O of g— ; thus, the symplectic leaves are called coadjoint
orbits. To understand this, consider the inclusion map ι : O ’ g— . The induced
commutative diagram
ι E g—
O‚
d  
d  
d  
ιd   id
d
‚  
©
g—
shows that the G-action on g— restricts to a G-action on O. Furthermore, this action
is transitive: at each µ ∈ g— , the {dhv | v ∈ g} span Tµ g— , so the corresponding




39
40 7 HAMILTONIAN ACTIONS


hamiltonian vector ¬elds {Xhv | v ∈ g} span the tangent space to the symplectic leaf
at µ. We conclude that each symplectic leaf O of g— is a symplectic homogeneous
™¦
space of G given as an orbit of the coadjoint action.
For a Poisson map • : M ’ g— , the diagram
• E g—
M
d  
d  
d  
•d   id
d
‚ ©
 
g—
shows that • is G-equivariant for the induced action of the corresponding Lie group
G on M and the coadjoint action on g— .
When g = R and G = R, the induced G-action of a map J : M ’ R is just the
hamiltonian ¬‚ow of J. In general, we say that a complete Poisson map J : M ’ g—
is a hamiltonian or momentum map for the resulting action of G on M .
In summary, a complete Poisson map

J : M ’’ g—

gives rise to a Lie algebra anti-homomorphism

ρ : g ’’ χ(M ) ,

which we integrate to a left action of G on M by Poisson automorphisms. The
original map J is G-equivariant with respect to this action and the coadjoint action
of G on g— .
Historical Remark. Much of the construction above is merely a modern formu-
lation of work done by Lie around 1890. Lie even refers to the “dual of the adjoint”
™¦
(see [106] and [164]).


7.2 First Obstruction for Momentum Maps
Given a Poisson map J : M ’ g— , we constructed in the previous section an action
on M by Poisson automorphisms for which J was the momentum map. Conversely,
given an action of a Lie group G by Poisson automorphisms on M , we would like
to ¬nd a corresponding momentum map.
The sets of Poisson vector ¬elds and of hamiltonian vector ¬elds on M will be
denoted χPoiss (M ) and χHam (M ).
An action of G on M by Poisson maps can be di¬erentiated to give an anti-
homomorphism ρ : g ’ χPoiss (M ). The ¬rst step in seeking a momentum map for
this G-action is attempting to lift ρ to a linear map J : g ’ C ∞ (M ) making the
following diagram commute:
C ∞ (M ) E χHam (M ) ‚E χPoiss (M )
s
d 
 
d  
d  
J? d  ρ
d  
g
7.3 Second Obstruction for Momentum Maps 41


The map ρ lifts to χHam (M ) if and only if its image is actually contained in
1
χHam (M ) ⊆ χPoiss (M ). Let HΠ (M ) be the ¬rst Poisson cohomology of M (de-
¬ned in Section 3.6). To measure the obstruction, we look at the exact sequence
1
χHam (M ) ‚E χPoiss (M ) E HΠ (M )
¨
B
¨¨
T  
 ¨
ρ 
? ¨¨
 
  ¨¨ ρ
¨
 ¨
¨
g
1 1
which induces a Lie algebra homomorphism ρ : g ’ HΠ (M ) (here we equip HΠ (M )
with the trivial Lie bracket). Clearly, ρ = 0 if and only if ρ lifts to χHam (M ). This
can be interpreted as a ¬rst characteristic class for the action of G on a manifold;
the vanishing of ρ is a necessary and su¬cient condition for G to act by hamiltonian
vector ¬elds.
Remark. Recall that in the symplectic case HΠ (M ) = H 1 (M ; R) with trivial
1

bracket, since the bracket of any two Poisson vector ¬elds X1 , X2 is hamiltonian:
[X1 , X2 ] = ’Xω(X1 ,X2 ) .
™¦
Even in this case, ρ can of course be non-trivial.
1
Question: Is the vanishing of HΠ (M ) necessary for all group actions to lift to
1
χHam (M )? More generally, are all elements of HΠ (M ) represented by complete
Poisson vector ¬elds? (Hint: see [172].)
1
Metaphorically speaking, HΠ (M ) is the algebra of vector ¬elds on the space
of symplectic leaves. It is as if the action of G on M induced an action of G on
1
the space of symplectic leaves, via the algebra homomorphism ρ : g ’ HΠ (M ).
The triviality of this action is a necessary and su¬cient condition for the lifting to
hamiltonian vector ¬elds. The following simple case illustrates that Poisson vector
¬elds are not necessarily tangent to the symplectic leaves.

Example. Take R2 with bracket {x, y} = x. The Poisson vector ¬eld ‚y preserves
the two open symplectic leaves (the half-planes {(x, y) | x < 0} and {(x, y) | x >
0}), but it is not tangent to the symplectic leaves on the y-axis (the points {(0, y)}),
and acts non-trivially on them. Thus it does not lift to χHam (M ), and hence this
1
™¦
Poisson manifold has HΠ = 0.


7.3 Second Obstruction for Momentum Maps
Assume that ρ = 0, so that there is a lift ρ : g ’ χHam (M ).
C ∞ (M ) E χHam (M ) ‚E χPoiss (M ) E HΠ (M )
1

¨¨
B
T ¨
s
d 
 
¨¨
ρ 
ρ
d ¨
d  
J? d ¨¨ ρ = 0
 
  ¨¨
d
¨
g
Because the map C ∞ (M ) ’ χHam (M ) is surjective, we can lift ρ to a linear map
J : g ’ C ∞ (M ), but J is not necessarily a Lie algebra homomorphism.
42 7 HAMILTONIAN ACTIONS


In any case, de¬ne the smooth map J : M ’ g— by

J(x), v = J (v)(x)

for all x ∈ M, v ∈ g. This is called a momentum map for the G action by
Kostant [99], Smale [152] and Souriau [153] (though our de¬nition above is more
restrictive). The map J is Poisson if and only if J is a Lie algebra homomorphism.
In that case, J is G-equivariant if G is connected. Conversely, we have the following
proposition:

Proposition 7.1 A G-equivariant momentum map is a Poisson map.

If J : M ’ g— is G-equivariant, then for any v ∈ g, the hamiltonian
Proof.
¬‚ow of J (v) on M is mapped by J to the hamiltonian ¬‚ow of hv on g— , since
J (v) = J — (hv ). By the last characterization in Section 6.1, the map J is Poisson
if, for all functions f ∈ C ∞ (g— ), J maps the hamiltonian ¬‚ow of J — (f ) to the
hamiltonian ¬‚ow of f . But it actually su¬ces to check this condition for the hv ™s
because the collection {dhv } spans the cotangent spaces of g— . We conclude that J
is a Poisson map. 2
What is the obstruction to constructing a lift J : g ’ C ∞ (M ) which is a Lie
algebra homomorphism? Here is a test to see whether a given J preserves the
Poisson bracket. For any v, w ∈ g, de¬ne

˜J (v, w) = {J (v), J (w)} ’ J ([v, w]) .

We would like to have ˜J (v, w) = 0 for any choice of v, w. Let β : C ∞ (M ) ’
χHam (M ) be the map β(f ) = Xf . Noting that both β and ρ = β —¦ J are anti-
homomorphisms, we compute

= β{J (v), J (w)} ’ βJ ([v, w])
β(˜J (v, w))
= ’[β(J (v)), β(J (w))] ’ ρ([v, w])
= ’[ρ(v), ρ(w)] + [ρ(v), ρ(w)]
=0.

So ˜J takes values in

ker β : C ∞ (M ) ’ χHam (M ) = HΠ (M ) .
0



Since ˜J is anti-symmetric, we regard it as a map
0
˜J : g § g ’’ HΠ (M ) ,

whose vanishing is equivalent to J being G-equivariant, as long as G is connected.

7.4 Killing the Second Obstruction
For a ¬xed ρ, the de¬nition of ˜J above does depend on J . As the lift J varies by
0
elements of HΠ (M ), the corresponding ˜J ™s can change. The question becomes: if
˜J is non-trivial, can we kill it by a di¬erent choice of lifting J ?
To answer this question, we start by evaluating

˜J (u, [v, w]) = J (u), {J (v), J (w)} ’ ˜J (v, w) ’ J ([u, [v, w]]) .
7.5 Obstructions Summarized 43


The cyclic sum

δ˜J (u, v, w) = ˜J (u, [v, w]) + ˜J (v, [w, u]) + ˜J (w, [u, v])

is called the coboundary, δ˜J , of ˜J .

Exercise 16
Prove that δ˜J (u, v, w) is 0. You should use the Jacobi identity and the fact
that ˜J (v, w) is a Casimir function.

0
Since δ˜J (u, v, w) = 0, ˜J is called a 2-cocycle on g with values in HΠ (M ).
0
Suppose that we replace J with J + K, where K : g ’ HΠ (M ) is a linear map.
The momentum map K : M ’ g— associated to K is constant on symplectic leaves.
0
Such a map K is called a 1-cochain on g with values in HΠ (M ). The 2-cocycle
˜J+K corresponding to J + K satis¬es

˜J+K (u, v) = ˜J (v, w) ’ K([v, w]) .

We de¬ne δK(v, w) = ’K([v, w]).

Exercise 17
0
Using the previous de¬nition of δ for 2-cochains on g with values in HΠ (M ),
2 K = 0.
show that δ


Let H 2 (g; HΠ (M )) be the second Lie algebra cohomology of g with coe¬cients
0
0
in HΠ (M ). We then conclude that the cohomology class

[˜J ] ∈ H 2 (g; HΠ (M ))
0


is independent of the choice of J and depends only on ρ. Furthermore, [˜J ] vanishes
if and only if a lift J exists which is a Lie algebra homomorphism.

7.5 Obstructions Summarized

Given an action of a Lie group G on a Poisson manifold M , there is an induced map
ρ : g ’ χPoiss (M ). The ¬rst obstruction to lifting ρ to a Lie algebra homomorphism
J : g ’ C ∞ (M ) is the map ρ : g ’ HΠ (M ). If HΠ (M ) is abelian, as for instance
1 1

in the symplectic case, then ρ is actually an element of H 1 (g; HΠ (M )). We think
1

of ρ as an action of g on the leaf space of M which needs to be trivial in order to
lift ρ.
When ρ = 0 there is a second obstruction in H 2 (g; HΠ (M )).
0



Exercise 18
Check that
H 2 (g; HΠ (M ))
0
H 2 (g) — HΠ (M ) .
0



0
Interpreting HΠ (M ) as the set of functions on the leaf space, we can view this
second obstruction as lying on “functions on the leaf space with values in H 2 (g)”.
Questions: Is there a variant for [˜J ] that makes sense even when ρ = 0? Is
it possible that the two objects ρ and [˜J ] be considered as parts of some single
geometric object related to the “action of G on the leaf space”? Can we integrate
44 7 HAMILTONIAN ACTIONS


cocycles on the Lie algebra into cocycles on the group? Perhaps some sense can be
made of these questions in the realms of Lie algebroid cohomology or equivariant
Poisson cohomology.
There is some terminology commonly used in these constructions. An action of
G by automorphisms of a Poisson manifold (M, Π) is called weakly hamiltonian if
there exists a momentum map J. If there is an equivariant momentum map J, then
the action is called hamiltonian. In some of the literature, weakly hamiltonian
actions are simply referred to as hamiltonian while hamiltonian actions as we have
de¬ned them are called strongly hamiltonian.
Remark. For a weakly hamiltonian action of a connected group G on a connected
symplectic manifold M , there is a modi¬ed Poisson structure on g— for which the
momentum map J : M ’ g— is a Poisson map. Consider the map
0
˜J : g § g ’’ HΠ (M )
as an element of (g— § g— ) — HΠ (M ), i.e. as a bivector ¬eld on g— with values in
0
0 0

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