18 1. Geometry of Coadjoint Orbits

The simply connected coverings of M and G are always suﬃcient.

2. Algebraic obstacle. The map X → fX of g to

C∞(M)

can always be

chosen to be linear. Indeed, let {Xi} be a basis in g. We choose functions

fi such that LXi = s-grad fi and put fX =

∑

i

cifi for X =

∑

i

ciXi. But in

general this is not a Lie algebra homomorphism: f[X,

Y ]

and {fX , fY } can

differ by a constant c(X, Y ).

Exercise 3. a) Show that the map c : g × g → R : (X, Y ) → c(X, Y )

is a 2-cocycle on g, i.e. satisfies the cocycle equation

c([X, Y ], Z) = 0 for all X, Y, Z ∈ g.

b) Check that the freedom in the choice of fX is not essential for the

cohomology class of c. In other words, the cocycle corresponding to a dif-

ferent choice of fX differs from the original one by a trivial cocycle (or a

coboundary of a 1-cycle b):

db(X, Y ) = b, [X, Y ]

where b ∈

g∗

is a linear functional on g. ♣

To cope with the algebraic obstacle we have to pass from the initial Lie

algebra g to its central extension g given by the cocycle c. By definition,

g = g ⊕ R as a vector space. The commutator in g is defined by

(22) [(X, a), (Y, b)] =

(

[X, Y ], c(X, Y )

)

.

We define the action of g on M by L(X,

a)

:= LX . So, practically it is the

action of g = g/R, which is of no surprise because the center acts trivially

in the adjoint and coadjoint representations.

We claim that this action is Poisson. Namely, we put

f(X,

a)

:= fX + a

and a simple computation shows that f[(X,

a), (Y, b)]

= {f(X, a), f(Y, b)}.

The final conclusion is

Proposition 4. Any symplectic action of a connected Lie group G on a

symplectic manifold (M, σ) can be modified to a Poisson action of a central

extension G of G on some covering M of M so that the following diagram

is commutative:

(23)

G × M −−−→ M

⏐

⏐

⏐

⏐

G × M −−−→ M

Here the horizontal arrows denote the actions and the vertical arrows denote

the natural projections.