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
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 =

cifi for X =

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
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,
:= 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
:= 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:
G × M −−−→ M

G × M −−−→ M
Here the horizontal arrows denote the actions and the vertical arrows denote
the natural projections.
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