Suppose that a Lie group $G$ acts freely and properly on a symplectic manifold $P$ via symplectomorphisms. Suppose further that we have at our disposal an $\text{Ad}^*$-equivariant momentum map $\mu:P\rightarrow\mathfrak{g}^*$.

In this setting, the poisson bracket on $P$, $\{\cdot,\cdot\}_P$, descends to a Poisson bracket on the orbit space $P/G$, $\{\cdot,\cdot\}_{P/G}$, that is uniquely determined by the formula $$ \pi^*\{F,G\}_{P/G}=\{\pi^*F,\pi^*G\}_P,$$
where $\pi:P\rightarrow P/G$ is the quotient map and $F,G\in C^\infty(P/G)$.

Because $P$ is symplectic, the bracket $\{\cdot,\cdot\}_P$ does not have any Casimirs. On the other hand, the reduced bracket $\{\cdot,\cdot\}_{P/G}$ has a natural class of Casimirs given in terms of the momentum map $\mu$ as follows. Let $C^\infty_G(\mathfrak{g}^*)\subset C^\infty(\mathfrak{g}^*)$ be the smooth functions on $\mathfrak{g}^*$ that are invariant under the coadjoint action. For each $A\in C^\infty_G(\mathfrak{g}^*)$, the function $A\circ\mu:P\rightarrow\mathbb{R}$ is $G$-invariant, which implies that it descends to a function $C_A$ on the quotient $P/G$ characterized by the equation $$\pi^*C_A=A\circ\mu.$$ It is straightforward to verify that $C_A$ is a Casimir of the reduced bracket for each $A\in C^\infty_G(\mathfrak{g}^*)$.

Now my question. Do the $C_A$ exhaust the (smooth) Casimirs of the reduced bracket $\{\cdot,\cdot\}_{P/G}$? I am half-way convinced that the answer is "yes" by a hand-wavy dimension counting argument. However, I'm especially interested in cases when $P$ and $G$ are infinite dimensional and dimension counting arguments can't be used; a proof that works in infinite dimensions (at least formally) would be great.

This post imported from StackExchange MathOverflow at 2015-05-29 19:53 (UTC), posted by SE-user Josh Burby