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  Casimirs of Poisson brackets obtained via Poisson reduction

+ 5 like - 0 dislike

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
asked Sep 6, 2014 in Mathematics by Josh Burby (120 points) [ no revision ]
retagged May 29, 2015

1. What is your argument in the finite-dimensional case?

2. Do you have a particular $\infty$-dimensional example in mind?

@Arnold Neumaier Here is the hand-wavy argument in finite dimensions. The number of independent functions that are invariant under the coadjoint action in a neighborhood of the coadjoint orbit $\mathcal{O}_{\mu_o}$ through $\mu_o\in\mathfrak{g}^*$ is $\dim(\mathfrak{g}^*)-\dim(\mathcal{O}_{\mu_o})$. I'll argue that this number is equal to the number of independent Casimirs of the reduced bracket. We know that the symplectic leaves in $P/G$ are diffeomorphic to the symplectic quotients $ \mu^{-1}(\mu_o)/G_{\mu_o}$, where $G_{\mu_o}$ is the stabilizer of $\mu_o$ under the coadjoint action. The dimension of a symplectic leaf is therefore $\dim(P)-\dim(\mathfrak{g}^*)-\dim(G_{\mu_o})$.  On the other hand, because a symplectic leaf can also be expressed as a common level set of a maximal number, $N$, of independent Casimirs, we have $N=\dim(G_{\mu_o})$. Finally, we note that $\dim(\mathcal{O}_{\mu_o})=\dim(G)-\dim(G_{\mu_o})=\dim(\mathfrak{g}^*)-\dim(G_{\mu_o})$.

@Arnold Neumaier One important example I have in mind is the Poisson bracket for the Vlasov-Poisson system. Here $P/G$ is the space of top-degree forms $f$ on $T^*Q$, where $Q$ is a finite-dimensional manifold (typically plasma physicists look at $Q=\mathbb{R}^3$.) The number of plasma particles in a region $U\subset T^*Q$ is given by $\int_U f$. The Poisson bracket on $P/G$ is $\{\mathcal{F},\mathcal{G}\}_{P/G}=\int f \{\frac{\delta \mathcal{F}}{\delta f},\frac{\delta\mathcal{G}}{\delta f}\}_{T^*Q}$, where $\{\cdot,\cdot\}_{TQ}$ is canonical bracket on $T^*Q$. I could describe the space $P$ and the group $G$ as well if you're interested.

Thanks; I'll think about a possible proof. 

By the way, to ping someone whose user name contains a blank you need to remove the blank.

I suspect that your conjecture might be invalid even if $P$ is finite-dimensional and $G$ is 1-dimensional. Though your argument indicates that there should be generically no other Casimirs, it is 

1 Answer

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I don't have a full answer. But I suspect that your conjecture might be invalid even if $P$ is finite-dimensional and $G$ is 1-dimensional.

Your argument indicates that there should be generically no other Casimirs. But it seems to me likely that particular, degenerate choices of the generator of $G$ lead to a kind of resonance situation that generates additional Casimirs.

Thus I recommend that you study this particular situation in more detail. I don't have the time for it but would be interested in the result of such an investigation.

answered Jun 4, 2015 by Arnold Neumaier (15,787 points) [ no revision ]

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