I'm a physicist who occasionally likes/needs to read math papers. In these, the definition of the a moment map is as follows: We have a symplectic manifold $(M,\omega)$ with a symplectic action of a Lie group $G$ on it. If it exists, a moment map is then defined as a map $\mu:M \to {\mathfrak g}^*$ where ${\mathfrak g}^*$ is the dual of the Lie algebra of $G$. I translate this into physics-speak by interpreting the definition to say that for each generator $\lambda_a$ of the Lie algebra $\mathfrak g$ (in which $[\lambda_a,\lambda_b]={f_{ab}}^c \lambda_c$) there is a function $\varphi_a:M\to \mathbb R$ (a "Hamiltonian") such that the group action flow field ${\bf v}_a$ is defined by $d\varphi_a = \omega({\bf v}_a,\_\, )$ and obeys
$$
[{\bf v}_a, {\bf v}_b]= {f_{ab}}^c {\bf v}_c.
$$
Because the Hamiltonians are only defined up to additive constants we can only expect the corresponding Poisson brackets
$$
\{\varphi_a, \varphi_b\}\stackrel{\rm def}{=} {\bf v}_a \varphi_b
$$
To obey
$$
\{\varphi_a, \varphi_b\}= {f_{ab}}^c \varphi_c+c_{ab}
$$
where the $c_{ab}$ (a central extension) are constants. In attempting to parse various other statements in the symplectic literature I gather that the nice case where we can redefine the $\varphi_a$ to get rid of the $c_{ab}$ and the Poisson algebra becomes
$$
\{\varphi_a, \varphi_b\}= {f_{ab}}^c \varphi_c
$$
is said to be an "equivariant" moment map. Is my interpretation correct? If so, $\mu$ is equivariant with respect to what? Can someone draw a commutative/intertwining diagram and explain why the vanishing of the central terms is equivalent to being equivariant.
Equivariant moment map. Equivariant with respect to what
classical-mechanics
Best Answer
I am really not used to how these things are treated in physics, but I think that I can at least say something about the equivariance and how it relates to the Poisson bracket.
Firstly, the equivariance required in the usual definition of moment map is equivariance of the symplectic action $\psi : G \times M \rightarrow M$ with respect to the coadjoint action $Ad^\sharp$ of $G$ on $\frak{g}^*$, which means that \begin{equation} \tag{1}\label{equiv} \mu \circ \psi_g = Ad^\sharp_g \circ \mu \end{equation} holds for every $g \in G.$
Well, as you said $\mu: M \rightarrow \mathfrak{g}^*$ associates hamiltonian functions to the infinitesimal generators of the action (for me these are those ${\bf{v}}_a$). Given such a map, one can define $\mu^*:{\frak{g}} \rightarrow C^\infty(M) $ by setting $\mu^*(X)(\, \cdot \, ) = \langle \mu(\, \cdot\, ), X\rangle,$ and this is usually called a comoment map (I think this is what gives those $\varphi_a$ that you have mentioned). Then, $\mu^*$ is a linear map between two Lie algebras, so it makes sense to ask if $\mu^*$ preserves the respective brackets. And here is the point where the equivariance of $\mu$ comes into play; indeed, one can prove that if $\mu$ is equivariant then $\mu^*$ is a Lie algebra morphism. Actually, the converse also holds when $M$ and $G$ are connected. Maybe it is worth to mention that if both $H^1(\frak{g}, \mathbb{R}) $ and $H^2(\frak{g}, \mathbb{R}) $ vanish then one can always fix $\mu^*$ (or, as you said, redefine $\varphi_a)$ in order to obtain Lie morphism; in this case, any symplectic action will be hamiltonian.
Then, as far as I understand, I believe that your interpretation of what equivariance means is correct.