I'm trying to learn about moment maps in symplectic topology (suppose our Lie group is $G$ with Lie algebra $\mathfrak g$, acting on the symplectic manifold $(M,\omega)$ by symplectomorphisms). I'm having a hard time, and I've realized this is because I don't have a good conceptual understanding of the Lie bracket, either on the Lie algebra $\mathfrak g$, or on the group of symplectomorphisms of $(M,\omega)$, or on the space of functions $\mathcal C^\infty(M,\mathbb R)$. Therefore I can't "visualize" the Hamiltonian condition, which requires that the linear map $\mathfrak g \rightarrow \mathcal C^\infty(M,\mathbb R)$, which exists when the action by $G$ is "exact," be a Lie algebra homomorphism.


Please tell me how you personally understand/intuit/conceptualize this situation, both the Lie bracket stuff and moment maps more generally! Any help is greatly appreciated.


**EDIT**: I didn't realize how non-standard some of this terminology is, so my question might be confusing. I call the action $\rho: G \rightarrow {\rm Symp}(M,\omega)$ "exact" if the image of the induced map $\rho: {\rm Lie}(G) \rightarrow {\rm Lie}({\rm Symp}(M,\omega))$ is contained in the sub-lie-algebra of Hamiltonian vector fields. The condition that was confusing me, I now realize, is just a technical point: that we choose a set of representative Hamiltonian functions for the image $\rho({\rm Lie}(G))$ which is a sub-Lie-algebra of $\mathcal C^\infty(M,\mathbb R)$ with its Poisson bracket. Thanks to all the helpful answers I think I understand this much better now.


In particular, if we present ${\rm Lie}(G)$ (assumed finite dimensional, semi-simple, etc) by Lie algebra generators (with some relations), then we can probably just choose appropriate elements in $\mathcal C^\infty(M,\mathbb R)$ for these generators, and then the rest of the map from ${\rm Lie}(G)$ to $\mathcal C^\infty(M,\mathbb R)$ is just forced on us, and this gives a Hamiltonian action? Is that right?



Understanding moment maps and Lie brackets
This question is (at least as I read it) about the Poisson bracket; the Poisson bracket is a Lie bracket structure on the functions on a symplectic manifold. 


So how should one think about Poisson bracket? Well, remember that for every function on a symplectic manifold, one has a Hamiltonian vector field $X\_f$. One way to think about this is that if your symplectic manifold is the phase space of a physical system (the space of possible positions and momenta), and $f$ is the energy function, then the resulting vector field is the derivative of the time evolution of the system.


The Poisson bracket $\{f,g\}$ is defined to be $X\_f(g)$, the derivative of $g$ along the vector field $X\_f$. That is **the Poisson bracket of $f$ and $g$ is the time derivative of $g$ if you use $f$ as the energy function**. Remarkably, this operation is anti-symmetric, and defines a Lie algebra structure.


As you mentioned, a moment map is equivalent to a Lie algebra homomorphism from $\mathfrak{g}$ to the space of functions on your manifold. I'm not sure how you're exactly supposed to visualize that, but let me explain how I think about it.


So, imagine you have your favorite G-action on a symplectic manifold, preserving the symplectic structure. Taking derivative, you get a map of Lie algebras from $\mathfrak{g}$ to vector fields on your manifold. It sounds from your question like how to think about such Lie algebra homomorphisms is actually what is confusing you. This just says that if you were to integrate the vector fields coming from $\mathfrak{g}$, you would get the group G (or maybe a finite cover). 


Now, each of these vector fields corresponds under the symplectic form to a 1-form. If your manifold has no $H^1$, then you can integrate these to functions, but of course, you can't do this uniquely; it's only unique up to a constant. So taking all of these lifts, you get a vector space of functions which is $\dim \mathfrak{g}+1$ dimensional (assuming $G$ acted faithfully). This is closed under Lie bracket, so it's a finite dimensional Lie algebra $\tilde {\mathfrak{g}}$ with a map $\tilde {\mathfrak{g}}\to {\mathfrak{g}}$.


It might be that $\tilde {\mathfrak{g}}\cong {\mathfrak{g}}\times \mathbb{R}$ as a Lie algebra, in which case you can get a moment map by picking a splitting of the map above, or it might not, in which case you don't have a moment map. If $\mathfrak{g}$ is semi-simple, then the latter case is impossible, so you always have a moment map.


