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?

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?

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?

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?

I'm trying to learn about moment maps in symplectic topology (suppose our Lie group is G with lie algebra g, acting on the symplectic manifold (M,w) 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 g, or on the group of symplectomorphisms of (M,w), or on the space of functions C^infty(M,R). Therefore I can't "visualize" the Hamiltonian condition, which requires that the linear map g --> C^infty(M,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 --> Symp(M,w) "exact" if the image of the induced map rho: Lie(G) ---> Lie(Symp(M,w)) 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(Lie(G)) which is a sub-Lie-algebra of C^inf(M) with its Poisson bracket. Thanks to all the helpful answers I think I understand this much better now.

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

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?