# Why ask for the co-moment map to preserve brackets?

Let $G$ be a Lie group and $(M, \omega)$ a symplectic manifold. An action of $G$ on $M$ is Hamiltonian if it is equipped with a co-moment map $\widetilde{\mu} : \mathfrak{g} \to C^\infty(M)$ which is a Lie algebra morphism and satisfies $X_{\widetilde{\mu}(\alpha)} = \alpha_M$ for any $\alpha \in \mathfrak{g}$, where $\alpha_M$ is the infinitesimal generator and $X_{\widetilde{\mu}(\alpha)}$ is the Hamiltonian vector field of $\widetilde{\mu}(\alpha)$.

This condition is sometimes relaxed by not requiring $\widetilde{\mu}$ to preserve brackets. In this setting, for instance, Noether's theorem still holds (if $f \in C^\infty(M)$ is $G$-invariant then the moment map $\mu$ is preserved along the flow of $X_f$).

For connected $G$, the condition that $\widetilde{\mu}$ preserves brackets is equivalent to $\mu : M \to \mathfrak{g}^*$ being equivariant (in general, the latter implies the former). This is important if one wants to do symplectic reduction.

What are other instances where this condition is useful? Why impose this condition instead of the stronger condition of $\mu$ being equivariant?

• I don't think there's any particularly good reason for this. Moment maps are a basically infinitesimal notion, so generally people don't worry a tremendous amount about non-connected groups. I also believe that if $G$ has finitely many components, one can just average to get an equivariant moment map from one equivariant on the connected component of the identity (just as one can fix a lack of infinitesimal equivariance by passing to a central extension). – Ben Webster Dec 26 '16 at 3:02