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For the proof of Noether's theorem, it seems like that the only thing that's important for the symmetry map $S_g : M \to M$ is that it conserves the Hamiltonian (which will then imply that the moment map is constant along hamiltonian integral curves), but I don't see that S_g conserving the symplectic form is being used in the proof. Is this true?

http://math.uchicago.edu/~may/REU2015/REUPapers/Spiegel.pdf (see last page)

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The conservation of the symplectic form is implicit in the definition of the momentum map (Def. 4.1 in the document you referenced); I recall the context here.

Let a Lie group $G$ act on a symplectic manifold $(M, \omega)$ and write for $\xi \in \mathfrak{g} = Lie(G)$ write $\xi_M$ for the induced vector field on $M$. Given a smooth function $\Phi : M \to \mathfrak{g}^*$ and defining for each $\xi \in \mathfrak{g}$ the smooth map $\Phi^{\xi} : M \to \mathbb{R} : \Phi^{\xi}(m) = \Phi(m)(\xi)$, we say that $\Phi$ is a moment map if it satisfies $ -\mathrm{d}\Phi^{\xi}(\cdot) = \omega(\xi_M, \cdot)$ for all $\xi \in \mathfrak{g}$.

The existence of a moment map associated to an action of $G$ on $M$ puts some implicit constraints on the action. For instance, the infinitesimal action of $\mathfrak{g}$ on $M$ preserves the symplectic form $\omega$, so to say; indeed, $$ \mathcal{L}_{\xi_M}\omega = i_{\xi_M} \mathrm{d}\omega + \mathrm{d}i_{\xi_M}\omega = i_{\xi_M}0 + \mathrm{d}(-\mathrm{d}\Phi^{\xi}) = 0+0 = 0 \, . $$ Any element $g$ in the connected component of $e \in G$ is the result of integrating a smooth path $c : [0,1] \to \mathfrak{g}$ starting from the identity element $e$ in $G$. Hence the elements in this connected component of $G$ act on $M$ as symplectomorphisms (in fact, Hamiltonian diffeomorphisms) of $\omega$.

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