# Where is it used that the symmetry conserves the symplectic form in noether's theorem.

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)

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 $$\omega$$.