# What does it mean for a Symplectic Form to be invariant under Group Action?

This should be a very basic question for people familiar with differential manifolds. I'm more or less new to the field so let me apologize in advance for ill-defined questions if arising. I split the question into 3 more specific ones.

Let $\mathcal{M}$ be a Differentiable Manifold and $\omega$ a Symplectic Form on $\mathcal{M}$.

I read the definition of a Symplectomorphism: A map $\phi:(\mathcal{M},\omega)\to(\mathcal{N},\omega ')$, s.t. $\phi^*\omega=\omega '$ where $\phi^{*}\omega(X,Y)=\omega(d\phi(X),d\phi(Y))$.

The Group Action $\Phi:(G\times\mathcal{M})\to\mathcal{M}$ of the group $G$ can be regarded as maps $\Phi_g:(\mathcal{M},\omega)\to(\mathcal{M},\omega)$ with $g\in G$.

1. Is the meaning of "invariant under Group Action $\Phi$" the same as "$\Phi_g$ is a Symplectomorphism for all $g\in G$"?

Example: $\mathcal{M} = \mathbb{R}^3$; $\omega=\varepsilon_{abc}x^c dx^a \land dx^b$, $\varepsilon$ being the Levi-Civita symbol. (I know that on $\mathbb{R}^3$, $\omega$ cannot be a Symplectic Form, but on $S^2 \subset \mathbb{R}^3$ it should be.)

1. How do I show this form is "invariant under SU(2)"?

I found this statement in literature. I guess the statement implies the action on $\mathbb{R}^3$ should be taken "naturally" as SO(3) matrix muliplication.(?) To really understand what's going on I would like to see two ways of showing this: a very abstract way and a coordinate-oriented "just-calculate" way (only if possible of course).

If $\omega$ is a Symplectic Form then there is a corresponding Poisson Bracket via $\{f,g\}:=\omega(X_f,X_g)$ where $X_f$ is a Hamiltonian Vector Field.

1. How does the invariance of a Symplectic Form relate to its corresponding Poisson Bracket? More specific: There should be an equivalent equation for the poisson bracket expressing the invariance of the symplectic structure. How does it look like?

Thanks, that's it.

2. If $\omega$ is a volume form on $M$ and $f:M\to M$ a diffeomorphism then $$\left(f^\ast\omega\right)_x=\det d_xf\cdot\omega.$$ In particular, if $\det d_xf=1$ for all $x\in M$, then $\omega$ is $f$-invariant. It's the case of the action of $SO(3)$ on $S^2$ by multiplication.
3. A diffeomorphism $\varphi : (M,\omega)\to(M,\omega)$ is a symplectomorphism iff it is a Poisson map, i.e. for all $f,g\in C^\infty(M)$: $$\{f,g\}\circ\varphi=\varphi^\ast\left(\{f,g\}\right)=\{\varphi^\ast f,\varphi^\ast g\}=\{f\circ\varphi,g\circ\varphi\}$$