I'm trying to understand the proof of Lemma 5.2.1 in McDuff-Salamon (3rd edition).
Let $(M, \omega)$ be a symplectic manifold, and $G$ a Lie group with a symplectic action on $(M, \omega)$. Let $X_\xi, X_\eta$ be the vector fields corresponding to $\xi,\eta \in \mathfrak{g}$, and $H_\xi,H_\eta$ functions that generate them. Let $\psi_g$ be a symplectomorphism where $g\in G$.
While showing that $H_{g^{−1}[\xi,\eta]g} = H_{[\xi,\eta]} \circ \psi_g$, the proof uses the following equality:
$$ \{H_\xi \circ \psi_g , H_\eta \circ \psi_g \} = \{H_\xi , H_\eta\} \circ \psi_g $$
At first I thought that the equality might be trivial, but now I'm not sure. I tried forgetting the fact that $H_\xi,H_\eta, \psi_g$ were defined using the Lie algebra, and instead show that $ \{F \circ \psi , G \circ \psi \} = \{F , G\} \circ \psi $, for general smooth functions $F,G$ and a symplectomorphism $\psi$:
$$
(\{F,G\} \circ \psi) (x)= \{F,G\}(\psi(x))= \omega_{\psi(x)}(X_F(\psi(x)), X_G(\psi(x))) = (\psi^*\omega)_x\big(\psi_*^{-1}X_F(\psi(x)), \psi_*^{-1}X_G(\psi(x))\big)
$$
and since $\psi$ is a symplectomorphism that leaves me with showing that $X_{H \circ \psi} = \psi_*^{-1}X_H \circ \psi$, for general $H$, and I'm not sure if it is true, or how to show it.
My questions are:
- Is this equality trivial and there's a much easier approach to show it?
- Was there any mistakes in my attempt to show it? Is it a good approach? Is it true that $X_{H \circ \psi} = \psi_*^{-1}X_H \circ \psi$ in general?
Edit: I believe that I've answered my question, but I'll wait a day or two before accepting it, to see if someone gives a nicer alternative proof.
Best Answer
I think I managed to complete my proof: $$\iota_{X_{H \circ \psi}}(\omega) = d(H \circ \psi) = d(\psi ^* H) = \psi^*(dH) = \psi^*(\iota_{X_H}(\omega)) = \iota_{\psi^*X_H}(\psi^*\omega) = \iota_{\psi^*X_H}(\omega) $$ Therefore, by the uniqueness of the Hamiltonian vector field, $X_{H \circ \psi} = \psi^*X_H = \psi_*^{-1}X_H \circ \psi$.
If there are any mistakes in my answer, please tell me. I am also still interested to see if there's an easier approach than mine.