I have a question regarding the following proof from Cannas da Silva – Introduction to Symplectic and Hamiltonian Geometry:
Let $(M, \omega)$ be a symplectic manifold, and let $\alpha$ be a 1-form such that
$\omega = −d\alpha$.
There exists a unique vector field $v$ such that its interior product with
$\omega$ is $-\alpha$, i.e., $\iota_v \omega = −\alpha$.Proposition 2.3 If $g$ is a symplectomorphism which preserves $\alpha$
(that is, $g^*\alpha = \alpha$), then $g$ commutes with the one-parameter group of
diffeomorphisms generated by $v$, i.e.,
$(\exp tv) \circ g = g \circ (\exp tv)$.Proof. Recall that, for $p \in M$, $(\exp tv)(p)$ is the unique curve in $M$
solving the ordinary differential equation
\begin{align*}
\frac{d}{dt}(\exp tv(p)) &= v(\exp tv(p))\\
(\exp tv)(p)|_{t=0} &= p
\end{align*}
for $t$ in some neighborhood of 0. From this is follows that $g \circ(\exp tv)\circ
g^{−1}$ must be the one-parameter group of diffeomorphisms generated
by $g_*v$. (The push-forward of $v$ by $g$ is defined by $(g_*v)_{g(p)} = dg_{p}(v_{p})$.)
Finally we have that $g_*v = v$, i.e., that $g$ preserves $v$.
I cannot figure out why this step holds:
From this is follows that $g \circ(\exp tv)\circ
g^{−1}$ must be the one-parameter group of diffeomorphisms generated
by $g_*v$.
I intuitively feel that the $g^{-1}$ should be there, otherwise the positioning on the manifold is wrong after pushing forward $v$, but haven't been able to formulate anything cohesive. I also cannot figure out how to show something along the lines of $g \circ \exp (tv) = \exp (t g_* v)$. In general this won't hold, so the strict conditions on $V$ and $g$ must be important.
Best Answer
I will use the notation $\rho_{t}:=\exp tv$. By assumption, $\rho_{t}$ is the one-parameter group of diffeomorphisms generated by $v$, i.e. \begin{equation}\tag{1}\label{1} \begin{cases} \left.\frac{d}{ds}\right|_{s=t}\rho_{s}(p)=v(\rho_{t}(p))\\ \rho_{0}(p)=p \end{cases} \end{equation} for all $p\in M$ and all $t\in\mathbb{R}$. To see that $g\circ\rho_{t}\circ g^{-1}$ is the one-parameter group of diffeomorphisms generated by $g_{*}v$, we need to show that \begin{equation}\tag{2}\label{2} \begin{cases} \left.\frac{d}{ds}\right|_{s=t}(g\circ\rho_{s}\circ g^{-1})(p)=(g_{*}v)\big((g\circ\rho_{t}\circ g^{-1})(p)\big)\\ (g\circ\rho_{0}\circ g^{-1})(p)=p \end{cases} \end{equation} for all $p\in M$ and $t\in\mathbb{R}$. For the first equality in \eqref{2}, we have \begin{align} \left.\frac{d}{ds}\right|_{s=t}(g\circ\rho_{s}\circ g^{-1})(p)&=(dg)_{\rho_{t}(g^{-1}(p))}\left(\left.\frac{d}{ds}\right|_{s=t}(\rho_{s}(g^{-1}(p))\right)\\ &=(dg)_{\rho_{t}(g^{-1}(p))}\left(v(\rho_{t}(g^{-1}(p)))\right)\\ &=\left(g_{*}v\right)\left((g\circ\rho_{t}\circ g^{-1})(p)\right). \end{align} Here the first equality is due to the chain rule, and the second equality uses the first equation of \eqref{1} at the point $g^{-1}(p)$.
As for the second equality in \eqref{2}, just use the second equation of \eqref{1} at the point $g^{-1}(p)$.