# Why does $g\circ \exp(tv) \circ g^{-1}$ give the one-parameter group of diffeomorphisms generated by $g_* v$?

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.

## 1 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 $$g^{-1}(p)$$.