Im trying to show that the Hofer metric on a symplectic manifold is bi-invariant but im struggling.
Firstly, given the flow $\rho_t$ of a hamiltonian $H_t$, the Hofer metric is $$ d(\rho_0,\rho_1)=\min_{\rho_t} \int^1_0(\max H_t- \min H_t).$$
where $\rho_t$ is a diffeo joining $\rho_0$ to $\rho_1$ with $H_t$ the corresponding Hamiltonian of the flow for each value of t.
Given a symplectomorphism $\phi$ on $(M,\omega)$ where $\phi^* \omega=\omega$, we want to show $d(\rho_0,\rho_1)=d(\rho_0(\phi),\rho_1(\phi))$.
We know that $dH_t=\omega(-,X_{H_t})$ where $\frac{d \rho_t}{dt}=X_{H_t} \circ \rho_t$.
So I assume we want to determine the vector field corresponding to flow $\rho_t \circ \phi$. Thus $\frac{d\rho_t \circ \phi}{dt}=X_{K_t} \circ (\rho_t \circ \phi)$. We would then want to determine $K_t$ in terms of $H_t$. By looking at it I would think $H_t=K_t$ as one can compose both sides of the equation by $\phi^{-1}$?
Apparantly its quite trivial, any help would be great.