I'm trying to prove this identity which is mentioned at the very beginning of this paper by Dostoglou and Salamon "Self-Dual instants and holomorphic curves".
Let $(M,\omega)$ be a closed symplectic manifold and let $\phi$ a symplectomorphims $\phi \colon M \to M$. Let $H \colon \Bbb R\times M\to M$ be a smooth time-dependant Hamiltonian function such that $H_s=H_{s+1}\circ \phi$. The Hamiltonian symplectomorphisms $\psi_s \colon M \to M$ generated by $H$ are defined by $$ \frac{d}{ds}\psi_s =X_s\circ \psi_s, \ \ \psi_0=Id, \ \ \omega(X_s, \cdot ) = dH_s(\cdot) $$
They satisfy $$\psi_{s+1}\circ \phi_H=\phi \circ \psi_s$$
where $\phi_H := \psi_1^{-1}\circ \phi$.
Equivalently we must show $$\psi_s\circ \phi =\phi \circ \psi_s$$ It's clear that the equation is satisfied at time $s=0$, and if we compute the derivative w.r.t. $s$ at $s=0$ we get on both sides $X_1\circ \phi$, where I used the fact that $H_s=H_{s+1}\circ \phi$. But from here I don't know what else can I say.
Can someone help me figuring out this equality?
Best Answer
Rewriting, you need to prove that $$ \psi_{s+1}\circ \psi_1^{-1} = \phi\circ\psi_s\circ\phi^{-1}. $$ This is not actually equivalent to what you wrote - I think you are assuming that $\psi_s$ is a one parameter group, and so $\psi_{s+1}\circ\psi_1^{-1} = \psi_s$, but this is not true for the flow of a time-dependent vector field.
Regardless, let $f_s = \psi_{s+1}\circ\psi_1^{-1}$, and $g_s = \phi\circ\psi_s\circ \phi^{-1}$. We want to show these two diffeomorphisms are the same. Firstly, both equal the identity at $s=0$. Now for any $p\in M$ $$ \frac{d}{ds}f_s(p) = \frac{d}{ds}\psi_{s+1}(\psi_1^{-1}(p)) = X_{s+1}(\psi_{s+1}(\psi_1^{-1}(p))) = X_{s+1}(f_s(p)). $$ Also, starting from $H_s = H_{s+1}\circ \phi$, it's not difficult to show that for any $p\in M$, $$ T_p\phi\,(X_s(p)) = X_{s+1}(\phi(p)). $$ It follows that $$ \frac{d}{ds}g_s(p) = \frac{d}{ds}(\phi\circ\psi_s\circ\phi^{-1})(p) = T_{\psi_s(\phi^{-1}(p))}\phi\, \left(X_s(\psi_s(\phi^{-1}(p)) \right) \\= X_{s+1}(\phi(\psi_s(\phi^{-1}(p))) = X_{s+1}(g_s(p)). $$ So both $f_s$ and $g_s$ satisfy the same 1st-order ODE with the same initial condition. Therefore they are identical.