Hamiltonian diffeomorphisms on a Poisson manifold

Question

Let $(M,\{-,-\})$ be a Poisson manifold.

An Hamiltonian isotopy is a smooth family of diffeomorphisms $\{\varphi^t:M\to M\}_{t\in [0,1]}$ such that $\varphi^0=\text{id}_M$ there exists a smooth family of functions $\{h_t:M\to \mathbb{R}\}_{t\in [0,1]}$ such that $$\frac{d\varphi^t(x)}{dt}=X_{h_t}|_{\varphi^t(x)}$$ where $X_{h_t}$ is the Hamilton vector field induced by $h_t$ (i.e. $X_{h_t}(g):=\{h_t,g\}$ for any $g\in C^\infty(M)$).

A diffeomorphism $\varphi:M\to M$ is a Hamiltonian diffeomorfism iff there exists an Hamiltonian isotopy $\{\varphi^t\}$ such that $\varphi^1=\varphi$.

I want to prove that the set of the hamiltonian diffeomorphisms $\text{Ham}(M)$ is a subgroup of the diffeomorphisms group. In order to do so my reference book ("Lectures on Poisson Geometry" by Crainic, Fernandes, Marcut) suggests proving that given two hamiltonian isotopies $\{\varphi^t\}$ and $\{\psi^t\}$, we have that $\{\varphi^t\circ \psi^t\}$ and $\{(\varphi^t)^{-1}\}$ are hamiltoniana isotopies, but I'm having some trouble in proving this.

My attempt Let $\{h_t\}$ and $\{k_t\}$ be smooth families associated to the hamiltonian isotopies $\{\varphi^t\}$ and $\{\psi^t\}$. I define the functions $\psi_x(t):=(t,\psi^t(x))$ and $\tilde{\varphi}(t,y):=\varphi^t(y)$. Clearly $$\frac{d}{dt}|_{t=t_0}(\varphi^t\circ \psi^t)(x)=\frac{d}{dt}|_{t=t_0} (\tilde{\varphi}\circ \psi_x)(t)=d_{t_0}(\tilde{\varphi}\circ \psi_x)\left(\frac{d}{dt}|_{t=t_0}\right)= $$ $$=\left(d_{\psi_x(t_0)}\tilde{\varphi} \circ d_{t_0}\psi_x\right)\left(\frac{d}{dt}|_{t=t_0}\right)=d_{\psi_x(t_0)}\tilde{\varphi}\left(1;X_{k_{t_0}}|_{\psi^{t_0}(x)}\right)=$$

And now I feel a bit stuck.

Answer 1

For the composition, you can check https://math.stackexchange.com/q/4248516 where the corresponding Hamiltonian vector field and its smooth family of functions are constructed.

You can now use this and the fact that $\varphi_t^{-1} \circ \varphi_t=id_M$ to show the result for inverses. You just need to plug $Z_t=0$ (the vector field associated to the Hamiltonian isotopy $\varphi_t=id_M$ for all $t$) along with its corresponding family of smooth functions being constant functions (so that the Poisson bracket is zero). If $h_t$ and $X_t$ correspond to $\varphi_t$, then $-h_t\circ\varphi_t$ and $-(\varphi_t^{-1})_* X_t(\varphi_t(x))$ would make $\varphi_t^{-1}$ a Hamiltonian isotopy from $\varphi_0^{-1}=id^{-1}=id$ to $\varphi_1^{-1}=\varphi^{-1}$.