An Hamiltonian diffeomorphism is also a Poisson diffeomorphism

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 diffeomorphism iff there exists an Hamiltonian isotopy $\{\varphi^t\}$ such that $\varphi^1=\varphi$.

I want to prove that an Hamiltonian diffeomorphism is also a Poisson diffeomorphism i.e.

$$\{f\circ \varphi,g\circ \varphi\}=\{f,g\}\circ \varphi$$

for any $f,g\in C^\infty(M)$.

My attempt

Let's prove that

$$\{f\circ \varphi^t,g\circ \varphi^t\}-\{f,g\}\circ \varphi^t=0$$

for any $t$. This trivially holds for $t=0$, so I just need to prove that the derivative of the expression above (with respect to $t$) is $0$:

$$\frac{d}{dt}|_{t=t_0}\{f\circ \varphi^t,g\circ \varphi^t\}-\{f,g\}\circ \varphi^t=$$ $$=\left\{f_\ast\left(\frac{d}{dt}|_{t=t_0}\varphi^t\right),g\circ \varphi^{t_0}\right\}+\left\{f\circ \varphi^{t_0},g_\ast\left(\frac{d}{dt}|_{t=t_0}\varphi^t\right)\right\}-\{f,g\}_\ast \left(\frac{d}{dt}|_{t=t_0}\varphi^t\right)=$$ $$=\left\{f_\ast\left(X_{h_{t_0}}|_{\varphi^{t_0}(\cdot)}\right),g\circ \varphi^{t_0}\right\}+\left\{f\circ \varphi^{t_0},g_\ast\left(X_{h_{t_0}}|_{\varphi^{t_0}(\cdot)}\right)\right\}-\{f,g\}_\ast \left(X_{h_{t_0}}|_{\varphi^{t_0}(\cdot)}\right)=$$ $$=\{\{h_{t_0},f\}\circ \varphi^{t_0},g\circ \varphi^{t_0}\}+\{f\circ \varphi^{t_0},\{h_{t_0},g\}\circ \varphi^{t_0}\}-\{h_{t_0},\{f,g\}\}\circ \varphi^{t_0}$$ And now I feel stuck.

Answer 1

Simple case

Consider first

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

1. Using $\varphi^{t'}\circ\varphi^{t} = \varphi^{t'+t}$ one obtains $$\{h, f\circ\varphi^{t'}\}\circ \varphi^{t} = \{h, f\} \circ\varphi^{t+t'} $$
2. Denote $F(f, g, t) = \{f\circ \varphi,g\circ \varphi\}-\{f,g\}\circ \varphi = F'(x, t).$
3. Then your last identity becomes \begin{align}\{\{h,f\}\circ \varphi^{t_0},g\circ \varphi^{t_0}\}+\{f\circ \varphi^{t_0},\{h,g\}\circ \varphi^{t_0}\}-\{h,\{f,g\}\}\circ \varphi^{t_0}= \\ \{\{h,f\circ \varphi^{t_0}\},g\circ \varphi^{t_0}\}+\{f\circ \varphi^{t_0},\{h,g\circ \varphi^{t_0}\}\}-\{h,\{f,g\}\circ \varphi^{t_0}\} \end{align}
4. Using Jacobi law one proves that $$ \{\{h,f\circ \varphi^{t_0}\},g\circ \varphi^{t_0}\}+\{f\circ \varphi^{t_0},\{h,g\circ \varphi^{t_0}\}\}-\{h,\{f,g\}\circ \varphi^{t_0}\} = \{h, F(f, g,t)\} $$
5. $$H: f \mapsto \{h, f\}:C^\infty(M)\to C^\infty(M)$$ is linear operator and we have that $F'$ satisfies: $$ \frac{\partial}{\partial t} F' = H (F') $$
6. Idea is to use $\exp(H)$ operator. That is to solve the above equation in operators. $\exp{H}$ is linear operator and $F'(x, t)=\exp{H}(F'(x, 0)),$ so $F'(x, t) = 0.$

Solution using $\exp{H}$ requires some continuity assumption on $H.$ Lang "Fundamentals of differential geometry" (pp. 76-77, prop.1.9-1.10) provides a result in the case $H$ is continuous linear map between Banach spaces (possibly depending on parameter $t$). It should be sufficient in this case as one can restrict attention to functions $f\in C^k(K),$ where $K\subseteq M$ is compact.

Without some continuity assumption on $H$ I have no hope to prove that $F'\equiv 0$ and it seems natural to assume $H$ is continuous. When $M$ is symplectic manifold, continuity of $H$ is easily satisfied for $C^3(K).$

Computation of derivative $\frac{d}{dt}\{f, g\} = \{\frac{d}{dt}f, g\} + \{f,\frac{d}{dt} g\}$ requires continuity of $\{, \}:C^1(M)\times C^1(M)\to C^1(M).$

General case

Use reduction of time-dependent vector fields to time-independent. There will be another linear operator, no new considerations are required: $$H: f \mapsto \frac{\partial f}{\partial t} + \{h_t, f\}:C^\infty(M\times [0,1])\to C^\infty(M\times [0,1])$$