Let $(\Sigma,\omega)$ be a closed surface equipped with an area form. For any $\phi:\Sigma \to \Sigma$ such that $\phi^{*}\omega = \omega$ (i.e., area-preserving), we define its mapping torus $M_{\phi}$ to be $[0,1] \times \Sigma $ by identifying $ (0,x) \sim (1,\phi^{-1}(x))$. Note that the pullback of $\omega$ to $[0,1] \times \Sigma$ descends to a two-form $\omega_{\phi}$ on $M_{\phi}$, by the area-preserving assumption.
Now, I want to compare the geometry between the mapping tori of two Hamiltonian isotopic maps $\phi, \phi'$. That is, say $\phi' = \phi \circ \varphi_{H}^1$, where $\varphi_{H}^{1}$ is the time-1 map of some time-dependent Hamiltonian. We can define a map
\begin{align*}
[0,1] \times \Sigma &\to [0,1] \times \Sigma\\
(t,x) &\to (t,(\varphi_{H}^t)^{-1}(x))
\end{align*}
which descends to a diffeomorphism $\Psi_H: M_{\phi} \to M_{\phi'}$. Now, I want to show that $\omega_{\phi'}$ pulled back to $\omega_{\phi} + dH \wedge dt$ under this diffeomorphism.
My thoughts so far: There is a vector field $\partial_t$ on $M_{\phi}$ where $t$ is the coordinate on $[0,1]$. This vector field is pushforward to $\partial_t – X_H$, where $X_H$ is the Hamiltonian vector field of $H$. Then, note that the tangent space $TM_{\phi}$ has a decomposition $\text{Span}(\partial_t) \oplus E$, where $E$ is the vertical tangent bundle. The linearized diffeomorphism preserves $E$ by the area-preserving assumption.
However, I don't know how to show that pullback form is given by $\omega_{\phi} + dH \wedge dt$. I confused myself when trying to write this in local coordinates. Any insight would be appreciated.
Best Answer
Let me work upstairs on $[0, 1]\times \Sigma$ before you take the quotients to form the mapping cylinders. So we have \begin{align} \psi_H : [0, 1]\times \Sigma&\to [0, 1]\times \Sigma;\\ (t, x)&\mapsto (t, (\varphi_H^t)^{-1}(x)). \end{align} Let me denote the pullback as $\tilde\omega = \pi_2^*\omega\in \Omega^2([0, 1]\times\Sigma)$. Now we want to show the corresponding statement $$ \psi_H^*(\tilde \omega) = \tilde \omega + dH\wedge dt. $$ By the decomposition $\Omega^2([0, 1]\times \Sigma) = \Omega^2(\Sigma) \oplus \Omega^1(\Sigma)\wedge \Omega^1([0, 1])$, we see that $$ \psi_H^*(\tilde \omega) = \sigma + \eta\wedge dt. $$
Note that $$ \psi_H^*(\tilde \omega)=\psi_H^*\pi_2^*(\omega)=\big((\varphi^t_H)^{-1}\big)^*\omega. $$
To compute $\sigma$, we restrict it to $\{t_0\}\times \Sigma$. Note that $$ \big((\varphi^t_H)^{-1}\big)^*\omega\big|_{\{t_0\}\times \Sigma}=\big((\varphi^{t_0}_H)^{-1}\big)^*\omega=\omega, $$ since $\varphi^t_H$ is Hamiltonian so symplectic. Therefore, $$ \sigma = \tilde \omega. $$
To compute $\eta$, we use interior product to get, up to a multiple of $dt$,
$$ \eta = -i_{\partial_t}\psi_H^*(\tilde \omega) = -\psi_H^*\big(i_{(\psi_H)_*\partial_t}\tilde \omega\big) = -\psi_H^*\big(i_{\partial_t-X_H} \tilde\omega\big) = \psi_H^*\big(i_{X_H} \tilde\omega\big) = \psi_H^*\big(dH\big) = dH, $$ since $H$ is constant along the flow lines of $\varphi_H^t$, i.e., $H\circ \psi_H = H$. (A similar calculation can also be done using $\big((\varphi^t_H)^{-1}\big)^*\omega$.)
So we are done. Since $\phi$ and $\phi'=\phi\circ \phi_H^1$ preserve the symplectic form, this result descends to the map $\Psi_H$ of mapping cylinders.
It may be fun to work out the example of $\Sigma=S^2$, and $H=z=\cos \phi$ (or even a time dependent $H=t\cos\phi$) in the spherical coordinates to check this.