Construct a nontrivial symplectomorphism of cotangent bundle

Question

I have tried to prove that exercise 4 on the page 20, Lectures on Symplectic Geometry, Ana Cannas da Silva, which is available on professor's website: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf

Let $X$ be an arbitrary $n$-manifold, and let $M=T^{*}X$ its cotangent bundle. Let h be a smooth function on $X$. Define $\tau_h : M \to M$ by setting $$\tau_h(x,\xi)=(x,\xi+dh_x).$$ Prove that $$\tau_h^{*} \alpha= \alpha + \pi^*dh$$ where $\pi$ is the projection map $\pi: M \to X$ defined by $(x,\xi) \to x$. Deduce that $$\tau_h^{*} \omega= \omega,$$ i.e., $\tau_h$ is a symplectomorphism.

I haven't discovered a possible pattern of proof yet. Any hints or suggestions?

Answer 1

$\alpha_{(x,\xi)}(u,v)=\xi(u)$, $d(\tau_h)_{(x,\xi)}(u,v)=(u,v+d^2h_x(u))$, this implies that $\tau_h^*\alpha_{(x,\xi)}(u,v)=\alpha_{(x,\xi+dh_x)}(u,v+dh^2_x.u)=(\xi+dh_x)(u)=\xi(u)+dh_x(u)$

$=\alpha_{(x,\xi)}(u)+ \pi^*dh_x(u)$.

We deduce that $\tau_h^*\alpha=\alpha+\pi^*dh$.

$\tau_h^*\alpha=\alpha+\pi^*dh$ implies that $(d\tau_h^*\alpha)=d\alpha+d(\pi^*dh)$

This is equivalent to say that: $\tau_h^*(d\alpha)=d\alpha+\pi^*(d(d(h))$, since $d^2=0$, and $-d\alpha=\omega$, we deduce that $\tau_h^*\omega=\omega$.

Answer 2

For any $p = (x ,\xi) \in M$ and $u \in T_pM$

$$\begin{aligned} (\tau_h \alpha)^*_p (u) &= \alpha_{\tau_h (p)} ((d\tau_h)_p(u))\\&= \alpha_{(x, \xi + dh_x)} ((d\tau_h)_p(u)) \\&= (\xi + dh_x)(d\pi_p \circ (d\tau_h)_p(u))\\&= (\xi + dh_x) (\underbrace{d(\pi \circ \tau)_p}_{d\pi_p} (u))\\&=\xi (d\pi_p (u)) + dh_{\pi(p)} (d\pi_p (u)) \\&= \alpha_p (u) + (\pi^*dh)_p (u) \end{aligned}$$ as required.

Finally, we see that

$$\tau_h^* \omega = \tau_h^* (- d\alpha) = - d(\tau^*\alpha) = -d (\alpha + \pi^*dh) = \underbrace{-d\alpha}_{\omega} - \pi^*\underbrace{d^2h}_0= \omega$$