symplectisation of a manifold

Question

I have a contact form $\alpha$ on a manifold $M$.

I have to show that the $2$-form $d(e^tp^*\alpha)$ is symplectic where $e^t:\mathbb{R}\times M\rightarrow \mathbb{R}:(tx)\mapsto e^t$ and $p:\mathbb{R}\times M\rightarrow M$.

Any hints to get me started?

closed is obvious but for degenerate less. The problem for me is that the calculations mix the two different views of 2-forms and that I can't seem to unify them. I.e. forms as maps from vector fields to maps and forms as sections of $\Lambda^rT^*M$.

Answer 1

Hint: $T(\Bbb R\times M)=\Bbb R \partial t \times \Bbb RR \times \ker \alpha$, where $R$ is a reeb vector field and let $\omega = e^t dt\wedge\alpha + e^t d\alpha$. Show that every vector $v$ which is purely in one of these factors, there exists a vector $v'$ with $\omega(v,v') \ne 0$.

Answer 2

I will suppose that $M$ is $2n-1$ dimensional.

Another approach would be to write $\omega = e^t dt \wedge \alpha + e^t d\alpha$ and then expand $\omega^n$. Recall that $\alpha$ is a contact form if and only if $\alpha \wedge (d\alpha)^{n-1}$ is a volume form.

When you expand $\omega^n$, you will find useful that $(dt \wedge \alpha)^2=0$.