$M$ is contact manifold with contact form $w$. I need to give a contact form on $M×\mathbb R^{2}$
The hints provided are:
- $M×\mathbb R$ has an exact symplectic form $d(e^{t}w) = dm$ say
- $M×\mathbb R^2$ has a contact form $m+ds$
Can someone elaborate these points? For example what exactly does $d(e^{t}w)$ mean?
Take $t$ to be the coordinate on the first factor of $\mathbb R$. Then $e^t$ is a nowhere vanishing function on $\mathbb R$ and on $M\times\mathbb R$. Similarly, you can pull back the contact form $w$ to $M\times\mathbb R$, and the pullback has been suppressed in the notation you use (which is no problem, since the exterior derivative commutes with pullbacks). So $e^tw$ is a one-form on $M\times\mathbb R$ and the claim in the end is that if $s$ denotes the coordinate on the second fact of $\mathbb R$ then $e^tw+ds$ (with implicit pullbacks as before) is a contact form on $M\times\mathbb R^2$. The proof will also show that $d(e^tw)$ is symplectic. Since $dds=0$, the exterior derivative of $e^tw+ds$ can be computed as $$ d(e^tw)=d(e^t)\wedge w+e^t dw=e^t(dt\wedge w+d w). $$ Assuming that $\dim(M)=2n-1$, we have to compute $d(e^tw)^n$. Since two forms commute, we can apply the binomial theorem. Taking into account that the wedge product of $dt\wedge w$ with itself is zero and that $dw^n$ is zero (since it is the pullback of a $2n$-form on $M$, we see that $$ d(e^tw)^n=ne^{nt}dt\wedge w\wedge (dw)^{n-1}, $$ which is nowhere vanishing since $w$ is a contact form. This shows that $d(e^tw)$ is symplectic on $M\times\mathbb R$. Moreover wedging with $(e^tw+ds)$ the first summand does not contribute (two factors of $w$) so we just get a wedge with $ds$, which again is clearly nowhere vanishing.