While reading to Geige's An Introduction to Contact Topology, I got stuck after the following statement:
Observe that $\alpha$ is a contact form precisely if $\alpha\wedge(d\alpha)^n$ is a volume form on $M$ (i.e. a nowhere vanishing top-dimensional differential form); in particular, $M$ needs to be orientable. The condition $\alpha\wedge(d\alpha)^n\not=0$ is independent of the specific choice of $\alpha$ and thus is indeed a property of $\xi=\ker\alpha$: any other 1–form defining the same hyperplane field must be of the form $\lambda\alpha$ for some smooth function $\lambda:M\rightarrow\mathbb R\setminus\{0\}$, and we have
$$(\lambda\alpha)\wedge d((\lambda\alpha))^n=\lambda\alpha\wedge(\lambda d\alpha+d\lambda\wedge\alpha)^n=\lambda^{n+1}\alpha\wedge(d\alpha)^n\not=0\text{.}$$
For context, $\alpha$ is a 1-form that doesn't vanish at any point of $M$, where $M$ is a $(2n+1)$-differentiable manifold, and $(d\alpha)^n$ denotes the $n$-fold wedge product $d\alpha\wedge\dots\wedge d\alpha$.
I don't know how to prove the second equality, that is, why do we have that
$$\lambda\alpha\wedge(\lambda d\alpha+d\lambda\wedge\alpha)^n=\lambda^{n+1}\alpha\wedge(d\alpha)^n?$$
This would be trivial if we somehow had that $d\lambda\wedge\alpha=0$, which certainly doesn't have to be the case.
Thanks in advance for your answer.
Best Answer
Expanding $(\lambda d\alpha + d\lambda \wedge \alpha)^n$, you should get $$ \sum_{k=0}^n c_k\cdot (\lambda d\alpha)^{n-k} \wedge (d\lambda \wedge \alpha)^k $$ where $c_k$ is a combinatorial constant. Notice that $c_0=1$. Now, any term with $k>0$ has some $\alpha$ term inside, and after a suitable permutation in the wedge products, one can write
$$ (\lambda d\alpha + d\lambda \wedge \alpha)^n = \lambda^ n (d\alpha)^n + \alpha \wedge \beta, $$ for $\beta$ a $(2n-1)$-form. Finally, $\alpha$ being a $1$-form, we have $\alpha\wedge \alpha = 0$, and therefore $$ \lambda \alpha \wedge (d(\lambda \alpha))^n = \lambda^{n+1}\alpha \wedge (d\alpha)^n + \lambda \underbrace{\alpha \wedge \alpha}_{=0} \wedge \beta = \lambda^{n+1} \alpha\wedge (d\alpha)^n. $$