I'm working on a qualifying exam question and I am stuck about how to even commence.
Let $M$ be a $3$ dimensional manifold. Suppose $\alpha$ is a $1$-form such that the $3$-form $\alpha \wedge d\alpha$ is nowhere zero. Show that there is a unique vector field $v$ such that $\alpha(v) = 1$ and $d\alpha(v,w) = 0$ for any other vector field $w$.
You may use, without proof, the fact that if a vector space has a non-degenerate skew-symmetric bilinear pairing then it must have even dimension.
First off, $\alpha \wedge d\alpha$ is a volume form on $M$. Let us work in local coordinates. Then $$\alpha = f dx + g dy + h dz$$ $$d\alpha = (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) dx\wedge dy + (\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z} ) dy\wedge dz + (\frac{\partial f}{\partial z}-\frac{\partial h}{\partial x}) dz \wedge dx$$ If we let $v = a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y} + c \frac{\partial}{\partial z}$, then the first equation boils down to $$af + bg + ch = 1$$ I wanted to use $a = \frac{1}{f}, b = \frac{1}{g}, c = \frac{1}{h}$ as a first guess but there's no guarantee yet that $f, g, h$ are never zero. I am not at all sure how to proceed from here.
I tried to understand the hint given, but assuming $d\alpha$ is the skew-symmetric form mentioned, we know it cannot be non-degenerate because it acts on a vector space of 3. So there does exist a degeneracy i.e. a vector field such that when paired with any $w$, gives a $0$.
I'd appreciate any help!
The form $\alpha$ is called a contact form and $v$ is the Reeeb vector field. The fact that $\alpha\wedge d\alpha\neq 0$ implies that for every $x$, $d\alpha_x$ is a non zero $2$-form so $U_x=\{w\in T_xM: d\alpha_x(w,.)=0\}$ has dimension $1$, take $v_x\in U_x$ with $\alpha_x(v_x)=1$.