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!

  • 1
    $\begingroup$ If $d\alpha(w, \cdot) = 0$ but $\alpha \wedge d\alpha$ is nonvanishing, you should be able to conclude $\alpha(w)$ is everywhere nonvanishing. Then just multiply $w$ by an appropriate function to normalize $\alpha(w) = 1$. $\endgroup$ – Anthony Carapetis Dec 12 '17 at 2:37
  • $\begingroup$ Not only does $w$ exist as you say, but it is unique up to multiples. Because if both $d\alpha(w,\cdot) = 0$ and $d\alpha(u,\cdot) = 0$ for indepdendent $w$ and $u$, then $\operatorname{span}\{w,u\}\cap \ker \alpha \neq \{0\}$ by counting dimensions. Picking an $x$ in this intersection, it follows that $\alpha \wedge d\alpha(x,\cdot, \cdot) = 0$, that is, $\alpha \wedge d\alpha = 0$. Together with Anthony Carapetis's comment, this gives you uniqueness of the $v$. $\endgroup$ – Jason DeVito Dec 12 '17 at 2:46

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 $\alpha_x(v_x)=1$.

  • $\begingroup$ Just for completeness sake, how do I know that there is a $v_x \in U_x$ such that $\alpha_x (v_x) = 1$? Is it because from the comments above, $\alpha$ is non-vanishing so just take a multiple of any $w$ that sets $\alpha_x$ equal to 1? $\endgroup$ – Osama Ghani Dec 12 '17 at 17:06
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    $\begingroup$ If ${\alpha_x}_{\mid U_x}=0$ then $\alpha_x \wedge d\alpha_x=0$. $\endgroup$ – Tsemo Aristide Dec 12 '17 at 17:09

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