Given an $(n-1)$-form $\varphi$ on a smooth orientable $n$-manifold, there is a vector field $v$ such that $i_v\varphi = 0$.

Question

I am working on the following problem.

Let $M$ be a smooth orientable $n$-manifold, $n \geq 2$, and let $\varphi$ be a smooth $(n-1)$-form on $M$. Show that there is a vector field $v$ on $M$ such that $$\varphi(v,\ \_\ , \dots, \_\ ) \equiv 0$$ and $v \neq 0$ at every point where $\varphi \neq 0$.

The condition can be rewritten as $i_v\varphi = 0$.

If I could show that $\varphi = i_w\omega$ where $w$ is a vector field and $\omega$ is an orientation form on $M$, then I could take $v = w$. Is it the case that every $(n-1)$-form on a smooth orientable $n$-manifold can be written in this way?

If this is not the case, I'd appreciate some hints on how to proceed.

Answer 1

Every differential form $\theta$ of degree n-1 has a corresponding 1 form $\eta$ with relation $\theta=i_\eta\omega$, it is called Hodge dual of $\theta$. To see this, assume $\theta\wedge\rho=\lambda\omega$, then $\theta$ can be seen as a linear functional $\theta:\Omega^1(M)\to\mathbb R$ by $\rho\mapsto \lambda$. By Rietz theorem, there exists a 1-form $\eta$ such that $\theta=\langle\eta,\cdot\rangle$.

Answer 2

It is the case. One way to see it is dimension count, spaces of vectors and $n-1$ exterior forms have the same dimension, and $v\mapsto i_v\omega$ is linear and injective, hence surjective, on every tangent space. Inverses to a smooth family of invertible linear maps also form a smooth family.