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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.

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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 $\theta=\langle\eta,\cdot\rangle$.

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  • $\begingroup$ You need a Reimannian metric to define the Hodge dual. Furthermore, if $\eta$ is a one-form, what does $i_{\eta}\omega$ mean? $\endgroup$
    – Michael Albanese
    Aug 20, 2014 at 3:22
  • $\begingroup$ @MichaelAlbanese Yes, a metric is required. And $i_\eta\omega=i_{\eta^\sharp}\omega$ $\endgroup$
    – Shuchang
    Aug 20, 2014 at 3:25
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    $\begingroup$ Riemannian metric does not restrict generality, one can be introduced on any manifold using partition of unity. Using Hodge star the inverse can be written as a formula $\varphi\mapsto(*\varphi)^\sharp$. However, proving that Hodge star is surjective requires essentially the same argument as for vectors/$(n-1)$-forms, and presence of metric obscures the fact that the correspondence only depends on the volume form, not metric. $\endgroup$
    – Conifold
    Aug 20, 2014 at 19:00
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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.

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