A question about a one-form on Riemannian manifold

Question

Let M be a Riemannian Manifold, $X$ is a smooth vector field on M with isolated zeros. Is there a one-form $\omega$ with isolated zeros such that $\omega(X)$ has nontrivial zeros? (nontivial zero means that the piont is neither in $X$'s zeros nor in $\omega$'s zeros.) If this $\omega$ exist, how to construct it?

Answer 1

Assuming the dimension of $M$ is at least 2 (otherwise it's false), you can do the following. Let $p_1,p_2,\dots$ be isolated points where $X$ does not vanish but where you want $\omega$ to vanish. In a neighborhood $U_i$ of each $p_i$, there are coordinates $(x^1,\dots,x^n)$ centered at $p_i$ on which $X$ has the coordinate representation $X = \partial/\partial x^1$. In each $U_i$, let $\omega_i = dx^2 + |x|^2 dx^1$. Then let $U_0$ be the complement of {$p_1,p_2,\dots$}, and let $\omega_0=X^\flat$ (the 1-form dual to $X$ via the metric). Let {$\phi_0,\phi_i$} be a partition of unity subordinate to the cover {$U_0,U_i$}, and let $\omega = \sum_{i\ge 0}\phi_i\omega_i$. The fact that $\omega_i(X)>0$ at points other than $p_i$ and zeros of $X$ ensures that $\omega(X)$ vanishes only at such points.

Answer 2

Yes; although my answer is so trivial that I'm not sure I'm understanding the problem well.

Given $X$, let $Y$ be any vector field perpendicular to $X$ relative to the metric and let $\omega = Y^\flat$ be the dual one-form. Then $\omega(X)=0$.

The only question here is to construct $Y$, but this can be done locally relative to a trivialisation of the frame bundle and then gluing the local $Y$s using a partition of unity subordinate to the trivialising cover.