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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?

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The case in which the tangent bundle of $M$ is trivial is not too difficult. The general case is not very different from this special case! – damiano Aug 13 '10 at 10:57
up vote 5 down vote accepted

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.

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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.

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Ehehe, we seem to have the same interpretation of the question! d – damiano Aug 13 '10 at 10:58
Thanks! I want to construct a $\omega$ with isolated nontivial zeros. But I don't known how to do it. – Chen Aug 13 '10 at 14:00
Just modify José's answer a little bit. For example over $\mathbb{R}^2$ if we take standard coordinates $x^1,x^2$, then $\omega = dx^1$ and $X = f(x^1,x^2) \frac{\partial}{\partial x^1} + \frac{\partial}{\partial x^2}$. Then $\omega(X)\vert_{(x^1,x^2)} = f(x^1,x^2)$, and neither $\omega$ or $X$ is ever zero for any choice of $f$. Just choose $f$ with any sort of isolated zeroes you want. – Otis Chodosh Aug 13 '10 at 18:09

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