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Let $M$ be an orientable 3-manifold. On $M$, fix a vector field $X$. The curl of $X$ relative to the Riemannien metric $g$ and the volume form $\mu$, $\nabla_{g,\mu}\times X$, is defined by the formula $$ di_X g=i_{\nabla_{g,\mu}\times X}\mu.$$

When is it possible to choose a metric and volume form such that $$ \nabla_{g,\mu}\times X=\lambda X,$$ where $\lambda$ is a nowhere vanishing function?

There are many $X$ for which such a metric and volume form can be found. In particular, $X$ that arise as Reeb vector fields relative to some contact 1-form on $M$ are all examples (http://www.math.upenn.edu/~ghrist/preprints/beltrami.pdf).

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  • $\begingroup$ Do you require that the volume form be that given by the metric (and orientation), or is it chosen separately? I'm guessing the former but your notation makes me unsure. $\endgroup$
    – Paul Reynolds
    Oct 25, 2012 at 23:35
  • $\begingroup$ @Paul Reynolds : Sorry for not being clear. I'm not requiring the volume form to be the one determined by the metric. $\endgroup$
    – Josh Burby
    Oct 26, 2012 at 0:00
  • $\begingroup$ In that case if you can find $g$ and $\mu$ then you can find them so that $\lambda = 1$. I was going to say something about the case $X$ is non-vanishing, but it's pretty much what is said in the paper you linked to. $\endgroup$
    – Paul Reynolds
    Oct 26, 2012 at 17:36

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