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I have a problem in which I have a vector field $u$ on a smooth manifold and want to find a Riemannian metric for which the vector field is geodesic, i.e. $$\nabla_u u = \sigma u,$$ and $\sigma$ is some smooth function. As a starting point, I would like to understand if there are some necessary conditions that the vector field has to satisfy for this to be true. That is, what are some properties of $u$ that, if they fail, rule out the possibility of finding a metric such that the above equation holds?

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    A geodesic is a parametrized curve, not a vector field, so you should rephrase. – Ted Shifrin Nov 18 '23 at 04:51
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    @TedShifrin each integral curve of such a vector field is a (pre-)geodesic, so this seems like a reasonable enough extension of terminology. Besides, using “geodesic” as an adjective in “the vector field is geodesic” isn’t that uncommon: geodesic triangle, geodesic ball/sphere etc. – peek-a-boo Nov 18 '23 at 16:52
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    Of course, the vector field must be everywhere non-zero. Locally, at least, the answer to this question is a start. – Ted Shifrin Nov 18 '23 at 18:20

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