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Is there a condition on vector fields $X$ and/or the underlying manifold $(M,g)$ such that $\nabla_X \omega$ is a closed differential form? That is, $d \nabla_X \omega = 0$. Let us assume for now that $\omega = df$, but if there is a more situations where a result holds then I would also be curious to know. Similar to this kind of question

When is the Hessian contracted with a vector field a closed form?

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    Have you tried to do the calculation yourself? – Deane Jun 01 '21 at 15:20
  • I've calculated a specific example I was interested in and got some truly horrendous relations. I am quite certain now that the answer to my question is a definitive "No", but I'll leave this up in case somebody has a lot of insight that I missed out and actually finds something. – Theo Diamantakis Jun 24 '21 at 18:56

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