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Let $M$ be a smooth manifold, $f$ a smooth function on $M$, and $X$ a vector field on $M$ that ascends $f$, i.e. $df_p(X_p)>0$ for all $p\in M$. Are there choices of $(M,f,X)$ such that there is no Riemannian metric $g$ on $M$ in which $X=\nabla_g f$, where $\nabla_g$ is the gradient operator on $(M,g)$?

I'm looking for either general obstructions or an explicit example (i.e. an example where no such metric can be found).

Note that I'm not looking for obstructions to $X$ being the gradient of some function, but the gradient of a function fixed in advance, of which we already know that $X$ ascends it (a necessary condition).

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Your condition implies that $f$ has no critical points and $X$ has no zeros. On the $(n-1)$-dimensional distribution $D$ defined by the kernel of $df$, choose any metric (in a way depending smoothly on the point). Then extend it to a metric on $M$ by declaring $X$ to be a unit vector orthogonal to $D$.

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  • $\begingroup$ Except that you don't want $X$ to be a unit vector. Rather, it should satisfy $X^2 = df(X)$. $\endgroup$
    – Igor Khavkine
    Jun 26, 2018 at 13:14
  • $\begingroup$ You are right, I think I want to weaken the condition $df(X)>0$. What about $df(X)\geq0$ and $X_p=0 \Leftrightarrow df_p=0$? $\endgroup$
    – S.Surace
    Jun 26, 2018 at 13:23
  • $\begingroup$ Right, @Igor, thanks. I am not sure about the case with zeros. $\endgroup$
    – Mikhail Katz
    Jun 26, 2018 at 13:34
  • $\begingroup$ Suppose that the zeros of $X$ and $df$ are isolated points. Can't we just use your construction on $M$ with those points removed? The extension to $M$ should not be a problem because on the zeros, the metric can be arbitrary. I'm probably overlooking something. $\endgroup$
    – S.Surace
    Jun 26, 2018 at 13:59
  • $\begingroup$ Well you have to check that the metric actually extends to give a smooth metric on all of $M$. Seems like it might be a delicate problem but I haven't thought about it carefully. $\endgroup$
    – Mikhail Katz
    Jun 26, 2018 at 14:01

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