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).



Obstructions for vector fields to be gradient fields of a fixed function
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 $D$.


