Let $M$ be a smooth manifold, and $f$ a smooth function with an isolated local minimum at $p$. Furthermore, let $X$ be a vector field vanishing at $p$ such that for some neighborhood $U$ of $p$, $df_q(X_q)<0$ for all $q\in U\setminus\{p\}$. Is it possible to find a Riemannian metric g on $U$ such that $X|_U=-\text{grad}_gf$? If not, what is a counterexample?

  • $\begingroup$ I think the answer should be no, in the same way not every 1-form is the differential of a function. However, since you allow to change the metric for each case... the answer is more difficult. $\endgroup$
    – Dog_69
    Jul 4, 2018 at 13:49
  • $\begingroup$ Do you know something about the order of vanishing at $p$? For example, if $f$ is Morse while $X$ vanishes at $p$ up to some high order, I believe your desired metric does not exist. However, if both $df$ and $X$ have a simple zero at $p$, there may be something you could do. $\endgroup$ Jul 4, 2018 at 15:19
  • $\begingroup$ No, I would like a statement in full generality. So if there is one, any explicit counterexample of the kind you mention would be helpful. $\endgroup$
    – S.Surace
    Jul 4, 2018 at 16:01
  • 1
    $\begingroup$ In the one-dimensional case this reduces to smooth divisibility of functions: $p=0,f=t^4, X=-t^2 \partial_t$ is a simple counterexample on $M=\mathbb R.$ $\endgroup$ Jul 5, 2018 at 9:37
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    $\begingroup$ Yeah, you're right, I forgot about the derivative when choosing the signs. $-t \partial_t$ is the right choice. $\endgroup$ Jul 5, 2018 at 11:50

1 Answer 1


The answer, in general, is no.

The following is a general fact: Let $M$ be a smooth manifold of dimension $n$, let $E\to M$ be a vector bundle of rank $n$, and let $s\in\Gamma(M,E)$ be a smooth section vanishing at $p\in M$. Let $\nabla$ be a linear connection on $E$. Then $s$ is transverse to the zero section at $p$ if and only if we have $$(\nabla_Xs)_p\neq0$$ for every $X\in T_pM$, $X\neq 0$, in other words, iff the endomorphism $X\mapsto(\nabla_Xs)_p$ is invertible.

Now, a Riemannian metric induces an isomorphism of vector bundles between the tangent and cotangent bundles. Transversality of sections is preserved under a bundle isomorphism. Suppose $f$ is Morse. In particular, the Hessian $\nabla^2f$ is non-degenerate at $p$, where $\nabla$ can be any linear connection on $TM$, or equivalently, the section $df$ of the cotangent bundle is transverse to the zero section at $p$. Take any vector field $X$ which vanishes at $p$ and which is not transverse to the zero section. Then no bundle isomorphism carries $X$ to $df$.

  • $\begingroup$ This looks like a very powerful and beautiful argument! I will study it in a bit more detail later. Thanks! $\endgroup$
    – S.Surace
    Jul 5, 2018 at 9:53

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