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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?

S.Surace
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  • 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. – Dog_69 Jul 04 '18 at 13:49
  • 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. – Amitai Yuval Jul 04 '18 at 15:19
  • 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. – S.Surace Jul 04 '18 at 16:01
  • What is your background in differential geometry? If you are comfortable with vector bundles and the notion of transversality, a fairly simple argument would help you construct many counterexamples. – Amitai Yuval Jul 04 '18 at 18:35
  • I‘m comfortable with bundles, but do not know how to get started. – S.Surace Jul 04 '18 at 18:36
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    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 $M=\mathbb R.$ – Anthony Carapetis Jul 05 '18 at 09:37
  • Thanks you @Anthony for the simple counterexample! However, on $t<0$ the vector field points in the wrong direction, so I suppose the counterexample works on $M=[0,\infty)$ instead. – S.Surace Jul 05 '18 at 09:56
  • Or take $X=-t\partial_t$. – S.Surace Jul 05 '18 at 10:02
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    Yeah, you're right, I forgot about the derivative when choosing the signs. $-t \partial_t$ is the right choice. – Anthony Carapetis Jul 05 '18 at 11:50

1 Answers1

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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 $df$.

S.Surace
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Amitai Yuval
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  • This looks like a very powerful and beautiful argument! I will study it in a bit more detail later. Thanks! – S.Surace Jul 05 '18 at 09:53