What is an example of a gradient vector field $X$ on a Riemannian manifold $(M,g)$ which cannot be converted to a divergence free vector field via the following processes:

  1. First we remove the singularities $S$ from $M$ then we set $M:=M\setminus S$

  2. We are allowed to reparameterize $X$ to $X:=fX$ for some positive function $f$

  3. We are allowed to change the initial Riemannian metric $g$ to a new metric $g'$ for computation of divergence of $X$ with respect to this new $g'$ to obtain a vector field $X$ on $M$ with $\operatorname{div}_{g'} X=0$.

In other words, with some abuse of terminology, we ask: "Is every gradient vector field a divergence free vector field?"

An obvious example is $X=x\partial_x+y\partial_y$ is a divergence free vector field on the punctured plane after rescaling $X:=\frac{1}{x^2+y^2}X$

Is the answer yes, at least in low dimensions?

  • 2
    $\begingroup$ Yes --- just choose a Riemannian metric that is invariant with respect to the flow generated by $X$. It exists since the singularities are removed. $\endgroup$ Commented Jan 29, 2021 at 19:57
  • $\begingroup$ @AntonPetrunin Yes I see you use gradient property here to ensure the flow is geodesible. $\endgroup$ Commented Jan 29, 2021 at 20:05
  • $\begingroup$ @AntonPetrunin Thanks for your attention I realize that the flow of X keep invariant the level sets of f so our frame is $X, \nabla f$. But the motivation for my question is another thing I am giving my question in my next post. $\endgroup$ Commented Jan 29, 2021 at 21:07
  • $\begingroup$ @AntonPetrunin My motivation for this question was the following: $\endgroup$ Commented Jan 29, 2021 at 21:52
  • $\begingroup$ mathoverflow.net/questions/382577/… $\endgroup$ Commented Jan 29, 2021 at 21:53

1 Answer 1


I think the answer is no as soon as your gradient vector field admits a saddle point where the divergence is non-zero.

Let $\omega$ denote the volume form associated to the Riemann metric. We have $$\mathrm{div}(X) \omega = X\cdot \omega$$ where $X\cdot \omega$ denotes the Lie derivative. The goal is to find positive functions $f$ and $g$ such that $$(fX)\cdot (g\omega) = X\cdot(fg)~\omega + (fg) \mathrm{div}(X)~\omega = 0~.$$ In other words, we want the function $h=\log(fg)$ to satisfy $$X\cdot h = -\mathrm{div}(X)~.$$

This is a dynamical question: we ask whether the function $\mathrm{div}(X)$ is a coboundary along the flow of $X$. Of course a gradient flow does not have very rich dynamics, but a saddle point is already too much for the following reason:

Assume $X$ has a saddle point. Then one can find sequences $(x_i)$ and $(y_i)$ which are bounded in $M\backslash S$ such that $y_i$ is on the trajectory of $x_i$ along the flow of $X$, and such that the trajectory from $x_i$ to $y_i$ is very long and spends most of its time very close to the saddle point $s$.

Assume now that we have $h:M\backslash S \to \mathbb R$ such that $\mathrm{div}(X)= X\cdot h$. Then $$\int_{x_i}^{y_i} \mathrm{div}(X) = h(y_i)-h(x_i)$$ is bounded independently of $i$. (Here $\int_{x_i}^{y_i} \mathrm{div}(X)$ denotes the integral of the function $\mathrm{div}(X)$ along the trajectory of $X$ from $x_i$ to $y_i$.)

But, on the other side, since this trajectory spends a very long time close to $s$, we have that $$\int_{x_i}^{y_i} \mathrm{div}(X)\underset{i\to +\infty}{\longrightarrow} \pm \infty$$ as soon as $\mathrm{div}(X)(s) \neq 0$. This is a contradiction.

  • $\begingroup$ Thank you for your answer. I think I am missing some thing: Put $f(x,y)=x^2-y^2$ then it has a saddle point at the origin and its divergence is zero: $\nabla f=2x\partial_x-2y\partial_y$. Right? $\endgroup$ Commented Feb 12, 2021 at 12:49
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    $\begingroup$ It is divergence free indeed, including at the saddle point $(0,0)$! My argument applies for instance if you take $g(x,y) = 2x^2 - y^2$, whose gradient has divergence $1$. $\endgroup$
    – Nicolast
    Commented Feb 12, 2021 at 16:10
  • $\begingroup$ Yes. But in your argument, there is no any restriction on saddle rotation $\lambda_1/\lambda_2$. So in principal it should work for every arbitrary saddle, including saddle ration=-1. So I think some thing is missing in your argument. Right? $\endgroup$ Commented Feb 12, 2021 at 17:56
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    $\begingroup$ To show that the integral of the divergence along a trajectory which spends a lot of time close to the saddle point diverges, I use that the divergence at the saddle point is non-zero (i.e. saddle ratio $\neq -1$ if you want). Your example shows that this condition is necessary. $\endgroup$
    – Nicolast
    Commented Feb 12, 2021 at 21:52
  • $\begingroup$ My sincere apology for not reading carefuly your answer. $\endgroup$ Commented Feb 12, 2021 at 21:57

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