3
$\begingroup$

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?

$\endgroup$
6
  • 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

2
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.