# Is every gradient vector field a divergence free vector field?

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?

• Yes --- just choose a Riemannian metric that is invariant with respect to the flow generated by $X$. It exists since the singularities are removed. Commented Jan 29, 2021 at 19:57
• @AntonPetrunin Yes I see you use gradient property here to ensure the flow is geodesible. Commented Jan 29, 2021 at 20:05
• @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. Commented Jan 29, 2021 at 21:07
• @AntonPetrunin My motivation for this question was the following: Commented Jan 29, 2021 at 21:52
• mathoverflow.net/questions/382577/… Commented Jan 29, 2021 at 21:53

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.

• 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? Commented Feb 12, 2021 at 12:49
• 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 $1$. Commented Feb 12, 2021 at 16:10
• 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? Commented Feb 12, 2021 at 17:56
• 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. Commented Feb 12, 2021 at 21:52
• My sincere apology for not reading carefuly your answer. Commented Feb 12, 2021 at 21:57