I'm concerned with the following

Proposition: If a compact manifold $M$ satisfies $$Rc + \textstyle\frac{1}{2}\mathcal{L}_Xg = \lambda g $$

where $\lambda$ is a constant (i.e. $M$ is a compact Ricci soliton), then in fact $$Rc + \nabla^2f = \lambda g $$

so $X = \nabla f + K$ where $K$ is a Killing vector.

(Quick note: I use $\nabla f$ to indicate grad $f$. I know, I know; technically $\nabla f=df$, but $\nabla f$ is too good as notation to waste on something that already has a name, and grad $f$ is unforgivably grotesque.)

One can see this fact quoted in, for example, the second paragraph of this paper: http://www.mathem.pub.ro/bjga/v19n1/B19-1-cd-924.pdf

The reference made there is to Perelman's first paper on the Ricci flow: https://arxiv.org/pdf/math/0211159.pdf

Perelman never comes out and states this theorem explicitly as far as I know, but the outline of his reasoning for it seems to occur in Remark 3.2 on page 9. Apparently it hinges on the monotonicity of his entropy functional, but I'm not much of an expert on this stuff.

I had the following idea for an alternate approach to this proposition:

First, solve the elliptic equation $$div(X)=\Delta f$$ for the function f. (In coordinates this is $\nabla_i X^i=g^{ij}\nabla_i\nabla_j f$, just so my conventions are clear.) Now write $$X=\nabla f+(X-\nabla f)=\nabla f+K$$

It is clear that $div(K)=0$. Given that we know that the above proposition is true, it must be the case that $K$ is a Killing vector. (This is because a Killing vector is divergence-free, and the decomposition of a vector field $X$ into a gradient and a divergence-free vector field is unique, which is easy to see.)

Unfortunately, I haven't been able to prove that this divergence-free vector field $K$ is a Killing field. One attempt I made begins by integrating: $$\int |\textstyle\frac{1}{2}\mathcal{L}_K g|^2 dV = \int <K^{\flat}, div(Rc+\nabla^2 f-\lambda g)>dV$$ $$=\int<K^{\flat}, div\nabla^2 f>dV$$ $$= \int <K^{\flat}, Rc(\nabla f)> dV$$ $$= -\textstyle\frac{1}{2}\int \mathcal{L}_K g(K, \nabla f) dV $$

(I've omitted details, but I used a bunch of integration by parts, $div(K)=0$, the second Bianchi identity, and the main equation. The details are for the reader if they are interested.).

So, for example, to prove $\mathcal{L}_K g=0$ it is sufficient to prove $\mathcal{L}_K g(K, \nabla f)=0$, though I can't see any reason why this might be true either.

My question is: Does anyone have any suggestion as to how we might prove $K$ is Killing? I would prefer that the proof be "elliptic" in the sense of relying on the equations as they are and not introducing the parabolic methods of the Ricci flow as Perelman did. Any thoughts at all would be welcome.


1 Answer 1


There is an "elliptic" proof in this paper by Eminenti, La Nave, Mantegazza, see Theorem 3.1. It does still use Perelman's $\mathcal{W}$ entropy (but it does not use the Ricci flow, it just uses a minimization argument).

The authors there ask in Problem 3.2 whether there is a proof that does not use any "arguments related to Ricci flow".

  • $\begingroup$ Thanks, I actually chanced upon this paper just a couple days ago. Given that the proof provided is "just" a calculus of variations argument, I suppose it fits into the parameters of an "elliptic" proof as I requested, though without knowing Perelman's developments in the Ricci flow it's hard to imagine (for a mortal like me) coming up with a proof like this. Andrzej Derdzinski also has a slightly more elementary proof in an unpublished manuscript which I obtained by emailing him this week. The key step is the surjectivity of a certain elliptic operator, but ... $\endgroup$ Apr 14, 2017 at 0:32
  • $\begingroup$ ... Derdzinski (in his own words) "claim[s] no credit for it [the idea]." This is altogether too humble in my assessment, but he seems to feel that the essential insight follows from the chain of logic in Perelman's first Ricci flow paper. $\endgroup$ Apr 14, 2017 at 0:39
  • $\begingroup$ are you talking about Theorem 6.1 here people.math.osu.edu/derdzinski.1/crs.pdf ? $\endgroup$
    – YangMills
    Apr 14, 2017 at 3:39
  • $\begingroup$ Yes, I looked for that document online and couldn't find it so I finally wrote to Prof. Derdzinski. Apparently my internet search skills need work. $\endgroup$ Apr 14, 2017 at 4:35

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