$\newcommand{\Ric}{\text{Ric}}$ Let $M$ be a smooth closed oriented Riemannian surface.

I am searching for a reference (or a sketch of proof) for the following inequality:

$$ \int_M | \nabla V|^2 \ge \int_{M} \Ric(V,V)=\int_{M} K|V|^2, \tag{1}$$

for every vector field $V \in \Gamma(TM)$, where $ \nabla$ is the Levi-Civita connection, and the integration is against the Riemannian volume form. ($K$ is the Gauss curvature).

I guess some kind of Bochner identity is needed. I am also interested to know if this inequality holds for manifolds of higher dimensions.

BTW, specializing for the case of the round $2$-sphere, we get

$$ \int_{\mathbb{S}^2} | \nabla V|^2 \ge \int_{\mathbb{S}^2} |V|^2. \tag{2}$$

A proof of this specific case can be found here.

  • $\begingroup$ Question: I wish there was $|K|$ instead of $K$: can we have a non-vacuous inequality when the curvature is nonpositive? $\endgroup$ – Seub Sep 21 '18 at 18:05
  • $\begingroup$ That is a very good question. I also wondered about it, but I don't know the answer. I guess we can ask this as a separate question. As a starting point, we need to find out if there are non-zero parallel vector fields on a surface of negative curvature. $\endgroup$ – Asaf Shachar Sep 21 '18 at 19:20
  • $\begingroup$ thank you. I'd really like to know, so maybe I'll ask the question. Regarding your starting point, unless I'm mistaken, a nonzero parallel vector field cannot exist on a surface, even locally, unless the curvature is zero. I guess you could see it as a consequence of this: mathoverflow.net/questions/16850/… $\endgroup$ – Seub Sep 21 '18 at 19:33
  • $\begingroup$ I just remembered that a closed surface with negative Ricci curvature does not have non-zero parallel vector fields (see e.g. math.stackexchange.com/questions/2607606/…). However, the flat Torus does have a non-zero parallel vector field. (e.g. $\frac{\partial}{\partial \theta_1}$) So, a Poincare-type inequality cannot hold on the flat Torus. However, we can still ask whether it holds on a surface with Ricci curvature that is everywhere negative...(Please let me know if you ask this) $\endgroup$ – Asaf Shachar Sep 21 '18 at 20:23
  • $\begingroup$ Yes, but cf my previous comment: "a nonzero parallel vector field cannot exist on a surface, even locally, unless the curvature is (everywhere) zero." $\endgroup$ – Seub Sep 21 '18 at 20:29

You probably figured it out by now but since it's such a nice inequality I'll just point out that it follows from the identity in OP's other question :

How to prove $\int_M-\text{Ric}(V,V)+ |\nabla V|^2 =\int_M \frac{1}{2}|L_Vg|^2-\big(\text{div}V\big)^2$?

Since $(\mathcal{L}_Vg)_{ij} = \nabla_iV_j + \nabla_jV_i$, on a manifold of dimension $n$ we find $$\frac12|\mathcal{L}_Vg|^2 \geq \frac{1}{2n}|\text{trace}(\mathcal{L}_Vg)|^2 = \frac{2}{n}(\text{div}V)^2.$$ Thus the RHS of that identity is bounded below by $(\frac{2}{n} - 1)(\text{div}V)^2$ which is non-negative when $n \leq 2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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