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$\newcommand{\Ric}{\text{Ric}}$ 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.

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  • $\begingroup$ 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
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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 $n \leq 2$.

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