# Converse to Bishop-Gromov Inequality

Is the converse to the Bishop-Gromov Inequality true?

In other words, if, for a complete $n$-dimensional Riemannian manifold $M$, there is $k \in \mathbb{R}$, such that defining $V_k(R)$ to be the size of a ball of radius $R$ in the standard space $S^n_k$, for any $p \in M$ and $R\geq 0$ we have that

$$vol_M(B_R(p)) \leq V_k(R)$$

then is it true that on $M$, $Ric \geq (n-1)k$?

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A very vague reason for this result not to hold is that, if it were true, this could have been used as a definition of Ricci curvature bounded form below for metric (or metric measured) spaces. In fact defining such bounds for Ricci curvature on metric spaces appears to need the machinery of optimal transport, as shown by Sturm, Lott and Villani, math.univ%2lyon1.fr/homes%2www/villani/Cedrif/P12.CIME.pdf and hal.archives-ouvertes.fr/docs/00/51/89/74/PDF/ohta.pdf are nice surveys on these questions. See also, arxiv.org/pdf/math/0701886. –  Thomas Richard Jan 20 '11 at 21:41
Thanks, this was my motivation for asking the question - I think that the B-G inequality is preserved under G-H convergence, so the space of metric spaces satisfying the B-H ineq. with a fixed n,k is compact, so I was confused as to why this didn't just solve the problem. –  Otis Chodosh Jan 21 '11 at 18:31

This is false for $n=3$, and for $k<0$. One can get a different pinching condition, $R(g)g-Ric \geq -4g$ which is weaker than $Ric \geq -2g$ ($R(g)$ is the scalar curvature), and which gives the same upper bound on volume as Gromov's inequality (for a manifold with negative curvature or no cut points). The idea is that in the proof of Gromov's inequality, one can integrate out the curvature tangent to the sphere of radius $r$ using the theorem egregrium and Gauss-Bonnet, giving a weaker curvature condition. There are metrics which satisfy the first curvature condition but not the Ricci lower bound. For example, there are homogeneous negatively curved metrics on $R^3$ which in one direction are foliated by totally geodesic hyperbolic planes of curvature $-K_1^2$, and in a perpendicular direction are foliated by totally geodesic hyperbolic planes of curvature $-K_2^2$ for $0 < K_1\neq K_2>0$ (this is a double warped product, and is Thurston's "ninth geometry" of the form $dr^2+e^{2K_1r}dx^2+e^{2K_2r}dy^2$). One obtains a counterexample for the choice of $K_1 =5/4,K_2=1/2$.

Added computations: The curvature operator of this metric may be normalized to have sectional curvatures $-K_1^2, -K_2^2, -K_1K_2$. If we assume that $K_1\geq K_2$, then the minimal curvature is $-K_1^2$. Under the same assumption, the minimal eigenvalue of the Ricci operator is $-K_1^2-K_1K_2$. The minimal eigenvalue of $Rg-Ric$ is $-2K_1^2-K_1K_2-K_2^2$. We then see that for $K_1=5/4, K_2=1/2$, the minimal eigenvalue of $Ric$ is $-35/16<-2$, and the minimal eigenvalue of $Rg-Ric$ is $-4$.

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Shouldn't it be even easier to construct counterexamples in higher dimensions, say $n = 4$, using a product metric? –  Deane Yang Jan 20 '11 at 20:16
It seems that your curvature condition survives under nice connected sum with thin necks. I.e. it can not give the same upper bound on volume as in Bishop inequality. It may hold locally and your counterexample seems to be OK, BUT you can not do global estimates with your curvature condition... –  Anton Petrunin Jan 20 '11 at 22:47
@ Deane: maybe, although I wouldn't expect a product to have as large volume growth, and I'm not sure how to estimate it. –  Ian Agol Jan 21 '11 at 0:59
@Anton: Nice observation - I wasn't sure if the estimate worked if there were cut points. How do you check that the curvature condition holds under connect sum? –  Ian Agol Jan 21 '11 at 1:50
@Agol - Thanks, this answers my question! @Deane Yang, could you provide such a counterexample if you know one? I'd be interested in the simplest possible such example. –  Otis Chodosh Jan 21 '11 at 18:34

If it is true for all $R,$ then the inequality follows from the infinitesimal calculation (as $R$ goes to $0.$)

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Igor, is this really right? It doesn't seem like there are enough variations to recover the full Ricci tensor at $p$. –  Deane Yang Jan 20 '11 at 15:44
It's a good calculation to check, but it sort-of suggests the opposite final answer. The first non-trivial derivative at $R = 0$ gives you the scalar curvature of the manifold, not the Ricci curvature. However, this is not a complete answer because we need a closed manifold by hypothesis. Possibly also the question is bad unless the manifold is simply connected. –  Greg Kuperberg Jan 20 '11 at 16:06