# Counterexample to volume comparison inequality assuming only scalar curvature bound?

The Gromov-Bishop volume comparison theorem says that if we have a lower bound for the Ricci curvature on $(M,g)$, then its geodesic ball has volume not greater than the geodesic ball with the same radius in the corresponding space form. Since the proof only uses the bound for the Ricci curvature in the radial direction, I am sure there is a counterexample if we merely assume a lower bound for the scalar curvature. However, I couldn't construct any counterexample, and couldn't find any reference either. Any idea?

• Take a product $\mathbb{H}^2 \times \mathbb{S}^2$, where the sphere has small radius so that the product has positive scalar curvature. This has infinite diameter, but the corresponding space form with the same scalar curvature has finite diameter. – Ian Agol Jun 9 '14 at 4:17
• Yes, I know that there are many non-compact manifolds with positive scalar curvature (such as long cylinder), but I want to compare the volumes of their geodesic balls, i.e. with radius smaller than the minimum of the rwo injectivity radii. – user50396 Jun 9 '14 at 4:52
• Volume of geodesic balls in one is uniformly bounded while in the other it is arbitrarily large. – Misha Jun 9 '14 at 18:02
• What I mean is, say, take $M=\mathbb{H}^2\times \mathbb{S}^2$, where $\mathbb{S}^2$ is the unit sphere. Since $\mathbb {S}^2$ has diameter $\pi$, any geodesic lying in a slice $\{p\}\times \mathbb {S}^2$ fails to be minimizing beyond $\pi$, so I do not consider the distance ball of radius $\ge \pi$ to be a geodesic ball. In other words, in this example, I only care about the volume of (geodesic) ball of radius less than $\pi$. – user50396 Jun 9 '14 at 22:59

Specifically, comparing a metric $g$ with the spherical metric $g_0$, which has $R_0 = 6$ and $Ric_0 = 2 g_0$, one requires $R_g \geq 6$ and $Ric_g \geq 2\epsilon g$, where $\epsilon<1$ is a given constant (which can be estimated).