# Volume comparison under Ricci curvature upper bounds

Say I have a Hadamard $d$-manifold $M$ with an upper Ricci curvature bound of $-b^2$. Write the volume form in polar exponential coordinates at $p\in M$ as $V(r,\theta) \, dr \, d\theta$, and similarly write the volume form of the simply connected $d$-manifold with constant curvature $-b^2$ as $V_{const}(r)\, dr\, d\theta$ in exponential coordinates.

Is it true that the ratio $V(r,\theta)/V_{const}(r)$ is bounded away from zero?

There's a true analogous statement for lower Ricci bounds, see e.g. Eq 4.2 of http://mail.math.ucsb.edu/~wei/paper/06survey.pdf.

• I would be genuinely surprised if such result were true, as Lokhamp showed ("Metrics of Negative Ricci Curvature", Ann.Math. 140 (1994) 655-683) that any (paracompact, Hausdorff, connected) smooth manifold of dimension $d\geq 3$ admits a Riemannian metric with negative Ricci curvature, scalar curvature bounded above by -1 and finite volume. On the other hand, lower bounds on Ricci curvature may have strong implications for the topology of the underlying manifold, which stem from Bishop-Gromov-type volume comparison estimates such as the one you stated. – Pedro Lauridsen Ribeiro May 17 '16 at 6:32
• @Pedro, strictly speaking your comment does not apply directly since the OP required the manifold to be Hadamard. I have no opinion at this point as to whether the answer is affirmative or not. – Mikhail Katz May 17 '16 at 7:12
• Thanks very much for the comments! It's a bit surprising to me that not much is really known here, although it's true that the Lokhamp paper does suggest caution. Tracing through the estimates used in proving the analogue of the above for Ricci lower bounds, it seems like the nonreversible one is $| II |^2 \geq m^2 / (d-1)$, where $II$ is the second fundamental form and $m$ is mean curvature, both of a metric sphere. I guess I'm not sure how to make examples of metrics where this is far from an equality, but I haven't thought about it much yet. – biringer May 18 '16 at 20:57