Geometry/Topology Seminar
Wednesday, November 22, 2000, 1:00pm, 120 Physics
Hubert Bray (M.I.T.)
Volume Comparison Theorems Involving Scalar Curvature
Abstract:
Bishop's theorem says that an n-manifold, $n \ge 2$, with Ricci curvature greater than that of the standard, round n-sphere has less total volume. It is then natural to ask whether or not this is true for scalar curvature as well. The answer, however, is no for $n \ge 3$, which can be seen by considering the cross product metric on $S^{n-1} \times {\bf R}$, which can be scaled to have arbitrarily large scalar curvature and yet infinite volume.

However, this counterexample metric is quite far from the standard, round metric on Sn, and it turns out that for metrics sufficiently close to the standard metric g0 on S3, greater scalar curvature does imply less total volume. To be precise, suppose (M3,g) has scalar curvature $R(g) \ge R(g_0)$ and Ricci curvature $Ric(g) \ge \epsilon Ric(g)$.(Notice that this condition is equivalent to the hypothesis of Bishop's theorem for $\epsilon = 1$ but is weaker for $\epsilon < 1$.) Then we prove that there exists an $\epsilon_0 \in (0,1)$ such that for all $\epsilon \in [\epsilon_0, 1],$ the total volume $Vol(g) \le Vol(g_0)$.Furthermore, using techniques involving isoperimetric surfaces (surfaces which minimize area given a volume constraint), we are able to derive an explicit formula for the optimal value of $\epsilon_0$, which we note is between 0.134 and 0.135. More generally, we find that for any $\epsilon \in (0,1)$, $Vol(g) \le \lambda(\epsilon) Vol(g_0)$, and we compute the $\lambda(\epsilon)$which makes the inequality sharp. It is interesting to note that for $\epsilon \in (0,\epsilon_0)$, the Riemannian manifold with maximal volume is best described as a ``football metric,'' which approaches a cylinder in the limit as $\epsilon$ goes to zero.

Generated at 5:46pm Tuesday, April 23, 2024 by Mcal.   Top * Reload * Login