Area of distance sphere in manifold with Ricci $\ge 0$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:49:03Z http://mathoverflow.net/feeds/question/77558 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77558/area-of-distance-sphere-in-manifold-with-ricci-ge-0 Area of distance sphere in manifold with Ricci $\ge 0$. unknown (google) 2011-10-08T21:23:26Z 2013-04-30T22:22:00Z <p>Let $M$ be a open complete manifold with Ricci curvature $\ge 0$. By a theorem of Calabi and Yau, the volume growth of $M$ is at least of linear. I am wondering whether the following statement is true:</p> <p>Let $p$ be any fixed point in $M$ and $B(p, r)$ be the distance ball of radius $r$ in $M$. Then for any given $R>0$, there exists a constant $c=c(p,R)>0$ such that $Area(\partial B(p, r))\ge c(p,R)$ for any $r>R$.</p> http://mathoverflow.net/questions/77558/area-of-distance-sphere-in-manifold-with-ricci-ge-0/77571#77571 Answer by Rbega for Area of distance sphere in manifold with Ricci $\ge 0$. Rbega 2011-10-09T00:55:39Z 2011-10-09T01:21:23Z <p>This is clearly false, just consider the cylinder </p> <p>$$R_t \times S_{\theta}$$</p> <p>with the product metric </p> <p>$$g_\alpha=dt^2+\alpha^2 d\theta^2.$$ </p> <p>This is a flat metric so $Ric_{g_\alpha} = 0$. On the other hand, for $r>>\alpha$, it is easy to see $Area(\partial B_r)&lt;8\pi \alpha$. Since $\alpha$ is arbitrary there is no uniform lower bound. </p> <p>Maybe you need a uniform lower bound on the injectivity radius? (I'm not an expert on comparison geometry so don't know off the top of my head if this would suffice) [Edit: Or maybe this can only happen if the metric splits off an isometric euclidean factor].</p> <p>[As an aside I can't seem to get math blackboard fonts to work anyone else have a problem with this?]</p>