Area of distance sphere in manifold with Ricci $\ge 0$.

2011-10-08T21:23:26Z

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:

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 $r>R$.

Answer by Rbega for Area of distance sphere in manifold with Ricci $\ge 0$.

2011-10-09T00:55:39Z

This is clearly false, just consider the cylinder

$$ R_t \times S_{\theta} $$

with the product metric

$$g_\alpha=dt^2+\alpha^2 d\theta^2.$$

This is a flat metric so $Ric_{g_\alpha} = 0$. On the other hand, for $r>>\alpha$, it is easy to see $Area(\partial B_r)<8\pi \alpha$. Since $\alpha$ is arbitrary there is no uniform lower bound.

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].

[As an aside I can't seem to get math blackboard fonts to work anyone else have a problem with this?]

Area of distance sphere in manifold with Ricci $\ge 0$. - MathOverflow

Area of distance sphere in manifold with Ricci $\ge 0$.

Answer by Rbega for Area of distance sphere in manifold with Ricci $\ge 0$.