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

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 $r>R$.

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Do you mean for $r$ sufficiently large? – Agol Oct 8 2011 at 21:52
Oh, yes. Thanks Agol, I've corrected my statement – unknown (google) Oct 8 2011 at 22:42
Don't have time to work out the details, but can't you use the Calabi-Yau lower bound on, say, $V(B(p, 4r))$ and the lower bound on Ricci to infer an isoperimetric inequality on all domains contained in $B(p, 2r)$. That and the Calabi-Yau lower bound on $V(B(p,r))$ implies a lower bound on the area of $\partial B(p,r))$. Not sure how the lower bound depends on $r$ though. – Deane Yang Oct 9 2011 at 0:15
Some evidence in favour: the analogue for horospheres (rather than distance spheres) is true. Specifically, for any Busemann function $b$ on $M$, the area of $b^{-1}(r)$ is eventually nondecreasing in $r$. See Lemma 20 in C. Sormani, "Busemann functions on manifolds with lower bounds on Ricci curvature and minimal volume growth," JDG 1998. (I don't think the theorem's hypothesis of exactly-linear volume growth is needed for this part of the conclusion.) – macbeth Dec 19 2011 at 1:15

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?]

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The constant $c=c(p,R)$ has to depend on the manifold of course. – unknown (google) Oct 9 2011 at 1:14