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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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    $\begingroup$ Do you mean for $r$ sufficiently large? $\endgroup$
    – Ian Agol
    Commented Oct 8, 2011 at 21:52
  • $\begingroup$ Oh, yes. Thanks Agol, I've corrected my statement $\endgroup$
    – user16750
    Commented Oct 8, 2011 at 22:42
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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Oct 9, 2011 at 0:15
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    $\begingroup$ 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.) $\endgroup$
    – macbeth
    Commented Dec 19, 2011 at 1:15

2 Answers 2

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The answer is YES.

Let $b\colon M\to \mathbb R$ be a Busemann function for a ray $\gamma$ from $p$, so that $b(p)=0$ and $b(x)\le 0$ for any $x\in \gamma$.

Set $$L_t=b^{-1}(t)\ \ \text{and}\ \ L^-_t=b^{-1}(-\infty,t].$$ Note that the sublevel sets $L_t^-$ are mean curvature concave for all $t$. In particular, any area minimizing hypersurface in $L_t^-$ with the boundary in $L_t$ lies in $L_t$.

Fix $t<0$. From above, $$\mathop{\rm area}\partial B(p,r)\ge\mathop{\rm area}(\partial B(p,r)\cap L_t^-)\ge \mathop{\rm area}( B(p,r)\cap L_t)\ge \mathop{\rm area}( B(p,R)\cap L_t);$$ i.e., the inequality holds for $c(p,R)=\mathop{\rm area}( B(p,R)\cap L_t)$.

It remains to choose $t$ and $R$ so that $\mathop{\rm area}( B(p,R)\cap L_t)>0$; $R=2$ and $t=-1$ will do the job.

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    $\begingroup$ P.S. I realized that there is yet simpler proof. Take a point $q$ on the distance $2{\cdot} r$ from $p$. Denote by $W_r$ the set of all minimizing geodesics from $q$ to the points in $B_\varepsilon(p)$. The standard comparison gives the lower bound on the area of $W_r\cap\partial B_r(p)$. $\endgroup$
    – Anton Petrunin
    Commented Aug 18, 2013 at 21:52
  • $\begingroup$ can you give more detail of how to estimate $W_r$? $\endgroup$
    – J. GE
    Commented Oct 6, 2013 at 22:11
  • $\begingroup$ @user3922, Consider the radial map $f\colon W_r\to \partial B_{2r}(q)$. The lower Ricci urvature gives a bound for the $(n-1)$-Jacobian of $f$. It remains to apply the coarea formula. $\endgroup$
    – Anton Petrunin
    Commented Oct 7, 2013 at 1:56
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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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    $\begingroup$ The constant $c=c(p,R)$ has to depend on the manifold of course. $\endgroup$
    – user16750
    Commented Oct 9, 2011 at 1:14

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