Let $M$ be a Riemannian manifold. For a point $p\in M$ and $r\ge0$, let $B_r(p)$ denote the closed ball of radius $r$ around $p$, i.e. the ball consisting of all points in $M$ that have distance less than or equal to $r$ from $p$ in the metric induced by the Riemannian metric. Do such balls always have finite volume? If not, what is a counter-example?

In case such a ball is compact, it necessarily has finite volume, because sufficiently small coordinate charts have finite volume and the ball could then be covered with finitely many of these. If $M$ is complete, the Hopf-Rinow theorem implies that these balls are compact, so a counter-example would have to be incomplete, but I don't know anything further.

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    $\begingroup$ This is interesting, I wonder if the bounds offered by Kodani at least shows that there is a lower bound? projecteuclid.org › euclid.kmj AN ESTIMATE ON THE VOLUME OF METRIC BALLS 1 ... - Project Euclid $\endgroup$ – Hyperkähler Jan 21 at 23:29

Consider the open unit ball in $\mathbb{R}^2$ with polar coordinates $(r,\theta)$ and a metric of the form $g=dr^2+f^2(r)d\theta^2$, where $f$ is a smooth positive function chosen so that $g$ is well behaved at the origin. The set $B_r(0)$ is the same as in the Euclidean metric, since radial geodesics are unchanged, but the volume differs; a simple computations shows $$ \operatorname{vol}(B_r(0))=2\pi\int_0^{\min(1,r)}f(\rho)d\rho $$ Thus $\operatorname{vol}(B_1(0))$ can be made to diverge by choosing a suitable $f$ which diverges as $r\to 1^-$.

I haven't computed it, but I suspect the curvature of such a metric also diverges as $r\to 1^-$, and that this specific behavior can be controlled by a lower bound on the curvature.


Without any kind of completeness assumptions, one can construct even more pathological counterexamples. Here's one which is flat and simply connected.

Let $S$ be the universal cover of the punctured Euclidean plane the with projection $\pi:S\to\mathbb{R}^2\setminus\{0\}$ (equipped with the pullback metric), and let $\rho:S\to\mathbb{R}$ be the "radial" function defined by $\rho(p)=\|\pi(p)\|$. The set $\{p\in S:\rho(p)<r\}$ is an open submanifold of infinite volume, since it is an infinite-sheeted cover of a punctured open ball. Additionally, one can show by constructing paths which loop arbitrarily tightly around the origin that $d(p,q)\le\rho(p)+\rho(q)$. Thus, any ball $B_r(p)$ in $S$ with $r>\rho(p)$ has infinite volume.

With this kind of local pathological behavior in mind, I don't think the size of geodesic balls can be controlled by bounding the curvature alone.

  • $\begingroup$ A lower bound on Ricci curvature would suffices (see here) $\endgroup$ – Arctic Char Jan 22 at 2:41
  • $\begingroup$ @ArcticChar: I am not so sure since in the BG inequality one usually assumes completeness. $\endgroup$ – Moishe Kohan Jan 22 at 3:31
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    $\begingroup$ @ArcticChar That inequality requires the existence of minimizing geodesics, so that $B_r(p)\subseteq \exp_p(B_r(0))$ (completeness is sufficient). I've added another counterexample which illustrates what can go wrong. $\endgroup$ – Kajelad Jan 22 at 5:03
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    $\begingroup$ @Thorgott $dr$ isn't defined at the origin, strictly speaking, but we can choose $f$ so that $g$ extends smoothly to the origin, e.g. by setting $f(r)=r^2$ for $r\in(0,\epsilon)$. You can write $g$ in Cartesian coordinates if you want to avoid this subtlety. $\endgroup$ – Kajelad Jan 23 at 1:14
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    $\begingroup$ @Thorgott $f$ only modifies lengths perpendicular to the radial geodesics, since $d\theta(\partial_r)=0$ and thus $g(\partial_r,\partial_r)=1$ independent of $f$. It's the length of the geodesic circles around the origin that diverges. $\endgroup$ – Kajelad Jan 23 at 4:37

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