# Do balls of finite radius have finite volume?

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.

• 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 – Hyperkähler Jan 21 at 23:29

## 1 Answer

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.

Edit:

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 $$\{p\in S:\rho(p) is an open submanifold of infinite volume, since it is an infinite-sheeted cover of a punctured open ball. Additionally, one can show by constructing paths which loop arbitrarily tightly around the origin that $$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.

• A lower bound on Ricci curvature would suffices (see here) – Arctic Char Jan 22 at 2:41
• @ArcticChar: I am not so sure since in the BG inequality one usually assumes completeness. – Moishe Kohan Jan 22 at 3:31
• @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. – Kajelad Jan 22 at 5:03
• @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. – Kajelad Jan 23 at 1:14
• @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. – Kajelad Jan 23 at 4:37