Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This may be embarassingly simple, but I can't see it.

Let $M$ be a Riemannian manifold of dimension $n$; fix $x \in M$, and let $B(x,r)$ denote the geodesic ball in $M$ of radius $r$ centered at $x$. Let $V(r) = \operatorname{Vol}(B(x,r))$ be the Riemannian volume of $B(x,r)$. It seems to be the case that for small $r$, $V(r) \sim r^n$, i.e. $V(r)/r^n \to c$ with $0 < c < \infty$. How is this proved, and where can I find it?

Given a neighborhood $U \ni x$ and a chart $\phi : U \to \mathbb{R}^n$, certainly $\phi$ has nonvanishing Jacobian, hence (making $U$ smaller if necessary) bounded away from 0. So $\operatorname{Vol}(\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))) \sim r^n$. But I do not see how to relate the pullback $\phi^{-1}(B_{\mathbb{R}^n}(\phi(x), r))$ of a Euclidean ball to a geodesic ball in $M$.

share|cite|improve this question
up vote 5 down vote accepted

A Google search for the title of the question finds this article where the first five coefficients in the series expansion of the volume (in powers of $r$) are computed. The first term is the same as the Euclidean volume (proportional to $r^n$, in other words); then come higher-order corrections depending on the curvature.

share|cite|improve this answer
By the way, concerning your idea for a proof: instead of mapping a ball from an arbitrary chart, maybe you could map a ball from the tangent space using the exponential map, where you know more about the Jacobian. (I'm going to bed now, and haven't thought this through properly, but someone else can surely say something wise about this.) – Hans Lundmark Nov 10 '10 at 21:58
The reference has a stronger result, which I think will be helpful as well. Thanks! – Nate Eldredge Nov 10 '10 at 22:45

It's simple, all right. As I realized not long after posting (and as Hans also suggested), the key is the exponential map. The tangent space $T_x M$ gets an inner product space structure from the Riemannian metric; we can isometrically identify it with $\mathbb{R}^n$. Now $\exp_x : \mathbb{R}^n \to M$ is a diffeomorphism on some small ball $B_{\mathbb{R}^n}(0,\epsilon)$; on this ball, straight lines map to length-minimizing geodesics (see Do Carmo, Riemannian Geometry, Proposition 3.6), and thus Euclidean balls map to geodesic balls of the same radius. Taking $\epsilon$ smaller if necessary, we can assume the Jacobian of $\exp_x$ is bounded away from $0$ and $\infty$ on $B_{\mathbb{R}^n}(0, \epsilon)$; thus for $r < \epsilon$ we have that $\operatorname{Vol}(B(x,r))$ is comparable to $\operatorname{Vol}(B_{\mathbb{R}^n}(0,r)) \sim r^n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.