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

I am given a smooth Riemannian submanifold of dimension $d$ embedded in $\mathbb{R}^D$ with condition number $1/\tau$ (a formal definition of condition number is on Page 3 of this paper

I would like to know how to calculate an upper bound on the $d$-dimensional volume of a ball (of radius $r \ll \tau$) around a point on the manifold intersected with the manifold. Lemma 5.3 in the paper referenced above gives a lower bound on this volume.

Proposition 6.3 gives a bound on the geodesic distance $d_M$ between any two points that are a distance $r$ apart in $\mathbb{R}^D$ as $$d_M(p,q) \leq \tau - \tau \sqrt{1 - \frac{2r}{\tau}} := d_{\max}$$ if $r < \frac{\tau}{2}$.

I think a valid upper bound is given by $$\mathrm{vol}(B(r,p) \cap M) \leq v_d d_{\max}^d$$ where $v_d$ is the volume of a unit ball in $\mathbb{R}^d$. How would I go about proving this (or an equivalent statement)?

share|cite|improve this question
That's a strange upper bound: why does not it involve the radius $p$? – user53153 Jan 8 '13 at 0:34
$p$ is the center (a point on $M$) of the ball and $r$ is the radius. – Sivaraman Jan 8 '13 at 3:14
OK, why does not it involve the radius $r$? – user53153 Jan 8 '13 at 3:16
It does. It involves $d_{\max}$ which depends on the radius. In particular if $r \ll \tau$ then $d_{\max} = O(r)$. – Sivaraman Jan 8 '13 at 6:09

Your Answer


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

Browse other questions tagged or ask your own question.