MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Is there a nice formula/method to find the convexity radius of a matrix Lie group (the manifold can be noncompact) ?

Edited based on comments:

Definition : Convexity Radius (Berger - Panoramic View of Riemannian Geometry) The convexity radius of a Riemannian manifold $M$ is the infimum of positive numbers $r$ such that the metric open ball $B(m,r)$ is convex for every $m ∈ M$.

share|cite|improve this question
It might be helpful to spell out the question in a little more detail, including motivation and a background source for the notion of convexity radius. I'm not a specialist in any of this but can easily find on MathSciNet one relevant-looking reference (maybe not what you are looking for and maybe not accessible to you): MR0458335 (56 #16538), Cheeger, Jeff; Ebin, David G., Comparison theorems in Riemannian geometry. North-Holland Mathematical Library,Vol. 9. North-Holland Publishing Co., Amsterdam-Oxford; American Elsevier Publishing Co., Inc., New York, 1975. viii+174 pp. – Jim Humphreys Feb 1 '11 at 14:25
@Jim: I have no clue what "convexity radius" is in this context (there is the convexity radius of a function, but not of a space), but a note: Cheeger and Ebin's book has been reprinted by AMS/Chelsea, at a fraction of the NH price. – Igor Rivin Feb 1 '11 at 15:29
According to Berger (Panoramic View of Riemannian Geometry) 'A set in a Riemannian manifold is (totally) $convex$ if for any pair of points in this set, $every$ segment connecting these two points belongs to this set'. – sam Feb 2 '11 at 1:05
The notion of convexity radius of a Riemannian manifold is well estabished, but since there is some confusion I recall that $r$ is the convexity radius if every $r$-ball $B_r(p)$ is convex in the sense that $B_r(p)$ contains the minimizing geodesic between any two points of $B_r(p)$. (This should not be confused with the notion of a totally convex set which is assumed to contain any not necessarily minimizing geodesic between its points). As stated the question makes no sense because a Lie group can admit many different metrics. – Igor Belegradek Feb 2 '11 at 16:42
I am not sure about convexity radius but the injectivity radius of symmetric spaces can be found at and and the answer is not simple. – Igor Belegradek Feb 2 '11 at 17:01

For simply connected Lie group with bi-invariant metric it is half of the distance to the cut locus.

share|cite|improve this answer
So, half the injectivity radius? – Igor Rivin Feb 1 '11 at 18:36
@ε-δ/Igor Thanks. Do you have some references regarding this. How about for a Lie group with left or right invariant metric? – sam Feb 2 '11 at 1:09
A simply connected Lie group with bi-invariant riemannian metric is necessarily of the form $K\times\mathbb{R}^n$, with $K$ compact. – BS. Feb 2 '11 at 11:11
I am also interested in a reference (or a proof) justifying that "for simply connected Lie group with bi-invariant metric it is half of the distance to the cut locus". – Igor Belegradek Feb 2 '11 at 16:44
I do not have a reference. But the curvature tensor is parallel, so the minimal curvature radius behaves as $\sin{2\pi x/l}$, where $l$ is the injectivity radius... – ε-δ Feb 3 '11 at 23:38

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.