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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 $m ∈ M$.

edited Feb 2 2011 at 1:15

asked Feb 1 2011 at 11:20

sam
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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 2011 at 14:25

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@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 2011 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 2011 at 1:05

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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 2011 at 16:42

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I am not sure about convexity radius but the injectivity radius of symmetric spaces can be found at arxiv.org/abs/math/0703521v1 and arxiv.org/abs/math/0609627v1 and the answer is not simple. – Igor Belegradek Feb 2 2011 at 17:01

ε-δ · Answer 1 · 2011-02-01 17:40:06Z UTC

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

answered Feb 1 2011 at 17:40

ε-δ
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So, half the injectivity radius? – Igor Rivin Feb 1 2011 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 2011 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 2011 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 2011 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 2011 at 23:38

Convexity radius of a Lie Group

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