Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a very stupid question: I often see that the volume entropy of a compact Riemmannian manifold with negative curvature coincide with the Hausdorff dim of the limit set or Patterson sullivan measure on the boundary.

But the volume entropy is not invariant under scaling of the measure, and I assume the dimension of limit set is. So what is the correct normalization to make the claim above true?

share|improve this question
    
"Hausdorff dim of" what "limit set"? $\:$ –  Ricky Demer May 17 '13 at 21:42
3  
In this context, limit set is the ideal boundary of the universal cover. Classically, one uses this in the case of Kleinian groups. Then the right formula is that volume entropy equals Hausdorff dimension of the conical limit set of the group (which could be less than dimension of the full limit set). See Nicholls' book "Ergodic theory of discrete groups". Normalization in variable curvature was worked out in papers by Besson, Courtois and Gallot (maybe also Hammenstadt) and you only have an inequality. Equality if I remember correctly is only in curvature -1 case. –  Misha May 17 '13 at 22:33
1  
See also math.univ-lyon1.fr/~remy/smf_sec_18_09.pdf and references therein. –  Misha May 17 '13 at 22:33
    
Thanks Misha, the answer is very helpful. While I'll try to look it up in the BCG papers, do you happen to remember in which direction the inequality is, between entropy and dimension, for non-constantly curved manifolds? –  Jerry May 17 '13 at 23:44
add comment

1 Answer 1

For the geodesic flow on the unit tangent bundle of a compact manifold with nonpositive curvature, a classical result of Manning asserts that the topological entropy of the geodesic flow equals the volume entropy of the universal cover. This is true in variable (nonpositive) curvature.

For the geodesic flow on the unit tangent bundle of a negatively curved manifold, the topological entropy is equal to the critical exponent of the fundamental group of the manifold (result due to Otal-Peigné in full generality. You just have to assume that the curvature is bounded between two negative constants).

The critical exponent of the group is always equal to the Hausdorff dimension of the conical (also called radial) limit set of the group acting on the boundary at infinity (in variable negative curvature). I don't remember if the result in this generality is due to Sullivan (he writes in curvature -1, but the arguments are often general), or maybe to Bishop-Jones.

For general noncompact manifolds, the critical exponent(=topological entropy) is strictly smaller than the dimension of the boundary. In this situation it is not reasonable to expect an equality with the volume entropy of the universal cover, which is the common value of the topological entropies of all geodesic flows of all compact quotients.

An interesting phenomenon arises in the finite volume case. One could expect that the result of Manning could still be true. In fact, if the curvature varies too much in the cusps, the equality fails to happen. See here a work of Dalbo Peigne Picaud Sambusetti http://www.lmpt.univ-tours.fr/~peigne/fichiers/volent-versionfinale2.pdf

For your question of normalization: I believe that when you scale the metric of the manifold, you scale also the natural distances on the boundary, and therefore Hausdorff dimension of limit sets.

share|improve this answer
add comment

Your Answer

 
discard

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.