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

Suppose $M$ is a hyperbolic $3$-manifold whose fundamental group has rank $r.$ What is the best (lower) bound on the volume of $M?$ Similar question for rank of $H_1.$ There is a bunch of papers of Culler and Shalen on related subjects, but they seem to care about "small" manifolds, whereas this question is more on the asymptotic dependence.

share|improve this question

1 Answer 1

up vote 10 down vote accepted

In Counting arithmetic lattices and surfaces, Mikhail Belolipetsky, Tsachik Gelander, Alex Lubotzky and Aner Shalev prove the following Theorem.

Let $H$ be a connected simple Lie group of real rank one. Then there is an effective computable constant $C=C(H)$ such that for any lattice $\Gamma < H$ we have $r(\Gamma)\le C\cdot \mathrm{vol}(\Gamma\backslash H)$, where $r(\Gamma)$ is the minimal number of generators of $\Gamma$.

Applying this to $SO(1,3)$ gives $\mathrm{vol}(M) \ge 1/C\cdot r(\pi_1(M))$ in your case.

share|improve this answer
    
I interpreted "rank" as "minimal number of generators", I hope this is what you meant. –  Aurel Jan 26 at 18:55
    
Thanks! This is obviously sharp (in terms of growth rate) in dimension $2,$ but less obviously so in dimension $3...$ –  Igor Rivin Jan 26 at 19:03
4  
The growth rate is sharp in dimension 3 as well. For example, take cyclic covers over one component of the Whitehead link complement. The rank (of $H_1$) and the volume both grow linearly. –  Ian Agol Jan 26 at 19:31
1  
Note that on 2204 of the published paper, a fairly explicit estimate is given in terms of the Margulis constant. The Margulis constant is estimated in a paper of Meyerhoff. However, when the rank of $H_1$ is at least $3$, one may assume the Margulis constant is $\geq \log(3)/2$. So this should give an improvement in this case. –  Ian Agol Jan 26 at 19:44
    
@IanAgol Ah, OK. I will need to ponder what this means in the grand scheme of things... –  Igor Rivin Jan 27 at 3:15

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.