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.
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asked Jan 26, 2014 at 15:28
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$\begingroup$ Is there an opposite inequality, i.e. a bound on the volume in terms of the rank (minimal number of generators)? I am a total novice in this field so perhaps I am unaware of some canonical counterexample. $\endgroup$– NWMTCommented Jun 27, 2016 at 19:31
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1 Answer
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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.
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$\begingroup$ I interpreted "rank" as "minimal number of generators", I hope this is what you meant. $\endgroup$– AurelCommented Jan 26, 2014 at 18:55
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$\begingroup$ Thanks! This is obviously sharp (in terms of growth rate) in dimension $2,$ but less obviously so in dimension $3...$ $\endgroup$ Commented Jan 26, 2014 at 19:03
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4$\begingroup$ 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. $\endgroup$– Ian AgolCommented Jan 26, 2014 at 19:31
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1$\begingroup$ 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. $\endgroup$– Ian AgolCommented Jan 26, 2014 at 19:44
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$\begingroup$ @IanAgol Ah, OK. I will need to ponder what this means in the grand scheme of things... $\endgroup$ Commented Jan 27, 2014 at 3:15