MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

For a orientable three manifold M with totally geodesic boundary, this inequality is true. Because the rank of (fundemantal group of boundary M)=rank (homology group of boundary M ) then we use the "half live half die" theorem to get the theorem. But in the orbifold case, we do not have such good things. After passing to a finite manifold cover. I can prove that rank(the fundamental group of boundary M)< 4 rank(fundmental group of M)

If I change the numeber 4 to 2,will this still be true???

share|cite|improve this question
What do you mean by "rank"? – Igor Rivin Oct 5 '11 at 17:05
rank(G) is the least number of elements in G that can generate G. – Lin Jianfeng Oct 5 '11 at 17:12
What if the boundary is disconnected? Do you want to take the sum of the ranks of the boundary components? This would work in the torsion-free case by your observation. – Ian Agol Oct 6 '11 at 3:41
I am very sorry, my question is a little misleading. I should not use the term "hyperbolic oblifold". But a "hyperbolic obrifold with totally geodesic boundary". That means the boundary is a totally geodeic 2-hyperbolic orbifold. we can double it to get a really closed orbifold. We can assume that the boundary is connected. – Lin Jianfeng Oct 6 '11 at 5:58
up vote 3 down vote accepted

One may obtain an estimate improving your factor of 4 to a factor of 3.

The ranks of hyperbolic 2-orbifolds were computed by Zieschang et al. If $\partial\mathcal{O}$ has genus $g$ and $p$ cone points, then they show that $rank(\pi_1\partial\mathcal{O})\leq 2g+p-1$, except in the case $p=0$, one has $rank(\partial\pi_1(\mathcal{O}))=2g$ (of course, one may deduce this estimate directly by thinking about the punctured case). The same argument (half-lives, half-dies) applies in that case (as Igor observes), so I'll assume $p>0$.

A theorem of Sullivan shows that the deformation space of geometrically finite structures on $\mathcal{O}$ is parameterized by the Teichmuller space of $\partial{\mathcal{O}}$. This follows from the theory of quasiconformal deformations of Kleinian groups. Now, one follows the proof of the Ahlfors finiteness theorem. If $rank(\pi_1\mathcal{O})=k$, then the space of deformations of representations of $\pi_1\mathcal{O}$ into $PSL_2(\mathbb{C})$ up to conjugacy has $\mathbb{C}$-dimension $\leq 3k-3$ (this follows by computing the dimension of the variety of representations, and using that the conjugacy action is faithful since the generators are non-commuting). This is also the dimension of the space of geometrically finite reps., since these are structurally stable.

On the other hand, the Teichmuller space of $\partial\mathcal{O}$ has complex dimension $3g-3+p$, so we get $3g-3+p\leq 3k-3$, or $g+p/3\leq k$. From the rank computation above, then we get $\frac13 rank(\pi_1\partial\mathcal{O})\leq 2g/3+p/3-1/3 \leq g+p/3 \leq k$. Obviously the worst estimate holds when $g=0$. One might be able to improve this result taking into account the relators.

share|cite|improve this answer
Thanks very much!!! I think 3 is a very elaborate estimate. I have no idea can this estimate be improved. I will think about it more carefully. Thank you very much! Your answer is very inspiring for me. – Lin Jianfeng Oct 6 '11 at 17:00

The statement is true when $\partial O$ is a surface without cone points, since the underlying topological space of an orientable 3-orbifold $O$ is a manifold $|O|,$ and the natural map from $O$ to $|O|$ induces a surjective homomorphism on fundamental groups. On the other hand, that map is identity on the boundary.

share|cite|improve this answer
Is it identity on the boundary?? I think if there are some singular points on the boundary, some torsion elements in the fundemantal group of orbifold will be quotiented? – Lin Jianfeng Oct 6 '11 at 0:34
I was assuming that the boundary had no orbifold points, correct. – Igor Rivin Oct 6 '11 at 9:36
WHy did someone downvote this??? – Igor Rivin Oct 6 '11 at 9:37
thanks very much! I also think that if the boundary has some orbifold point, it may be a little complicated. – Lin Jianfeng Oct 6 '11 at 13:00
@Agol: perhaps remarking this would have been a more appropriate course of action. – Igor Rivin Oct 6 '11 at 17:33

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.