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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???

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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.

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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.

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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

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