I am doing some rank control about the fundamental group of a 3 dim hyperbolic orbifold. After cuting out the regular neighborhood of all the singularity, I get a manifold with imcompressible boundary. I can control the rank of this manifold from below using homology of the boundary and "half live half die theorem". But then I glue the singluarity back, this is equal to gluing a disk along n (n>1) times of simple cloesed curve on the imcompressible boundary. Will this operation reduce the rank of the fundamental group? Thanks very much!
1 Answer
I would guess that the rank could go down in general, but that something like the following should be true (and may well be, but my memory is a little foggy).
Let $N$ be your $3$-manifold obtained by deleting a neighborhood of the singular locus.
If all the boundary components of $N$ are tori, then I would expect that in a neighborhood of infinity in the hyperbolic Dehn filling space (maybe after deleting some hyperplanes) all of the fillings $N(\alpha)$ will have fundamental groups with the same rank as $N$. You should be able to make some sort of geometric limit argument. I think there is a theorem of Rubinstein to this effect concerning the Heegaard genus.
If $N$ has higher genus boundary components, you probably need to assume some things about how you are gluing in the singular handlebody. Here I would suggest looking at the work of Juan Souto on ranks of hyperbolic $3$-manifold groups.
Maybe Agol will come around and fill in my gaps.
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1$\begingroup$ That Agol might get a bit of a rep if he keeps coming around and doing that... $\endgroup$ Commented Oct 5, 2011 at 20:22