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Oct 18, 2010 at 16:40 history edited HJRW CC BY-SA 2.5
Acknowledged mistake, added something hopefully useful
Oct 18, 2010 at 15:56 comment added Greg Kuperberg He's not allowing gluing with ridges, only gluing along boundary.
Oct 18, 2010 at 15:32 comment added Bruno Martelli It might be worth saying that if $S= \{M_1,\ldots , M_k\}$ consists of manifolds with toric boundaries, then any hyperbolic manifold generated by $S$ has volume not bigger than the maximum Gromov norm of the $M_i$'s. Therefore you cannot get all hyperbolic 3-manifolds with finitely many cusped ones. Things however are more complicate if we admit (as we do) higher-genus boundary.
Oct 18, 2010 at 15:26 comment added Bruno Martelli You are right that $M_8$ is universal, but its coverings are not obtained by gluing copies of $M_8$. If you construct a manifold by gluing copies of $M_8$ then you never get a hyperbolic manifold, because it contains many incompressible tori. (On the other hand, all coverings of $M_8$ are hyperbolic.)
Oct 18, 2010 at 15:19 history answered HJRW CC BY-SA 2.5