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## Closed hyperbolic manifold with right-angled fundamental domain

What is an example (as simple as possible, please!) of a closed hyperbolic three-manifold with a right-angled polyhedron as fundamental domain?

If we allow cusps then the Whitehead link or the Borromean rings are good answers (fundamental domains have not too many sides and the gluings can be understood). If we allow orbifolds then the (4,0) filling on all components of the Borromean rings is a good answer. (This is carefully explained in the NotKnot video.) I tried to answer the original question, using SnapPea (see also SnapPy), to build a cyclic, four-fold, manifold cover of the Borromean orbifold. This was not successful. It is "obvious" that the resulting manifold has a right-angled fundamental domain, but the domain is huge (two octagons, eight hexagons, 16 pentagons) and it is hard (for me) to describe the face pairings or check that I haven't messed up in some way.

More generally, I guess that one could use Andreev's theorem to build as "simple as possible" right-angled polyhedra and then look for low index torsion free subgroups of the resulting reflection groups. However I don't know how large an index we'd have to sacrifice or even if the resulting manifold will have a right-angled domain...

Edit: I've accepted bb's answer below, because the first paper cited gives the required construction. However, I didn't understand this until I asked an expert off-line, who puckishly told me that this was equivalent to the four color theorem! Here is the construction:

Suppose that $R$ is a right-angled three-dimensional hyperbolic polyhedron. Note that the edges of $R$ form a cubic graph in $\partial R$. Let $G_R$ be the subgroup of isometries of $H^3$ generated by reflections in the sides of $R$. Now, by the four-color theorem there is a four coloring of the faces of $R$ (so no two adjacent faces have the same color). This defines surjective homomorphism from $G_R$ to $(Z/2)^4$. Let $\delta = (1,1,1,1) \in (Z/2)^4$ and let $\Delta$ be the preimage in $G_R$ of the subgroup generated by $\delta$. Note that $\Delta$ has index eight in $G_R$ and is torsion-free. Finally, a fundamental domain for $\Delta$ is a obtained by gluing eight copies of $R$ around a vertex.

Other cryptic tidbits I was fed: There is such a hyperbolic manifold in dimension four, coming from the 120-cell. This is the only example known. There are no such examples in dimensions five and higher. See Bowditch and Mess, referring to Vinberg and Nikulin.

Further edit: In fact the argument using the four-color theorem can be found in the second paper of Vesnin, as cited by bb.

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Seifert-Weber dodecahedral space is pretty simple. How do you want to measure "simple"? Volume? SWDS has pretty large volume. – Ryan Budney Mar 10 2010 at 22:17
Doesn't the Seifert-Weber dodecahedral space have a dodecahedral fundamental domain with a 72 degree dihedral angle? I tried to cut that into right-angled pieces, but I haven't succeeded yet. – Douglas Zare Mar 10 2010 at 22:20
Ah, right. Why is 90 degrees special to Sam Nead? – Ryan Budney Mar 10 2010 at 23:30
@Ryan - I was roughly measuring complexity by the number of faces, edges, and vertices on the boundary of the polyhedron. I vaguely guess that this is proportional to the volume when the polyhedron is right-angled? 90 degrees is important for a potential application (which I don't really understand at all yet). – Sam Nead Mar 11 2010 at 13:09

A. Yu. Vesnin has some articles on these Lobell manifolds. The first one describes how to construct arbitrarily many non-isometric closed hyperbolic manifolds from one right-angled polyhedron.

1. "Three-dimensional hyperbolic manifolds with a common fundamental polyhedron" Math. Notes 49 (1991), no. 5-6, 575--577

2. "Three-dimensional hyperbolic manifolds of Löbell type" Siberian Math. J. 28 (1987), no. 5, 731--733

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The smallest volume right-angled polyhedron is the dodecahedron. But by an observation of Best, this cannot be the fundamental domain for a 3-manifold: any face pairing would have to have orbits of edges in fours, but there are 30 edges. The next thing to try would be the second Lobell polyhedron L(6), which has two opposite hexagonal faces, and 12 pentagonal faces. The two hexagonal faces get glued together, and then one must glue the resulting 6 hexagonal faces together (each one coming from two right-angled pentagons). But there are only 12 dihedral corners left, which won't work since there must be 8 dihedral angles per vertex. So I don't think this works either (although this is not a proof, since the paired pentagons don't have to be preserved). This has an index 8 manifold cover called the Lobell manifold.

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