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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.

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.

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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...