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I have a specific concrete example of a complete, finite volume, cusped hyperbolic 3-manifold $M$, and I am trying to determine whether or not $M$ is a cover of the figure-8 knot complement (call this $F_8$).

The manifold $M$ is described combinatorially, and all the numbers work out; for instance I know $M$ and $F_8$ are commensurable. But $M$ is big: it is built out of 630 regular ideal cubes, or equivalently, 3150 regular ideal tetrahedra, so it would have to be a 1575-sheeted cover. This is of course a finite problem, but too big for SnapPea (or any other software, I imagine) to handle. By the way, I know that $M$ is not a link complement in $S^3$.

I am looking for ideas for (a) how to prove $M$ IS NOT a cover of $F_8$, if indeed it is not; and (b) how to prove $M$ IS a cover of $F_8$, if indeed it is.

For (b) I am looking for symmetries of $M$ to mod out by, and I am making slow progress, but for (a) I have no ideas.

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What descriptions of the manifold do you have? If you have the triangulation in SnapPea form, perhaps you could post it on-line somewhere? –  Ryan Budney Jan 17 '11 at 5:37
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@Ryan: I have essentially one description. (The others I know follow from it.) Namely, start with the Cartesian product of 3 copies of the complete graph $K_7$. This is a cube complex. Then take the subcomplex consisting of the (closed) cubes that are disjoint from the fat diagonal (in the configuration space sense). What you have now is the "3-point discretized configuration space of $K_7$." This is a manifold away from the vertices. The final step is to delete the vertices; the result is $M$. There are 630 cubes because ${7 \choose 2}{5 \choose 2}{3 \choose 2} = 630.$ –  aaron Jan 17 '11 at 6:16
    
When you say "commensurable", what do you mean? Do you mean the volume of $M$ is a multiple of the volume of $F_8?$ –  Igor Rivin Jan 17 '11 at 15:14
    
One simple way that might prove that it doesn't cover $F_8$ would be to show that the links of the vertices do not cover the link of a vertex of the figure 8 knot. These have a well-defined conformal type, and you can estimate the index of the cover. Then construct the corresponding covers of the figure 8 knot cusp, and try to show that they are not conformally equivalent to the cusps of your example. –  Ian Agol Jan 17 '11 at 18:18
    
@Igor: I think in general being a rational (or integral) multiple of the volume does not ensure commensurability, but in this case (as Bill described) there is a tiling of $H^3$ by regular ideal tetrahedra whose symmetry group contains both an index 2 subgroup with quotient $F_8$ and an index 3150 subgroup with quotient $M$. So $F_8$ and $M$ are commensurable. @Ian: I thought of that, actually. Turns out the cusp shapes are compatible. –  aaron Jan 17 '11 at 19:04
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up vote 18 down vote accepted

Edited, in light of description of manifold in comments (at end)

Added #2: The answer is no, details at the end

In principle this is doable, but caution is necessary. Given a manifold tiled by ideal simplices, its fundamental group is a subgroup $H$ of the full group $G$ of isometries of finite index in the tiling, as is the fundamental group $H_0$ of the figure eight knot complement. You can determine whether $H$ is conjugate in $G$ to a subgroup of $H_0$ by a finite check: label the two simplices of the figure eight knot complement, assigning them vertex orderings, and just try developing this pattern into the pattern for triangulation of $H$. If you transcribe the information suitably into a computer program, the check should be instantaneous (on human timescale). Snappea has a good system for handling the combinatorics of how simplices are glued together; this could be copied or used. (or you could do it by hand).

But a caution: $H_0$ might be conjugate within $PSL(2,\mathbb C)$ but not within $G$. To start, the tiling of $H^3$ by regular ideal cubes has two subdivisions into tilings by regular ideal simplices, obtained by inscribing a tetrahedron in the cube in one of two ways. More generally, since these particular groups are arithmetic, they have a large commensurability group, which is $PGL(2, \mathbb Q(\sqrt{-3}))$. If your manifold is not a covering of the figure eight knot, to prove it you need to consider conjugation within this larger group. It's known how to understand all the maximal lattices commensurable with an arithmetic group. This translates into a countable sequence of patterns, one for each prime ideal in the Eisenstein lattice $O_{-3}=\mathbb Z[1/2 + \sqrt{-3}/2$, given the triangulation of $H^3$ by ideal regular tetrahedra, of grouping them into a repeating pattern of larger polyhedra which can be retriangulated a second way. The retriangulation of the cube is the one associated with 2.

Whether or not these retriangulations are compatible to your given manifold depends on congruence conditions. I.e. find generators for the group in $PSL(2,O_{-3})$, and look what subgroup they generate mod various prime ideals. This gives a finite set of ways to retriangulate --- then you need to continue, obtaining a finite list of subgroups of $PSL(2,O_{-3})$ that are conjugate to $H$ within $PSL(2, \mathbb Q(\sqrt(-3)))$. For each of these, you would need to check whether they are contained in $H_0$.

It's reasonably likely your manifold is a finite-sheeted covering of the figure eight knot complement, in which case, most of this isn't necessary---you'll find the covering without such an elaborate search. If it's not a covering, you might be able to prove that it's not more easily by looking at the length spectrum.

If you have a reasonably conceptual description of your manifold, then it might also be possible to work out the answer by pure thought, without having to do a lot of combinatorial searching.

Added I didn't notice your comment giving a concrete description when I wrote the above. It's a nice manifold. To recap the description: for each triple of disjoint edges on the complete graph K7, there is a cube. The faces of the cube correspond to the ends of edges; each face is glued to the cube obtained by moving the corresponding endpoint to the unoccupied vertex. It's easy to check that the edges have order 6. There are $7!/2^3$ cubes.

First question: can the cubes be consistently divided into 5 tetrahedra, by choosing for each cube a set of 4 vertices at distance 2 apart (on the 1-skeleton of the cube), so that the subdivisions of the faces match up? No: think of the operation of reflecting a cube in a pair of opposite faces. This corresponds to walking a single edge around a triangle in K7. It comes back after 3 times, with its ends reversed; the reflected subdivision does not match. This means there is a 2-fold cover of the manifold where the subdivision would work consistently. This 2-fold covering space appears more likely to be a covering of the figure eight knot complement, but it requires checking.

I think it should be possible to prove from this that your manifold does not cover the figure eight knot complement, by showing that your group is conjugate to a subgroup of the stabilizer of an edge but not the stabilizer of a vertex in the action of $PSL(2, \mathbb Q(\sqrt{-3}))$ on the 5-way-branching tree (Tits building) corresponding to the prime $\left <2\right >$, but I haven't thought through the details.

Additional details The prime 2 does not split in the ground field, so $O_{-3}/\left <2\right >$ is the field with 4 elements. The cusps of the Bianchi group are at elements of the projective line of $\mathbb Q(\sqrt{-3})$; reduction mod $\left <2\right >$ maps them to the 5 elements of the projective line of $F_4$. You can think of the structure of the corresponding Tits building or Bass-Serre tree as follows: from a tiling by regular ideal tetrahedra, if you pick a tetrahedron, it and its four facial neighbors make an ideal cube; by reflections, this extends to a grouping of the tiling by tetrahedra into cubes. There are 5 ways to do this, depending which tetrahedron in the original cube you pick to be the central one, and these can be distinguished algebraically by naming which of the 5 elements of $\mathbb P^1F_4$ is omitted form the vertices of the central tetrahedron.

For any tiling by cubes, you can subdivide it into regular ideal tetrahedra in two ways. Thus, the full group of automorphisms of the cubical tesselation intersects the full group of automorphisms of the tetrahedral tesselation in index 2 in the first, and index 5 in the second (a parabolic subgroup mod $<2>$). To get from a parabolic subgroup mod $\left <2\right >$ to the edge stabilizer, you add an element of $PGL(2,O_{-3})$ that is not conjugate to $PSL$, something like diagonal matrix $(2,1)$.

So: it's clear that the group as described stabilizes an edge of this tree, but does not stabilize a vertex. Since the full commensurability group, $PGL_2(\mathbb Q(\sqrt{-3}))$, acts on the tree, this implies it is not conjugate to a subgroup that stabilizes a vertex. Since the figure eight knot group stabilizes a vertex, it is not conjugate to a subgroup of the figure eight knot group, i.e. it is not a covering space.

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@Bill: I hadn't noticed that the division into simplices can't be done consistently. That seems helpful. I will attempt to decipher your suggestion, although I don't yet understand the action you refer to. Is this a Bass-Serre tree? –  aaron Jan 17 '11 at 19:16
    
Yes, it's the same thing as a Bass-Serre tree. –  Bill Thurston Jan 17 '11 at 20:58
    
@Bill: I've said it before and I'll say it again: you rock. Thanks a lot -- I was definitely not going to think of this argument. –  aaron Jan 18 '11 at 11:29
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