Detecting a cover of the figure-8 knot complement 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.
 A: 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.
