Finding hyperbolic metrics by approximation

Given a presentation $< X ; R >$ of a group $G$. Suppose we know for some reason that $G$ is the fundamental group of a three-dimensional finite volume manifold.

Then there is a injective group homomorphism $G\rightarrow Isom(\mathbb{H}^3)$. Mostows rigidity theorem then tells us that it is unique up to composition with inner automorphisms of $\mathbb{H}^3$ from the left (and probably automorphisms of $G$ from the right).

As usual to speficy a homomorphism from a group with a specific presentation to another group we have to pick a image for each generator, such that all relators get mapped to the neutral element. My hope is that there might be an algorithm that starts with some first choice of the images of the generators. Note that $Isom(H^3)$ can be viewed as a subgroup of $GL_4(\mathbb{R})$. Then I would like to iteratively minimize the sum of some norm of the images of the relators (hoping that it will converge to a homomorphism). Of course there are also noninjective homomorphisms (like the trivial one) so I cannot hope that the sequence always converges to an injective homomorphism. But maybe it does with high probability for a reasonably random first choice of generators.

Has someone already done this ? If so I am interested in the convergence properties. They might also depend on the given presentation. So one could also hope that any fundamental group of a hyperbolic 3-manifold has a nice presentation.

To my knowledge this hasn't been done in theory (although see Harriet Moser's thesis http://www.math.columbia.edu/~moser/). But it certainly has been done in practice by Jeff Weeks' program SnapPea. Note that $\mbox{Isom}(\mathbb{H}^3) = \mbox{PSL}(2, \mathbb{C}) = \Gamma$. So your source group $G = \pi_1(M^3)$ already has a very nice matrix group as a target. SnapPea assumes that the three-manifold $M$ is given as a triangulation. (I am sure that it is not easy to go backwards from a presentation of $G$ to a triangulation of $M$.) After tidying the triangulation (and drilling out a curve if necessary - but lets suppose that $M$ has a single torus boundary component) SnapPea gives all of the tetrahedra in the triangulation the same shape, namely that of the regular ideal tetrahedron. "Developing" in $\mathbb{H}^3$ turns shapes of tetrahedra into matrices, one per generator. Naturally this is not yet a representation of $G$ into $\Gamma$. The failure to be a representation is measured by the failure of the shapes to satisfy the Thurston gluing equations. SnapPea uses a multivariate version of Newton's method to find new shapes for the tetrahedra, hopefully converging to the discrete and faithful representation of $G$ into $\Gamma$.

Details, references, and more can be found in Moser's thesis. However, I'll add a final remark - the convergence properties of SnapPea's method certainly do depend sensitively on the initial triangulation. There are certain triangulations where SnapPea will consistently produce wrong answers. This is not a problem in practice -- you randomize the triangulation a few times and SnapPea typically starts to behave much better. But finding an actual algorithm appears to be difficult. So something mysterious going on.

The problem (of finding the discrete and faithful representation) is easy to solve even for very respectfully sized manifolds (say up to 100 tetrahedra). But why? I certainly don't know. You'd have to ask Thurston, Weeks, Hodgson or some other expert for an opinion.

• @Sam: I think you need a triangulation or something equivalent to it, like Heegaard diagram. If you start with an arbitrary presentation then you would not be find an algorithm for the word problem (even though, you know that one exists). For instance, take the free product of a hyperbolic 3-manifold group (with a nice presentation) with the trivial group that is given an "unrecognizable" presentation. Thus, in this case, one cannot hope for an algorithm for finding a discrete and faithful representation. – Misha Apr 16 '12 at 14:37
• Dear Misha - Hence the parenthetical remark near the beginning of my post. (If you will kindly read "not easy" to mean "impossible" ;) – Sam Nead Apr 16 '12 at 14:47
• @Sam: Dear Sam, yes, of course. By the way, to the best of my knowledge, nobody really knows why Snappea works so well and so fast. Casson's (unsuccessful) approach to the geometrization conjecture was an attempt to explain why. Maybe, if there is a combinatorial Ricci flow, it would explain it, but so far there was very little real progress in this direction. – Misha Apr 16 '12 at 21:10

You might be interested in this paper of Jason Manning. He proves (assuming the solution to the word problem in $G$---basic undecidability results tell you that it's necessary to assume something like this) that there is an algorithm to construct the discrete faithful hyperbolic representation of $G$.

I don't think the strategy is at all what you suggest, though. Instead, he builds the character variety

$\chi(G)=\mathrm{Hom}(G,SL_2(\mathbb{C}))//SL_2(\mathbb{C})$

and computes its decomposition into irreducibles. Mostow Rigidity implies that a discrete faithful representation is contained in a 0-dimensional component. This gives a finite list of representations to check. Now, for each of these, his algorithm attempts to either construct a fundamental domain using hyperbolic geometry, or to find a proof that the representation is not faithful or discrete.

So one could also hope that any fundamental group of a hyperbolic 3-manifold has a nice presentation.

This sounds extremely optimistic to me. (Well, it depends what you mean by 'nice'.)

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• Restating my comment more directly: If I give you $G$, written as a presentation of finite index supergroup of a surface bundle group, then there at least two reasons to hope one can find the hyperbolic structure algorithmically. For example, see the work of Helling and Menzel. Also, Thurston's proof that bundles (with pA monodromy) are hyperbolic is an iterative proof -- this is called the "double limit theorem". However... – Sam Nead Apr 16 '12 at 16:48