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The Mostow Rigidity Theorem is phrased in terms of a relationship between isometries and isomorphisms of fundamental groups, which raises an obvious question. Given the fundamental group of a complete finite-volume hyperbolic manifold of dimension $> 2$, is it possible to reconstruct the hyperbolic manifold?

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    $\begingroup$ Yes, but it's quite hard and is not algorithmic to the best of my knowledge. By the way, you should specify what you mean by "reconstruct." $\endgroup$ – Moishe Kohan Apr 24 at 16:37
  • $\begingroup$ What do you mean by reconstruct? Do you have an algorithmic question in mind? If so, how do you encode the group, and how do you encode the manifold? (a possible output is a triangulation but it doesn't say what the metric is) $\endgroup$ – YCor Apr 24 at 16:37
  • $\begingroup$ I was thinking more generally about a mathematical construction rather than an effective algorithm, so any answer is fine. I don't have any particular application in mind; I was more thinking about why rigidity theorems are often stated in isomorphism form. $\endgroup$ – Cameron Zwarich Apr 24 at 16:49
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    $\begingroup$ At a theoretical level the Mostow rigidity theorem precisely say that $\Gamma$ determines the manifold $X_\Gamma$ up to isometry. If I understand correctly, the question is to have a more "constructive" description of $\Gamma\mapsto X_\Gamma$? Maybe asking the manifold only should be more reasonable, because if you can output the metric easily, basically you reprove Thurston's hyperbolization conjecture (in this case in dim 3, from my very rough understanding the metric is output by the Hamilton-Perelman Ricci flow procedure). $\endgroup$ – YCor Apr 24 at 17:07
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    $\begingroup$ ... reading Toffee's answer, I understand that something is doable without reproving Thurston-Hamilton-Perelman... I understand (again, roughly) that if we know beforehand that hyperbolization exists, then it can be done in some kind of effective way, by "retrieving" the relevant representation $\Gamma\to\mathrm{PSL}_2(\mathbf{C})$. $\endgroup$ – YCor Apr 24 at 17:33
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How is your fundamental group $\Gamma$ given to you? As a presentation in terms of generators and relations? Here is an answer for hyperbolic $3$-space that can probably be generalized to $\mathbb{H}^n$, $n \ge 4$, with some effort.

In short, you compute the $\mathrm{SL}_2(\mathbb{C})$ character variety. The relevant ideas and facts I use below are in Culler and Shalen's famous paper [CS].

Here is an algorithm.

  1. Compute the algebraic set $\mathrm{Hom}(\Gamma, \mathrm{SL}_2(\mathbb{C}))$ from your presentation. This is naturally an affine algebraic set in $\mathbb{C}^{4 d}$, where $\Gamma$ has $d$ generators.
  2. Then, there is an explicit set of elements $\sigma_1, \dots, \sigma_n \in \Gamma$ for which the trace function $$ t(\rho) = \left(\mathrm{tr}(\rho(\sigma_1)), \dots, \mathrm{tr}(\rho(\sigma_n))\right) \in \mathbb{C}^n $$ has image the $\mathrm{SL}_2(\mathbb{C})$ character variety, $X(\Gamma)$. See Proposition 1.4.1 in [CS].
  3. Compute this algebraic set in $\mathbb{C}^n$ and find its irreducible components (maybe in Macaulay?).
  4. Mostow rigidity tells you that a discrete and faithful representation $\rho : \Gamma \to \mathrm{SL}_2(\mathbb{C})$ determines an isolated point of $X(\Gamma)$. You might get several isolated points in $X(\Gamma)$ from different lifts from $\mathrm{PSL}_2$ to $\mathrm{SL}_2$ and from different Galois conjugates of the discrete and faithful representation, not to mention possibly other random isolated points. (Small nitpick: In the finite volume, noncompact case you need to also cut out the equations that traces of peripheral elements are all $2$ in order to get isolated points. You need to cut away deformations related to Dehn filling.)
  5. There is a standard way described in [CS] to algorithmically lift a point on $X(\Gamma)$ back to a representation $\rho$. Basically you choose a lift so that your first generator is upper-triangular with the right trace and $1$ in the upper right entry. You make the second generator lower-triangular of the right trace, where now the lower left entry is going to depend on the trace of the product of the two generators, etc. Compute such a $\rho$ for each isolated point of $X(\Gamma)$.
  6. Now, compute a Dirichlet or Ford domain in $\mathbb{H}^3$ for each $\rho(\Gamma)$ from your list of isolated points. One of these will terminate to give you a fundamental domain for the complete structure.

Now you have the representation associated with the complete structure and a fundamental domain. So if that is your desired notion of "reconstructing" the manifold, there you go.

A last remark on the finite volume case. Your presentation oracle also needs to enumerate the conjugacy classes of $\mathbb{Z}\oplus \mathbb{Z}$ subgroups - these are precisely the peripheral subgroups. (The manifold is homeomorphic to the interior of a compact manifold with boundary a union of essential tori.) As described in [CS], Thurston proved that the dimension of the character variety at the complete structure equals the number of cusps, and the complete structure is cut out by making each peripheral subgroup unipotent (in fact, setting the traces of these elements all equal to $2$ cuts out the desired point). You can probably make finding these algorithmic in just the presentation. The rank of the abelianization of $\Gamma$ is an upper bound for the number of peripheral subgroups (by Poincaré duality), so one knows when enough subgroups have been found. This point is irrelevant in higher dimensions.

[CS] Culler, Marc; Shalen, Peter B. Varieties of group representations and splittings of 3-manifolds. Ann. of Math. (2) 117 (1983), no. 1, 109–146.

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  • $\begingroup$ Thanks for this great answer. It's more than what I was expecting to get. $\endgroup$ – Cameron Zwarich Apr 25 at 15:17
  • $\begingroup$ the rigidity result which gives discrete+faithful => isolated in character variety would be local (Calabi--Weil) rigidity rather than strong (Mostow) rigidity. $\endgroup$ – Jean Raimbault Apr 26 at 14:56
  • $\begingroup$ Have the isolated points that do not come from the monodromy representation or its Galois conjugates ever been looked at? I can't even think of an example. $\endgroup$ – Jean Raimbault Apr 26 at 15:03
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    $\begingroup$ @JeanRaimbault That's right about local rigidity, except for the fact that local rigidity fails for nonuniform lattices in $\mathrm{SL}_2(\mathbb{C})$. Local rigidity only holds once you deem the parabolic subgroups must stay parabolic (i.e., traces of peripheral elements are $\pm 2$). $\endgroup$ – Toffee Apr 26 at 16:13
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    $\begingroup$ @JeanRaimbault As for other isolated points, I don't know an example off the top of my head. You could probably build one with a dominant map between closed hyperbolic $3$-manifolds that isn't a covering. Such an example is probably sufficiently complicated that cranking out the character variety isn't practical though. There are known examples of dominant maps between knot complements, so you see the canonical component of the target knot complement in the character variety of the other one. I can't remember one off-hand, but papers by Boyer and Zhang are a likely source. $\endgroup$ – Toffee Apr 26 at 16:13

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