How do we know there are no more Deligne–Mostow/Thurston lattices?

Question

In the context of hypergeometric functions, Deligne and Mostow enumerated several lattices in complex hyperbolic space/the rank 1 Lie group $\operatorname{PU}(1,n)$ (see [1] and [2]). Thurston used the perspective of flat structures to get the same results [3] (and apparently corrected the Deligne–Mostow list by slightly extending it, according to another paper of Mostow). In the Thurston perspective, the idea is to enumerate the choices of cone angles (or equivalently, singular curvatures) for flat cone metrics on the sphere such that the completion of the relevant moduli space is an orbifold. In his appendix, Thurston stated

These examples were enumerated by a routine computer program, which checks all possibilities having a given least common denominator $q$. The enumeration was not rigorously verified (even though it should not be hard to do so and search more ‘intelligently’ at the same time) but was a simple check of all denominators through 999 in a few minutes of computer time.

How do we know that this enumeration is complete? Is there some condition I'm missing that tells us we won't find orbifold moduli spaces past denominator $q=999$?

How did Thurston do this enumeration so quickly? I've tried to run a similar computer search myself. As best I can figure, the enumeration requires generating all partitions of integers. There are many, many partitions of integers (apparently the number of partitions of $n$ grows like $\exp(\sqrt n)/n$). Thurston was doing this in the 1980's. In my computer search on a modern day laptop, I wasn't able to get past denominators in the 30's even with many hours of home computer time (though I'm certainly no expert programmer).

Additional thought: One way to cut down the search time is to realize that all of the boundary strata, which correspond to collision of two cone points, must necessarily be smaller moduli spaces that are also orbifolds. (This idea appears for intance as Lemma 2.4 in Mostow's later paper [4].) So e.g. you could try to enumerate all 5-part partitions, then examine only the 6-part partitions which are refinements of those, etc. I imagine this could speed up the search.

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[1] Deligne, P.; Mostow, G. D., Monodromy of hypergeometric functions and non-lattice integral monodromy, Publ. Math., Inst. Hautes Étud. Sci. 63, 5-89 (1986). ZBL0615.22008.

[2] Mostow, G. D., Generalized Picard lattices arising from half-integral conditions, Publ. Math., Inst. Hautes Étud. Sci. 63, 91-106 (1986). ZBL0615.22009.

[3] Thurston, William P., Shapes of polyhedra and triangulations of the sphere, Rivin, Igor (ed.) et al., The Epstein Birthday Schrift dedicated to David Epstein on the occasion of his 60th birthday. Warwick: University of Warwick, Institute of Mathematics, Geom. Topol. Monogr. 1, 511-549 (1998). ZBL0931.57010.

[4] Mostow, G. D., On discontinuous action of monodromy groups on the complex n-ball, J. Am. Math. Soc. 1, No. 3, 555-586 (1988). ZBL0657.22014.

Answer 1

Note that Thurston says

Mostow has rigorously enumerated examples by hand, so this table can be regarded as just a check.

That is effectively a claim that it suffices to check up to denominator $42$.

This answer is a sketch as to how Mostow might have done it.

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The denominators make things unnecessarily long-winded. Normalise by dividing the curvatures by $2\pi$: we have $\mu_1, \ldots, \mu_n$ each in interval $(0, 1)$ and summing to $2$. Let $r_{i,j} = (1 - \mu_i - \mu_j)^{-1}$. We further require that for all $1 \le i < j \le n$ such that $\mu_i + \mu_j < 1$ either $r_{i,j} \in \mathbb{N}$ or $(\mu_i = \mu_j) \wedge 2r_{i,j} \in \mathbb{N}$.

Wlog $\mu_1 \ge \cdots \ge \mu_n$, and then obviously $\mu_i \le \frac2i$ with equality only if $i = n$ and all the parts are equal.

Observe that for all $3 \le i < j$ we must have $\mu_i + \mu_j \le 1$ with equality only if $n = 4$ and $\mu_1 = \mu_2 = \mu_3 = \mu_4 = \frac12$. Proof is by contradiction: since $\mu_1 \ge \mu_i$ and $\mu_2 \ge \mu_j$ the total sum would otherwise exceed $2$.

By inverting the definition of $r_{i,j}$ we get $\mu_i + \mu_j = \frac{r_{i,j} - 1}{r_{i,j}}$.

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Case: $n = 5$. From the assumption of descending size of $\mu_i$ we get

• $\mu_5 \le \mu_4$ so $\frac{\mu_4 + \mu_5}2 \le \mu_4$
• $\mu_4 \le \mu_3$
• $\mu_3 \le \frac13(2 - \mu_4 - \mu_5)$

which can be combined to give $\mu_4 + \mu_5 \le \frac45$. Then $1 - \mu_4 - \mu_5 \ge \frac15$ so $r_{4,5} \le 5$.

By substitution, we get $\mu_3 \le \frac{r_{4,5} + 1}{3r_{4,5}}$ and can follow through to $r_{3,5} \le \frac{6r_{4,5}}{r_{4,5} + 1}$.

If $r_{4,5} > 2$ the constraint $\mu_4 \le \mu_3$ allows us to bound $\mu_3 + \mu_4$ away from $1$ and get $r_{3,4} \le \frac{3r_{4,5}}{r_{4,5} - 2}$. We have a finite number of values for $r_{4,5}, r_{3,5}, r_{3,4}$ and each element of the Cartesian product gives three simultaneous linear equations. From $\mu_2 \le \frac12(2 - \mu_3 - \mu_4 - \mu_5)$ we can filter out some of them because they violate the assumption $\mu_2 \ge \mu_3$. This leaves the following possibilities:

$$\begin{array}{ccc} \mu_3 & \mu_4 & \mu_5 \\ 3/10 & 3/10 & 3/10 \\ 11/30 & 3/10 & 3/10 \\ 9/20 & 3/10 & 3/10 \\ 1/3 & 1/3 & 1/3 \\ 5/14 & 5/14 & 13/42 \\ 3/8 & 3/8 & 7/24 \\ 7/18 & 7/18 & 5/18 \\ 2/5 & 2/5 & 4/15 \\ 9/22 & 9/22 & 17/66 \\ 5/12 & 5/12 & 1/4 \\ 11/26 & 11/26 & 19/78 \\ 3/7 & 3/7 & 5/21 \\ 13/30 & 13/30 & 7/30 \\ 7/16 & 7/16 & 11/48 \\ 15/34 & 15/34 & 23/102 \\ 4/9 & 4/9 & 2/9 \\ 5/12 & 1/3 & 1/3 \\ 53/120 & 43/120 & 37/120 \\ 5/14 & 5/14 & 5/14 \\ 11/28 & 5/14 & 5/14 \\ 3/8 & 3/8 & 3/8 \\ 7/18 & 7/18 & 13/36 \\ 2/5 & 2/5 & 7/20 \\ 9/22 & 9/22 & 15/44 \\ 5/12 & 5/12 & 1/3 \\ 7/18 & 7/18 & 7/18 \\ 2/5 & 2/5 & 2/5 \\ \end{array}$$

For each of these, the bound on $\mu_2$ turns out to give $\mu_3 + \mu_2 < 1$ and we could follow through to bound $r_{2,3}$ and enumerate and check all possible cases.

The remaining subcases of $n=5$ are $r_{4,5} \in \{\frac32, 2\}$.

If $r_{4,5} = \frac32$ then since it's non-integral we have $\mu_4 = \mu_5$, and we can solve to find $\mu_5 = \frac16$, $\mu_3 \in \{\frac16, \frac13, \frac12\}$. $\mu_5 + \mu_2$ is bounded away from $1$, so we can bound $r_{2,5}$ and enumerate and check all possible cases.

$r_{4,5} = 2$ is the tricky one. We have $\mu_4 + \mu_5 = \frac12$ and $\mu_3 \le \frac12$. That gives $\mu_5 \le \frac14$ and $\mu_2 \le \frac58$ so that $r_{2,5} \le 8$, but we need to bound $\mu_3 + \mu_4$ away from $1$ to pin the values down. $r_{3,5} \le 4$ so either $\mu_3 = \mu_5 = \frac14$ or $r_{3,5} \in \{2,3,4\}$. Take the non-equal subcases one by one in order of difficulty:

• $r_{3,5} = 4$. Then $\mu_3 + \mu_5 = \frac34$ so $\mu_5 \ge \frac14$, giving $\mu_5 = \mu_4 = \frac14$, $\mu_3 = \mu_2 = \mu_1 = \frac12$.

• $r_{3,5} = 3$. Then $\mu_3 + \mu_5 = \frac23$ so $\mu_5 \ge \frac16$, whence $\mu_4 \le \frac13$ and $\mu_3 + \mu_4 \le \frac56$.

• $r_{3,5} = 2$. Then $\mu_3 = \mu_4$. We must have $\mu_3 + \mu_4 < 1$ since otherwise $\mu_5 = 0$. Then $2r_{3,4} = \frac{2}{1 - 2\mu_4} \in \mathbb{N}$. Rearranging, $\mu_4 = \frac12 - \frac{1}{2r_{3,4}}$, so $\mu_5 = \frac{1}{2r_{3,4}}$ is an Egyptian fraction. Let $k = 2r_{3,4}$.

  If $r_{2,5} = 2$ then $\mu_2 = \mu_4$. We have a family parameterised by $4 \le k \in \mathbb{N}$: $\frac{k+4}{2k}, \frac{k-2}{2k}, \frac{k-2}{2k}, \frac{k-2}{2k}, \frac{2}{2k}$. When $k > 6$ we need $r_{1,5} = \frac{2k}{k-6} \in \mathbb{N}$, so we only need to consider $k \in \{4,5,6,7,8,9,10,12,18\}$.

  Otherwise we have variables $k$ and $2 < r_{2,5} \le 8$ which between them determine all of the parts. $\mu_2 + \mu_5 = 1 - \frac{1}{r_{2,5}}$. But $\mu_1 = 2 - \mu_2 - \mu_3 - \mu_4 - \mu_5 \ge \mu_2$, so $k \le \frac{3r_{2,5}}{r_{2,5} - 2}$ and we again have a finite number of cases to check.

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For $n \ge 6$ the bound $\mu_i \le \frac2i$ suffices to bound $r_{5,6}, r_{4,6}, r_{4,5}$, so it should be easier than for $n=5$. But it's even more straightforward to consider possible refinements of the finite list of solutions for $n=5$, and so on. If we choose to split $\mu_i$ then the smaller part will be at most $\tfrac12 \mu_i$; we choose $j \neq i$ to maximise $\mu_j + \tfrac12 \mu_i$ subject to $\mu_j + \tfrac12 \mu_i < 1$ for the most efficient search. This approach is implemented in this Python code, which runs in less than a second on a modernish desktop computer and produces the 94 configurations of Thurston's appendix.

Note also that $n \le 12$, since we get the bound $r_{12,13} \le \frac{78}{53} < \frac 32$.