Classifying spaces and geometric obstructions for Artin groups

The explanation

Take two generators, a and b, with one rule: aba = bab. This is the braid group on three strands. Swap strands 1 and 2, then 2 and 3, then 1 and 2 again. You get the same braid as swapping 2 and 3, then 1 and 2, then 2 and 3. An Artin group allows such a rule for each pair of generators. A label m says both alternating words of length m are equal; the label ∞ imposes no ruleSource 1. Adding the rule that each generator squared is 1 gives the Coxeter groupSource 1. For a and b with label 3, the Coxeter group has 6 elements, like the symmetries of a triangle.

The Salvetti complex is a space with one cell of dimension |T| for each spherical subset TSource 2. A subset is spherical when its Coxeter group is finite. For the braid example, the subsets ∅, {a}, {b} and {a, b} are all spherical. So the space has 1 point, 2 loops and 1 disc. Its Euler characteristic is 1 − 2 + 1 = 0. With label ∞ instead, {a, b} is not spherical. The space is two loops at a point, with Euler characteristic 1 − 2 = −1. The K(π,1) conjecture asks that this space be aspherical. That means its universal cover can shrink to a point, so all higher homotopy groups vanish.

Deligne proved the spherical case in 1972Source 3. Charney and Davis covered FC type and dimension at most twoSource 4. Paolini and Salvetti covered affine typesSource 5. Huang and Przytycki covered dimension at most threeSource 6. The general case stayed open. The first paper proves it for every Coxeter matrix on a finite setSource 7. Labels may be finite or infinite, and diagrams may be disconnectedSource 7.

The second paper asks about geometry. A CAT(0) space is one whose triangles are at least as thin as flat ones. A geometric action is proper and cocompact on such a space. Haettel recorded the expectation that every finite-rank Artin group has oneSource 8. Brady and Crisp showed an obstruction in two dimensions can vanish in threeSource 9. So a counterexample must work in every dimension. The paper proves there is an Artin group on 116 generators, with labels in {2, 3, ∞}, that has no geometric actionSource 10.

The third paper studies parabolic subgroups, which are conjugates of subgroups generated by some of the generatorsSource 11. It proves that the intersection of any two parabolic subgroups is parabolicSource 12. A corollary is that arbitrary intersections are parabolic, and every subset has a smallest parabolic subgroup containing itSource 13. The paper also claims a decidable word problem for every finite-rank Artin groupSource 14.

How sure is this? A computer proof checker called Lean has formal proofs of the contractible Salvetti cover, the 116-generator counterexample, and the parabolic intersection results. The Lean note leaves out later corollaries in the manuscripts. No specialist review of these papers has been published.

What it doesn't show. This does not cover Artin groups with infinitely many generators, and the Lean note leaves out the later corollaries in the manuscripts.