2018
DOI: 10.4171/ggd/442
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Subspace arrangements, BNS invariants, and pure symmetric outer automorphisms of right-angled Artin groups

Abstract: We introduce a homology theory for subspace arrangements, and use it to extract a new system of numerical invariants from the Bieri-Neumann-Strebel invariant of a group. We use these to characterize when the set of basis conjugating outer automorphisms (a.k.a. the pure symmetric outer automorphism group) of a right-angled Artin group is itself a right-angled Artin group.

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Cited by 9 publications
(35 citation statements)
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“…Then Aut (AnormalΓ;scriptGt) is the pure symmetric automorphism group and Out (AnormalΓ;scriptGt) is the pure symmetric outer automorphism group . These groups have been studied by Day–Wade , Koban–Piggott and Toinet .…”
Section: Computation and Examplesmentioning
confidence: 99%
“…Then Aut (AnormalΓ;scriptGt) is the pure symmetric automorphism group and Out (AnormalΓ;scriptGt) is the pure symmetric outer automorphism group . These groups have been studied by Day–Wade , Koban–Piggott and Toinet .…”
Section: Computation and Examplesmentioning
confidence: 99%
“…. , r 20 } in the free group F = F (a 1 , a 2 ), we denote by Γ(S) the group with generators a 1 , a 2 , a 3 , a 4 Note that this presentation is of the type described in Theorem 1.1. The group Γ(S) is an HNN extension of F × F with a stable letter t that commutes with the fibre product P < F × F associated to the presentation P = a, b | S .…”
Section: Proof Of Theorem 11mentioning
confidence: 99%
“…In more detail, there is no algorithm that, given 22 words u i in the free group F (a 1 , a 2 , a 3 , a 4 ) can determine whether or not the group with presentation a 1 , a 2 , a 3 , a 4 is a RAAG. Nor is there an algorithm that can determine whether or not such a group is commensurable with a RAAG or quasi-isometric to a RAAG.…”
Section: Introductionmentioning
confidence: 99%
“…It says that when you have a SIL (x1,x2|x3), there is a subset of normalΓ that appears as a connected component of each of the three sets Γ st (x1), Γ st (x2), and Γ( lk false(x1false) lk false(x2false)). It follows from [, Lemma 4.5], and we also refer the reader to [, Section 2.3], however we include a short proof for completeness.…”
Section: Separating Intersections Of Linksmentioning
confidence: 99%