On the subgroups of right-angled Artin groups and mapping class groups

Bridson, Martin R.

doi:10.4310/mrl.2013.v20.n2.a1

Cited by 21 publications

(35 citation statements)

References 32 publications

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“…Recall that the isomorphism problem asks if, given a class C of finitely presented groups, whether there exists an algorithm which on an input of two members A, B ∈ C, halts and determines whether or not A ∼ = B. In [9], Bridson uses right-angled Artin subgroups of mapping class groups to show that the isomorphism problem is unsolvable when C is the class of finitely presented subgroups of a sufficiently high genus surface mapping class group. Theorem 1 allows us to replace the surface mapping class group with a braid group: Corollary 6.…”

Section: Corollarymentioning

confidence: 99%

Anti-trees and right-angled Artin subgroups of braid groups

Kim

Koberda

2015

Geom. Topol.

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We prove that an arbitrary right-angled Artin group $G$ admits a quasi-isometric group embedding into a right-angled Artin group defined by the opposite graph of a tree. Consequently, $G$ admits quasi-isometric group embeddings into a pure braid group and into the area-preserving diffeomorphism groups of the 2--disk and the 2--sphere, answering questions due to Crisp--Wiest and M. Kapovich. Another corollary is that a pure braid group contains a closed hyperbolic manifold group as a quasi-isometrically embedded subgroup up to dimension eight. Finally, we show that the isomorphism problem, conjugacy problem, and membership problems are unsolvable in the class of finitely presented subgroups of braid groups.Comment: The condition p>2 is included for the result on symplectomorphisms of the sphere. To appear in Geometry and Topolog

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Section: Corollarymentioning

confidence: 99%

Anti-trees and right-angled Artin subgroups of braid groups

Kim

Koberda

2015

Geom. Topol.

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“…Such groups are now known to be an extremely rich class, following the theory of special cube complexes. See [Wis09], [HW08], [Ago13], [Bri13] and [Bri17]. Stable commutator length may serve as an invariant to distinguish virtually special from special cube complexes.…”

Section: Introductionmentioning

confidence: 99%

Gaps in scl for Amalgamated Free Products and RAAGs

Heuer

2019

Geom. Funct. Anal.

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We develop a new criterion to tell if a group G has the maximal gap of 1/2 in stable commutator length (scl). For amalgamated free products G = A C B we show that every element g in the commutator subgroup of G which does not conjugate into A or B satisfies scl(g) ≥ 1/2, provided that C embeds as a left relatively convex subgroup in both A and B. We deduce from this that every non-trivial element g in the commutator subgroup of a right-angled Artin group G satisfies scl(g) ≥ 1/2. This bound is sharp and is inherited by all fundamental groups of special cube complexes.We prove these statements by constructing explicit extremal homogeneous quasimorphisms φ : G → R satisfyingφ(g) ≥ 1 and D(φ) ≤ 1. Such maps were previously unknown, even for non-abelian free groups. For these quasimorphismsφ there is an action ρ : G → Homeo + (S 1 ) on the circle such that [δ 1φ ] = ρ * eu R b ∈ H 2 b (G, R), for eu R b the real bounded Euler class.

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“…Kim-Koberda [12] proved that for any tree T , there exists a pure braid group PB n such that A(T ) is embedded in PB n . Bridson [3] proved that the isomorphism problem for the mapping class group of a surface whose genus is sufficiently large is unsolvable by using right-angled Artin subgroup in mapping class groups. See Koberda [14] for other researches about right-angled Artin groups and their subgroups.…”

Section: Introductionmentioning

confidence: 99%

Right-angled Artin groups on finite subgraphs of disk graphs

Kuno

2017

Topology and its Applications

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Koberda proved that if a graph $\Gamma$ is a full subgraph of a curve graph $\mathcal{C}(S)$ of an orientable surface $S$, then the right-angled Artin group $A(\Gamma)$ on $\Gamma$ is a subgroup of the mapping class group ${\rm Mod}(S)$ of $S$. On the other hand, for a sufficiently complicated surface $S$, Kim-Koberda gave a graph $\Gamma$ which is not contained in $\mathcal{C}(S)$, but $A(\Gamma)$ is a subgroup of ${\rm Mod}(S)$. In this paper, we prove that if $\Gamma$ is a full subgraph of a disk graph $\mathcal{D}(H)$ of a handlebody $H$, then $A(\Gamma)$ is a subgroup of the handlebody group ${\rm Mod}(H)$ of $H$. Further, we show that there is a graph $\Gamma$ which is not contained in some disk graphs, but $A(\Gamma)$ is a subgroup of the corresponding handlebody groups.Comment: 14 pages, 9 figure

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On the subgroups of right-angled Artin groups and mapping class groups

Cited by 21 publications

References 32 publications

Anti-trees and right-angled Artin subgroups of braid groups

Anti-trees and right-angled Artin subgroups of braid groups

Gaps in scl for Amalgamated Free Products and RAAGs

Right-angled Artin groups on finite subgraphs of disk graphs

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