# Combinatorial Methods for Detecting Surface Subgroups in Right-Angled Artin Groups

Authors: Robert W. Bell
Journal: ISRN Algebra
 (2011)

---

## Abstract

We give a short proof of the following theorem of Sang-hyun Kim: ifA(Γ) is a right-angled Artin group with defining graph Γ, then A(Γ) contains a hyperbolic surface subgroup if Γ contains an induced subgraph C¯n for some n≥5, where C¯n denotes the complement graph of an n-cycle. Furthermore, we give a new proof of Kim's cocontraction theorem.


---

## Body

## 1. Introduction and Definitions

Suppose thatΓ is a simple finite graph with vertex set VΓ and edge set EΓ. We say that Γ is the defining graph of the right-angled Artin group defined by the presentation(1.1)A(Γ)=〈VΓ;[v,w]:=vwv-1w-1=1,∀{v,w}∈EΓ〉.
Right-angled Artin groups are also called graph groups or partially commutative groups in the literature. All graphs in this paper are assumed to be simple and finite.Right-angled Artin groups have been studied using both combinatorial and geometric methods. In particular, it is well known that these groups have simple solutions to the word and conjugacy problems. Moreover, each right-angled Artin group can be geometrically represented as the fundamental group of a nonpositively curved cubical complexXΓ called the Salvetti complex. For these and other fundamental results, we refer the reader to the survey article by Charney [1].LetΓ be a graph, and suppose that W⊂VΓ. The induced subgraph ΓW is the maximal subgraph of Γ on the vertex set W. A subgraph Λ⊂Γ is called an induced subgraph if Λ=ΓVΛ. In this case, the subgroup of A(Γ) generated by VΛ is canonically isomorphic to A(Λ). This follows from the fact that f:A(Γ)→A(Λ) given by f(v)=v if v∈VΛ and f(v)=1 if v∉VΛ defines a retraction. Therefore, we identify A(Λ) with its image in A(Γ).In this paper, we study the following problem: find conditions on a graphΓ which imply or deny the existence of hyperbolic surface subgroup in A(Γ). Herein, we say that a group is a hyperbolic surface group if it is the fundamental group of a closed orientable surface with negative Euler characteristic.Servatius et al. proved that ifΓ contains an induced n-cycle, that is, the underlying graph of a regular n-gon, for some n≥5, then A(Γ) has a hyperbolic surface subgroup [2]. In fact, they construct an isometrically embedded closed surface of genus 1+(n-4)2n-3 in the cover of the Salvetti complex XΓ corresponding to the commutator subgroup of A(Γ).Kim and, independently, Crisp et al. gave the first examples of graphs without inducedn-cycles, n≥5, which define a right-angled Artin groups which, nonetheless, contain hyperbolic surface subgroups [3, 4]. We give one such example here to illustrate the main lemma of this paper.Consider the graphs in Figure1. The map ϕ:A(P′)→A(P) sending v1 to v12, vi to vi, and vi′ to vi for i>1 defines an injective homomorphism onto an index two subgroup of A(P) (see the discussion below). Since P′ contains an induced circuit of length five (the induced subgraph on the vertices v2,…,v5, and v6′), A(P) contains a hyperbolic surface subgroup; however, P does not contain an n-cycle for any n≥5.The group given by the graphP on (b) injects into the group given by the graph P′ on (a). The vertex labeled by i is referred to as vi in the discussion below.
(a)(b)That the mapϕ is injective can be seen from several perspectives. The approach of Kim and Crisp et al. is to use dissection diagrams; these are collections of simple closed curves which are dual to van Kampen diagrams on a surface over the presentation A(Γ). The method was introduced in this context by Crisp and Wiest [5] and used with much success by Kim [3] and Crisp et al. [4].The purpose of this paper is to demonstrate that classical methods from combinatorial group theory offer another perspective and simplify some of the arguments in the aforementioned articles. We will use the Reidemeister-Schreier rewriting process to give a direct proof that the mapϕ, above, is injective, and we also indicate how this can be proven using normal forms for splittings of groups. This in turn will lead to a short proof of Kim's theorem on cocontractions (in [3, Theorem  4.2]) alluded to in the abstract; see Theorem 3.5 in this paper.Lemma 1.1.
Suppose thatA(Γ) is a right-angled Artin group, and let n be a positive integer. Choose a vertex z∈V(Γ), and define ϕ:A(Γ)→〈x;xn=1〉≅ℤ/nℤ by ϕ(v)=1 if v≠z and ϕ(z)=x. Then kerϕ is a right-angled Artin group with defining graph Γ′ obtained by gluing n copies of Γ∖st(z) to st(z) along lk(z), where st and lk are the star and link, respectively. Moreover, the vertices of Γ′ naturally correspond to the following generating set:
(1.2){zn}∪lk(z)∪{u:u∉st(z)}∪{zuz-1:u∉st(z)}∪⋯∪{zn-1uz1-n:u∉st(z)}.The proof is a straightforward computation using the Reidemeister- Schreier method. The details are given in Section2. Applying Lemma 1.1 to the graphs in Figure 1 proves that A(P′) injects into A(P): if ϕ:A(P)→ℤ/2ℤ maps z=v1 to 1 mod 2, then A(P′)=kerϕ.Another way to prove Lemma1.1 is to take advantage of “visual” splittings of the groups A(Γ) and A(Γ′) as an HNN extension or amalgamated free product. This second approach is stated in the article by Crisp et al. (see [4, Remark  4.1]).We illustrate the utility of Lemma1.1 by giving a short proof of the following theorem of Kim.Theorem 1.2 (see Kim [3, Corollary  4.3(2)]).
LetC¯n denote the complement graph of an n-cycle. For each n≥5, the group A(C¯n) contains a hyperbolic surface subgroup.In fact, we give a new proof of Kim's more general theorem (in [3, Theorem  4.2]) on cocontractions of right-angled Artin groups in Section 3. Kim's proof used the method of dissection diagrams. Kim has also discovered a short proof using visual splittings (personal communication).In preparing this paper, we found that the Reidemeister-Schreier method has been used previously to study certain Bestvina-Brady subgroups of right-angled Artin groups (see [6, 7]).This work was inspired by a desire to better understand Crisp et al.'s very interesting classification of the graphs on fewer than nine vertices which define right-angled Artin groups with hyperbolic surface subgroups. We hope that this paper will help to clarify some aspects of the general problem.


## 2. The Reidemeister-Schreier Method and Proof of Lemma1.1

The Reidemeister-Schreier method solves the following problem: suppose thatG is a group given by the presentation 〈X;R〉, and suppose that H⊂G is a subgroup; find a presentation for H. The treatment below is brisk; see [8] for details and complete proofs.LetF=F(X) be free with basis X, and let π:F→G extend the identity map on X. Consider the preimage P=π-1(H). Let T⊂F be a right Schreier transversal for P in F, that is, T is a complete set of right coset representatives that is closed under the operation of taking initial subwords (of freely reduced words over X). Given w∈F, let [w] be the unique element of T such that Pw=P[w]. For each t∈T and x∈X, let s(t,x)=tx[tx]-1. Define S={s(t,x):t∈T,x∈X,ands(t,x)≠1}. Then S is a basis for the free group P. Define a rewriting process τ:F→P on freely reduced words over X by(2.1)τ(y1y2⋯yn)=s(1,y1)s([y1],y2)⋯s([y1⋯yn-1],yn),
where y∈X∪X-1. Then τ(w)=w[w]-1 for every reduced word w∈F, and(2.2)H=〈S;τ(t-1rt)=1,∀t∈T,r∈R〉.
This rewriting process together with the resulting presentation for the given subgroup H of G=〈X;R〉 is called the Reidemeister-Schreier Method.Proof of Lemma1.1.
LetΓ be a graph, G=A(Γ) the corresponding right-angled Artin group, and z a distinguished vertex of Γ. Let ϕ:G→〈x;xn〉 be given by ϕ(v)=1 if v≠z and ϕ(z)=x.
LetF be free on X=VΓ, and let R be the set of defining relators corresponding to EΓ. Let H=kerϕ, and let P be the inverse image of H in F under the natural map F→G. The set T={1,z,…,zn-1} is a right Schreier transversal for P<F. One verifies (directly) that the following equations hold:
(2.3)s(zk,v)=zkvz-k,ifv≠z,k=0,…,n-1,s(zk,z)=1,ifk=0,…,n-2,s(zk,z)=zn,ifk=n-1.
Thus, we have a setS of generators for kerϕ; however, many of these generators are redundant. Again, one verifies (using τ(w)=w[w]-1) that the following equations hold:
(2.4)τ(zk[u,v]z-k)=[s(zk,u),s(zk,v)],ifu,v≠z,k=0,…,n-1,τ(zk[z,v]z-k)=s(zk+1,v)⋅s(zk,v)-1,ifv≠z,k=0,…,n-2,τ(zk[z,v]z-k)=zn⋅s(1,v)⋅z-n⋅s(zn-1,v)-1,ifv≠z,k=n-1.
Therefore, if[z,v]=1 is a relation in A(Γ), then
(2.5)v=s(1,v)=s(z,v)=⋯=s(zn-1,v),[zn,v]=1
hold in kerϕ. It follows that kerϕ is generated by zn, the vertices adjacent to z in Γ, and n copies (u,zuz-1,…,zn-1uz1-n) of each vertex u≠z and not adjacent to z in Γ. Moreover, the relations are such that kerϕ is presented as a right-angled Artin group where the defining graph is obtained from Γ by taking the star of z and n copies of the complement of the star of z and gluing these copies along the link of z. This completes the proof of Lemma 1.1.


## 3. A Short Proof of Two Theorems of Kim

Suppose thatΓ is a graph. The complement graph Γ¯ is the graph having the same vertices as Γ but which has edges complementary to the edges of Γ. Recall that an n-cycle Cn is the underlying graph of a regular n-gon.Theorem1.2 follows from Kim's cocontraction theorem (see Theorem 3.5 below); however, we present a short independent proof here.Proof of Theorem1.2.
Suppose thatΓ is a graph which contains an induced C5. Then A(Γ) contains a hyperbolic surface subgroup by [2]. Since C5≅C¯5, Theorem 1.2 follows from Lemma 3.1 below.Lemma 3.1 (see Kim [3, Corollary  4.3(1)]).
For eachn≥4, A(C¯n-1)<A(C¯n).Proof.
LetVCn={x1,…,xn}=VC¯n. Define ϕ:A(C¯n)→〈a,a2〉 by ϕ(xn)=a and ϕ(xi)=1 for i≠n. By Lemma 1.1, the defining graph Γ of kerϕ has vertex set VΓ={z2}∪{x1,…,xn-1}∪{y1,yn-1}, where z=xn and yi=zxiz-1. Let S={y1,x2,…,xn-1}. Consider the induced subgraph ΓS. The vertices y1 and xi are not adjacent if and only if i∈{2,n-1}. The vertices xi and xj are not adjacent if and only if |i-j|≤1. Therefore, ΓS≅C¯n-1.Kim proved a more general theorem about subgroups of a right-angled Artin groupA(Γ) defined by “cocontractions.” Let S⊂VΓ, and let S′=VΓ∖S. If ΓS is connected, then the contraction CO(Γ,S) of Γ relative to S is defined by taking the induced subgraph ΓS′ together with a vertex vS and declaring vS to be adjacent to w∈S′ if w is adjacent in Γ to some vertex in S. The cocontraction CO¯(Γ,S) is defined as follows:(3.1)CO¯(Γ,S)=CO(Γ¯,S)¯.Kim insists thatΓS be connected whenever he considers the contraction CO(Γ,S). This assumption is not necessary. Moreover, the following lemma shows that the structure of ΓS is immaterial; the proof follows directly from the definitions.Lemma 3.2.
Suppose thatΓ is a graph and S⊂VΓ. Let Γ′ be the graph obtained from Γ by removing any edges joining two elements of S. Then CO(Γ,S)=CO(Γ′,S).Corollary 3.3.
Suppose thatΓ is a graph and S⊂VΓ. Let Γ′ be a graph obtained from Γ by adding or deleting any collection of edges with both of their vertices belonging to S. Then CO(Γ,S)=CO(Γ′,S) and CO¯(Γ,S)=CO¯(Γ′,S).Lemma 3.4.
Suppose thatΓ is a graph and n≥2. Suppose that S={s1,…,sn}⊂VΓ. Let Λ=CO¯(Γ,{s1,…,sn-1}) and S′=S∖{sn}. Then CO¯(Γ,S)=CO¯(Λ,{vS′,sn}).Proof.
It suffices to compare the collection of vertices which are adjacent tovS in Γ1=CO¯(Γ,S) and Γ2=CO¯(Λ,{vS′,sn}); in the latter case, we are identifying the vertex vS′∪sn with vS.
A vertexw in Γ1 not belonging to S is adjacent to vS if and only if w is adjacent to every si in Γ. A vertex w in Γ2 not equal to vS′ nor sn is adjacent to vS if and only if w is adjacent to vS′ and sn in Λ; but this, in turn, means that w is adjacent to every si in Γ. (Note that the case of n=2 is trivial since CO(Γ,{s1})=Γ.)A collection of verticesS⊂VΓ is said to be anticonnected if Γ¯S is connected. (Note that (Γ¯)S=(ΓS)¯.)Theorem 3.5 (see Kim [3, Theorem  4.2]).
Suppose thatΓ is a graph and S⊂V(Γ) is an anticonnected subset. Then A(CO¯(Γ,S)) embeds in A(Γ).Proof.
First consider the case whenS consists of two nonadjacent vertices z,z′∈VΓ. Define ϕ:A(Γ)→〈x;x2〉 by ϕ(z)=x, and ϕ(v)=1 if v≠z. Let A(Γ′)=kerϕ. Let
(3.2)T=(V(Γ)∖{z2,z′})∪{zz′z-1}⊂V(Γ′).
We claim that ΓT′≅CO¯(Γ,S) via v↦v if v≠zz′z-1 and zz′z-1↦vS.
Ifv and w are distinct from z and z′, then v and w are adjacent in CO¯(Γ,S) if and only if they are adjacent in Γ.
On the other hand, a vertexw is adjacent to vS in CO¯(Γ,S) if and only if w is adjacent to z and z′, whereas a vertex w is adjacent to zz′z-1 in ΓT′ if and only if w belongs to the link of z and to the link of z′, that is, w is adjacent to z and z′. Therefore, ΓT′≅CO¯(Γ,S) and, hence, A(CO¯(Γ,S)) embeds in A(Γ).
Now we prove the general statement by induction on|S|. Suppose that S={s1,…,sn} is anticonnected, and suppose that we have chosen the ordering so that S′={s1,…,sn-1} is also anticonnected. (This is always possible: choose sn so that it is not a cut point of Γ¯S.) Let Λ=CO¯(Γ,S′). Suppose that A(Λ) embeds in A(Γ). By the case of the two vertices above, A(CO¯(Λ,{vS′,sn})) embeds in A(Λ). (Note that vS′ and sn are not adjacent in Λ for, otherwise, sn would be adjacent to every si, i=1,…,n-1, which would contradict the hypothesis that S is anticonnected.) This proves the inductive step. The proof of the theorem is completed by applying Lemma 3.4.



---

*Source: 102029-2011-08-23.xml*