We give a short proof of the following theorem of Sang-hyun Kim: if 𝐴(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 𝐶𝑛 for some 𝑛≥5, where 𝐶𝑛 denotes the complement graph of an 𝑛-cycle. Furthermore, we give a new proof of Kim's cocontraction theorem.

1. Introduction and Definitions

Suppose that Γ 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𝑉𝐴(Γ)=Γ;[]𝑣,𝑤∶=𝑣𝑤𝑣−1𝑤−1=1,∀{𝑣,𝑤}∈𝐸Γ.(1.1) 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 complex 𝑋Γ called the Salvetti complex. For these and other fundamental results, we refer the reader to the survey article by Charney [1].

Let Γ 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 graph Γ 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 if Γ 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 induced 𝑛-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 Figure 1. The map 𝜙∶𝐴(𝑃′)→𝐴(𝑃) sending 𝑣1 to 𝑣21, 𝑣𝑖 to 𝑣𝑖, and ğ‘£î…žğ‘– 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 ğ‘£î…ž6), 𝐴(𝑃) contains a hyperbolic surface subgroup; however, 𝑃 does not contain an 𝑛-cycle for any 𝑛≥5.

That the map 𝜙 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 map 𝜙, 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 that 𝐴(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: {𝑧𝑛}∪lk(𝑧)∪{𝑢∶𝑢∉st(𝑧)}∪𝑧𝑢𝑧−1𝑧∶𝑢∉st(𝑧)∪⋯∪𝑛−1𝑢𝑧1−n.∶𝑢∉st(𝑧)(1.2)

The proof is a straightforward computation using the Reidemeister- Schreier method. The details are given in Section 2. Applying Lemma 1.1 to the graphs in Figure 1 proves that 𝐴(ğ‘ƒî…ž) injects into 𝐴(𝑃): if 𝜙∶𝐴(𝑃)→ℤ/2ℤ maps 𝑧=𝑣1 to 1 mod 2, then 𝐴(ğ‘ƒî…ž)=ker𝜙.

Another way to prove Lemma 1.1 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 Lemma 1.1 by giving a short proof of the following theorem of Kim.

Theorem 1.2 (see Kim [3, Corollary  4.3(2)]). Let 𝐶𝑛 denote the complement graph of an 𝑛-cycle. For each 𝑛≥5, the group 𝐴(𝐶𝑛) 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 Lemma 1.1

The Reidemeister-Schreier method solves the following problem: suppose that 𝐺 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.

Let 𝐹=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 by𝜏𝑦1𝑦2⋯𝑦𝑛=𝑠1,𝑦1𝑠𝑦1,𝑦2𝑦⋯𝑠1⋯𝑦𝑛−1,𝑦𝑛,(2.1) where 𝑦∈𝑋∪𝑋−1. Then 𝜏(𝑤)=𝑤[𝑤]−1 for every reduced word 𝑤∈𝐹, and𝑡𝐻=𝑆;𝜏−1.𝑟𝑡=1,∀𝑡∈𝑇,𝑟∈𝑅(2.2) 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 Lemma 1.1. Let Γ 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.
Let 𝐹 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: 𝑠𝑧𝑘,𝑣=𝑧𝑘𝑣𝑧−𝑘𝑠𝑧,if𝑣≠𝑧,𝑘=0,…,𝑛−1,𝑘𝑠𝑧,𝑧=1,if𝑘=0,…,𝑛−2,𝑘,𝑧=𝑧𝑛,if𝑘=𝑛−1.(2.3)
Thus, we have a set 𝑆 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: 𝜏𝑧𝑘[]𝑧𝑢,𝑣−𝑘=𝑠𝑧𝑘𝑧,𝑢,𝑠𝑘𝜏𝑧,𝑣,if𝑢,𝑣≠𝑧,𝑘=0,…,𝑛−1,𝑘[]𝑧𝑧,𝑣−𝑘𝑧=𝑠𝑘+1𝑧,𝑣⋅𝑠𝑘,𝑣−1𝜏𝑧,if𝑣≠𝑧,𝑘=0,…,𝑛−2,𝑘[]𝑧𝑧,𝑣−𝑘=𝑧𝑛⋅𝑠(1,𝑣)⋅𝑧−𝑛𝑧⋅𝑠𝑛−1,𝑣−1,if𝑣≠𝑧,𝑘=𝑛−1.(2.4)
Therefore, if [𝑧,𝑣]=1 is a relation in 𝐴(Γ), then 𝑧𝑣=𝑠(1,𝑣)=𝑠(𝑧,𝑣)=⋯=𝑠𝑛−1,𝑧,𝑣𝑛,𝑣=1(2.5) 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 that Γ 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.

Theorem 1.2 follows from Kim's cocontraction theorem (see Theorem 3.5 below); however, we present a short independent proof here.

Proof of Theorem 1.2. Suppose that Γ is a graph which contains an induced 𝐶5. Then 𝐴(Γ) contains a hyperbolic surface subgroup by [2]. Since 𝐶5≅𝐶5, Theorem 1.2 follows from Lemma 3.1 below.

Lemma 3.1 (see Kim [3, Corollary  4.3(1)]). For each 𝑛≥4, 𝐴(𝐶𝑛−1)<𝐴(𝐶𝑛).

Proof. Let 𝑉𝐶𝑛={𝑥1,…,𝑥𝑛}=𝑉𝐶𝑛. Define 𝜙∶𝐴(𝐶𝑛)â†’âŸ¨ğ‘Ž,ğ‘Ž2⟩ by 𝜙(𝑥𝑛)=ğ‘Ž 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≅𝐶𝑛−1.

Kim proved a more general theorem about subgroups of a right-angled Artin group 𝐴(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:CO(Γ,𝑆)=CO.Γ,𝑆(3.1)

Kim insists that Γ𝑆 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 that Γ 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 that Γ 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 that Γ 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 to 𝑣𝑆 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 vertex 𝑤 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 vertices 𝑆⊂𝑉Γ is said to be anticonnected if Γ𝑆 is connected. (Note that (Γ)𝑆=(Γ𝑆).)

Theorem 3.5 (see Kim [3, Theorem  4.2]). Suppose that Γ 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 when 𝑆 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 𝑧𝑇=𝑉(Γ)⧵2,ğ‘§î…žâˆªî€½î€¾î€¸ğ‘§ğ‘§î…žğ‘§âˆ’1î€¾î€·Î“âŠ‚ğ‘‰î…žî€¸.(3.2) We claim that Î“î…žğ‘‡â‰…CO(Γ,𝑆) via 𝑣↦𝑣 if ğ‘£â‰ ğ‘§ğ‘§î…žğ‘§âˆ’1 and ğ‘§ğ‘§î…žğ‘§âˆ’1↦𝑣𝑆.
If 𝑣 and 𝑤 are distinct from 𝑧 and ğ‘§î…ž, then 𝑣 and 𝑤 are adjacent in CO(Γ,𝑆) if and only if they are adjacent in Γ.
On the other hand, a vertex 𝑤 is adjacent to 𝑣𝑆 in CO(Γ,𝑆) if and only if 𝑤 is adjacent to 𝑧 and ğ‘§î…ž, whereas a vertex 𝑤 is adjacent to ğ‘§ğ‘§î…žğ‘§âˆ’1 in Î“î…žğ‘‡ if and only if 𝑤 belongs to the link of 𝑧 and to the link of ğ‘§î…ž, that is, 𝑤 is adjacent to 𝑧 and ğ‘§î…ž. Therefore, Î“î…žğ‘‡â‰…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 ğ‘†î…ž={𝑠1,…,𝑠𝑛−1} is also anticonnected. (This is always possible: choose 𝑠𝑛 so that it is not a cut point of Γ𝑆.) Let Λ=CO(Γ,ğ‘†î…ž). 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.


The author would like to thank several colleagues for advice and feedback: Jon Hall, Ian Leary, Ulrich Meierfrankenfeld, and especially Mike Davis and Tadeusz Januskiewicz. Furthermore, the author owes many of the ideas herein to Kim, Crisp et al., as this paper is the product of his attempts to better understand their body of work on this problem.