Abelian subgroups of two-dimensional Artin groups

We classify abelian subgroups of two-dimensional Artin groups.


Introduction
Let S be a finite set and for all s = t ∈ S let m st = m ts ∈ {2, 3, . . . , ∞}. The associated Artin group A S is given by generators and relations: Assume that A S is two-dimensional, that is, for all s, t, r ∈ S, we have 1 m st + 1 m tr + 1 m sr ≤ 1.
We say that A S is of hyperbolic type if the inequality above is strict for all s, t, r ∈ S. In [MP19, Thm D] we classified explicitly all virtually abelian subgroups of A S of hyperbolic type. In this article we extend this, using different techniques, to all twodimensional A S . Charney and Davis [CD95,Thm B] showed that two-dimensional Artin groups satisfy the K(π, 1)-conjecture, in particular they are torsion-free and of cohomological dimension 2. Thus all noncyclic virtually abelian subgroups of A S are virtually Z 2 . The first step of our classification involves the modified Deligne complex Φ (see Section 2) introduced by Charney and Davis [CD95], and generalising a construction of Deligne for Artin groups of spherical type [Del72].
Theorem A. Let A S be a two-dimensional Artin group, and let H be a subgroup of A S that is virtually Z 2 . Then: (i) H is contained in the stabiliser of a vertex of Φ, or (ii) H is contained in the stabiliser of a standard tree of Φ, or (iii) H acts properly on a Euclidean plane isometrically embedded in Φ.
The stabilisers of the vertices of Φ, appearing in (i), are cyclic or conjugates of the dihedral Artin groups A st (see Section 2), which are virtually Z × F for a free group F , see e.g. [HJP16,Lem 4.3(i)]. The stabilisers of the standard trees of Φ, appearing in (ii), are also Z×F [MP19, Lem 4.5] and were described more explicitly in [MP19,Rm 4.6].
In the second step of our classification, we will list all H ∼ = Z 2 satisfying (iii). Note that the statement might seem daunting, but in fact it arises in a straightforward way from reading off the labels of the Euclidean planes obtained in the course of the proof. We use the following notation. Let S be an alphabet. If s ∈ S, then s • denotes the language (i.e. set of words) of form s n for n ∈ Z − {0}. We treat a letter s ∈ S ± as a language consisting of a single word. If L, L are languages, then LL denotes the language of words of the form ww where w ∈ L, w ∈ L . If L is a language, then L * denotes the union of the languages L n for n ≥ 1.
Theorem B. Let A S be a two-dimensional Artin group. Suppose that Z 2 ⊂ A S acts properly on a Euclidean plane isometrically embedded in Φ. Then Z 2 is conjugated into: (a) w, w , where w ∈ A T , w ∈ A T and m tt = 2 for all t ∈ T, t ∈ T , for some disjoint T, T ⊂ S, or Z 2 is conjugated into one of the following, where s, t, r ∈ S: It is easy to check directly that the above groups are indeed abelian. Since A S is torsion-free [CD95,Thm B], the only other subgroups of A S that are virtually Z 2 are isomorphic to the fundamental group of the Klein bottle. They can be also classified, see Remark 5.8.
In the proof of Theorem B we will describe in detail the Euclidean planes in Φ stabilised by Z 2 ⊂ A S . Huang and Osajda established properties of arbitrary quasiflats in the Cayley complex of A S , and one can find similarities between our results and [HO17, §5.15.2 and Prop 8.3]. Organisation of the article. In Section 2 we describe the modified Deligne complex Φ of Charney and Davis and we prove Theorem A. In Section 3 we prove a lemma on dihedral Artin groups fitting in the framework of [AS83]. In Section 4 we introduce a polarisation method for studying Euclidean planes in Φ. We finish with the classification of admissible polarisations and the proof of Theorem B in Section 5.

Modified Deligne complex
Let A S be a two-dimensional Artin group. For s, t ∈ S satisfying m st < ∞, let A st be the dihedral Artin group A S with S = {s, t} and exponent m st . For s ∈ S, let A s = Z.
Let K be the following simplicial complex. The vertices of K correspond to subsets T ⊆ S satisfying |T | ≤ 2 and, in the case where |T | = 2 with T = {s, t}, satisfying m st < ∞. We call T the type of its corresponding vertex. Vertices of types T, T are connected by an edge of K, if we have T T or vice versa. Similarly, three vertices span a triangle of K, if they have types ∅, {s}, {s, t} for some s, t ∈ S.
We give K the following structure of a simple complex of groups K (see [BH99, §II.12] for background). The vertex groups are trivial, A s , or A st , when the vertex is of type ∅, {s}, {s, t}, respectively. For an edge joining a vertex of type {s} to a vertex of type {s, t}, its edge group is A s ; all other edge groups and all triangle groups are trivial. All inclusion maps are the obvious ones. It follows directly from the definitions that A S is the fundamental group of K.
We equip each triangle of K with the Moussong metric of an Euclidean triangle of angles π 2mst , π 2 , (mst−1)π 2mst at the vertices of types {s, t}, {s}, ∅, respectively. As explained in [MP19,§3], the local developments of K are CAT(0) and hence K is strictly developable and its development Φ exists and is CAT(0). See [CD95] for a detailed proof. We call Φ with the Moussong metric the modified Deligne complex. In particular all A s and A st with m st < ∞ map injectively into A S (which follows also from [vdL83, Thm 4.13]). Vertices of Φ inherit types from the types of the vertices of K.
Let r ∈ S and let T be the fixed-point set in Φ of r. Note that since A S acts on Φ without inversions, T is a subcomplex of Φ. Since the stabilisers of the triangles of Φ are trivial, we have that T is a graph. Since Φ is CAT(0), T is convex and thus it is a tree. Thus we call a standard tree the fixed-point set in Φ of a conjugate of a generator r ∈ S of A S . , and since the action is without inversions, we can take this point to be a vertex as required in (i).
Secondly, suppose that Γ has both an element γ that fixes a point of Φ and an element that is loxodromic. Then Min(Γ) is not a single point, so it contains an edge e. Since Min(Γ) ⊂ Fix(γ), we have that γ fixes e. Thus γ is a conjugate of an element of S and so Min(Γ) is contained in a standard tree T . For any h ∈ H we have that the intersection h(T ) ∩ T contains Min(Γ) ⊃ e and thus by Remark 2.1 we have h ∈ Stab(T ), as required in (ii).
Finally, suppose that all elements of Γ are loxodromic. By [BH99, Thm II.7.20(1,4)] we have Min(Γ) = Y × R n with H preserving the product structure and Γ acting trivially on Y . As before H/Γ fixes a point of Y and so H stabilises R n isometrically embedded in Φ. By [BH99, Thm II.7.20(2)], we have n ≤ 2, but since H acts by simplicial isometries, we have n = 2 and the action is proper, as required in (iii).

Girth lemma
and only if up to interchanging s with t, and a cyclic permutation, it is of the form: Proof. The 'if' part follows immediately from Figure 1. We prove the 'only if' part by induction on the size of any reduced (van Kampen) diagram M of the word w in question, where we prove the stronger assertion that, up to interchanging s with t, M is as in Figure 1.
Let R be a strip of M with boundary labelled by rb, where r labels ∂R ∩ ∂M and so w = rw . Assume also that R shares no regions with some other strip (such a pair of strips exists by [AS83, Lem 2]). By [AS83, Lem 5], we have that the syllable lengths satisfy ||r|| ≥ ||b|| + 2 and so by [AS83, Lem 6], we have ||r|| ≥ m + 1, hence ||w || ≤ m + 1. In fact, since the outside boundary of the other strip has also syllable length ≥ m + 1, we have ||r|| = m + 1, and hence ||b|| = m − 1. Let M be the diagram with boundary labelled by b −1 w obtained from M by removing R. By the induction hypothesis, M is as in Figure 1. If R is a singleton strip, then there is only one way of gluing R to M to obtain ||w|| = 2m and it is as in Figure 1. If R is a compound strip, then by the induction hypothesis R is also as in Figure 1. Moreover, since all regions of R share exactly one edge with M , up to interchanging s with t, and/or Since m > 2, we have that b cannot be a subword of the boundary word of M , unless M is a mirror copy of R, contradiction.

Polarisation
Definition 4.1. Let F be a Euclidean plane isometrically embedded in Φ. Then for each vertex v in F of type {s, t} there are exactly 4m st triangles in F incident to v. We assemble them into regular 2m st -gons, and call this complex the tiling of F . We say that a cell of this tiling has type T if its barycentre in Φ has type T .
For a Coxeter group W , let Σ denote its Davis complex, i.e. the complex obtained from the standard Cayley graph by adding k-cells corresponding to cosets of finite T for T ⊂ S of size k. For example for W the triangle Coxeter group with exponents {3, 3, 3}, the complex Σ is the tiling of the Euclidean plane by regular hexagons. Proof. The 2-cells of the tiling are regular polygons with even numbers of sides, hence their angles lie in [ π 2 , π). If there is a vertex v of F incident to four 2-cells, then all these 2-cells are squares. Consequently, any vertex of F adjacent to v is incident to at least two squares, and thus to exactly four squares. Then, since the 1-skeleton of F is connected, the tiling of F is the standard square tiling.
If v is incident to three 2-cells, which are 2m, 2m , 2m -gons, then since 1 m +   Definition 4.5. Suppose Z 2 ⊂ A S acts properly and cocompactly on Σ ⊂ Φ. For an edge e of type {s} in Σ, its vertices correspond to elements g, gs k ∈ A S for k > 0. We direct e from g to gs k . By Lemma 3.1, the boundary of each 2-cell σ is subdivided into two directed paths joining two opposite vertices. The induced polarisation of Σ assigns to each σ the longest diagonal l(σ) joining these two vertices.
Lemma 4.6. An induced polarisation is admissible.
Indeed, suppose that we have v ∈ l(σ), l(τ ). Without loss of generality suppose that the edge e = σ ∩ τ is directed from v. Then the other two edges incident to v are also directed from v. We will now prove by induction on the distance from v that each edge of Σ is oriented from its vertex closer to v to its vertex farther from v in the 1-skeleton Σ 1 (they cannot be at equal distance since Σ 1 is bipartite).
For the induction step, suppose we have already proved the induction hypothesis for all edges closer to v than an edge uu , where u is closer to v than u. Let u be the first vertex on a geodesic from u to v in Σ 1 . Let σ be the 2-cell containing the path uu u . By [Ron89, Thms 2.10 and 2.16], σ has two opposite vertices u 0 closest to v and u max farthest from v. By the induction hypothesis, the edge u u is oriented from u to u , and both edges of σ incident to u 0 are oriented from u 0 .
Thus if the edge uu was oriented to u we would have that u is opposite to u 0 , so u = u max , contradiction. This finishes the induction step.
As a consequence, v is the unique vertex of Σ with all edges incident to v oriented from v. This contradicts the cocompactness of the action of Z 2 on Σ and proves Step 1.
Step 2. For each v there is at least one l(σ) containing v.
Among the edges incident to v there are at least two edges directed from v or at least two edges directed to v. The 2-cell σ containing such two edges satisfies l(σ) v. Note that if the conclusion of Proposition 5.1 holds, then the translation ρ mapping one hexagon containing e to the other preserves l.

Classification
Remark 5.2. It is easy to prove the converse, i.e. that if each l(γ) has endpoints on edges of γ parallel to e, and if l is ρ-invariant, then l is admissible. This can be used to classify all admissible polarisations, but we will not need it.
Case 2. All the l(σ) are parallel.
In this case it suffices to take any edge e intersecting some l(σ). An edge e of Σ lies either in two octagons σ, σ or there is a square with two parallel edges e, e in octagons σ, σ . The translation of Σ mapping σ to σ is called an e-translation. Note that if the conclusion of Proposition 5.4 holds, then an e-translation preserves l. Proof. Suppose that ♦ holds for all octagons γ in the ρ-orbit of an octagon σ. We can assume e ⊂ σ. Suppose first that e lies in another octagon σ . Then let τ be an octagon outside the ρ-orbit of σ adjacent to some ρ k (σ), say σ. Let , ρ( ) be the two squares adjacent to both σ and τ (see Figure 3). By the admissibility of l, we have that l( ), l(ρ( )) contain the two vertices of σ ∩ τ . Consequently, l(τ ) intersects the edge τ ∩ ρ(τ ), and so γ = τ satisfies ♦. It is easy to extend this to all the octagons γ. σ σ τ Figure 3.
It remains to consider the case where there is a square with two parallel edges e, e in octagons σ, σ . Let τ be an octagon adjacent to two of them, say to σ and ρ(σ). Let v = e ∩ τ, x = e ∩ τ . Since ♦ holds for γ = σ, σ , each of v, x lies in one of l(σ), l( ), l(σ ). Thus by the admissibility of l, we have v, x / ∈ l(τ ). Any of the two remaining choices for l(τ ) satisfy ♦ for γ = τ . It is again easy to extend this to all the octagons γ.
Proof of Proposition 5.4. Note that we fall in one of the following two cases.
Let v be the vertex of f distinct from u = l(τ ) ∩ f . By the admissibility of l, the vertex v is contained in l( ) for the square incident to v. Let x be the vertex in τ ∩ distinct from v, and let σ be the octagon incident to x distinct from τ . By the admissibility of l, the vertex x is contained in l(σ ). Hence ♦ holds for e = σ ∩ and γ = σ , τ (see Figure 4). Let ρ be the translation mapping σ to σ .  Now let be the square incident to u, and let z be the vertex in σ ∩ distinct from u. Note that by the admissibility of l, we have z ∈ l( ), and consequently ρ −1 (x) ∈ l(σ) and ρ −1 (v) ∈ l(ρ −1 ( )). Hence l(ρ −1 (τ )) cannot contain neither z, nor ρ −1 (v), nor ρ −1 (x) ∈ l(σ). There is only one remaining choice for l(ρ −1 (τ )), and it satisfies ♦ for γ = ρ −1 (τ ).
We can now argue exactly as in the proof of Proposition 5.1 that that ♦ holds for γ = ρ n (σ), ρ n (τ ) for all n ∈ Z. It then remains to apply Lemma 5.5. Case 2. For each edge e in octagons σ, σ with l(σ) ∩ e = ∅ we have l(σ ) ∩ e = ∅.
Let σ be any octagon and e an edge contained in another octagon σ and intersecting l(σ). Let ρ be the translation mapping σ to σ . One can show inductively that ♦ holds for octagons γ = ρ n (σ) for all n ∈ Z. It then remains to apply Lemma 5.5.
Note that for l satisfying ♦ for all octagons γ, the values of l on octagons determine its values on squares.
Proposition 5.6. Let Σ be the Davis complex for W T the triangle Coxeter group with exponents {2, 3, 6} with an admissible polarisation l. Then there is an edge e such that such that each 12-gon γ of Σ satisfies ♥: the diagonal l(γ) has endpoints on edges of γ parallel to e.
Let e be an edge. An e-translation is the translation of Σ mapping σ to σ in of the two following configurations. In the first configuration we have a square with two parallel edges e, e in 12-gons σ, σ . In the second configuration we have four parallel edges e, e , e , e such that e , e lie in a square, e, e in a hexagon φ and e , e in another hexagon, and we consider 12-gons σ ⊃ e, σ ⊃ e . Again, if the conclusion of Proposition 5.6 holds, then an e-translation preserves l. To see this in the configuration with hexagons it suffices to observe that l(φ) (and similarly for the other hexagon) is not parallel to e: otherwise l(φ) would intersect l( ) for the square containing e ∩ l(σ).
Lemma 5.7. Let e be an edge and ρ an e-translation. If ♥ holds for all 12-gons γ in one ρ-orbit, then it holds for all γ.
The proof is easy, it goes along the same lines as the proofs of Lemmas 5.3 and 5.5 and we omit it.
Proof of Proposition 5.6. We adopt the convention that if we label the vertices of an edge in a 12-gon σ by v 0 v 1 , then all the other vertices of σ get cyclically labelled by v 2 · · · v 11 . Let τ be a 12-gon and suppose that l(τ ) contains a vertex v 1 of an edge v 0 v 1 ⊂ τ for a square = v 0 v 1 u 1 u 0 . Let σ be the 12-gon containing u 0 u 1 . Then l( ) = u 1 v 0 and furthermore l assigns to the hexagon and square containing u 1 u 2 , u 2 u 3 , respectively, the longest diagonal containing u 2 , u 3 , respectively. Thus the only three remaining options for l(σ) are the diagonals u 0 u 6 , u 5 u 11 , and u 4 u 10 . Thus we fall in one of the following three cases. Case 0. There is such a τ with l(σ) = u 4 u 10 .
Let φ be the hexagon containing u 3 u 4 . Then l(φ) is parallel to u 3 u 4 and there is no admissible choice for l in the square containing u 4 u 5 (see Figure 5). This is a contradiction. It is easy to see that l agrees with Figure 6 on the hexagon containing v 0 v 11 and the square containing v 11 v 10 . Thus the only 2-cell φ with l(φ) containing v 10 may be (and is) the hexagon containing v 10 v 9 . Consequently the only 2-cell with l( ) containing v 9 may be (and is) the square containing v 9 v 8 . Denote by σ the 12-gon adjacent to both φ and at the vertex x = v 9 . Then x may lie (and lies) only in l(σ ). Denote by ρ the translation mapping σ to σ .
It is also easy to see that the 2-cells surrounding σ and ρ −1 (τ ) depicted in Figure 7 have l(·) as indicated. This leaves only two choices for l(ρ −1 (τ )), where one of them leads to Case 0, and the other satisfies ♥. Replacing repeatedly σ and τ in the above argument by τ and σ or by ρ −1 (τ ) and σ gives that ♥ holds for γ = ρ n (σ), ρ n (τ ) for all n ∈ Z. It remains to apply Lemma 5.7. Case 2. There is no such τ as in Case 0 or 1.
It is then easy to see that e = v 0 v 1 and ρ mapping σ to τ satisfy the hypothesis of Lemma 5.7.
Note that for l satisfying ♥ for all 12-gons γ, the values of l on 12-gons determine its values on squares and hexagons.
We are now ready for the following. First consider the case where the tiling of F is the standard square tiling. We can partition the set of edges into two classes horizontal and vertical of parallel edges. Let T be the set of types of horizontal edges and T be the set of types of vertical edges. Since for each square the type of its two horizontal (respectively, vertical) edges is the same, we have m tt = 2 for all t ∈ T, t ∈ T . Moreover, by Remark 3.2, if one of the edges is of form g, gt k , then the other is of form h, ht k . Thus, up to a conjugation, the stabiliser of F in A S is generated by a horizontal translation w ∈ A T and a vertical translation w ∈ A T . This brings us to Case (a) in Theorem B.
It remains to consider the case where the tiling of F is the one of Σ. Consider its induced polarisation l from Definition 4.5. By Lemma 4.6, l is admissible. By Propositions 5.1, 5.4, and 5.6, there is an edge e such that each for each γ a maximal size 2-cell, the diagonal l(γ) has endpoints on edges of γ parallel to e, and there is a particular translation ρ in the direction perpendicular to e preserving l.
For an edge f of type {s} in Σ, its vertices are of form g, gs k for k > 0, directed from g to gs k . If k > 1, then we call f k-long. By Lemma 3.1 and Remark 3.2, if f is k-long, then so is its opposite edge in both of the 2-cells that contain f . Consequently all the edges crossing the bisector of f are k-long. Moreover, all such bisectors are parallel, since otherwise the 2-cell where they crossed would have four long edges, so would be a square by Lemma 3.1. Analysing l in the 2-cells adjacent to leaves then no admissible choice for l( ).
Furthermore, by Lemmas 3.1, 5.3, 5.5 and 5.7, if f is a long edge, then we can assume that f is parallel to e.
Suppose first that W T is the triangle Coxeter group with exponents {3, 3, 3} and T = {s, t, r}. Let ω be a combinatorial axis for the action of ρ on Σ. Since none of the edges of ω are parallel to e, by the definition of the induced polarisation we see that they are all directed consistently (see Figure 8). Thus, up to replacing F by its translate and interchanging t with r, the element strstr preserves ω, and coincides on it with ρ 3 . In fact, ρ 3 not only preserves the types of edges, but also by Lemma 3.1 their direction and k-longness. Thus strstr preserves F . The second generator of the type preserving translation group of Σ in W T is tstr. Note that the path representing it in Σ ⊂ Φ corresponds to a word in t • str ⊂ A T . That word depends on whether the second edge of the path is long and on the polarisation. Since Z 2 acts cocompactly, there is a power of tstr such that its corresponding path in Σ ⊂ Φ reads off a word in (t • str) * ⊂ A T that is the other generator of the orientation preserving stabiliser of F in A S . This brings us to Case (b) in Theorem B. One similarly obtains the characterisations of orientation preserving stabilisers of F for the two other W (see Figures 9 and 10). Remark 5.8. Analysing the full stabilisers of F in A S one can easily classify also the subgroups of A S acting properly on Φ isomorphic to the fundamental group of the Klein bottle. For example, suppose that the second generator of the Z 2 in Case (b) of Theorem B has the form g = (t k str)(t −k str). Then our Z 2 is generated by strstr and g = g(strstr) −1 = t k sr −k s −1 . Note that str normalises our Z 2 with (str) −1 g (str) = (g ) −1 . Thus str, g is isomorphic to the fundamental group of the Klein bottle. We do not include a full classification, since it is not particularly illuminating.