Abelian subgroups of two-dimensional Artin groups
AABELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTINGROUPS
ALEXANDRE MARTIN †∗ AND PIOTR PRZYTYCKI ‡∗ Abstract.
We classify abelian subgroups of two-dimensional Artin groups. Introduction
Let S be a finite set and for all s (cid:54) = t ∈ S let m st = m ts ∈ { , , . . . , ∞} . Theassociated Artin group A S is given by generators and relations: A S = (cid:104) S | sts · · · (cid:124) (cid:123)(cid:122) (cid:125) m st = tst · · · (cid:124) (cid:123)(cid:122) (cid:125) m st (cid:105) . Assume that A S is two-dimensional , that is, for all s, t, r ∈ S , we have m st + 1 m tr + 1 m sr ≤ . 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 ofhyperbolic type. In this article we extend this, using different techniques, to all two-dimensional A S . Charney and Davis [CD95, Thm B] showed that two-dimensionalArtin groups satisfy the K ( π, -conjecture, in particular they are torsion-free andof cohomological dimension . Thus all noncyclic virtually abelian subgroups of A S are virtually Z . The first step of our classification involves the modified Delignecomplex Φ (see Section 2) introduced by Charney and Davis [CD95], and general-ising 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 subgroupof A S that is virtually Z . 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 ofthe dihedral Artin groups A st (see Section 2), which are virtually Z × F for a freegroup 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 explicitlyin [MP19, Rm 4.6].In the second step of our classification, we will list all H ∼ = Z satisfying (iii).Note that the statement might seem daunting, but in fact it arises in a straightfor-ward way from reading off the labels of the Euclidean planes obtained in the courseof the proof.We use the following notation. Let S be an alphabet. If s ∈ S , then s • denotesthe language (i.e. set of words) of form s n for n ∈ Z − { } . We treat a letter s ∈ S ± † Partially supported by EPSRC New Investigator Award EP/S010963/1. ‡ Partially supported by NSERC, FRQNT, and National Science Centre, Poland UMO-2018/30/M/ST1/00668. ∗ This work was partially supported by the grant 346300 for IMPAN from the Simons Foun-dation and the matching 2015–2019 Polish MNiSW fund. a r X i v : . [ m a t h . G R ] J u l A. MARTIN AND P. PRZYTYCKI as a language consisting of a single word. If L , L (cid:48) are languages, then LL (cid:48) denotesthe language of words of the form ww (cid:48) where w ∈ L , w (cid:48) ∈ L (cid:48) . If L is a language,then L ∗ denotes the union of the languages L n for n ≥ . Theorem B.
Let A S be a two-dimensional Artin group. Suppose that Z ⊂ A S actsproperly on a Euclidean plane isometrically embedded in Φ . Then Z is conjugatedinto:(a) (cid:104) w, w (cid:48) (cid:105) , where w ∈ A T , w (cid:48) ∈ A T (cid:48) and m tt (cid:48) = 2 for all t ∈ T, t (cid:48) ∈ T (cid:48) , for somedisjoint T, T (cid:48) ⊂ S ,or Z is conjugated into one of the following, where s, t, r ∈ S :(b) (cid:104) strstr, w (cid:105) , where w ∈ ( t • str ) ∗ and m st = m tr = m sr = 3 .(c) (cid:104) strt, w (cid:105) , where w ∈ ( r • t − s • t ) ∗ and m st = m tr = 4 , m sr = 2 .(d) (cid:104) stsrtr, w (cid:105) , where w ∈ ( t • str ) ∗ and m st = m tr = 4 , m sr = 2 .(e) (cid:104) ststsrtstr, w (cid:105) , where w ∈ ( t • ststr ) ∗ and m st = 6 , m tr = 3 , m sr = 2 .(f ) (cid:104) tststr, w (cid:105) , where w ∈ ( s • tstrt − ) ∗ and m st = 6 , m tr = 3 , m sr = 2 . It is easy to check directly that the above groups are indeed abelian. Since A S istorsion-free [CD95, Thm B], the only other subgroups of A S that are virtually Z are isomorphic to the fundamental group of the Klein bottle. They can be alsoclassified, see Remark 5.8.In the proof of Theorem B we will describe in detail the Euclidean planes in Φ stabilised by Z ⊂ A S . Huang and Osajda established properties of arbitraryquasiflats in the Cayley complex of A S , and one can find similarities between ourresults and [HO17, § Organisation of the article.
In Section 2 we describe the modified Deligne com-plex Φ of Charney and Davis and we prove Theorem A. In Section 3 we prove alemma on dihedral Artin groups fitting in the framework of [AS83]. In Section 4we introduce a polarisation method for studying Euclidean planes in Φ . We finishwith the classification of admissible polarisations and the proof of Theorem B inSection 5. 2. 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 (cid:48) with S (cid:48) = { 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 tosubsets T ⊆ S satisfying | T | ≤ and, in the case where | T | = 2 with T = { s, t } ,satisfying m st < ∞ . We call T the type of its corresponding vertex. Verticesof types T, T (cid:48) are connected by an edge of K , if we have T (cid:40) T (cid:48) or vice versa.Similarly, three vertices span a triangle of K , if they have types ∅ , { s } , { s, t } forsome 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 vertexis of type ∅ , { s } , { s, t } , respectively. For an edge joining a vertex of type { s } to avertex of type { s, t } , its edge group is A s ; all other edge groups and all trianglegroups are trivial. All inclusion maps are the obvious ones. It follows directly fromthe definitions that A S is the fundamental group of K .We equip each triangle of K with the Moussong metric of an Euclidean triangleof angles π m st , π , ( m st − π m st at the vertices of types { s, t } , { s } , ∅ , respectively. Asexplained in [MP19, § K are CAT(0) and hence K isstrictly developable and its development Φ exists and is CAT(0) . See [CD95] for adetailed 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 BELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTIN GROUPS 3 also from [vdL83, Thm 4.13]). Vertices of Φ inherit types from the types of thevertices 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 trianglesof Φ are trivial, we have that T is a graph. Since Φ is CAT(0) , T is convex andthus it is a tree. Thus we call a standard tree the fixed-point set in Φ of a conjugateof a generator r ∈ S of A S . Remark 2.1 ([MP19, Rm 4.4]) . Each edge of Φ belongs to at most one standardtree. Proof of Theorem A.
Let Γ ⊂ H be a finite index normal subgroup isomorphicto Z . By [Bri99], Γ acts on Φ by semi-simple isometries. Let Min(Γ) = (cid:84) γ ∈ Γ Min( γ ) ,where Min( γ ) is the Minset of γ in Φ . By a variant of the Flat Torus Theo-rem not requiring properness [BH99, Thm II.7.20(1)], Min(Γ) is nonempty. By[BH99, Thm II.7.20(4)] we have that H stabilises Min(Γ) .Suppose first that each element of Γ fixes a point of Φ . Then Γ acts trivially on Min(Γ) . By the fixed-point theorem [BH99, Thm II.2.8(1)] the finite group H/ Γ fixes a point of Min(Γ) , and since the action is without inversions, we can take thispoint to be a vertex as required in (i).Secondly, suppose that Γ has both an element γ that fixes a point of Φ and anelement that is loxodromic. Then Min(Γ) is not a single point, so it contains anedge e . Since Min(Γ) ⊂ Fix( γ ) , we have that γ fixes e . Thus γ is a conjugate of anelement of S and so Min(Γ) is contained in a standard tree T . For any h ∈ H wehave that the intersection h ( T ) ∩ T contains Min(Γ) ⊃ e and thus by Remark 2.1we 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 actingtrivially on Y . As before H/ Γ fixes a point of Y and so H stabilises R n isomet-rically embedded in Φ . By [BH99, Thm II.7.20(2)], we have n ≤ , but since H acts by simplicial isometries, we have n = 2 and the action is proper, as requiredin (iii). (cid:3) Girth lemma
Lemma 3.1.
Let S = { s, t } with m st ≥ . A word with m syllables (i.e. ofform s i t j · · · s i m t j m with all i k , j k ∈ Z − { } ) is trivial in A S if and only if up tointerchanging s with t , and a cyclic permutation, it is of the form: • s k t · · · s (cid:124) (cid:123)(cid:122) (cid:125) m − t − k s − · · · t − (cid:124) (cid:123)(cid:122) (cid:125) m − for m odd, • s k ts · · · t (cid:124) (cid:123)(cid:122) (cid:125) m − s − k t − s − · · · t − (cid:124) (cid:123)(cid:122) (cid:125) m − for m even,where k ∈ Z − { } .Proof. The ‘if’ part follows immediately from Figure 1. We prove the ‘only if’ partby induction on the size of any reduced (van Kampen) diagram M of the word w inquestion, where we prove the stronger assertion that, up to interchanging s with t , M is as in Figure 1. A. MARTIN AND P. PRZYTYCKI aaaaa ss s a = ts · · · (cid:124) (cid:123)(cid:122) (cid:125) m − Figure 1.
The diagram M . The top arrows are either all la-belled s or all labelled t , depending on the parity of m .We use the vocabulary from [AS83], where the -cells of M are called regions and the interior degree i ( D ) of a region is the number of interior edges of ∂D (after forgetting vertices of valence ). For example the two extreme regions inFigure 1 have interior degree . A region D is a simple boundary region if ∂D ∩ ∂M is nonempty, and M − D is connected. For example, the two extreme regions inFigure 1 are simple boundary regions, but the remaining ones are not. A singletonstrip is a simple boundary region with i ( D ) ≤ . A compound strip is a subdiagram R of M consisting of regions D , . . . , D n , with n ≥ , with D k − ∩ D k a singleinterior edge of R (after forgetting vertices of valence ), satisfying i ( D ) = i ( D n ) =2 , i ( D k ) = 3 for < k < n and M − R connected.Let R be a strip of M with boundary labelled by rb , where r labels ∂R ∩ ∂M andso w = rw (cid:48) . Assume also that R shares no regions with some other strip (such apair of strips exists by [AS83, Lem 2]). By [AS83, Lem 5], we have that the syllablelengths satisfy || r || ≥ || b || + 2 and so by [AS83, Lem 6], we have || r || ≥ m + 1 ,hence || w (cid:48) || ≤ m + 1 . In fact, since the outside boundary of the other strip has alsosyllable length ≥ m + 1 , we have || r || = m + 1 , and hence || b || = m − . Let M (cid:48) bethe diagram with boundary labelled by b − w (cid:48) obtained from M by removing R . Bythe induction hypothesis, M (cid:48) is as in Figure 1. If R is a singleton strip, then thereis only one way of gluing R to M (cid:48) to obtain || w || = 2 m 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 (cid:48) , up to interchanging s with t , and/or b with b − , we have b = s k ts · · · (cid:124) (cid:123)(cid:122) (cid:125) m − or b = · · · st (cid:124) (cid:123)(cid:122) (cid:125) m − s k , where k ≥ .Since m > , we have that b cannot be a subword of the boundary word of M (cid:48) ,unless M (cid:48) is a mirror copy of R , contradiction. (cid:3) Remark 3.2.
Let S = { s, t } with m st = 2 . A word with syllables is trivial in A S = Z if and only if up to interchanging s with t it is of the form s k t l s − k t − l ,where k, l ∈ Z − { } . 4. Polarisation
Definition 4.1.
Let F be a Euclidean plane isometrically embedded in Φ . Thenfor each vertex v in F of type { s, t } there are exactly m st triangles in F incidentto v . We assemble them into regular m st -gons, and call this complex the tilingof 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 obtainedfrom the standard Cayley graph by adding k -cells corresponding to cosets of finite (cid:104) T (cid:105) for T ⊂ S of size k . For example for W the triangle Coxeter group withexponents { , , } , the complex Σ is the tiling of the Euclidean plane by regularhexagons. BELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTIN GROUPS 5
Lemma 4.2.
Let F be a Euclidean plane isometrically embedded in Φ . Then thetiling of F is either the standard square tiling, or the one of the Davis complex Σ for W , where W is the triangle Coxeter group with exponents { , , } , { , , } or { , , } .Proof. The -cells of the tiling are regular polygons with even numbers of sides,hence their angles lie in [ π , π ) . If there is a vertex v of F incident to four -cells,then all these -cells are squares. Consequently, any vertex of F adjacent to v isincident to at least two squares, and thus to exactly four squares. Then, since the -skeleton of F is connected, the tiling of F is the standard square tiling.If v is incident to three -cells, which are m, m (cid:48) , m (cid:48)(cid:48) -gons, then since m + m (cid:48) + m (cid:48)(cid:48) = 1 , we have { m, m (cid:48) , m (cid:48)(cid:48) } = { , , } , { , , } or { , , } . Moreover, avertex u of F adjacent to v is incident to two of these three -cells, and this impliesthat the third -cell incident to u has the same size as the one incident to v . Thisdetermines uniquely the tiling of F as the one of Σ . (cid:3) Henceforth, let Σ be the Davis complex for W , where W is the triangle Coxetergroup with exponents { , , } , { , , } or { , , } . Lemma 4.3.
Suppose Σ is the tiling of a Euclidean plane isometrically embeddedin Φ . Then the natural action of W on Σ preserves the edge types coming from Φ . In particular, W = W T for some T ⊂ S with | T | = 3 . Proof.
Chose a vertex v of Σ , and let { s } , { t } , { r } be the types of edges incidentto v . Let u be a vertex of Σ adjacent to v , say along an edge e of type { r } . Hencethe -cells in Σ incident to e have types { s, r } and { t, r } . Consequently, the typesof the remaining two edges incident to u are also { s } and { t } , and in such a waythat the reflection of Σ interchanging the endpoints of e preserves the types of theseedges. This determines uniquely the types of the edges of Σ , and guarantees thatthey are preserved by W . (cid:3) Definition 4.4. A polarisation of Σ is a choice of a longest diagonal l ( σ ) in each -cell σ of Σ . A polarisation is admissible if every vertex of Σ belongs to exactlyone l ( σ ) . Definition 4.5.
Suppose Z ⊂ A S acts properly and cocompactly on Σ ⊂ Φ . Foran edge e of type { s } in Σ , its vertices correspond to elements g, gs k ∈ A S for k > . We direct e from g to gs k . By Lemma 3.1, the boundary of each -cell σ is subdivided into two directed paths joining two opposite vertices. The inducedpolarisation of Σ assigns to each σ the longest diagonal l ( σ ) joining these twovertices. Lemma 4.6.
An induced polarisation is admissible.Proof.
Step 1.
For each vertex v of Σ , there is at most one l ( σ ) containing v . Indeed, suppose that we have v ∈ l ( σ ) , l ( τ ) . Without loss of generality supposethat 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 -skeleton Σ (they cannot be at equal distance since Σ is bipartite).For the induction step, suppose we have already proved the induction hypothesisfor all edges closer to v than an edge uu (cid:48) , where u (cid:48) is closer to v than u . Let u (cid:48)(cid:48) be the first vertex on a geodesic from u (cid:48) to v in Σ . Let σ be the -cell containingthe path uu (cid:48) u (cid:48)(cid:48) . By [Ron89, Thms 2.10 and 2.16], σ has two opposite vertices u closest to v and u max farthest from v . By the induction hypothesis, the edge u (cid:48) u (cid:48)(cid:48) is oriented from u (cid:48)(cid:48) to u (cid:48) , and both edges of σ incident to u are oriented from u . A. MARTIN AND P. PRZYTYCKI
Thus if the edge uu (cid:48) was oriented to u (cid:48) we would have that u (cid:48) is opposite to u , so u (cid:48) = u max , contradiction. This finishes the induction step.As a consequence, v is the unique vertex of Σ with all edges incident to v orientedfrom v . This contradicts the cocompactness of the action of Z on Σ and provesStep 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 orat least two edges directed to v . The -cell σ containing such two edges satisfies l ( σ ) (cid:51) v . (cid:3) Classification
Proposition 5.1.
Let Σ be the Davis complex for W T the triangle Coxeter groupwith exponents { , , } with an admissible polarisation l . Then there is an edge e such that each hexagon γ of Σ satisfies ♣ : the diagonal l ( γ ) has endpoints on edges of γ parallel to e . Note that if the conclusion of Proposition 5.1 holds, then the translation ρ map-ping one hexagon containing e to the other preserves l . Remark 5.2.
It is easy to prove the converse, i.e. that if each l ( γ ) has endpointson edges of γ parallel to e , and if l is ρ -invariant, then l is admissible. This can beused to classify all admissible polarisations, but we will not need it.To prove Proposition 5.1 we need the following reduction. Lemma 5.3.
Let e be an edge and ρ a translation mapping one hexagon contain-ing e to the other. If ♣ holds for all hexagons γ in one ρ -orbit, then it holds forall γ .Proof. Suppose that ♣ holds for all hexagons γ in the ρ -orbit of a hexagon σ . Let τ be a hexagon adjacent to two of them, say to σ and ρ ( σ ) . Let v = σ ∩ ρ ( σ ) ∩ τ .Since ♣ holds for γ = σ and γ = ρ ( σ ) , by the admissibility of l , v belongs to one of l ( σ ) , l ( ρ ( σ )) . Thus v / ∈ l ( τ ) and hence ♣ holds for γ = τ . Proceeding inductively,by the connectivity of Σ , we obtain ♣ for all γ . (cid:3) Proof of Proposition 5.1.
Case 1.
There are adjacent hexagons σ, τ with non-parallel l ( σ ) , l ( τ ) .Let f = σ ∩ τ . Without loss of generality l ( σ ) ∩ f = ∅ , l ( τ ) ∩ f (cid:54) = ∅ . Let v be thevertex of f outside l ( τ ) . By the admissibility of l , v is contained in l ( σ (cid:48) ) for the thirdhexagon σ (cid:48) incident to v . Hence ♣ holds for e = σ ∩ σ (cid:48) and γ = σ, σ (cid:48) , τ (see Figure 2).Let ρ be the translation mapping σ to σ (cid:48) . Replacing the pair σ, τ with τ, σ (cid:48) andrepeating inductively the argument shows that ♣ holds for γ = ρ n ( σ ) , ρ n ( τ ) for all n > (note that e gets replaced by parallel edges in this procedure). BELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTIN GROUPS 7 σ σ (cid:48) v τ
Figure 2.
Furthermore, by the admissibility of l , since l ( ρ − ( τ )) is disjoint from l ( σ ) and l ( τ ) , it leaves us only one choice for l ( ρ − ( τ )) , and it satisfies ♣ for γ = ρ − ( τ ) .Replacing the pair σ, τ with ρ − ( τ ) , σ and repeating inductively the argument showsthat ♣ holds for γ = ρ − n ( σ ) , ρ − n ( τ ) for all n > . It remains to apply Lemma 5.3. Case 2.
All the l ( σ ) are parallel.In this case it suffices to take any edge e intersecting some l ( σ ) . (cid:3) Proposition 5.4.
Let Σ be the Davis complex for W T the triangle Coxeter groupwith exponents { , , } with an admissible polarisation l . Then there is an edge e such that each octagon γ of Σ satisfies ♦ : the diagonal l ( γ ) has endpoints on edges of γ parallel to e . An edge e of Σ lies either in two octagons σ, σ (cid:48) or there is a square with twoparallel edges e, e (cid:48) in octagons σ, σ (cid:48) . The translation of Σ mapping σ to σ (cid:48) is calledan e -translation . Note that if the conclusion of Proposition 5.4 holds, then an e -translation preserves l . Lemma 5.5.
Let e be an edge and ρ an e -translation. If ♦ holds for all octagons γ in one ρ -orbit, then it holds for all γ .Proof. Suppose that ♦ holds for all octagons γ in the ρ -orbit of an octagon σ . Wecan assume e ⊂ σ . Suppose first that e lies in another octagon σ (cid:48) . Then let τ bean octagon outside the ρ -orbit of σ adjacent to some ρ k ( σ ) , say σ . Let (cid:3) , ρ ( (cid:3) ) bethe two squares adjacent to both σ and τ (see Figure 3). By the admissibility of l ,we have that l ( (cid:3) ) , l ( ρ ( (cid:3) )) 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 toall the octagons γ . A. MARTIN AND P. PRZYTYCKI σ σ (cid:48) τ Figure 3.
It remains to consider the case where there is a square with two parallel edges e, e (cid:48) in octagons σ, σ (cid:48) . Let τ be an octagon adjacent to two of them, say to σ and ρ ( σ ) . Let v = e ∩ τ, x = e (cid:48) ∩ τ . Since ♦ holds for γ = σ, σ (cid:48) , each of v, x lies inone of l ( σ ) , l ( (cid:3) ) , l ( σ (cid:48) ) . Thus by the admissibility of l , we have v, x / ∈ l ( τ ) . Any ofthe two remaining choices for l ( τ ) satisfy ♦ for γ = τ . It is again easy to extendthis to all the octagons γ . (cid:3) Proof of Proposition 5.4.
Note that we fall in one of the following two cases.
Case 1.
There is an edge f in octagons σ, τ with l ( σ ) ∩ f = ∅ , l ( τ ) ∩ f (cid:54) = ∅ .Let v be the vertex of f distinct from u = l ( τ ) ∩ f . By the admissibility of l , thevertex v is contained in l ( (cid:3) ) for the square (cid:3) incident to v . Let x be the vertex in τ ∩ (cid:3) distinct from v , and let σ (cid:48) be the octagon incident to x distinct from τ . Bythe admissibility of l , the vertex x is contained in l ( σ (cid:48) ) . Hence ♦ holds for e = σ ∩ (cid:3) and γ = σ (cid:48) , τ (see Figure 4). Let ρ be the translation mapping σ to σ (cid:48) . σ σ (cid:48) τu vz x Figure 4.
Now let (cid:4) be the square incident to u , and let z be the vertex in σ ∩ (cid:4) distinctfrom u . Note that by the admissibility of l , we have z ∈ l ( (cid:4) ) , and consequently ρ − ( x ) ∈ l ( σ ) and ρ − ( v ) ∈ l ( ρ − ( (cid:3) )) . Hence l ( ρ − ( τ )) cannot contain neither z ,nor ρ − ( v ) , nor ρ − ( x ) ∈ l ( σ ) . There is only one remaining choice for l ( ρ − ( τ )) ,and it satisfies ♦ for γ = ρ − ( τ ) .We can now argue exactly as in the proof of Proposition 5.1 that that ♦ holdsfor 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 (cid:48) with l ( σ ) ∩ e (cid:54) = ∅ we have l ( σ (cid:48) ) ∩ e (cid:54) = ∅ .Let σ be any octagon and e an edge contained in another octagon σ (cid:48) and inter-secting l ( σ ) . Let ρ be the translation mapping σ to σ (cid:48) . One can show inductively BELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTIN GROUPS 9 that ♦ holds for octagons γ = ρ n ( σ ) for all n ∈ Z . It then remains to applyLemma 5.5. (cid:3) Note that for l satisfying ♦ for all octagons γ , the values of l on octagonsdetermine its values on squares. Proposition 5.6.
Let Σ be the Davis complex for W T the triangle Coxeter groupwith exponents { , , } with an admissible polarisation l . Then there is an edge e such that such that each -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 (cid:48) in ofthe two following configurations. In the first configuration we have a square withtwo parallel edges e, e (cid:48) in 12-gons σ, σ (cid:48) . In the second configuration we have fourparallel edges e, e (cid:48) , e (cid:48)(cid:48) , e (cid:48)(cid:48)(cid:48) such that e (cid:48) , e (cid:48)(cid:48) lie in a square, e, e (cid:48) in a hexagon φ and e (cid:48)(cid:48) , e (cid:48)(cid:48)(cid:48) in another hexagon, and we consider 12-gons σ ⊃ e, σ (cid:48) ⊃ e (cid:48)(cid:48)(cid:48) . Again, if theconclusion of Proposition 5.6 holds, then an e -translation preserves l . To see thisin the configuration with hexagons it suffices to observe that l ( φ ) (and similarly forthe other hexagon) is not parallel to e : otherwise l ( φ ) would intersect l ( (cid:3) ) for (cid:3) the square containing e ∩ l ( σ ) . Lemma 5.7.
Let e be an edge and ρ an e -translation. If ♥ holds for all -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 5.3and 5.5 and we omit it.
Proof of Proposition 5.6.
We adopt the convention that if we label the vertices ofan edge in a 12-gon σ by v v , then all the other vertices of σ get cyclically labelledby v · · · v .Let τ be a 12-gon and suppose that l ( τ ) contains a vertex v of an edge v v ⊂ τ for a square (cid:3) = v v u u . Let σ be the 12-gon containing u u . Then l ( (cid:3) ) = u v and furthermore l assigns to the hexagon and square containing u u , u u ,respectively, the longest diagonal containing u , u , respectively. Thus the onlythree remaining options for l ( σ ) are the diagonals u u , u u , and u u . Thus wefall in one of the following three cases. Case 0.
There is such a τ with l ( σ ) = u u .Let φ be the hexagon containing u u . Then l ( φ ) is parallel to u u and thereis no admissible choice for l in the square containing u u (see Figure 5). This is acontradiction. σ τφ u u v v Figure 5.
Case 1.
There is such a τ with l ( σ ) = u u .It is easy to see that l agrees with Figure 6 on the hexagon containing v v andthe square containing v v . Thus the only -cell φ with l ( φ ) containing v maybe (and is) the hexagon containing v v . Consequently the only -cell (cid:4) with l ( (cid:4) ) containing v may be (and is) the square containing v v . Denote by σ (cid:48) the 12-gonadjacent to both φ and (cid:4) at the vertex x (cid:54) = v . Then x may lie (and lies) onlyin l ( σ (cid:48) ) . Denote by ρ the translation mapping σ to σ (cid:48) . σ σ (cid:48) τφu u v v v v v v Figure 6.
It is also easy to see that the -cells surrounding σ and ρ − ( τ ) depicted inFigure 7 have l ( · ) as indicated. This leaves only two choices for l ( ρ − ( τ )) , whereone of them leads to Case 0, and the other satisfies ♥ . σ τ Figure 7.
Replacing repeatedly σ and τ in the above argument by τ and σ (cid:48) or by ρ − ( τ ) and σ gives that ♥ holds for γ = ρ n ( σ ) , ρ n ( τ ) for all n ∈ Z . It remains to applyLemma 5.7. Case 2.
There is no such τ as in Case 0 or 1.It is then easy to see that e = v v and ρ mapping σ to τ satisfy the hypothesisof Lemma 5.7. (cid:3) Note that for l satisfying ♥ for all 12-gons γ , the values of l on 12-gons determineits values on squares and hexagons.We are now ready for the following. BELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTIN GROUPS 11
Proof of Theorem B.
Let F be a Euclidean plane isometrically embedded in Φ witha proper (and thus cocompact) action of Z . By Lemma 4.2, the tiling of F is eitherthe standard square tiling, or the one of the Davis complex Σ for W , where W isthe triangle Coxeter group with exponents { , , } , { , , } or { , , } .First consider the case where the tiling of F is the standard square tiling. Wecan partition the set of edges into two classes horizontal and vertical of paralleledges. Let T be the set of types of horizontal edges and T (cid:48) be the set of types ofvertical edges. Since for each square the type of its two horizontal (respectively,vertical) edges is the same, we have m tt (cid:48) = 2 for all t ∈ T, t (cid:48) ∈ T (cid:48) . Moreover, byRemark 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 horizontaltranslation w ∈ A T and a vertical translation w (cid:48) ∈ A T (cid:48) . 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 . Considerits induced polarisation l from Definition 4.5. By Lemma 4.6, l is admissible. ByPropositions 5.1, 5.4, and 5.6, there is an edge e such that each for each γ a maximalsize -cell, the diagonal l ( γ ) has endpoints on edges of γ parallel to e , and there isa 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 > , directedfrom g to gs k . If k > , 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 -cells that contain f .Consequently all the edges crossing the bisector of f are k -long. Moreover, all suchbisectors are parallel, since otherwise the -cell (cid:3) where they crossed would havefour long edges, so (cid:3) would be a square by Lemma 3.1. Analysing l in the 2-cellsadjacent to (cid:3) leaves then no admissible choice for l ( (cid:3) ) .Furthermore, by Lemmas 3.1, 5.3, 5.5 and 5.7, if f is a long edge, then we canassume that f is parallel to e .Suppose first that W T is the triangle Coxeter group with exponents { , , } and T = { s, t, r } . Let ω be a combinatorial axis for the action of ρ on Σ . Since none ofthe edges of ω are parallel to e , by the definition of the induced polarisation we seethat they are all directed consistently (see Figure 8). s ss t ttt • r rr Figure 8.
Dashed lines indicate possible polarisations.Thus, up to replacing F by its translate and interchanging t with r , the element strstr preserves ω , and coincides on it with ρ . In fact, ρ not only preserves thetypes 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 wordin t • str ⊂ A T . That word depends on whether the second edge of the path is longand on the polarisation. Since Z acts cocompactly, there is a power of tstr suchthat its corresponding path in Σ ⊂ Φ reads off a word in ( t • str ) ∗ ⊂ A T that is theother generator of the orientation preserving stabiliser of F in A S . This brings us toCase (b) in Theorem B. One similarly obtains the characterisations of orientationpreserving stabilisers of F for the two other W (see Figures 9 and 10). s sss t ttt ttt r rrrs • t • r • Figure 9. s sss s sss st t t ttt ttttt t rrr rrs • t • Figure 10.
BELIAN SUBGROUPS OF TWO-DIMENSIONAL ARTIN GROUPS 13 (cid:3)
Remark 5.8.
Analysing the full stabilisers of F in A S one can easily classify alsothe subgroups of A S acting properly on Φ isomorphic to the fundamental groupof the Klein bottle. For example, suppose that the second generator of the Z inCase (b) of Theorem B has the form g = ( t k str )( t − k str ) . Then our Z is generatedby strstr and g (cid:48) = g ( strstr ) − = t k sr − k s − . Note that str normalises our Z with ( str ) − g (cid:48) ( str ) = ( g (cid:48) ) − . Thus (cid:104) str, g (cid:48) (cid:105) is isomorphic to the fundamental group ofthe Klein bottle. We do not include a full classification, since it is not particularlyilluminating. References [AS83] K. I. Appel and P. E. Schupp,
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