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ISSN (print): 2251-7650, ISSN (on-line): 2251-7669 Vol. 4 No. 3 (2015), pp. 77-117.
c
⃝2015 University of Isfahan
www.ui.ac.ir
CONJUGACY IN RELATIVELY EXTRA-LARGE ARTIN GROUPS
ARYE JUH ´ASZ
Dedicated to the memory of my friend David Chillag
Communicated by Patrizia Longobardi
Abstract. In this work we consider conjugacy of elements and parabolic subgroups in details, in a new class of Artin groups, introduced in an earlier work, which may contain arbitrary parabolic
subgroups. In particular, we find algorithmically minimal representatives of elements in a conjugacy class and also an algorithm to pass from one minimal representative to the others.
1. Introduction
Let A be an Artin group with standard generators X = {x1, . . . , xn}, n ≥ 1 and defining graph ΓA. (See 2.1 for definition.) Astandard parabolic subgroup of A is a subgroup generated by a subset of X. For elements u and v of A we say (as usual) that u is conjugate to v by an element h of A if h−1uh=v holds inA. Similarly, ifK andLare subsets of AthenK is conjugate to Lby an element
h of Aifh−1Kh=L.
In this work we consider the conjugacy of elements and standard parabolic subgroups of a certain type of Artin groups. Results in this direction occur in [4, 6, 7, 8, 10, 14, 13, 15], for example: Of particular interest are centralisers of elements, and of standard parabolic subgroups, normalisers of standard parabolic subgroups and commensurators of parabolic subgroups. In this work we consider similar problems in a new class of Artin groups, introduced in [11] where the word problem is solved, among other things. Also, intersections of parabolic subgroups and their conjugates are considered. To define this class of Artin groups, letNAbe the adjacency matrix of ΓA, such that ifvi, vj ∈V(ΓA) and the vertices vi and vj are connected with an edge labelled by a natural number λ(vi, vj) = ni,j
MSC(2010): Primary: 20F02; Secondary: 20F36.
Keywords: Artin groups, Relative presentations, Small cancellation theory, Conjugacy in groups. Received: 13 March 2015, Accepted: 30 September 2015.
then the (i, j) entry ofNA isnij. Ifvi andvj are not connected by an edge then the (i, j) entry ofNA is 0.
Let H be a standard parabolic subgroup of A, generated by {x1, . . . , xk}. We say that A is extra-large relative to H if NA decomposes into
(
NH B
C D
)
, where NH is the defining submatrix of H,
B and C are block matrices of size l×k and k×l, respectively, in which all the non-zero entries are at least 4 and D is a block matrix of size l×l in which all the non-zero entries are at least 3, where k+l=n. There are no assumptions on NH.
We denote by E0,1 the edges of ΓA which have one endpoint in {v1, . . . , vk} and the other end-point is in {vk+1, . . . , vn}. Similarly, we denote by E1 the set of edges of ΓA with both endpoints in
{vk+1, . . . , vn} and by E0 the set of all the edges with both endpoints in {v1, . . . , vk}. As usual,vi is
the vertex of ΓA which corresponds to the generatorxi,i= 1, . . . , n.
To formulate our results we have to introduce some classes of words which describe minimal elements and their conjugacy, (see below). Thus, let 1̸=u be an element of A represented by a reduced word U. Since we are interested in conjugacy, it is convenient to consider the cyclic word Ub, rather than the linear word U and we shall do so. As mentioned, we are interested, as usual, to represent Ub by a word which is minimal in some sense. In this work we allow three operations to reduce a word W, W =xα1i1 · · ·xαm
im : ζ,κ and L, where
(1) ζ(W) - the freely reduced form of W
(2) κ(W) - the sequence of cyclic conjugates of W, starting withxα1i1 thenxα2i2 and so on
(3) L(W) - the sequence of words obtained from W := W1W2W3 reduced as written, where W2
is a subword of a relator R having length ≥ 12|R|, by replacing W2 with its complement W2′,
W2W ′−1 2 =R.
Notice that ζ,κ, and Lare well defined.
For a set Y of words let ζ(Y) = {ζ(Y) | Y ∈ Y}, κ(Y) = {κ(Y) | Y ∈ Y}, and define L(Y) =
{L(Y) | Y ∈ Y}. Apply (ζ ◦ L ◦ζ ◦κ)m, m = 0,1,2, . . . on W. We get a set of words of length at most |W|. Therefore, there is a uniquely defined number m0 such that (ζ ◦ L ◦ζ ◦κ)m0(W) =
(ζ◦ L ◦ζ◦κ)m0+1(W) =· · ·
Denote the set of words obtained byσ(W). Thus σ(W) = m0∪
i=1
(ζ◦ L ◦ζ◦κ)i and σ(W) is invariant
under ζ,κ and L.
To define the minimal representatives, we first construct a directed rooted tree Λ0(W), the vertices
of which are labelled by the words in σ(W) The root of Λ0(W) is the vertex v0, labelled withW. In
application of L on the words in the third level. Continue this way (to at most 2m0 levels) until no
new words are obtained.
We obtain Λ(W) from Λ0(W), as follows:
(*) If a word in level iis shorter than its immediate predecessor, then erase all its predecessors.
The different words in each tree (connected component) form an equivalence class, which is the orbit under ⟨K,L⟩ on any one of them. Denote the equivalence class of a word U by [U] and let U1, . . . , Um be the different representatives of the classes, (one representative for each class). Then we defineρ0(W) =
m
∪
i=1
[Ui]. Notice thatρ0(W) is well defined and can be constructed algorithmically.
Notice that all the words in the same component have equal lengths, henceL, when applied, replaces a subrelator by a subrelator of the same length.
In Theorem A we describe a modificationρ(W) ofρ0(W) and the interrelation between its elements.
Thus, let W =xa1i1 · · ·xam
im be a word in F :=⟨X|−⟩,xij ̸=xij+1,j= 1, . . . , m−1. We define now the possible sets of words forρ(U).
W1 = {cW |W =U1W1U2W2· · ·UkWk, k = 1,2,3, . . . |
Ui∈F(V0), Wi ∈F(Vi)} W′
2(a, b, c) = {cW |W =W1U1· · ·WkUk, k= 1,2,3, . . . , Ui ∈F(V0) and either
Wi=aαibβiaγiorWi=aαicβiaγi,and Ui=U′
i inH, whereUi′ =cδi inF in the first case andU′
i =bδi orcδibξi orbδiaξi inF in the second case, for all i,where {a, b, c} ⊆X, λ(a, b) =λ(a, c) = 4 and λ(b, c) = 2 and at most one of |αi|,|βi|,|γi|,|δi|is greater than 1, where
αi, βi, γi, δi ∈Z\ {0}.}
W′
3(a, b, c) = {cW |W =W1U1W2U2· · ·WkUk, k = 1,2,3, . . . , Ui∈F(V0), and
Ui=Ui′ inH, Wi =aαi and Ui′ =bβicγi inF forieven and Ui′=cγibβi fori odd. whereλ(b, c) = 2 and λ(a, c) =λ(a, b) = 4 and if one of
|βi|and|γi|is greater than 1, where|αi|= 1, where αi, βiγi ∈Z\ {0}}
W4(a, b, c) = {cW |W =W1· · ·Wk, k= 1,2, . . . , Wi=aαibβi fori≡1 mod 3,
Wi=cγiaαifori≡2 mod 3 andWi =bβicγi fori≡0 mod 3, where {a, b, c} ⊆X and λ(a, b) =λ(b, c) =λ(a, c) = 3 and
αi, βi, γi ∈Z\ {0} and at most one of|αi|,|βi|and|γi|are greater than 1.}
W5 = {am |a∈V1 , m= 1,2, . . .}
To obtain ρ(U) from ρ0(U) we replace Ui by Ui′ in W2′(a, b, c) and W3′(a, b, c). We need to make
Next, we describe the elements which conjugate two words in ρ(U). For elements of Wi we denote the set of these words by W⊥
i ,i= 1,2,3,4,5.
W⊥
1 = {an |a∈X, n∈Z} W⊥
2 = {W |W =W1· · ·Wk, k = 1,2, . . . , Wi=cαibβiaγi, where {a, b, c} is as in W2 and at most two of|αi|,|βi|,|γi|is greater than 1}
W⊥
3 = {W |W =W1· · ·Wk, k = 1,2, . . . , Wi =aαibβiaγicδi, {a, b, c} as inW3 and no consecutive exponents in (αi, βi, γi, δi, αi) may be both with absolute value greater than 1}
W4⊥ = {W |W =W1· · ·Wk, k = 1,2, . . . , Wi=aαicβibγi, {a, b, c}
as inW4 and as inW⊥
3 no two consecutive exponents may have
absolute value at least 2}
To define W5⊥ we need to introduce some notations:
Let ΓAbe the defining graph ofA. In what follows we identify the vertices of ΓAwith the generators they represent. Let N(v) = {w ∈ V(ΓA), w ̸= v | (w, v) ∈ E(ΓA)}. Then N(v) = No(v)∪Ne(v) whereNo(v) ={w∈N(v)|λ(v, w) is odd}andNe(v) ={w∈N(v)|λ(v, w) is even}. Foru∈Ne(v) letW(u) = (uv)λ(u,v)−1u and letNbe(v) =⟨W(u)|u∈Ne(v)⟩. Now let µ=v
1µ1v2µ2· · ·vm be a path
in ΓA,vi∈V(ΓA),µi ∈E(ΓA).
Call µ admissible if vi+1 ∈ No(vi), i = 1, . . . , m−1. Denote by Z(v1, vm) the set of all the
admissible paths from v1 tovm. Forµ∈Z(v1, vm) letW(µ) ={W1· · ·Wm |Wi ∈Nbe(vi)} and finally
let J(v1, vm) = ∪
µ∈Z(v1,vm)
W(µ). Thus,J(v1, vm) is the set of all words which conjugate v1 tovm and
are products of m subrelators of lengthsni−1, whereni =λ(vi, vi+1), i= 1, . . . , m−1.
Define
W5⊥(a, b) =J(a, b) {a, b} ⊆X
From now on ρ, Wi and Wi⊥, i= 1,2,3,4,5 will denote sets of words, as introduced above. Our main results are Theorems A, B and C.
Let A be an Artin group and let H be a standard parabolic subgroup of A. Suppose that A is extra-large relative to H.
Theorem A Let W be a cyclically reduced word inF. One of the following holds (a) ρ(W)⊆H
(b) ρ(W)∩F(V0) =∅, ρ(W) = [U]for a cyclically reduced word U. In this case all the elements
of ρ(W) have the same length and if U1 andU2 are inρ(W) thenU2 can be obtained fromU1
via the reduction procedure described above.
(c) ρ(W)∩F(V0) =∅ and there are U1 and U2 in ρ(W) with [U1]̸= [U2]. In this case there is a
forming a(2,4,4)or(3,3,3)triangle inΓAsuch that for certain wordsU1∗ ∈[U1]andU2∗∈[U2]
the following hold (i) U∗
1, U2∗∈ Wm
(ii) U1∗ is conjugated to U2∗ by an element of W⊥
m
Theorem B deals with conjugacy of elements in H, while Theorem C deals with conjugacy of elements inA\H.
Theorem B Assume u, v∈H and g∈A.
(a) If g−1ug =v and g ̸∈H then u is conjugate in H to xm1 for some x ∈V
0 and m1 ∈Z\ {0}
and v is conjugate inH to ym2, for somey ∈V
0 and m2∈Z\ {0}.
(b) IfP ⊆H is a non-cyclic parabolic subgroup of AthenCA(P) =CH(P),NA(P) =NH(P) and the commensurator ofP in A is its commensurator in H.
(c) Let K be a subgroup ofH such that⟨z⟩ ∩K={1}, for every z∈V0. IfK is malnormal inH
thenK is malnormal in A.
Theorem C
(a) Let P be a non-cyclic parabolic subgroup ofA, P ̸⊆H.
ThenCA(P) =CH(P)Z(P), where Z(P) is the centre of P.
(b) Let P and Q be non-cyclic parabolic subgroups of A, let t ∈ A\H and let Z = t−1P t∩Q. Assume P ̸⊆H and Q̸⊆H.
Then one of the following holds (i) Z={1}
(ii) Z is cyclic (iii) t∈ ⟨P, Q⟩
(c) Let P be a non-cyclic parabolic subgroup of A, P ̸⊆H. Letg∈A\H and let u, v be elements of P which are not conjugate to the power ofxε∈X, ε∈ {−1,1}.
Ifg−1ug=vtheng∈P. In particular,NA(P) =P CH(P)andP is its own commensurator.
The work is organised as follows.
Section 2 contains preliminary results on Artin groups and their diagrams, which are the main tools of the work. In particular, we recall results on a small cancellation condition (V(6)) which is satisfied by certain modified diagrams ( ˜M′). We also discuss the ways conjugacy of elements is expressed in these modified diagrams.
Section 3 recalls known results from [1] on diagrams of Artin groups generated by two elements and also consider a special kind of diagram, needed for the proof of the main results.
Section 4 contains the basic structure theorems of simply connected and of annular diagrams recalled from [10] and [11] and some consequences needed in the sequel.
In Section 5 using a result of S. Orevkov we show that in any Artin group on at least three standard generators the product of powers of three generators is a shortest representative of the element it represents.
2. Preliminary Results
2.1. Artin groups, notation and definitions. (a) Let X={x1, . . . , xn}, n≥2, be a set and let
F =F(X) be the free group, freely generated by X. For 1 ≤i < j ≤n let nij be natural numbers, let Rij be the empty word if nij = 0 and let Rij be the word (relator) inF defined by Rij =UijVij, whereUij is the head of lengthnij of (xixj)nij andV is a head of lengthnij of (x−1
j x−1i )nij, ifnij >0. Thus Rij is (xixj)mijxi(x−1
j x
−1
i )mijx
−1
j if nij is odd, nij = 2mij + 1 andRij is (xixj)mij(x
−1
i x
−1
j )mij if nij is even, nij = 2mij. Let N be the n×n matrix defined by (N)(i,j) = nij if i < j, nji = nij,
nii= 0. The groupA defined by the presentation
(2.1) ⟨x1, . . . , xn |Rij, nij ̸= 0⟩
is called the Artin group defined by N. The xi are called the standard generators of A and the presentation (1) is called the standard presentation ofA defined by N. The subgroups ofA generated by subsets of X are calledstandard parabolic subgroups. An alternative way to define Ais by a graph Γ with vertex set V(Γ) = {v1, . . . , vn} in one to one correspondence with x1, . . . , xn and the edge set
E(Γ) ={(vi, vj) |nij ̸= 0} such that the edge (vi, vj) is labelled withnij. We shall identifyvi withxi and denote nij =λ(vi, vj), except in the definition ofJ(a, b) in the introduction.
Artin groups for whichnij ≥4 for all non-zeronij are called extra-largeand those for whichnij ≥3 for all non-zero nij are called large. An Artin group in which all the nonzero nij are 2 is called a right-angled Artin group.
To each Artin group given by (1) there corresponds a Coxeter group obtained from (1) by adding the relationsx21, x22, . . . , x2n. An Artin group is calledof finite type if the corresponding Coxeter group is finite.
We shall often use the following general results.
Theorem 2.1 ([13] (Parabolics embed)). Let A be an Artin group given by (1) and let S be a subset of X. Let ΓS be the subgraph of the defining graph Γ of A. Let P =⟨S⟩. Then P is an Artin group with defining graph ΓS.
Theorem 2.2 ([3] (Parabolics are convex)). Let notation be as in Theorem 2.1 and let W ∈F(S) be a word. If U is a shortest representative for W in A thenU ∈ ⟨S⟩.
Finally, when it causes no ambiguity we shall not distinguish between elements of A and their representatives inF and similarly between elements ofH and words of F(x1, . . . , xk) which represent
them.
2.2. Diagrams and small cancellation conditions. For diagrams in general we follow [12, Ch. V]. We denote by Φ the labelling function of a diagram M over a free group F. (Thus, Φ :M →F). We denote by Reg(M) the set of all the regions of M. We shall also use the following notation.
Let W be a reduced word in F(x1, . . . , xn). Denote, as usual, by ||W|| its syllable length and
denote by l(W) its word length. For a word W =xα1i1 · · ·xαm
{xi1, . . . , xim}. Thus, if W = x
2
1x32x−11 x44 then ||W|| = 4, l(W) = 2 + 3 + | −1|+ 4 = 10 and
Supp(W) ={x1, x2, x4}. Ifµis a path inM we write l(µ) forl(Φ(µ)) and for a pathµwrite Supp(µ)
forSupp(Φ(µ)). Also, for a region Din M writeSupp(D) forSupp(Φ(∂D)). For a path µdenote by h(µ) and t(µ) the endpoints ofµ,h for head and tfor tail.
It was observed already in [1] that Artin groups on more than two standard generators do not satisfy small cancellation conditions. To overcome this difficulty, a modification has been carried out on the diagrams which declares certain subdiagrams as new regions - which we will call in this work derived regions - such that the obtained diagrams - which we will call in this workderived diagrams - satisfy the small cancellation conditon C(6), for large Artin groups (see [1], for the definition of the C(6) condition see [12, Ch. V]). It turns out that for our work we have to carry out a further modification on the derived diagrams, as described in [11], in order to get a diagram which satisfies a more involved small cancellation condition. We describe both modifications below.
2.2 (a) The underlying small cancellation condition
The main tool we use in this work is the small cancellation condition V(6) and its theory which we recall in Section 4. In this subsection we present the main notions involved.
Definition 2.3 ([11]). Let M be a map. M satisfies the small cancellation condition V(6) if each of the following holds:
(i) Each inner region has at least 4 neighbours;
(ii) If an inner region has less than 6 neighbours then every vertex on its boundary has valency at least 4.
Notice that theC(4)&T(4)and also the C(6) small cancellation conditions are special cases of the V(6) condition.
Definition 2.4. Let D be a boundary region of M. Call D a k-corner region of M, k∈ {1,2,3} if the following hold
(1) ∂D∩∂M is connected; (2) D has k neighbours in M
Definition 2.5 (one layer maps). (a) Let M be a connected, simply connected regular map (i.e. every edge is on the boundary of a region). M is called a one layer map if its dual is a line segment.
(b) Let A be an annular regular map. A is called a one layer map if its dual is a circle.
Finally, we describe the basic properties of V(6) diagrams, as we shall need this later.
The main result of [11] on annular diagrams A with the small cancellation condition V(6) is that A splits into annular subdiagrams with boundaries homotopic to the outer boundary ω(A) and the inner boundary τ(A) of A, respectively and the regions inside each layer are controlled and behave ”nicely”.
Denote by dM(D), as usual, the number of neighbours (neighbouring regions) of a region D inM. For a vertex v inM denote, as usual, bydM(v) the valency of v inM.
We denote for a boundary path µthe number of segments it contains by |µ|, where a segment is a path with endpoints having valency at least 3 and every inner vertex has valency 2.
We have the following:
Theorem 2.6 ([11], Theorem 1). Let A be a regular annular map with connected interior which satisfies the condition V(6). Let L0 be an annular submap with boundaries homotopic to the boundary
of A, with minimal possible number of regions. DefineS0 =L0 and for i >0 define
Li ={D̸∈Si−1 | ∂D∩∂Li−1 ̸=∅}
and define
Si =Si−1∪Li
Similarly, for i <0 define
Li ={D̸∈Si+1 | ∂D∩∂Li+1 ̸=∅}
and Si=Si+1∪Li.
(a) Then, Si is annular and A is the union of all the Si. Either Li is an annular one-layer map, or Li is the union of simply connected one-layer maps. Also, for i >0 |∂Si| ≤ |∂Si+1|, and
similarly fori <0.
For i≥0 we denote by ω(Li), (ω(Si)), the outer boundary of Li (Si) and denote by τ(Li) (τ(Si)) the inner boundary of Li (Si).
Similar notations used for i <0, accordingly.
(b) Let D be a region in Li, i≥0. Then D has at most 2 neighbours in Li. If dM(D) <6 then Dhas at most 2 neighbours in Li−1 and if it has exactly two neighbours then there is a region
E in Li with ∂E∩ω(Li−1) ={v},v a point. We call such regions 3-exceptional. If d(D)≥6
then D has at most 3 neighbours in Li−1 and if it has exactly three neighbours in Li−1 then
there is a region E in Li which has at most one neighbour in Li−1. We call such regions
2-exceptional. Similar statement holds for D∈Li, i <0.
(c) Letv be a vertex in ω(Li). ThendLi(v)≤4and dLi(v) = 4 if and only ifv is on the boundary of a region D in Li with dA(D) = 4and has two neighbours in Li−1. Similar statements hold
for v∈τ(Li), i <0.
For further application for a region Din Li,i≥1, denote
A(D) the set of neighbours ofD inLi−1, and a(D) =|A(D)|, B(D) the set of neighbours ofD inLi, and b(D) =|B(D)|,
C(D) he set of neighbours ofD inLi+1 (Li−1 ifi <0), andc(D) =|C(D)|.
Theorem 2.7 ([11], Theorem 2). Suppose M is simply connected with connected interior containing at least two regions. If M satisfies the small cancellation condition V(6), then for every vertex v M has a layer structure with L0 ={v}, as described above, with the following properties:
(1) Every vertex v on ω(Li) has valency ≤3 in Li.
(2) The last layer contains at least two corner regions of M.
(2’) The last layer contains a corner region withl(∂M ∩∂∆)≥ 12l(∂∆)
2.2 (b) Derived diagrams
Let M be an R-diagram, R a set of Artin relators. Say that two regions are neighbours or are adjacent if they have a common edge. Say that two adjacent regions D1 and D2 are friends, if
Supp(D1) =Supp(D2). Let ′′∼′′ be the transitive closure of friendness. Then ′′∼′′ is an equivalence
relation on Reg(M). For a regionD∈Reg(M) denote by [D] the equivalence class of Dand let ∆(D) =Int (∪E¯)
E ∈[D] , where ¯E denotes the closure of E inE
2.
It is standard to see that if M is simply connected than ∆(D) is homeomorphic to an open disc, hence we may consider ∆(D) as a region. We denote byM′the diagram obtained by considering ∆(D),
for D ∈Reg(M), as regions and call M′ the derived diagram of M . We call {∆(D)|D∈ Reg(M)}
the derived regions of M′. Observe that l(∂M′) =l(∂M) and Φ(∂M′) = Φ(∂M), since by definition ∂M =∂M′. The main property of these diagrams is given by the lemma below.
Lemma 2.8 ([1], Lemma 3). Let notation be as above and let ∆1 and ∆2 be derived regions of M′.
Then ||∂∆1∩∂∆2|| ≤1.
As we shall see, combining Lemma 2.8 with Lemma 3.1 (a) implies that derived diagrams of Artin groups of large type satisfy the small cancellation condition C(6).
2.2 (c) Derived Howie Diagrams
The presentation (1) is usually called afree presentation, becauseA is presented as a homomorphic image of a free group. In the sequel we shall also need a free-product presentation of A, sometimes also called a relative presentation. See [2]. Thus, let V0 be a subset ofV and letV =V0
·
∪V1. LetR0
be the symmetric closure (s.c.) of the set of all the relations involving letters ofV0 only, letR1 be the
s.c. of the set of all the relations involving letters of V1 only and let R0,1 be the s.c. of the set of all
the relations Rij in which i∈V0 and j∈V1.
Clearly,R=R0∪ R0· ,1 ·
∪ R1, whereRis the s.c. of {Rij}in (1). ThenF(V0)/R0 is isomorphic to
the subgroupHofAgenerated by the images ofV0 by the natural projectionπ:F →Acorresponding
to (1) andA is isomorphic to
(2.2) ⟨H∗F(V1) | R0,1∪ R1⟩
obtained from M′ by shrinking the R0-subdiagrams to a point. (See [11]). Then the regions of ˜M′
are RegE1( ˜M′) ∪. RegE0
,E1( ˜M′), where RegE1( ˜M′) is the set of regions of ˜M′ with boundary label from R1 and RegE0,E1 is the set of regions of ˜M′ with boundary label from R0,1, after shrinking the
edges with labels in V0 to vertices and introducing them as corner labels. (This notion has nothing
to do with corner regions.) See Figure 1. We call such vertices, with corner labels from V0, coloured
vertices. The main properties of ˜M′ for relatively extra-large M are summarised below.
D
D
D
D
D
D
D
1
4 5
6 2
3 a
b
b a
b a
c c
c c
c c
⇝
D
D
D
1
5 3
a c c
a c c
a c
c b
b b
c, a∈V0 , b∈V0,1
Figure 1
Lemma 2.9 ([11], Lemma 2.1). Let ∆′ be an inner region of M′,∆′ simply connected. Denote by∆˜′ the corresponding region in M˜′, for ∆′ ∈RegE1(M′)∪RegE0,E1(M′).
(a) If ∆˜ ∈RegE1( ˜M′) then the corner label of ∆˜′ at each boundary vertex is 1
(b) If∆˜′∈RegE0,E1( ˜M′)withSupp( ˜∆′) ={ai, aj},ai ∈V
0,aj ∈V1 then at every boundary vertex
of ∆˜′ the corner label is aαi
i , αi ∈Z\ {0}.
In particular, every boundary vertex of∆˜′ is a coloured vertex.
(c) If v is an inner coloured vertex of M˜′, then d ˜
M′(v)≥4.
We also have the following
Lemma 2.10 ([11], Lemma 2.2). Let ∆˜ ∈Reg( ˜M′), ∆simply connected.
(a) If ∆˜ ∈RegE1( ˜M′) and is an inner region of M˜′ then dM˜′(∆)≥6.
(b) Let ∆˜ ∈RegE0,E1( ˜M′). Then
(i) Every inner vertex of ∂∆˜′ has valency at least 4. (ii) If ∆˜ is an inner region of M˜′ thend
˜
M′(∆)≥4.
dM˜′(∆)≥4 and every inner vertex has valency at least 4.
(c) Every region in M˜′ either belongs to RegE1( ˜M′) or to RegE0
,E1( ˜M′) and M˜′ satisfies the condition V(6).
2.3. Conjugacy diagrams. We follow [12]. LetA be a group.
such that Φ(τ) = Φ(σ) = T, Φ(µ) = U and Φ(ν) = W. Since Φ(σ) = Φ(τ), we can glue (identify) σ toτ to get a map M, without distorting Int(M0) and without leaving the plane. See Figure 2 for
possible outcomes.
..
µ
. τ
.
σ
. ν
. M0
..
⇝
..ν .
σ =τ
.
µ
.
M
(a)
.. τ
. u2
. ν1
. ν2
.
σ
.
u1
.
µ2
.
v
.
µ1
..
⇝
..µ1
.
u1
.
µ2
.
σ =τ . u.2
ν1
. ν2
(b)
.. τ
. ν
.
σ
.
µ
..
⇝
..µ
. σ=τ
. ν
(c)
Figure 2
In Figure 2(a) Int(M) is not connected. As a result ν is not a simple closed curve. This means thatW contains a non-empty subwordW0 withW0 = 1 inA. To avoid this situation we shall require
that
In Figure 2(b) µ∩ν ̸=∅. As a result we can subdivide µ intoµ=µ1wµ2u1 and subdivide ν into
ν =u2ν1wν2 such that in M Φ(µ2u1µ1) = Φ(ν2u2ν1). Hence, if we let Ui = Φ(µi) andWi = Φ(νi),
i= 1,2 then U2U1 =A W2W1. SinceU2U1 =U1−1U U1 and W2W1 =W1−1W W1, we see that a cyclic
conjugate of U equals in A to a cyclic conjugate of W. Concerning the cyclic words ˆU and ˆW this means that ˆU =AWˆ.
Since we deal with cyclic words ˆU and ˆV, in what follows we shall assume, without loss of generality, (H1) and the following
(H2) U andW have no cyclic conjugates U∗ and W∗ with U∗ =AW∗.
(H3) U andW are cyclically reduced.
Under these assumptionsInt(M) is an annular diagram with boundary labelsU and W, respectively. See Figure 2(c).
Now letAbe a relatively extra-large Artin group and letM be an annularR-diagram with connected interior. When formingM′ fromM then in contrast to the simply connected case it may happen that ∆(D) is not homeomorphic to an open disc. Due to a standard argument, using the fact that every 2-generated Artin group is torsion free, it follows that in this caseInt(∆(D)) is annular with boundary components homotopic to the boundary components of M. It will turn out (see Section 6) that the structure of derived diagrams with such a derived region is different from the structure of annular derived diagrams with simply connected derived regions.
Next, when forming ˜M′ from M′, we shrink some of the edges of the annular diagram M′. As a
result, we get
Lemma 2.11. Let notation be as above. Then M˜′ satisfies one of the following:
(a) For every D∈Reg(M),∆(D) is simply connected.
Subcase 1M˜′ contains an annular subdiagramMv overF(V
0), maximal with respect to being
over F(V0), which is homotopic to the boundary components of M˜′. See Figure 3(a).
ThenM˜′ is the union of two simply connected diagrams M˜′
1 and M˜2′ withM˜1′ ∩M˜2′ ={v},
pointv corresponds toMv in M′. See Figure 3(a).
Subcase 2 M′ contains no subdiagram M as in Subcase 1. Then one of the following holds:
(i) There is no R0-subdiagram N′ in M′ with ∂N′∩ω ̸= ∅ and ∂N′∩τ ̸=∅, where ω and τ are the boundary components of M. Then M˜′ is an annular diagram with connected
interior.
(ii) There is anR0-subdiagram N′ in M′ with ∂N′∩ω ̸=∅ and ∂N∩τ ̸=∅. Then Int( ˜M1′)
is simply connected, whenM˜′
1 = ˜M′\ {v}. See Figure 3(b).
(b) ∆(D) is annular for someD∈Reg(M). See Figure 3(d).
. .
w .
τ .
N′
..
⇝
..ω
. ˜ M1′
.vN′
.
˜ M2′
.
τ
..
⇝
..w
. ˜ M′
1
. vN′
. τ .
˜ M2′
.
vN′
(a)
.. ω
. τ . N′
..
⇝
..vN′
.
τ .
ω .
˜ M′
(b)
.. ω
. ∆(D)
. τ
.
˜ M′
. .
ω
. ∆(D)
.
τ .
˜ M′
(c) (d)
Proof. Clearly all the possibilities are covered. The results follow immediately from the construction
of ˜M′. □
Lemma 2.12. Let ⟨X | R⟩ be a presentation of a group and let M0 be a minimal simply connected R-diagram with connected interior and boundary cyclev1µv2ηv3νv4θ, such that
(1) Φ(µ) =U, Φ(ν) =W, Φ(η) =T and Φ(θ) =T−1
(2) U, W and T are shortest cyclically reduced representatives.
Since Φ(η) = Φ(θ)−1, we can glue θ alongη to get an annular diagram M1. Let M be the annular
diagram obtained from M1, after cancelling out all cancelling pairs. Then
(1) ω(M) =ω(M1) and τ(M) =τ(M1)
(2) M contains a path ξ with endpointsv1 and v4 such that Φ(ξ) =AT.
.. ξ . v1
.
v2
.
v3
. v4
.
µ
.
ν .
θ .
η
(a)
..
E
.
D
.
ω
.
ω
.
v2
.
η
.
z1
.
ηE
.
ξE
.
z2
.
τ .
τ .
ξD
.
ηD
.
v3
.
η0
.. z1
.
ξE
. ξD
. η0
. z2
.
⇝
.
ξD
.
ξE
.
z2
.
z1
.. ↓
z1
.
z2
.
ξD =ξE .
←−
.
z2
. z1
.
ηE .
ηD
(c)
Figure 4
Proof. Let (D, E) be a cancelling pair. Since M0 is minimal, D and E are boundary regions, which
became a cancelling pair due to the identification of η with θ. Hence, after the identification ∂D∩
∂E∩η ̸=∅. Let η0 be a connected component of ∂D∩∂E∩η. Let ηD be the complement ofη0 in
∂D and let ηE be the complement of η0 in ∂E. See Figure 4(b). Since (D, E) is a cancelling pair,
Φ(ηD) = Φ(ηE). To carry out the cancellation, we add two simple paths inside D and E, ξD and ξE respectively, with labels Φ(ηD) and Φ(ηE), respectively. Now we replace η0 with ξD and identify ξD withξE. See Figure 4(c). (We can do this because Φ(ηD) = Φ(ηE) and Int(D) is homeomorphic to an open disc.) This way we replaced η0 by ξD(= ξE). Finally, we identify ηD with ξD and ηE
with ξE. See Figure 4(b). This ends the cancellation of the pair (D, E) towards getting M from M1.
Since Φ(ηD) =A Φ(η0), after carrying out this procedure for all cancelling pairs we did not change
Φ(η) mod N, where N is the normal closure of R in F. Consequently, after carrying out all the cancellations of all the cancelling pairs we get a path η′ with Φ(η′) ≡Φ(η) modN. Since the above procedure does not change ω and τ (though may distort the inner structures by identifying certain
edges or vertices in course of the procedure), our claim holds. □
3. Two-generated Artin groups and their diagrams
Two-generated parabolic subgroups of Artin groups are very important ingredients of the study of Artin groups via diagrams. (See [1].) The following two results from [1] are fundamental.
Lemma 3.1 ([1], Lemma 6). Let A be given by the presentation ⟨x1, x2 | R12⟩ where R12 is the
symmetric closure of R1,2 in F, and let 1 ≠ W be a cyclically reduced word in F(x1, x2) which
represents 1 in A. Then
(b) Every R12-diagram satisfies the small cancellation conditionC(4) &T(4). (See [12, Ch. V].)
We shall need a more detailed version of part (a) of the Lemma for the special case when m :=
||R1,2||/2 = 4. For this we need the Lemma below.
Lemma 3.2 ([1], Lemma 7). Let 1̸=W be a cyclically reduced word inF(x1, x2) which represents 1
in A. LetW =W1W2, reduced as written.
(a) If ||W1|| ≤n1,2
(
= ||R1,2||
2
)
then|W1| ≤ |W2|.
(b) If ||W1||< n1,2 then |W1|<|W2|.
Corollary 3.3. LetAbe an Artin group given by (1). Let xα11 xα22 W = 1inA,W an arbitrary reduced word. Then |W| ≥ |α1|+|α2|.
Proof. Without loss of generality we may assume that xα11 xα22 V is cyclically reduced, as written. By Lemma 3.2 |xα11 xα22 | ≤ |Z|, for every Z∈ ⟨x1, x2⟩ with Z−1=Axα11 xα22 .
Let U be a shortest representative of xα11 xα22 in A. Then by Theorem 2.2 U ∈ ⟨x1, x2⟩, hence |xα11 xα22 | ≤ |U|. But clearly |U| ≤ |xα11 xα22 |by choice of U, hence xα11 xα22 is a shortest representative
forxα11 xα22 and in particular|W| ≥ |α1|+|α2|. □
Corollary 3.4. Let x1 x2 x3 ∈X. Then xα1x
β
2x
γ
3 ̸= 1 in A, if not all of α, β, γ are zero.
Proof. Supposexα1xβ2xγ3 = 1 inA. Then due to Corollary 3.3 we may assume
|{x1, x2, x3}|= 3
Again from Corollary 3.3 we get |α| = |β|+|γ|, |β| = |α|+|γ| and |γ| = |α|+|β|. Denote C =
|α|+|β|+|γ|. Then adding up the equations yieldsC= 2C, i.e. C= 0, a contradiction. □
Definition 3.5. (a) Let D be a region in an R1,2-diagram M. ThenU V−1 is a boundary label of
D, where U and V are as in subsection 2.1. Let µ be the boundary path with Φ(µ) =U and let ν be the boundary path of D with Φ(ν) =V. Then vµwν is a boundary cycle of D. The vertices v and w are called separating vertices and the line segment connecting them is called separating segment. (It does not belong to the diagram.)
(b) Let v be an inner vertex in M with valency 2m and let D1, . . . , D2m be the regions which contain v on their boundary, in this order clockwise.
Call v alternating if v is a separating vertex for all Di with i odd (even) and v is not a separating vertex for allDi withi even (odd).
Lemma 3.6. LetM be a reduced van KampenR1,2 diagram with connected interior and with maximal
number of vertices with valency 2, among all van Kampen diagrams with the same boundary label. Then every inner vertex with valency 4 in M is alternating.
separating vertex. For each inner vertex v with valency 4 let θ(v) be the length of a longest common edge of two adjacent regions which contain v on their boundary. Let w be a vertex with valency 4 with maximal θ(v) among all vertices with valency 4 in all diagrams with boundary label, like M, satisfying the assumption of the Lemma.
Let u1 andu2 be the neighbours of w,u1 ∈∂D1 and u2∈∂D2, where θ(w) =|∂D1∩∂D2|. Carry
out a diamond move T around w which identifies u1 with u2. See Figure 5. Then T splits w into
v1 and v2, such that v1 ∈ ∂D3 ∩∂D4 and v2 ∈ ∂D1∩∂D2, both vertices with valency 2. By the
maximality assumption,u1 andu2 inM should have valency 2. Thereforeu=u1 =u2 (after carrying
outT) has valency
(dM(u1) + 1) + (dM(u2) + 1)−2 =dM(u1) +dM(u2) = 4
and u is not alternating. Therefore u is an inner vertex (after carrying out T) on the common boundary of D1 and D2 with valency 4, which is not alternating. But this contradicts the maximality
of θ(w), since θ(u) ≥ θ(w) + 1, (namely, the edge connecting v2 with u is added to ∂D1 ∩∂D2), a
contradiction. Hence M contains no non-alternating inner vertices with valency 4. □
.. D1
.
D2
.
D3
. D4
. u1
.v .
u2
. θ(v)
..
⇝
..
v2
.
v1
.
u1
.
u2
. θ(v)
(a) (b)
↙
.. D1
. D2
.
D3
.
D4
. v2
.
v1
.
u=u1 =u2
. θ(v)
(c) Figure 5
Corollary 3.7. Let M be a reduced R1,2-diagram. Then there exists a reduced R1,2-diagramM1 with
the same boundary labels in which every inner vertex with valency 4 is alternating.
Definition 3.8. LetM be anR0-van Kampen diagram, R0 the symmetric closure of R(x1, x2) =U V,
nx1,x2 ≥ 3. Let D be a corner region. Call D complete if Φ(∂D∩∂M) ∈ {U±1, V±1}. Call D
alternatingif∂D∩∂M contains a separating vertex ofD. Finally, callDlarge alternatingif∂D∩∂M contains a separating vertex of D and also Φ(∂D∩∂M) contains U±1 or V±1.
Lemma 3.9 (Induction of corner regions). Let Λ = (L0, . . . , Lp) be a layer structure of M,L0 ={v}.
Let D be a corner region in Li.
(a) If D is complete thenD induces in Li+1 either a complete corner region with the same
orien-tation or a large alternating corner region with the same orienorien-tation.
(b) If D is alternating then D induces in Li+1 either an alternating corner region or a large
alternating region.
(c) If D is large alternating then D induces a complete corner region, oriented as ∂D∩ωi, and an alternating (possibly large) region.
Proof. (a) Consider ∆ :=C(D). Case 1 Int(C(D)) is connected.
If ∆ is a single region then since|α|< na,b,β andα are oriented in the same direction, hence ∆ is a k-corner region, k ∈ {1,2} such thatγ is either complete or large alternating, with complete part oriented as ∂D. If ∆ is not a single region, then ∆ =⟨E1, . . . , Er⟩,r≥2.
If ∂Er∩∂D is a vertex then Er is a one-corner region, and the result follows as above. Since M satsifies the conditionT(4), ∂Er∩∂D is a vertex
IfEr−1 has no left neighbour thenEr−1is a 1-corner region and the result follows. Assume therefore
thatEr−1 has a left neighbour E. If∂E∩∂D is a vertex then one ofE orEr−1 is a complete or large
alternating, for otherwiseE and Er−1 cancel each other.
Finally, if ∂E∩∂D is not a vertex then the common vertex v of ∂E, ∂Er−1, ∂Er and ∂D is an
inner vertex with valency 4 hence by Lemma 3.6 we may assume it is alternating. But then Er−1 is
either large alternating with complete part oriented as DorEr−1 is complete, oriented asD.
(b) Let C1 be a connected component of C(D) and letv be the separating vertex on∂D∩wi.
Case 1 C1 containsv.
If v has valency 4, then C1 contains a single region E and due to Lemma 3.6 E is an alternating 2-corner region. If v has valency≥ 6 then by one of the regions, say E1 with ∂E∩∂D = {v} is an
alternating 2-corner region. Case 2 C1 does not contain v.
Then eithervis a boundary vertex ofM in which caseLi+1has two consecutive oppositely directed
complete 2-corner regions or it contains an alternating 2-corner region.
(c) Follows from parts (a) and (b).
□
Proof. Let Λ = (L0, . . . , Lp) be the layer structure of M with L0 ={v}. Then L1 contains at least 3
complete and 3 alternating corner regions in an alternating manner. Hence
||∂L1|| ≥3nx1,x2 + 3·2−6·1 = 3nx1,x2 ≥2nx1,x2 + 2
Now, a successive application of the Lemma implies that the same inequality holds for ∂M. □
The proof of the refinement of Lemma 3.2 below closely follows the way in which Lemma 3.9 is proved, hence we omit it.
Lemma 3.11. Let M be an R0-van Kampen diagram with connected interior, |R0|= 2m. Suppose that M has a boundary cycle µν such that ||µ|| ≤ m. Then M contains a complete corner region D with ∂D∩∂M =∂D∩ν.
Lemma 3.12. Let M be a van Kampen R0-diagram which contains an inner vertex v. Ifnx1,x2 ≥4 then ||∂M|| ≥2nx1,x2 + 2.
Proof. Due to Corollary 3.10 we may assume that every inner vertex has valency 4 and hence by Lemma 3.6 every inner vertex is alternating, Let v be an inner vertex of M and consider the layer structure Λ ={L0, . . . , Lp}, whereL0 ={v}. Then every region inL1 is a 2-corner region and
||∂L1||= ¯a+ ¯b+ ¯c+ ¯d+ 2nx1,x2−4 = 6nx1,x2−(a+b+c+d+ 4), where ¯x=nx1,x2 −x But a+d=nx1,x2 and b+c=nx1,x2, hence||∂L1||= 4nx1,x2 −4. See Figure 6.
Since nx1,x2 ≥4, by assumption, we get ||∂L1|| ≥2nx1,x2 + 2. Now apply the induction argument
of Lemma 3.9. □
..
a .
b
. c
.
d . ¯
a
. ¯b
.
¯ c .
¯ d .
nx1,x2
.
nx1,x2
. v
Figure 6
a,b,c and ddesignate syllable lengths.
Lemma 3.13. Let M be a van KampenR0-diagram with||∂M||= 8andnx1,x2 ≥4. Thennx1,x2 = 4,
M is a one-layer diagram in which each common piece has length 3, and M has a boundary cycle ω with Φ(ω) =xm1 x2x1x2x1−mx−12 x−11 x−12 , m≥1.
Proof. By Lemma 3.12nx1,x2 ≤4, hence nx1,x2 = 4.
LetM∗ be the dual diagram of M. It follows from Lemma 3.12 thatM contains no inner vertices.
henceM∗has at most two boundary vertices of valency 1 and consequently,M is a one-layer diagram. If one of the regions has a common edge of length less than 3 then||∂M|| ≥9. □
4. Diagrams with the small cancellation condition V(6)
In Section 2.2(a) we introduced the underlying small cancellation condition V(6) and in Theorem 2.6 and Theorem 2.7 described basic properties of V(6) diagrams. We need some further results, which we develop below.
Definition 4.1. LetM be a diagram and let Dbe a corner region ofM. CallDeffectiveif∂D∩∂M has at least one endpoint with valency 3.
The reason for introducing this special type of corner regions is the following
Remark 4.2. Let ∆˜′ be an effective corner region of M˜′ with nx,y ≥4, where Supp(∆) ={x, y}, let µ =∂∆∩∂M and let ν be the complement of µ on ∂∆. Then |µ| ≥ |ν|, since then ||ν|| ≤ 4, hence
|ν| ≤ |µ|, by Lemma 3.1.
Lemma 4.3. Let M be a simply connected C(6)-map with connected interior which contains an inner vertex v with valency ≥3. Then M has at least three effective k-corner regions, k≤3.
Proof. LetL0 ={v}and let Λ ={L0, L1, . . . , Lp}be the corresponding layer structure. SincedM(v)≥
3,L1 contains at least three regions D1,D2,Dk,k≥3 such thatA(Di) ={v} and b(Di) = 2. Hence
each Di is a 2-corner region. Due to Theorem 2.7 each boundary vertex of ∂L1 = ∂S1 has valency
3 in L1, hence each Di is an effective corner region. Now, by a standard induction argument on i,
1 ≤ i ≤ p, it follows along the lines of the proof of Lemma 3.9 that each Si, 2 ≤i ≤ p contains at
least 3 effective corner regions. □
Following the same method we have the C(4) & T(4) and the V(6) analouge of the previous Lemma, the proof of which we omit.
Lemma 4.4. Let M be a simply connected map with connected interior. Asumme that M contains an inner vertex with valency at least 4.
(a) If M satisfies the condition C(4) & T(4) then M has at least 4 effectivek-corner regions. (b) If M satisfies the condition V(6) then M has at least two effective corner regions.
.. Li+1
.
Li .
Li−1
..
Li
.. Li+1
. Li
. Li−1
(c) Figure 7
Definition 4.5 (See Figure 7.). (a) Let M be an annular R-diagram with layer decomposition Λ = (L−l, . . . , L0, . . . Lp). Call M full ifLi is annular, for every i, l≤i < p.
It follows from the definitions of Li and Si that M is full if and only if L−l and Lp are annular.
(b) Let M0 be a full annular V(6) diagram. Call Li 2-homogeneous if every vertex v ∈ ωi has
valency 3 in Li and valency 3 in Li+1, l≤i≤p−1 and every region D in Li has a(D) = 1,
b(D) = 2 and c(D) = 1. See Figure 7(a).
(c) Call Li 3-homogeneous if every vertex v ∈ ωi has valency 3 in Li and valency 2 in Li+1, if
l≤i≤p−1 and a(D) =b(D) = 2. See Figure 7(b).
(d) Call Li 4-homogeneous if every vertex v ∈ωi has valency 4 in Li and valency 2 in Li+1 and
a(D) = 0, b(D) = 2 for every region D in Li. See Figure 7(c). (e) Call M r-homogeneous, r = 2,3,4, if every Li is.
Lemma 4.6. Let notation be as above. Suppose that Lp is not full. Then Lp contains an effective corner region.
Proof. LetC1, . . . , Cr,r≥1, be the connected components ofInt(Lp) occuring in this order, clockwise.
Since Lp is not 4-homogenous, there is a connected componentCi such that either ∂Ci−1∩∂Ci =∅ or∂Ci+1∩∂Ci =∅(i−1 andi+ 1 are counted modulor). Without loss of generality we may assume
that ∂Ci−1∩∂Ci =∅. Let v be the left endpoint of ∂Ci∩ω(Lk−1) and let {E1, . . . , Es} be the set
of regions in Lk−1 which contain v on their boundary. By Theorem 2.6 1 ≤ s ≤ 3. Consider the
casess= 1,2,3 in turn. Let D1 be the leftmost region of Ci. Then b(D1)≤1, hence ifa(D1) = 0 or
Ci ={D1}then D1 is an effective corner region, due to Theorem 2.6. Hence we may assume
(∗) (i) |Ci| ≥2 and (ii) α(D1)≥1
..
D1
.
E1
. v
..
w
. E1
.
E2
.
D1
.
D2
.
Ci
. Li
(a) (b)
.. v
.
w
.
E1
.
E2
.
D1
.
D2
(c)
..
w
. v
.
E1
.
E2
.
D1
.
D2
(d)
..
v=w
. E1
.
E2
.
D1
.
D2
(e)
Figure 8
One of the following holds
(1) E1 has no right neighbour inLi−1. See Figure 8(a).
ThenD1 is an effective 2-corner region.
(2) E1 has a right neighbour E2 inLi−1 and w:=∂E1∩∂E2∩ω(Li−1) has valency 4 in Li−1.
ThenD2 is an effective 2-corner region, due to Theorem 2.7. See Figure 8(b).
(3) E1 has a right neighbour E2 and whas valency 3.
ThenD1 is an effective 2-corner region or an effective 3-corner region. See Figures 8(c) and
8(d), respectively.
Case 2 s= 2
Then D1 is an effective 2-corner region, (see Figure 8(e)), since b(D1) = 1 and a(D1) = 1, by (∗).
Case 3 s= 3
Then by Theorem 2.6 E1 is the left companion of the exceptional 2-corner region E2, hence D1
Corollary 4.7. Suppose that M contains no corner regions. Then M is full.
Proof. Clearly Lp does not contain an effective corner region. Therefore, by the contraposed version of the Lemma, either Lp is annular or Lp is 4-homogeneous. Since Lp contains no corner regions, Lp
cannot be 4-homogenous, by definiton. Therefore M is full, by the Lemma. □
Definition 4.8. Let M be a diagram over on Artin presentation. Call a boundary path µ of a region ∆ of M′ with||µ||= 1 a split edge if ∆has more than one neighbour which intersects µ by an edge.
In the next Lemma M is an R-diagram, as defined in the Introduction. We consider ˜M′, as introduced in Subsection 2.2(c). For simplicity, we denote the regions of ˜M′ by ∆, rather than ˜∆′. This causes no ambiguity.
Lemma 4.9. Let M˜′ be an annular V(6) diagram and let ∆ be a region in Li, 0 ≤i < p. Assume that M˜′ is full.
(a) If ∆is a corner region of Li thenC(∆) contains a corner region ofLi+1.
(b) In each of the following cases Li+1 contains a corner region ofLi+1.
(i) ∆ is an exceptional region. (ii) d(∆)≥7
(iii) d(∆)∈ {4,5} and one of the boundary vertices has valency ≥5. (iv) d(∆) = 6 and one of the vertices on ∂∆has valency ≥4.
(v) ∆ has a split edge.
As a corollary, we have the following
Corollary 4.10. Suppose thatM˜′ is an annular V(6) diagram with more than one layer in which all
the regions are simply connected and which contains no corner regions. Then M˜′ is full and none of the assumptions of the Lemma holds. In particular, M˜′ is k-homogeneous for some k∈ {2,3,4}.
..
Li+1
.
Li .
∆
.
v
.
∆1
..
Li+1
.
Li .
u
.
v
. ∆
.
∆1
.
∆2
..
∆
.
∆1
.
∆ℓ
.
∆r
.
Li+1
.
Li
.
v
..
v
. v1
.
∆1
.
∆2
.
∆ℓ
.
∆r
. Li .
Li+1
c(1) c(2)
Figure 9
Proof. Due to Corollary 4.7 we may assume that M is full. Hence the result follows from the last
Lemma, by inspection. □
Proof of the Lemma. (a) Suppose first that ∆ is a 2-corner region ofLi,i < p. IfdM(∆)≤5 then all the boundary vertices of ∆ have valency at least 4, hence in particular the inner vertex v in Figure 9(a).
But thenC(∆) contains a region ∆1 with∂∆1∩ωi ={v}, sincedLi(v) = 2 and hencedLi+1(v)≥4. Hence, ∆1 is a 2-corner region in C(∆). Next, suppose that ∆ is a 3-corner region of Li. Then
dM(∆) ≥6, hence||∂C(∆)∩ωi|| ≥3. In particular, there is a region ∆1 inLi+1 which contains the
adjacent vertices u and v on its boundary. See Figure 9(b).
If one ofuandvhas valency at least 4 inM then it has valency at least 4 inLi+1, sincedLi(w) = 2, w∈ {u, v}. HenceC(∆) contains a 2-corner region ∆i which contains it. Finally, if bothu andv have valency 3 in Li+1 thendM(∆)≥6, and ∆1 is a corner region. This also proves the assertion when ∆
is a 2-corner region and dM(∆)≥6.
(b) (i) Let ∆ be a 2-exceptional region with companion 2-corner regions ∆r and ∆l. See Figure 9(c). Consider L:=C(∆r∪∆l∪∆).
If there is a unique region ∆1 in Lwhich contains v on its boundary, then ∆1 is a 2-corner region
by definition, since then dLi(v1) = 2, hencedLi+1(v1) ≥ 4, see Figure 9(c1). Assume therefore, that there are f regions, f ≥ 2, {∆1, . . . ,∆f} which contain v on their boundary. If f ≥3 then ∆2 is a
2-corner region. Therefore, we may assume that f = 2, see Figure 9(c2). If d(∆1) ≤5 then v1 has
valency at least 4 in Li+1, again L contains a corner region. Henced(∆1)≥6. But thena(∆1) = 1,
hence again ∆1 is a 3-corner region. Consequently,C(L) contains a corner region.
.. u1
.
u2
.
u3
.
µ1
.
µ2
.
w
.
∆1
.
∆2
.
∆
..
∆
.
∆1
.
∆1
.
∆2
.
∆3
. u1
.
u2
.
u3
(a) (b)
..
∆ .
∆1
..
∆ .
∆1
. Li .
Li+1
(c) (d)
Figure 10
(b) (ii) See Figure 10. We may assume that a(∆) = 2, otherwise ∆ is already a 3-corner region. Let
A(∆) = (∆1,∆2), let µi =∂∆∩∂∆i,i= 1,2. Ifh(µ1) ort(µ2) have valency at least 4, then ∆ has a
neighbour in Li which is a 2-corner region, hence Li+1 contains a corner region, by part (a). Assume
therefore, that they both have valency 3 in M. Clearly, the vertex w:=∂∆1∩∂∆2∩∂∆ has valency
3 inM. Therefore, one of the verticesu1,u2, andu3 has valency 4. See Figure 10(a). If this vertex is
u2 thenC(∆) contains a region Θ with ∂Θ∩ωi =u2, hence Θ is a 2-corner region. Assume therefore
that u2 has valency 3. See Figure 10(b). If this is one ofu1 and u3 then the region ∆1 ofLi+1 or ∆3
of Li+1 has a(∆1) = 1 or a(∆3) = 1, hence d(∆1)≥6 ord(∆3) ≥6. Consequently, one of them is a
3-corner region of Li+1.
(b) (iii) By Theorem 2.6 a(∆)≤2,hence
c(∆) =d(∆)−(a(∆) +b(∆))≥7−(2 + 2) = 3
LetC(∆) = (∆1, . . . ,∆f),f ≥3. Letv=∂∆1∩∂∆2∩∂∆. If dM(v)≥4 thendLi+1(v)≥4. But then one of ∆1 and ∆2 is a 2-corner region ofLi+1. IfdM(v) = 3 then letv=∂∆2∩∂∆3∩∂∆. IfdM(v)≥4 then one of ∆2 and ∆3 is a 2-corner region and if dM(v) = 3 then d(∆2)≥6 and a(∆2) = 1. Hence,
∆2 is a 3-corner region.
(b) (iv) Assume first that d(∆) = 5. By Theorem 2.6a(∆)≤1, hence
Let C(∆) = (∆1,∆2, . . . ,∆f), f ≥2. Letv =∂∆1∩∂∆2∩∂∆. Then dM(v) ≥4, by assumption,
and since dLi(v) = 2,dLi+1(v)≥4. Consequently either ∆1 or ∆2 is a 2-corner region inLi+1. Assume now that dM(∆) = 4 and one of the boundary vertices of ∆ has valency ≥5. By Theorem 2.6 0≤a(∆)≤2. Ifa(∆) = 0 then ∆ is a 2-corner region. Ifa(∆) = 1 then either ∆ has a neighbour ∆1 in Li which is a 2-corner region (see Figure 10(c)) or C+(∆) contains a region ∆1 which is a
2-corner region. See Figure 10(d). (HereC+(∆) isC(∆) together with the regions inLi
+1which touch
∂∆ by a vertex only.) Hence, by part (c)Li+1 contains a corner region.
(b) (v) The number of unsplit edgesn(∆) of ∆ is at least 4. Suppose ∆ has a split edge. Ifn(∆) = 4 then ∆ has at least 5 edges and all vertices with valency at least 4. Hence by part (b) (iii) Li+1
contains a corner region. If n(∆) = 5 then ∆ has at least 6 edges and each boundary vertex has valency at least 4, hence by part (b) (iv) the result follows. Finally, if n(∆) ≥ 7 again the result
follows by part (b) (ii). □
5. Three-generated parabolic subgroups in Artin groups
Theorem 5.1. Let x1, x2, x3 ⊆X. If xα1x
β
2x
γ
3W = 1, where W is a reduced word in A, then |W| ≥ |α|+|β|+|γ|.
Let U be a shortest representative of xα
1xβ2x
γ
3 in A. Then |U| ≤ |α|+|β|+|γ|. By Theorem 2.2
U ∈ ⟨x1, x2, x3⟩.Hence if we show that
xα1xβ2xγ3 is a shortest representative of itself in ⟨x1, x2, x3⟩ (∗)
then the result will follow. We may assume without loss of generality that αi ̸= 0, i = 1,2,3 and xα
1x
β
2x
γ
3U−1 is cyclically reduced, as written and A = ⟨x1, x2, x3⟩. We prove (∗) by checking cases.
The first case is when⟨x1, x2, x3⟩ is of finite type.
The proof of this case is due to Professor S. Orevkov. I am grateful to Professor Orevkov for the proof and for his permission to quote it here.
Let Gbe an Artin group of finite type on 3 standard generators u,v and w.
Proposition 5.2. If uavbwc =W in G where W is a reduced word on u, v andw and a, b andc are non-zero integers, then the length of W is at least as that of uavbwc.
S. Orevkov. (For undefined terms and general background for the case of braid groups see [5].) We consider the standard Garside structure on G with the Garside element denoted by ∆. Each element Y of G can be presented in a unique way in the form Y =N−1P (resp. Y =P N−1) where N and P are positive elements whose left (resp. right) gcd is 1. If, moreover, N and P are given in the left (resp. right) normal form, then we call this presentation of Y the left (resp. right) balanced normal form. The balanced left normal form is used by default in GAP’s package Chevie, version
Lemma 5.3. Leta, b, cbe non-zero integers one of which is negative (we denote it−m) and the others are positive (we denote them k, l). Then inf(uavbwc) =−m.
Proof. We consider only the case whenGis an irreducible Artin group. Thus,Gis of the typeA3,B3
or H3, i.e. G is generated by x, y, z subject to the relations xyx =yxy, xz =zx and ⟨yz⟩µ =⟨zy⟩µ
for µ = 3,4, or 5; here we use Brieskorn-Saito’s notation ⟨yz⟩µ = yzyz . . . (µ alternating letters), so the Coxeter graph ofG is
..
x... y µ z
Figure 11
When presenting the left (right) normal forms, we use the notation A1 ·A2· · · which means that
the Ai’s are simple elements and each pair of consecutive elements is left (right) weighted. If an expressionAnoccurs in the left (right) normal form, it stands forA·A· · ·A(ntimes),⟨A·B⟩nstands forA·B·A·B· · · (n alternating factors), andtalways denotes the last letter of the previous simple element. For example, (yzy)2· ⟨yz·zy⟩3t3 stands for
yzy·yzy·yz·zy·yz·z·z·z
To prove the desired result, it is enough to compute the left or right balanced normal form ofuavbwc for all permutations (u, v, w) of (x, y, z) and for all permutations of (k, l,−m). The right balanced normal forms are:
xkylz−m=xk−1·xy·yl−1·z−m, zkylx−m=zk−1·xy·yl−1·x−m,
xkzly−m=xk−1·(xz)l·y−m, k≥l
xkzly−m=zl−k·(xz)k·y−m, k≤l
xky−mzl=xk−1·xzy· ⟨yz·zy⟩m−1·tl−m· ⟨yz·zy⟩−m, l≥m, A
3
xky−mzl=xk−1·xzyz·(zyz)m−1·zl−m·(yzy)−m, l≥m, B3
xky−mzl=xk−1·xzyzy· ⟨yzyz·zyzy⟩m−1·tl−m· ⟨yzyz·zyzy⟩−m, l≥m, H
3
zky−mxl=zk−1·zxy· ⟨yx·xy⟩m−1·tl−m· ⟨yx·xy⟩−m, l≥m xky−mzl=xk−1·xzy· ⟨yz·zy⟩l−1·(⟨yz·zy⟩l·tm−l)−1, l≤m, A
3
xky−mzl=xk−1·xzyz·(zyz)l−1·((yzy)l·ym−l)−1, l≤m, B3
xky−mzl=xk−1·xzyzy· ⟨yzyz·zyzy⟩l−1·(⟨yzyz·zyzy⟩l·tm−l)−1, l≤m, H 3
zky−mxl=zk−1·zxy· ⟨yx·xy⟩l−1·(⟨yx·xy⟩l·tm−l)−1, l≥m
We see that the infimum is equal to −min all listed cases. All other cases are reduced to these ones. Indeed, if a=−m (the first power), then we compute the left balanced normal form. If x−m orz−m occurs in the middle, then we exchange it with the commuting letter.
The fact that the expressions in the right hand sides are right weighted is evident. The fact that
the equalities hold can be easily proved by induction. □
that two of the numbersa, b, c are positive (we denote them bykand l) and the third one is negative (we denote it by −m).
Let p and n be the number of positive and negative letters inW respectively. Since the relations in the Artin groups are homogeneous, we havep−n=k+l−m. We have inf(W)≥ −n. By Lemma 5.3, this meansm≤n. Thus, p+n= (p−n) + 2n≥(p−n) + 2m= (k+l−m) + 2m=k+l+m. □
We turn now to the remaining cases. Let Γ be the defining graph of Awith edgesE =E(Γ), with respect to the standard generators x1,x2 and x3.
If E = ∅ then A is a free group and the result is clear. If E consists of a single edge, say (x1, x2)
thenA=⟨x3⟩ ∗ ⟨x1, x2 |R1,2⟩and the result follows from the Normal Form Theorem for free products (see [12]) and Corollary 3.3 and Corollary 3.4. If E consists of two edges, say (x1, x2) and (x2, x3),
then
A=⟨x1, x2 |R1,2(x1, x2)⟩ ∗ ⟨t2, x3 |R2,3(t2, x3)⟩
x2 =t2
and again the result follows from the Normal Form Theorem for amalgamated free products (see [12]) and Corollary 3.3 and Corollary 3.4.
So assume that Γ is a triangle. Let xα
1x
β
2x
γ
3U−1 = 1 in A = (⟨x1, x2, x3⟩), where U is a shortest representative of xα1x
β
2x
γ
3 in A.
Then there exists a van Kampen diagram M with boundary label xα
1x
β
2x
γ
3U−1. We propose to show
that Reg(M) = ∅, i.e. M contains no regions. This implies U = xα1xβ2xγ3 in F. Assume by way of contradiction that Reg(M) ̸= ∅. M has a boundary cycle uµvν, u and v vertices, such that Φ(µ) =xα
1x
β
2x
γ
3 and Φ(ν) =U. Thus, we shall assume
(H1) W is a shortest word for which Reg(M) ≠ ∅, where w = xα1i1 xα2i2 xα3i3 , |αi| ̸= 0 {i1, i2, i3} = {1,2,3} and M is the corresponding van Kampen diagram.
(H2) U is a shortest representative ofW.
(H3) M contains minimal number of regions among all diagrams with boundary cycle uµvν, where
Φ(µ) =W andΦ(ν) is a shortest representative ofW, for which (H1) and (H2) hold.
We shall derive a contradiction to one of (H1), (H2) and (H3), showing that Reg(M) =∅. (For if Reg(M)̸=∅ then there is a diagram which satisfies all of (H1), (H2) and (H3), this however leads to a contradiction, as we show.)
First observe that if µ contains a subpath µ0 ̸= ∅ such that Φ(µ0) = 1 in A then by ||µ0|| ≥ 4,
a contradiction, since ||µ|| = 3. Hence, µ cannot contain such a subpath. Also, ν may not contain a subpath ν0 ̸= ∅ with Φ(ν0) = 1 in A, since then (H2) is violated. Consequently, if M1, . . . , Mr
, r ≥ 1, are the connected components of Int(M) with Reg(Mi) ̸= ∅, then µ = µ1δ1µ2δ2· · ·µr,
ν =ν1δ1ν2δ2· · ·νr, where µi =∂Mi∩µ,νi =∂Mi∩ν and δi are (possibly empty) paths. See Figure
.. µ1
. M1
.
ν1
. δ1
.
µ2
. M2
.
ν2
. δ2
.
µr
. Mr
.
νr
Figure 12
We concentrate onMr and show thatReg(Mr) =∅. This contradiction implies that Reg(M) =∅, hence we may write M forMr without loss of generality.
Case 1 Γ is a (c1, c2, c3) triangle withci ≥3,i= 1,2,3.
In this caseM′ is of large type, hence due to Lemma 3.1 it satisfies the condition C(6). Hence, ifM′ contains an inner vertex with valency greater then 2 thenM′ contains at least three (effective) corner
regions, due to Lemma 4.3. SinceM′ is planar,u andv may occur on the boundary of at most one of them (each), hence at least one of the corner regions, say ∆, satisfies θ:=∂∆∩∂M =∂∆∩ξ, where ξ ∈ {µ.ν}. We claim that this is not possible. Let η be the complement of θ on ∂∆. Then due to Lemma 3.2|η| ≤ |θ|, hence ifξ=ν then we can remove ∆ fromM′and get a smaller diagram with the
required properties, thereby violating (H3). On the other hand, ifξ=µthen Φ(µ) contains a subword xβ1i1xβ2i2xβ3i1,xi1 ̸=xi2, βi ̸= 0, which is not possible, since Φ(µ) =xα
1x
β
2x
γ
3. Consequently, M′ has no
inner vertices, hence its dual is a tree. Therefore, all the corner regions of M′ are 1-corner regions. Hence, if ∆ is a corner region then∂∆∩∂M again cannot be a subpath ofµor ofν. Consequently,M′
is a one layer diagram, M′ =⟨∆1, . . . ,∆t⟩,t≥1. We concentrate on ∆t and show thatReg(∆t) =∅.
This proves our claim. Assumet≥2. Letρ=∂∆t∩∂∆t−1. Then due to Lemma 2.8||ρ||= 1. Clearly ||µt|| ≤2, hence||µ−1t ρ|| ≤3. Butµ−1t ρνtis a boundary cycle of ∆t, hence by Lemma 2.8 and Lemma 3.2|µ−1t ρ| ≤ |νt|. Consequently, we may remove ∆tand replace νtby µ−1t ρ, violating (H3). Therefore t= 1. But then, the above argument withρ=∅ leads to a violation of (H3). HenceReg(M′) =∅, as required.
Case 2 Γ is a (2, c1, c2) triangle, c1, c2 ≥4.
In this case we considerM as a relative diagram withH=⟨x1, x2⟩,R0=x1x2x−11 x−12 . ThenM′ is
extra-large relative toH. Consequently, it follows from Lemma 2.9 that ˜M′ is a C(4) & T(4) diagram, hence if ˜M′ contains an inner vertex, then by Lemma 3.2 ˜M′ has at least 4 effectivek-corner regions, k ∈ {1,2}. Let ∆ be such a corner region and assume θ := ∂∆∩∂M = ∂∆∩ξ, ξ ∈ {µ, ν}. Let η be the complement of θ on ∂∆. Then ||η|| ≤ 4, since ∆ is effective. Consequently, |η| ≤ |θ| and
||θ|| ≥4, since c1, c2 ≥4. This however, violates (H3), if ξ=ν and the fact that ||W||= 3, ifξ =µ.
Consequently, ˜M′ has no inner vertices and as in Case 1, the dual of ˜M′ is a tree. From this it follows easily, as in Case 1, that ˜M′ is a one layer diagram ⟨∆
1. . . . ,∆t⟩. Again, concentrate on ∆t. Assume
first that t≥2 and let ρ= ∂∆t∩∂∆t−1. Then||ρ||= 1. Let z1 and z2 be the endpoints of ρ, such
that z1 ∈µ and z2 ∈ν. Since ˜M′ is planar, at least one of z1 and z2 has valency 3 in ˜M′. Assume
..
v
. µt
. ξt
. MZ1.
ξt−1
. ρ
.
∆t
.
νt
. MZ2
.
κt
(a)
.. ξt
. ρ
.
κt .
κt−1
. M′
Z2
(b)
Figure 13
Then||ξt−1||=||ξt||= 1, hence||∂Mz1∩µ|| ≥ ||ξt−1ξt||= 2, i.e. ||∂Mz1∩µ|| ≥2. NowSupp(Mz1) =
{x1, x2}, while µt starts with x±13 , hence ||µt|| = 1. Consequently ||µ−1t ξt−1ρκt|| ≤ 4, where κt = ∂Mz2∩∂∆. But µ−1t ξt−1ρκtνt is a boundary cycle of ∆t. Therefore |µ−1t ξt−1ρκt| ≤ |νt|and as in Case 1, we may remove ∆tand violate (H2). Ift= 1 then the same argument works withρ=∅.
Assume now thatdM˜′(z2) = 3. See Figure 13(b).
Then ||κtκt−1|| = 2, hence |κtκt−1| ≤ |∂Mz2′ ∩ν|, by Lemma 3.2. But then the removal of ˜Mz2′ violates (H3). Consequently, Reg(Mz2′ ) = ∅. But then |µt−1ξt−1ρ| ≤ |νt|, as above, since ||µt|| ≤ 2,
||ξt||=||ρ||= 1. Hence we may replace νt with µ−1t ξtρ and get a smaller diagram with the required properties, again violating (H3). A similar argument applies if t= 1. Consequently Reg(∆t) =∅, as required.
Case 3 Γ is a (2,3, c) triangle,c≥6.
In this case we take H = ⟨x2, x3 | x2x3x2 = x3x2x3⟩. The relative diagram ˜M′ obtained is not
relatively extra-large, however, we may use the ideas used in Case 2 in order to show thatReg(M′) =∅.
Now, Reg( ˜M′) consists of two types of regions: n-gons ∆i with n ≥ 6 having Supp(∆i) = {x1, x3}
and bigons Λj having Supp(Λj) = {x1, x2}, (x1x2 =x2x1). Since ||∂Mv|| ≥6, for every vertex v for
whichReg(Mv)̸=∅, every inner vertex in ˜M′ has valency at least 6. (See proof of Lemma 2.3 in [11]).
..
∆i+1
. Λ1
. Λ2
.
w .
∆i .
x1
.
x2
.
x3
.
x2
.
x1
Figure 14
Let w be a boundary vertex of ˜M with valency 3 and consider Mw. Let ∆i, ∆i+1 and Λj be the regions of ˜M′ containing w on their boundary. Let ξi, ξi+1 and ξj be he common edges of ∆i, ∆i+1
and Λj, respectively, with ∂Mw, and let η=∂Mw∩∂M˜′. See Figure 15.
..
η
.
Mw
.
∆i
.
∆i+1
.
Λj
.
ξi
.
ξj
.
