The conjugacy problem in right-angled Artin groups and their subgroups
aa r X i v : . [ m a t h . G R ] F e b The conjugacy problem in right-angled Artin groupsand their subgroups
John Crisp, Eddy Godelle, Bert Wiest
Institut de Math´emathiques de Bourgogne, CNRS UMR 5584Universit´e de Bourgogne, B.P. 47870, 21078 Dijon cedex, France [email protected] andLaboratoire de Math´ematiques Nicolas Oresme, CNRS UMR 6139Universit´e de Caen, 14032 Caen cedex, France [email protected] andIRMAR, Campus de Beaulieu, CNRS UMR 6625Universit´e de Rennes 1, 35042 Rennes, France [email protected]
Abstract
We prove that the conjugacy problem in right-angled Artingroups (RAAGs), as well as in a large and natural class of subgroups ofRAAGs, can be solved in linear-time. This class of subgroups contains,for instance, all graph braid groups (i.e. fundamental groups of configura-tion spaces of points in graphs), many hyperbolic groups, and it coincideswith the class of fundamental groups of “special cube complexes” studiedindependently by Haglund and Wise.
AMS Classification
Keywords right-angled Artin group, partially commutative group, graphgroup, graph braid group, conjugacy problem, cubed complex, special cubecomplex.
It is well known that the conjugacy problem in free groups can be solved inlinear-time by a RAM (random access memory) machine. This result has beengeneralized in two different directions. On the one hand, Epstein and Holt [14]have shown that the conjugacy problem is linear in all word-hyperbolic groups.On the other hand, Liu, Wrathall and Zeger have proved the analogue result forall right-angled Artin groups ([23], based on [29]). Note that these groups arealso called “partially commutative groups” or “graph groups” in the literature.The aim of the present paper is to extend the second approach, in order toprove linearity of the conjugacy problem in a large class of subgroups of right-angled Artin groups. Very roughly speaking, the subgroups in question are1undamental groups of cubical complexes, sitting inside the right-angled Artingroup in a convex fashion. This class of groups has previously been studiedby Crisp and Wiest [11, 12], and independently by Haglund and Wise [19],as fundamental groups of so-called special cube complexes (or, more precisely, A -special cube complexes).The class of groups considered in this paper contains in particular all graphbraid groups [1, 2, 15, 16, 24] and more generally all state complex groups[3, 17]. These classes of groups have attracted considerable interest recently,which stems partially from their close relations to robotics [2, 17]. Indeed, ourresults can be interpreted as giving very efficient algorithms for motion planningof periodic robot movements. However, our results also apply to the variousword-hyperbolic groups discussed in [11, 12] – in particular, to all surface groupsexcept the three simplest non-orientable ones.The present paper raises the stakes on the conjecture of Haglund and Wise [19]that all Artin groups (e.g. braid groups) are virtually fundamental groups ofspecial cube complexes. If this conjecture was known to be true, then our workwould imply that Artin groups have finite index subgroups where the conjugacyproblem can be solved in linear time.The plan of the paper is as follows. In the second section we present an alterna-tive approach to the conjugacy problem in right-angled Artin groups, differentfrom the one of Liu, Wrathall and Zeger, but rather close in spirit to the methodsof Lalonde and Viennot [22, 28]. In the third section we prove that isometricallyembedded subgroups of right-angled Artin group inherit a linear-time solutionto the conjugacy problem from their supergroups. We recall that a right-angled Artin group is a group given by a finite presenta-tion, where every relation states that some pair of generators commutes. Graph-ically, a right-angled Artin group A can be specified by a simple graph Γ A ,where the generators of A correspond to the vertices of Γ A , and a pair of gen-erators commutes if and only if the corresponding vertices are not connected byan edge. Note that the opposite convention (connecting commuting generatorsby an edge) is also very common, but in the present paper we shall stick to thisconvention.Right-angled Artin groups have been widely studied in the last decades – see [10]for an excellent survey. Several solutions to the word and conjugacy problem2ave been found. It seems to be difficult to have a complete bibliography ofthe large number of articles on these two problems. The first solutions to theword and the conjugacy problem was obtained by Servatius in [25]. In [27], VanWyk constructed a normal form in right-angled Artin groups and proved thatthese groups are biautomatic. Indeed, even thought our point of view is verydifferent from Van Wyk’s, the normal form constructed in the present paperis the very similar to his. One of the main papers regarding the algorithmiccomplexity of these two problems is [23] (based on [29]) by Liu, Wrathall andZeger, which proves that they are both of linear complexity.The word problem in partially commutative monoids has also been widely stud-ied and numerous papers appeared on that topics. Several approaches appearedto be successful. In [9], Cartier and Foata constructed a normal form on par-tially commutative monoids, and then obtained the first solution to the wordproblem. This normal form is the restriction of the normal form obtained in [27].More recently, Viennot introduced in [28] a new tool, the so-called Viennot’spiling , based on a geometrical representation of partially commutative monoids.Several works deal with this tool (see for instance [13] and [22]). The Viennotpiling method associates a piling to each element of a partially commutativemonoid and thereby provides a linear-time solution to the words problem insuch a monoid. As remarked by Krob, Mairesse, and Michos in [21], this pilingis canonically related to the normal form constructed in [9]. In [22], Lalondeintroduces and uses the notion of a pyramid in order to study the conjugacyproblem in partially commutative monoids. In the present paper, we are goingto extend the notions of a piling and of a pyramid to the context of right-angledArtin groups , and use them in order to obtain a linear-time solution (to the wordproblem and) to the conjugacy problem. This leads us to introduce the notionof a cyclic normal form.In order to get an intuition for the nature of the conjugacy problem in right-angled Artin groups, let us first consider the relatively easy case of free groups.Given two cyclic words of length ℓ and ℓ respectively, there is a two stepalgorithm which can be performed in time O ( ℓ + ℓ ) on a RAM machine:first each word can be cyclically reduced in time O ( ℓ ) and O ( ℓ ), respectively.If the reduced words have different lengths, then they are not conjugate. Ifthey have the same length ℓ , then they can be compared in time O ( ℓ ) usingstandard pattern matching algorithms, like the Knuth-Morris-Pratt algorithm,the Boyer-Moore algorithm, or algorithms based on suffix-tree methods – see[20, 7, 4, 18, 26]. It should be stressed that on a Turing machine these algorithmstake time O ( ℓ log( ℓ )).In the sequel, we assume that A is a fixed right-angled Artin group given3y a fixed presentation. We denote by { a , · · · , a N } the generating set of A associated with this presentation.The aim of this section is to provide an algorithm which does, very roughlyspeaking, the following: given a word w , another word w ′ with smaller or equallength is created in linear time such that w and w ′ represent conjugate elementsof A . Furthermore, the word w ′ depends only on the conjugacy class in A of theelement represented by w , up to a cyclic permutation of its letters. This yieldsa linear-time solution to the conjugacy problem in A because, given words w and v we can compute the canonical cyclic words w ′ and v ′ representing theirconjugacy classes, and compare those by one of the algorithms mentioned above. We start by recalling the following classical lemma.
Lemma 1 [25] Any element of A can be represented by a reduced word (onewhich does not contain a subword of the form a ± i xa ∓ i , where all letters of x commute with a i ). Moreover, any two reduced representatives of the sameelement are related by a finite number of commutation relations – no inser-tions/deletions of trivial pairs are needed. Now we introduce our main tool, the notion of a piling . Definition 2 An abstract piling is a collection of N words, one for each gen-erator a i of A , over the alphabet with three symbols { + , − , } .The word associated with the generator a i will be called the a i -stack of theabstract piling. The product of two abstract pilings is defined as the pilingobtained by concatenation of the corresponding stacks.We define a function π ⋆ on the set { a ± , . . . , a ± N } ∗ of words on the 2 n let-ters a , a − , . . . , a N , a − N that associates an abstract piling to every word in thefollowing way: starting with the empty piling, we read the word from left toright. When a letter a ǫi is read, we check what the last letter of the a i -stackof the piling is. If this letter is different from − ǫ (the no-cancellation cases:the a i -stack is empty, or finishes either with 0 or ǫ ), then we append a letter +or − at the end of the a i -stack of the piling (the sign of ǫ ). Moreover, wealso append a letter 0 at the end of each of the a j -stacks associated with agenerator a j that does not commute with the generator a i . On the other hand,4f the last letter of the a i -stack is − ǫ (the cancellation case), then we erasethis last letter, and we also erase the terminal letter of each of the a j -stacksof the piling associated with a generator a j that does not commute with thegenerator a i – note that the terminal letter of the a j -stack is necessarily “0”. Definition 3 A piling is an abstract piling in the image of the function π ⋆ .The set of pilings is denoted Π.We observe that the number of letters + and − occuring in the piling π ⋆ ( w ) isat most equal to the length of the word w . Moreover, it is immediate from thedescription of the function π ⋆ that, given a word w of length ℓ , the piling π ⋆ ( w )can be calculated in time O ( ℓ ) (linear-time).It may be helpful to keep in mind the following physical interpretation of apiling: we have N vertical sticks, labelled by the generators a , . . . , a n , withbeads on it; the beads are labelled by +, − or 0 such that when readingfrom bottom to top the sequence of labels of the beads on the a i -stick, weobtain the a i -stack of the piling. A letter a i or a − i of the word w correspondsto a set of beads (which we call a tile ), consisting of one bead labelled +or − on the corresponding stick, and one bead labelled 0 on each of the stickscorresponding to generators of A which do not commute with a i ; each 0 labelledbead is connected to the ± labelled bead by a thread. The rule is: on a stick,adjacent 0-beads can commute with (“slide through”) each other, but 0-beadsdo not commute with ± -beads. In this physical model, we construct the imageof a word by adding beads from the top, and removing opposed tiles when oneobtains on a stick two adjacent ± -beads with opposite signs. In fact, when weare dealing with the word problem we can forget about the threads betweenthe beads, but they are helpful for thinking about the conjugacy problem. Example 4
In the group A with group presentation h a , a , a , a | a a = a a ; a a = a a ; a a = a a i we can calculate the piling p of the word a − a − a a a a a a − a a − as indi-cated in Figure 1.The map π ⋆ induces a well-defined function π : A → Π because words repre-senting the same element of A have the same image under π ⋆ : the image ofa word is unchanged by applying a commutation relation, and by inserting ordeleting a trivial pair a i a − i or a − i a i . Now, from the definitions it is immedi-ate that no cancellation occurs during the construction of the piling π ⋆ ( w ) ofa reduced word w . Then, the identity of A is the unique element of A whose5 a aa a a aa a a aa p= Figure 1: The pilings of the prefixes a − a − a and a − a − a a , and of the fullword a − a − a a a a a a − a a − image by π is the trivial piling, and therefore the word problem is solved inlinear-time: a word w represents the identity if and only if its piling π ⋆ ( w ) istrivial; this piling can be built in linear-time.The following notion will be extremely useful in the next section when weconsider the conjugacy problem. Definition 5
Let w be a reduced word.(i) We say that w is initially normal when w is trivial or when the index of itsfirst letter is greater or equal to the index of the first letter of any equivalentreduced word.(ii) We say that w is normal when all its suffixes are initially normal.We remark that all the factors of a normal word are normal words. Proposition 6
Any element of A has a unique normal reduced representativeword. Proof
For any reduced word w = a ε i · · · a ε k i k , where ε j = ±
1, we setΩ( w ) = { ( r, s ) | ≤ r < s ≤ k and i r < i s } . Let a be in A . In order to prove that a has normal reduced representativeword, we choose, among all words representing a , a word w for which thenumber w ) is as small as possible (possibly equal to zero). This word w isminimal.We shall prove uniqueness of the normal representative by induction on thelength. If a is of length 1, i.e. if a = a εi for ǫ = ±
1, then uniqueness is obvious.Now suppose that a has two normal reduced representatives w = a ε i · · · a ε k i k and w ′ = a ε ′ i ′ · · · a ε ′ k i ′ k . Since the the suffixes of length k − w and w ′ are6gain normal, it is, by induction hypothesis, sufficient to prove that a ε i = a ε ′ i ′ .Since w and w ′ are normal, we have i = i ′ . Now, the exponents also haveto be equal by Lemma 1: we can not transform the word a ε i · · · a ε k i k into theword a − ε i a ε ′ i ′ · · · a ε ′ k i ′ k by using commutation relations only: starting from thereduced w , no word of the form ua ε i a − ε i u ′ can appear by any sequence ofcommutation relations.In the sequel, we call this unique normal reduced word representing a the normal form of a . Proposition 7
There is a linear-time algorithm that associates to each pil-ing p a normal word σ ⋆ ( p ) such that π ⋆ ( σ ⋆ ( p )) = p . Furthermore, for anyelement a of A the word σ ⋆ ( π ( a )) is the normal form of a . Example 8
Using the notation of Example 4, the word σ ⋆ ( p ) is equal to a − a a − a a a − a a . The calculation is shown in Figure 2. a a a extract a aaa a a a aa aaa extract a extract a a extract extract a extract a
21 32 a aaa
14 1 extract a a a aaa a aa a a aa aa a a extract a a a Figure 2: The word σ ⋆ ( p ) associated to a piling p Proof of Proposition 7
Let p be a piling. By definition, this means thatthere exists an element a of A such that π ( a ) = p . In order to prove Proposi-tion 7, it suffices to find an algorithm for constructing in linear time a word σ ⋆ ( p ),and to prove that σ ⋆ ( p ) is a normal reduced representative of a .We start with the observation that the element a has a reduced representativestarting with the letter a ± i if and only if the a i -stack of the piling is nonemptyand starts with the letter + or − , respectively (not with the letter 0).7e associate to p a normal reduced word σ ⋆ ( p ) by induction on the number ofletters + and − in p in the following way. If p is empty then σ ⋆ ( p ) is the emptyword. Otherwise, let i be the largest index with the property that the a i -stackof p is nonempty and starts with the letter + or − , not with 0. Then, accordingto this sign, we define the first letter of σ ⋆ ( p ) to be a i or a − i , respectively. Thenwe remove the tile consisting of the first letter (+ or − ) of the a i -stack, and ofthe initial letter (which has to be 0) of each of the a j -stacks associated with agenerator a j that does not commute with a i . What remains is a piling p withstrictly fewer letters. Thus the word σ ⋆ ( p ) is already defined, by inductionhypothesis, and we define the word σ ⋆ ( p ) by concatenation σ ⋆ ( p ) = a ± i σ ⋆ ( p ).We claim that the word σ ⋆ ( p ) is a normal reduced representative of a ; indeed,in the above construction we see that the first letter of σ ⋆ ( p ) is also the firstletter of some reduced representative of a . By induction, the whole word σ ⋆ ( p )is a reduced representative of a . Moreover, the word σ ⋆ ( p ) is initially normal,by construction, and by induction its suffix σ ⋆ ( p ) is normal. Hence the wholeword σ ⋆ ( p ) is normal. We are now ready to attack the conjugacy problem.
We recall that a cycling of a reduced word w is the operation of removing thefirst letter of the word, and placing it at the end of the word. A word is called cyclically reduced if it is reduced and if any word obtained from it by a sequenceof cyclings and commutations is still reduced – in other words, if it is not of theform x a ± i x a ∓ i x , where all the letters of x and x commute with a i . Asfar as we know, all known solutions to the conjugacy problem in RAAGs arebased on the following lemma. Lemma 9
Two cyclically reduced words represent conjugate elements of A ifand only if they are related by a sequence of cyclings and commutation relations. Therefore two reduced words w , w with letters in { a ± , . . . , a ± n } representconjugate elements of A if and only if there is a sequence of words w −→ v ↔ v ←− w ↔ represents a finite sequenceof cyclings and commutation relations. Definition 10
If, in a piling p , the a i -stack starts ( resp. finishes) with aletter + or − , the bottom a i -tile ( resp. the top a i -tile ) of p is the sub-piling formed by the first ( resp. last) letter of the a i -stack and the first ( resp. last)letter of the a j -stacks such that a i and a j do not commute in A . Example 11
With the notation of Example 4, Figure 3 gives an example oftop and bottom tiles of a piling. a aaa cyclic reduction a aaa Figure 3: a top a -tile and a bottom a -tile, and the associated cyclic reduction Definition 12
If in a piling p the a i -stack starts with the letter + and endswith − , or vice versa, a cyclic reduction is the act of removing both top andbottom a i -tiles. We say that the piling is cyclically reduced if no cyclic reductionis possible.Note that cyclically reducing a piling yields again a piling. We remark that thereis an obvious linear-time algorithm for transforming any piling into a cyclicallyreduced one by a finite sequence of cyclic reductions. We also observe that fora reduced word w ∈ { a ± , . . . , a ± N } ∗ , cycling of w corresponds to a cycling ofits piling, and that w is cyclically reduced if and only if the piling π ⋆ ( w ) is.Now we have a fast algorithm for cyclically reducing words and pilings. Incontrast to the case of free groups, however, the reduced words which we canobtain are not unique up to cyclic permutation. In order to circumvent thisproblem, we shall introduce in the sequel the notion of a cyclic normal form . Our first objective is to restrict the conjugacy problem to the case of non-splitcyclically reduced words ( or pilings ). We recall that a graph Γ A is associatedto the right-angled Artin group A . 9 efinition 13 Let w be a reduced word different from 1, and let p be itsimage by π ⋆ . Consider ∆( p ) (or ∆( w )) the full subgraph of Γ A whose verticesare those whose correponding stacks contain at least one bead different from 0(in other words, the letters a i such that a ± i occurs in w ). Then, the word w and the piling p are said to be non-split when the graph ∆( p ) is connected.In other words, w is non-split if and only if its set of letters cannot be separatedin two disjoint subsets such that every letter of one of the subset commutes in A with every letter of the other subset. Clearly, it takes linear-time to obtain theset of vertices of the graph ∆( p ), and constant time (which depends on thegraph Γ A ) to decide if ∆( p ) is connected. If it is not, it takes still constant timeto determine the connected components ∆ ( w ) , . . . , ∆ k ( w ) of ∆( w ). Figure 4(which still uses the notation of Example 4) contains examples of both split andnon-split pilings. a a aa a a cyclic reduction aa Figure 4: The word a − a a a a − is not split, but cyclic reduction yields aword which is split: a ( a a − ) = ( a a − ) a Now, if w is a cyclically reduced word that is split, then it is equivalent to aproduct w · · · w k of non-split cyclically reduced words, one for each connectedcomponent ∆ i ( w ) of the graph ∆( w ); the graph ∆ i ( w ) is equal to ∆( w i ).Furthermore, once that the connected components ∆ ( w ) , . . . , ∆ k ( w ) of ∆( w )are computed, appropriate words w , . . . , w k can be obtained in linear-time. Remark 14
The following observation will be crucial: if v is another cyclicallyreduced word, then then w and v represent conjugate elements if and only iftwo conditions are satisfied: firstly the graph ∆( v ) is equal to ∆( w ); secondly,if v , . . . , v k are words such that ∆( v i ) = ∆ i ( v ) and such that v is equivalentto the product v · · · v k , then for each index i the words w i and v i representconjugate elements.Therefore, in order to obtain a solution to the conjugacy problem in linear-timeit is enough to consider the case of cyclically reduced non-split words.10 .2.3 Pyramidal piling and cyclic normal form To solve the conjugacy problem, we associate in the sequel a cyclic normal word to each cyclically reduced non-split word. We first do the analogue of this in theframework of pilings: to each non-split cyclically reduced piling, we associate a pyramidal piling.
Definition 15
Let p be a non-empty piling, and denote by i the smallestindex such that the a i -stack contains an a ± i -bead. We say that the piling p is pyramidal if the first bead of every a j -stack except the a i -stack is either emptyor starts with the letter 0. In that case, we say that a i is the apex of thepyramidal piling.Note that a pyramidal piling has to be non-split. Lemma 16 (i) Let p is a non-empty piling and denote by i the smallest indexsuch that the a i -stack of p contains an a ± i -bead; then there exists a uniquedecomposition p · p of p such that p is a pyramidal piling with a i as apex,and p is a piling without a ± i -beads. Furthermore, one has the equality ofwords σ ⋆ ( p ) = σ ⋆ ( p ) σ ⋆ ( p ) .(ii) The above decomposition p · p can be computed in linear-time on thenumber of ± -beads of the piling p . Example 17
Using the notation of Example 4, Figure 5 gives an example ofa decomposition of a piling.
432 1 a a aa a aaa
Figure 5: Decomposition of a piling as p · p Proof of Lemma 16
We start by exhibiting a linear-time algorithm for find-ing such a decomposition of a given non-empty piling p . Let p be the emptypiling. Reading all the stacks (in the index order), obtain in linear-time thesmallest index i for which the a i -stack contains a bead distinct from 0. Then,11pply iteratively the following recipe: consider the largest index j (necessar-ily greater than i ) for which the a j -stack starts with a letter + or − ; thenremove all the beads in the bottom a j -tile, and add them to the top of thepiling p . When no more beads can be extracted from the bottom of the pil-ing p , then the construction of the factor p is complete, and what remainsis the piling p . This proves the existence part of (i), as well as part ( ii ).The formula σ ⋆ ( p ) = σ ⋆ ( p ) σ ⋆ ( p ) is now immediate by construction. For theuniqueness part of (i), we notice that in any decomposition p = p · p , thefactor p has to contain exactly those tiles that can be extracted on the bottomfrom p without extracting any apex bead.We call the piling p the 0 -factor of p . Thus the piling p is pyramidal if andonly if its 0-factor is empty.In our physical interpretation, if i is the smallest index such that the a i -stackcontains a a ± i -bead, we can lift up the first a ± i -bead along its stick to the firstfloor. Then some part of the piling stays on the ground, while some beads arelifted up. Here it is essential to keep in mind that each 0-bead is connected bya thread to a ± -bead, and that adjacent 0-beads on a stick can slide througheach other. The factor that stays down is p , the factor that is lifted up is p .This latter factor has the structure of an upside-down pyramid supported byone of the apex-beads, hence the names.If in a cyclically reduced piling, the a i -stack starts with a letter + or − , thenone can perform a cycling of the bottom tile containing that bead to the topof the piling, i.e., one can move the initial letter + on the a i -stack, and theinitial letters 0 on the stacks corresponding to letters that do not commutewith a i , to the end of their respective stacks. A physical interpretation (seeFigure 6) of this procedure is obtained by replacing the sticks by concentrichula hoops. A cycling of a bottom tile corresponds to the operation of cyclingthe corresponding tile along the hula hoops. Proposition 18
There is an algorithm which takes as its input any non-split cyclically reduced piling p and which outputs a pyramidal piling that isobtained from the input piling by a finite sequence of cyclings. If the pilinghas ℓ beads, then the algorithm requires O ( ℓ ) cyclings, so its computationalcomplexity is O ( ℓ ) . Proof
The basic procedure of the algorithm is in two steps; given a cyclicallyreduced piling p , first determine the 0-factor p of the canonical decomposition12 reductioncyclic a a operationfirst cycling a a hold This piling is pyramidal p p a a a a a a a a aa a operationsecond cycling a Figure 6: The calculation of a pyramidal piling(by the method of Lemma 16). Secondly, cycle all the tiles belonging to p inorder to obtain a new piling. This procedure takes time O ( ℓ ). The algorithm issimply to iterate this basic procedure until the factor p is empty. It remains toprove that there is a bound on the number of iterations which depends only onthe group A , not on the piling p . In fact, if we denote i the smallest index suchthat p contains an a i -tile, and ∆( p ) the full subgraph of the defining graph Γdefined above, we claim that max a j ∈ ∆( p ) dist ∆ ( a i , a j ) is an upper bound onthe number of iterations, where each edge of ∆( p ) has length 1. This quantityis finite, because ∆( p ) is connected, and is bounded above by N , the number ofgenerators of the group A (which does not depend on the piling p ). This fact isobvious from the geometrical representation, and the proof is a straightforwardinduction: after the first iteration of the basic procedure, no a ± j -beads suchthat a j is at distance 1 from a i in ∆( p ) appear in the 0-factor; after a seconditeration no a ± j -beads such that a j is at distance at most 2 from a i in ∆( p )appear in the 0-factor, and so on.Now, if w is a non-split reduced word, we can apply the above algorithm tothe piling π ( w ) to obtain a pyramidal piling p . Then, the words σ ⋆ ( p ) and w represent conjugate elements. Definition 19
Let w be a word in { a ± , . . . , a ± n } ∗ that is reduced and cycli-cally reduced. We say that the word w is a cyclic normal form if it is normaland all its cyclically conjugate words are normal.Intuitively, if we regard w as a cyclic word, and we start reading anywhere in theword, then the first letter that we read must always be the largest-index letterthat can be extracted on the left. For instance, with the notation of Example 4,the word a − a a − a a a − a a is not a cyclic normal form: starting from the13ast letter and reading cyclically, we read out a a − . . . , which is already illegal,because the letters commute, and a − has a larger index than a , so a − shouldcome first. Another example: the word a a a − a a − a is a cyclic normal form.Our linear-time solution to the conjugacy problem is based on the two followingresults. Proposition 20 If p is a non-split cyclically reduced pyramidal piling, then σ ⋆ ( p ) is a cyclic normal form. Proposition 21
Two cyclic normal forms represent conjugate elements if andonly if they are equal up to a cyclic permutation.
Proof of Proposition 20
Firstly, we remark that a consequence of Lemma 1is the following fact: if a ǫ , b η are letters ( ǫ, η = ±
1) and w is a reduced wordsuch that b − η w and wa ǫ are both reduced ( i.e. no word equivalent to w starts and finishes with b η and a − ǫ , respectively) but the word b − η wa ǫ is notreduced ( i.e. wa ǫ is equivalent to some word that starts with b η ), then a ǫ = b η and all the letters of w commute with a . Now, we know that σ ⋆ ( p ) is anormal cyclically reduced word. For a cyclically reduced word w , the word ww is cyclically reduced (this follows directly from the above fact, or from thepiling representation), and all the words cyclically conjugate to the former aresubwords of the latter. Therefore, in order to prove the result, it is enoughto prove that the word σ ⋆ ( p ) σ ⋆ ( p ) is normal. Assume that this is not thecase. Since σ ⋆ ( p ) is normal, we can then write σ ⋆ ( p ) = w a ηj w = v a ǫi v such that a ηj w v is initially normal but a ηj w v a ǫ is not. In particular, thereexists a νk , with k > j , such that a − νk a ηj w v is reduced but a − νk a ηj w v a ǫi isnot. Since a ηj w v a ǫi is a subword of σ ⋆ ( p ) σ ⋆ ( p ), it is reduced. Using the abovefact, we get that a ǫi = a νk , and a νk commute with all the letter of a ηj w v . Inparticular, the word σ ⋆ ( p ) is equivalent to a νk v v . This is impossible because k is greater that j , and p is pyramidal. Therefore, σ ⋆ ( p ) σ ⋆ ( p ) is normal. Proof of Proposition 21
The “if” implication is obvious, we have to provethe “only if” part.Let w and w ′ be two cyclic normal forms that represent conjugate elements.Let i be the smallest index that appears in w and choose a distinguishedletter a εi in w . As the words w and w ′ are cyclically reduced, there exists asequence of words w = w → w → · · · w r = w ′ that transforms w into w ′ , suchthat w i +1 is obtained from w i by a commutation or a cycling transformation.14e can keep track of the distinguished letter a εi along the transformations:write w j = w ′ j a εi w ′′ j . Assume the number ℓ of commutations that involve thedistinguished letter is positive. Since w is a normal word, the first commuta-tion w j → w j +1 that involves a εi is “from left to right”, i.e. it is of the followingform: w j = w ′ j +1 a ε ′ i ′ a εi w ′′ j and w j +1 = w ′ j +1 a εi a ε ′ i ′ w ′′ j with i ′ > i .Now, consider the last operation w p → w p +1 such that a letter a ηk is exchangedwith the distinguished letter a εi from left to right: we have w p = w ′ p +1 a ηk a εi w ′′ p and w p +1 = w ′ p +1 a εi a ηk w ′′ p . We can also keep track of the distinguished letter a ηk .As long as the two letters do not cross each other again in the opposite direction,we have w ′′ q w ′ q = y q a ηk z q such that all the letters of y q commute with a k (where q satisfies q > p ). In particular, a εi w ′′ q w ′ q is not initially normal. But w ′ isnormal, so the two distinguished letters have to cross each other again in theopposite direction: there exists s , with p < s < r , such that w s = w ′ s a εi a ηk w ′′ s +1 and w s +1 = w ′ s a ηk a εi w ′′ s +1 . Hence, we have a sequence w p = w ′ p +1 a ηk a εi w ′′ p → v ′ p +1 a ηk a εi v ′′ p +1 → · · · → v ′ s − a ηk a εi v ′′ s − → v ′ s a ηk a εi v ′′ s → w s +1 such that each word v ′′ q v ′ q is equal to the word y q z q w ′ q . Thus we obtain anew sequence from w to w ′ with only ℓ − a εi .It follows that we can assume that no commutation involves the distinguishedletter a εi along the sequence w = w → w → · · · w r = w ′ . But this implies thatthe words a εi w ′′ w ′ and a εi w ′′ r w ′ r are equivalent. As they are both cyclic normalforms, they are normal words. Therefore they are equal by Proposition 6.Hence, the words w and w ′ are equal up to a cyclic permutation.Summing up, in order to decide whether two nonsplit cyclically reduced wordsrepresent conjugate elements, it suffices to decide whether their cyclic normalforms are equal (as cyclic words), and these cyclic normal forms can be calcu-lated in linear time. More formally, we have Theorem 22
The conjugacy problem in a right-angled Artin group A is linear-time on the sum of the lengths of the two input words. Proof
Here is a summary of the algorithm:Given any two words w and v ,(i) produce the piling π ⋆ ( w ), and then by cyclic reduction a cyclically reducedpiling p ; similarly for the word v produce first the piling π ⋆ ( v ), and cyclicallyreduce it to a piling q ; 15ii) factorize each of the pilings p and q into non-split factors. If the collectionof subgraphs ∆ i ( p ) and ∆ i ( q ) of the defining graph Γ A do not coincide, output“NO, w and v do not represent conjugate elements” and stop. Otherwise,(iii) if p = p (1) · . . . · p ( k ) and q = q (1) · . . . · q ( k ) are the factorizations found instep (ii), then for i = 1 , . . . , k do the following(a) transform the non-split cyclically reduced pilings p ( i ) and q ( i ) into pyra-midal pilings e p ( i ) and e q ( i ) , using a sequence of cyclings. Then producethe words in cyclic normal form σ ⋆ ( p ( i ) ) and σ ⋆ ( q ( i ) );(b) decide whether the words in cylic normal form found in the previous stepsare the same up to cyclic permutation (in linear-time, using a standardalgorithm). If they are not, answer “NO” and stop.(iv) answer ”YES”. The centralizer of a cyclically reduced element of A has a canonical finite gen-erating set: suppose that w is a cyclically reduced word, written as a productof cyclically reduced non-split words w = w · · · w k , c.f. Section 2.2.2. Then,according to [6], for each i in { , . . . , k } there exists a unique maximal infinite-cyclic subgroup of A containing [ w i ], generated by some cyclically reducedelement [ z i ], and by [25] the centralizer of [ w ] in A is generated by(1) the elements [ z i ], and(2) the generators of A which commute with all the generators occurringin w .In the next section we will need to algorithmically determine explicit represen-tatives of these generators, in the special case where the words w i are cyclicnormal forms. Proposition 23
There is a linear-time algorithm which takes as its input acyclically reduced word w , decomposed as a product of words in cyclic normalform w = w · · · w k , and which outputs the canonical generating set of thecentralizer of w . roof It takes linear time to determine the graph ∆( w ), and then constanttime to deduce from this the generators of type (2).Now we turn to the generators of type (1), i.e. the minimal roots [ z i ] of theelements [ w i ]. As a first step, we claim that periodicity of elements is visible intheir cyclic normal form. More precisely, if one of the words w i is equivalent toa word of the form e z ri for some word e z i and some integer r , then the word w i itself is of the form z ri , for some word z i . In order to prove this claim, weobserve that e z i is equivalent to a word z i in cyclic normal form (because the0-factor of p ( e z i ) must divide the 0-factor of p ( w i ), which is the trivial word).Now the word z ri is still in cyclic normal form (c.f. the proof of Proposition 20),and it is equivalent to w i . Therefore we have z ri = w i .We claim that for each of the factors w i , the desired minimal root z i of w i isdetectable in linear-time: we can calculate a pair ( z i , r ), where z i is a wordand r an integer with z ri = w i , and r is maximal among all such pairs. Indeed,this algorithm works as follows: consider the word w ∗ i obtained by removingthe first letter from the word w i w i . Then find the starting point of the firstoccurrence of w i as a subword of w ∗ i – this can be done by standard algorithms,like the Boyer-Moore algorithm, in time O ( ℓ i ), where ℓ i denotes the length of w i . If this starting point is at the ℓ i th letter of w ∗ , then there is no periodicity.If on the other hand the starting point is at the t th letter with t < ℓ i , thenlet z i be the prefix of w i of length t . By construction we have an equality ofwords z i w i = w i z i . This implies that the words w i and z i have a common root.By the choice of z i , this root has to be z i itself and for r := ℓ i /t we have anequality of words w i = z ri . Finally, by the choice of t , no prefix of w i of lengthless than t can be a root of w i , so z i is indeed the minimal root. In the previous section we saw that the conjugacy problem in a fixed right-angled Artin group can be solved in linear-time on a RAM-machine with con-stant that depends only on the group. In this section we shall prove analogueresults for a large class of subgroups of right-angled Artin groups, namely thoseconsidered in the papers [11, 12], as well as in [19].17 .1 A class of subgroups of RAAGs
Every right-angled Artin group A admits a finite K ( A, A , which we shall denote Y and which can be constructed explicitelyfrom the presentation of A . It is a cubed complex which has one single vertex,and one edge of length 1 for every generator of A . Moreover, for every n -tupleof mutually commuting generators of A , there is one ( n + 1)-torus in Y . Weequip every cell, of any dimension, of this complex with the flat metric, in thesense that in the universal cover e Y every cell is a Euclidean cube of sidelength 1.Then the complex is locally CAT(0), and its universal cover e Y is CAT(0). Forinstance, for the group A = Z = h a , a | [ a , a ] = 1 i , the complex Y is atorus, constructed out of one vertex, two edges, and one square which glued tothe 1-skeleton according to the commutation relation. See [11] for details. Thereader should note that as soon as an orientation is chosen on each edge ( i.e. simple loop) of Y , one obtains an explicit isomorphism between A and π ( Y )such that the image of each generator a i of A is represented by the simple looplabelled by a i traversed in the positive direction.Now, suppose that X is a finite locally CAT (0) cubed complex, and considera cubical map Φ : X → Y , sending each open cube of X bijectively and locallyisometrically to a cell of the same dimension in Y . (Here Y still denotes theSalvetti complex of some right-angled Artin group.) If one of the vertices of X is designated as its basepoint, then such a mapping induces a homomorphismΦ ∗ : π ( X ) → π ( Y ). See Figure 7 for an example where X and Y are 1-dimensional complexes.We need some more notation: for any vertex x of X , we denote by Φ lk : lk ( x, X ) → lk (Φ( x ) , Y ) the induced map from the link of x in X to the link of Φ( x ) in Y We shall be interested in the following two properties which our map Φ mayhave: • The convexity property: for any vertex x of X , and any two verticesof lk (Φ( x ) , Y ) which belong to the image Φ lk ( lk ( x, X )) and which areconnected by an edge, the connecting edge belongs to the image Φ lk ( lk ( x, X )),as well. • The injectivity property: the map of universal covers e Φ : e X → e Y isinjective. In particular, Φ ∗ : π ( X ) → π ( Y ) is a monomorphism.We remark that a map Φ satisfying the two hypotheses is a local isometry.Now, the subgroups of the right-angled Artin group A ∼ = π ( Y ) for which weshall solve the conjugacy problem are the fundamental groups π ( X ) of cubical18omplexes X which admit a cubical map Φ : X → Y with the convexity andinjectivity property. Remark If X and Y are both known to be CAT (0) cube complexes thenthe convexity property implies the injectivity property – cf. [11], Theorem 1and the remark following. Conversely, the two conditions, together with theknowledge that Y is CAT (0), imply that the complex X is itself CAT (0).The reader unfamiliar with the geometrical language used in stating the condi-tions should remember that the convexity and injectivity properties are satisfiedby all the subgroups of right-angled Artin groups discussed in Theorem 1 of [11].So some typical examples to keep in mind are those given in this paper. Moregenerally, in order to get a mental image of the class of subgroups satisfyingthe two hypotheses, one can think of a subgroup whose Cayley graph sits in theCayley graph of A in a “flat” way. Moreover, as proven by Haglund and Wise([19], Theorem 4.2), for a cubed complex X , the property of admitting map Φto a RAAG with the convexity and injectivity property can be characterizedpurely in terms of certain combinatorial conditions on the complex X – theycall such complexes special . General Notation and Conventions for the rest of the section • We fix once and for all a right-angled Artin group A given by a presenta-tion with generators a , . . . , a N , and we denote by Y the cubed complexassociated with A . We fix an orientation on every edge of Y and iden-tify A with π ( Y ), using the chosen orientations. • We also fix a finite cubed complex X and Φ : X → Y a cubical mapsatisfying the convexity and injectivity condition. Finally, we fix a label x , x , x , . . . for each vertex of X .Roughly speaking, our main result is the following Claim: Using the General Notation and Conventions of this section, the conju-gacy problem in the group π ( X ) , with respect to any finite set of generators of π ( X ) , is solvable in linear-time. Phrased in this way, however, this statement is somewhat dissatisfying, becausewe have not even stated how the generators of π ( X ) are specified. A moreprecise statement will be given in Theorem 25 below.In fact, we will not directly solve the conjugacy problem in the fundamentalgroup of X , but a more general problem, namely the conjugacy problem in the19undamental groupoid of X , in linear-time. First, we explain what preciselythat means.Let us fix a (positive) orientation for each edge of the complex X by pullingback along Φ the orientation of edges in Y . An element of the fundamentalgroupoid is, by definition, a homotopy class of paths (with fixed endpoints)from some vertex x i to some vertex x j . Such an element of the fundamentalgroupoid can be represented by a finite sequence of successive directed edges,which may be traversed in the positive or in the negative direction. We shallcall such a sequence an edge path from x i to x j . Similarly in Y we have ananalogue notion of an edge path as a homotopy class of path specified as asequence of positively or negatively directed edges.We shall use the following very convenient way of coding edge paths in X and Y : in Y , we shall simply identify closed edge paths with words in theletters a ± , . . . , a ± N . As for X , the map Φ gives rise to a coding of edge pathsin X by based words . Definition 24 A based word is a word of the form x i wx j , where x i and x j are vertices of X , and w is the image under Φ of an edge path in X startingat x i and ending at x j . The vertex x i is called the base vertex of the basedword.In other words, the edge path x i wx j is by definition the pullback to X ofthe path w in Y which starts at x i and ends at x j . Notice that not everyword of the form x i wx j , with x i and x j vertices of X and w a word withletters in { a ± , . . . , a ± N } , is a based word. However, when it is, then it uniquely determines an edge path in X , because of the injectivity property. For instance,if x i wx j is a based word, and if the word w can be written as a concatenation w = w w , then there exists a unique vertex x k such that x i w x k and x k w x j are based words. For an example of based words, see again Figure 7.Two elements of the fundamental groupoid of X can be multiplied if the termi-nal vertex of the first coincides with the initial vertex of the second. In termsof based words, ( x i w x j ) · ( x j w x k ) = x i w w x k . Two loops in X are freelyhomotopic if and only if they represent conjugate elements of the fundamentalgroupoid. If the loops are represented by based words x wx and x vx , thenthis equivalent to the existence of a based word x ux such that the elementsof the fundamental groupoid represented by x uvu − x and x wx coincide.Our main result can now be stated precisely. The proof will occupy the wholerest of the paper: 20 heorem 25 Using the General Notation and Conventions of this section,given two based words x wx and x vx , one can decide whether they representfreely homotopic loops in X . Moreover, if w and v have length ℓ and ℓ ,respectively, the decision can be performed by an algorithm which takes time O ( ℓ + ℓ ) on a RAM machine, where the linear constants depend on X , Y and Φ only. X Why did we pass to the fundamental groupoid, rather than sticking to thefundamental group? In other words, why do we pay so much attention tobasepoint issues? By the way of motivation, let us look at a wrong “proof” ofTheorem 25, and see how how we get into trouble if we don’t make basepointsexplicit at every step.
Wrong Claim
Let α , β be two closed edge paths in X based at a commonvertex x . Then the loops α and β represent conjugate elements of π ( X ) ifand only if the words Φ( α ) and Φ( β ) represent conjugate elements of A . Wrong proof of the Wrong Claim
The implication “ ⇒ ” is obvious. For“ ⇐ ”, we suppose that the words Φ( α ) and Φ( β ) represent conjugate elementsof π ( Y ), so the loops Φ( α ) and Φ( β ) in Y are freely homotopic. Thus we canapply sequences of free reductions, cyclings, and commutation relations (homo-topies across squares) in Y to each of the two loops so as to transform bothof them into some loop Γ in Y . By the injectivity- and convexity hypothe-sis, these transformations can be pulled back to free homotopies of the originalloops α and β in X . Therefore α and β are both freely homotopic to someloop γ in X , i.e. they are freely homotopic.This proof is almost correct, and our real proof of Theorem 25 shall follow thisoutline. The mistake, however, is the conclusion in the very last sentence: wecan only conclude that α and β are freely homotopic to some loops γ and γ ′ ,respectively, where Φ( γ ) = Γ = Φ( γ ′ ). Intuitively, the loops γ and γ ′ in X maylook like two different “liftings” of Γ, we did not pay attention to basepoints!An explicit counterexample to the Wrong Claim illustrating the base pointproblem is given in Figure 7.In order to prepare the proof of Theorem 25, let us study what homotopies ofpaths in X look like.If α is an edge path in X giving rise to a based word x i wx j , and if x i e wx j is abased word obtained from x i wx j by one application of a commutation relation21 e X = x x = Y a = Φ( e )basepoint Φ e a = Φ( e ) = Φ( e ) Figure 7: π ( Y ) is the free group on two generators a = Φ( e ) = Φ( e )and a = Φ( e ). The loops e and e e e − are not conjugate as elements of π ( X ), whereas their images in π ( Y ) are. In order to describe the loops in X it is better to use the based words x a x and x a a a − x . The latter isconjugate to x a x .(corresponding to a homotopy of a path in Y across a square) then there existsan edge path e α in X , starting from the same vertex as α and homotopic to α ,which gives rise to the based word x i e wx j – this is an immediate consequenceof the convexity condition. Similarly, free cancellations in w can be realised bycancellations of backtracking path segments in α .Let us summarize the situation in even more geometric language. Given avertex x of X , it is in general not true that every loop in Y is the imageunder Φ of a path in X starting at x . However, when such a pullback of theloop exists, then it is unique. Moreover, in that case all homotopies of the loopin Y , except length-increasing ones, can be pulled back to based homotopies ofthe path in X .Let us now look more generally at free homotopies of loops in X , i.e., homo-topies that move the basepoint. Definition 26
Suppose that xwx is a based word. A parallel transport of xwx is a replacement of the vertex x by a vertex x ′ , where x ′ is obtained from x by walking along an oriented edge e with the property that the element Φ( e )of A commutes with all the generators of A occurring in the word w .Geometrically, this move corresponds to replacing a closed path based at x by aparallel one based at x ′ , where x and x ′ are joined by an edge e . The two pathstogether bound an annulus-shaped region of X . Notice that, under Φ, the twopaths have the same image w in Y . Another way of moving the basepoint of aloop is to push it along the loop: Definition 27
Suppose that w = xy y . . . y ℓ x is a based word, and denoteby e the unique edge of X that has one of its extremities equal to x and suchthat Φ( e ) = y . A based cycling of the based word w is its replacement by22he word x ′ y . . . y ℓ y x ′ , such that the vertex x ′ is the second extremity of theedge e .Geometrically, if α is a loop in X based at a vertex x , and described by abased word xy y . . . y ℓ x , and if we apply a cycling operation (in the sense ofsection 2) to the word y y . . . y ℓ , then this cycling can be pulled back to X to a based cycling of the based word, yielding a loop e α , which looks exactlylike α , except that it based at a different vertex x ′ , “one notch further alongthe loop”. Example 28
In the example of Figure 7, we can apply a based cycling tothe based word x a a a − x , yielding x a a − a x . After a cancellation, weobtain the based word γ = x a x . We note that this is different from thebased word γ = x a x , which was also discussed in that example – in fact,the based words x a x and x a x are not even related by parallel transport(because Φ( e ) = a does not commute with a ). As we shall see in Lemma 31,this implies that the two loops e and e e e − are not freely homotopic in X .Also note that a cyclic reduction of a word on the generators of A and theirinverses can be decomposed as a cycling, followed by a usual cancellation ofletters, and each of these operations can be pulled back to operations on theloop in X . Summarizing the last few paragraphs, we have the following Key Observation 29 If α is a loop in X then all non-length-increasing freehomotopies of the loop Φ( α ) in Y can be pulled back to free homotopies of α .Thus for a based word xwx , all cancellations, applications of commutation re-lations, cyclings, and cyclic reductions of the word w can be pulled back toanalogue cancellations, commutation relations, and based cyclings of the basedword. Similarly, if x wx is a based word, and if the word w can be trans-formed into a word w ′ by applying cancellations and commutation relations,then x w ′ x is again a based word. The aim of this subsection is to prove Theorem 25. We recall that we areconsidering two based words x wx and x vx representing two loops in X traversing ℓ and ℓ edges, respectively. A necessary condition for these loopsbeing conjugate in the fundamental groupoid of X is that the words w and w represent conjugate elements of the right-angled Artin group A . In geometric23erms, for the two loops to be freely homotopic in X , their images under Φin Y must be freely homotopic. This is a condition which we can check in time O ( ℓ + ℓ ) by the results of Section 2. However, this condition is not sufficient,as seen in Example 28. So let us now try to refine this approach. Proposition 30
There is an algorithm with running time O ( ℓ + ℓ ) whoseinput consists of two based words x wx and x vx of lengths ℓ and ℓ , andwhich outputs (1) either the information that they do not represent freely homotopic loopsin X , or (2) two based words x ′ e w . . . e w k x ′ and x ′ e w . . . e w k x ′ , representing two loopsin X which are respectively freely homotopic to the original two, andwhere the e w i are mutually commuting cyclic normal forms. Proof of Proposition 30
As seen in Section 2 we can decide in linear-timewhether w and v represent conjugate elements of A . If they do not, thenthe two based words do not represent conjugate elements of the fundamentalgroupoid either, and it suffices to output this information (case (1)).For the rest of the proof we have to deal with the case where w and v dorepresent conjugate elements of A .We already know from Section 2 that the word w can, by a sequence of can-cellations, commutation relations and cyclings be transformed into a word w ′ with the required decomposition w ′ = w ′ . . . w ′ k . Moreover, we know how tocalculate the word w ′ in linear-time.We also know from the Key Observation 29 above that the transformation ofthe word w into the word w ′ can be pulled back to a transformation of thebased word x wx into a based word x w ′ x . Our next task is to determinethe corresponing base vertex x in linear-time.We shall fulfill this task by “carrying along information about the base vertexin X during the algorithm”. While running the algorithm of Section 2, theonly steps that affect the base vertex are the cyclings of pilings (includingcyclic reductions of pilings, which can be decomposed as cyclings, followed bycancellations of tiles): when we cycle an a ± j -tile, we have to determine howthe base vertex is affected. However, this can be done simply by a lookup in afinite, precalculated list: for every vertex x of X , for every generator a j of A ,and for every ǫ ∈ {− , } , this list must tell us at which vertex of X we arriveif we pull back the loop a ǫj ∈ A = π ( Y ) to a path in X starting at x (if that is24ossible). Since the algorithm of Section 2 performs a linearly bounded numberof cyclings, we can calculate the new base vertex x in time O ( ℓ ).In a similar manner we can algorithmically transform the based word x vx into a word x ′ e wx ′ , where e w is equipped with an analogue decomposition e w = e w · · · e w j .But since w and v represented conjugate elements of A , we have, by the resultsof Section 2, that the words w ′ and e w are in fact the same, at least after areordering of the factors of w ′ and a linearly bounded number of cyclings of eachfactor w ′ i ; in particular, we have j = k . Moreover, the Boyer-Moore algorithmtells us how many letters from each factor we have to cycle in order to achievethis. Thus we can transform the based word x w ′ x into the based word x ′ e wx ′ for some vertex x ′ , using a reordering of the factors (which does not affect thebase vertex) and a linearly bounded number of based cyclings.Thus in order to prove Theorem 25, it is enough to prove it for the specialcase v = w = e w . . . e w k , where the words e w , . . . , e w k are mutually commutingcyclic normal forms. (For instance, this is the situation of Example 28, where weneed to decide if the based words x a x and x a x represent freely homotopicloops in X .) For the rest of the proof of Theorem 25 we fix such a word e w ,with such a decomposition.Suppose a based word x e ux is such that x e u e w e u − x and x e wx represent thesame element of the fundamental groupoid. Then in particular the elementsof A represented by e u and e w commute: we have [ e u ][ e w ][ e u ] − = [ e w ] in A .As seen in Section 2.3, and using the notation of this section, the word e u isequivalent to another word u of the form u = z p . . . z p k k ζ where p , . . . , p k are integers and ζ is a word whose letters are generators of A which commute with, but are different from, all the generators occurring in w ,and their inverses. We shall call such a word u a word in preferred form . Wedefine the norm k u k of u by k u k = k X i =1 | p i | + length( ζ )We are now ready state an algorithmically checkable criterion for x e wx and x e wx representing conjugate elements (i.e. representing freely homotopic loopsin X ): 25 emma 31 The two based words x e wx and x e wx represent conjugate ele-ments in the fundamental groupoid if and only if there exists a based word x ux such that u is a word in preferred form with k u k { vertices of X } (1) Proof
We first suppose that an edge path x ux exists, where u is a wordin preferred form. Then the word u e wu − can be transformed into the word e w by a finite number of commutation relations and cancellations (but no length-increasing transformations). By Key Observation 29, this homotopy can bepulled back to X , to yield a based homotopy between the paths in X repre-sented by the based words x u e wu − x and x e wx . In other words, the elements x e wx and x e wx are conjugate, with conjugating element x ux .Conversely, let us suppose that a conjugating element in the fundamentalgroupoid exists, and is represented by a based word x e ux . This means thatthere exists an edge path in X from x to x such that reading out the edgelabels along the path yields the word e u . As seen before, [ e u ] belongs to thesubgroup of A generated the elements [ z ] , . . . , [ z k ] and [ a j ] , . . . , [ a j m ]. Thusthere is a word u in preferred form which can be obtained from e u by a sequenceof reductions and commutation relations. By Key Observation 29, x ux is alsoa based word, i.e. it also represents an edge path in X .We have shown the existence of a based word x ux with u a word in preferredform, and without loss of generality we can suppose that u is chosen so that k u k is minimal among all such based words.Now for t in { , . . . , k u k} let us denote by x ( t ) the vertex of X obtained by awalk in X starting at x and following the edges of X according to the t firstsubwords. Now, if this function { , , . . . , k u k} −→ { vertices of X } , t x ( t )is not injective (for instance, if k u k is larger than the number of vertices of X ),then there exists a strictly shorter edge path in X represented by a basedword x u ′ x with u ′ also in preferred form, obtained by cutting out somesegment of the previous edge path (c.f. the paragraph following Definition 24).This is in contradiction to the choice of u , and we can conclude that we have k u k { vertices of X } Let us now prove that the condition of Lemma 31 can be checked algorithmicallyin linear-time, i.e. in time O ( ℓ ), where ℓ is the length of the word w .26irstly, recalling that the centralizer of [ e w ] is generated by a finite numberof elements (some of them represented by the words z , . . . , z k and the othersequal to certain generators of A ), we observe that there is a universal upperbound on the number of generators, namely the number of generators of A .Moreover, as seen in Proposition 23, words representing these generators canbe determined in linear time.Now there is a very simpleminded linear-time algorithm to check for the ex-istence of a conjugating element: for all words u in preferred form satisfyingcondition (1) check whether x ux is a based word, i.e. whether there existsan edge path in X represented by the based word x ux . Indeed, there is auniversal bound on the number of words to be checked, and for each word u the check takes linear time (since the length of the words z i can grow linearlywith the length of e w ).Here is a summary of the whole algorithm: given two based words x ∗ wx ∗ and x ∗ vx ∗ representing loops α and β in X ,(1) Apply steps (i) and (ii) of the algorithm of Section 2.2, always carrying alongthe base vertex, to find graphs ∆ j ( w ) ( j = 1 , . . . , k ), ∆ j ( v ) ( j = 1 , . . . , k ′ ), basevertices x , x , and based words x w . . . w k x and x v . . . v k ′ x representingloops that are freely homotopic to α and β .(2) If k = k ′ , or if the collections of full subgraphs ∆ j ( w ) and ∆ j ( v ) ⊂ Γ arenot the same, or if for some j between 1 and k the words v j and w j do nothave the same length ℓ j , return “NO”.(3) Apply step (iii)(a) of the algorithm of Section 2.2 to each of the k factors,always carrying along the base vertices, to transform x w . . . w k x into a basedword x w ′ . . . w ′ k x and similarly x v . . . v k x into x ′ e w . . . e w k x ′ , where allwords w ′ i and e w i are cyclic normal forms.(4) For each factor, use a standard pattern matching algorithm to decide if w ′ i = e w i as cyclic words. If no, return “NO”. If yes, keep in mind how manycyclings of each factor w ′ i are required to achieve equality w ′ i = e w i as (non-cyclic) words.(5) Perform the required based cyclings of x w ′ . . . w ′ k x to obtain a based wordof the form x ′ e w . . . e w k x ′ .(6) Calculate the minimal roots z i of the words e w i , as explained in Section 2.3.Also determine the set of generators that commute with all the letters occurringin the words e w i , but do not occur in any of them.(7) Check, for all words u in preferred from satisfying condition (1), whetherthere exists an edge path in X represented by the based word x ′ ux ′ . If for one27f the words u the answer is affirmative, then return “YES”. Otherwise return“NO”. Acknowledgement
We thank Sam Sang-Hyun Kim, Tim Hsu, Lucas Sabalka,and Michah Sageev for interesting conversations.
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