aa r X i v : . [ m a t h . R A ] J a n Presentations for Temperley-Lieb algebras
James East
Centre for Research in Mathematics and Data Science,Western Sydney University, Locked Bag 1797, Penrith NSW 2751, Australia.
J.East @ WesternSydney.edu.au
Abstract
We give a new and conceptually straightforward proof of the well-known presentation for theTemperley-Lieb algebra, via an alternative new presentation. Our method involves twisted semi-group algebras, and we make use of two apparently new submonoids of the Temperley-Lieb monoid.
Keywords : Temperley-Lieb algebras, Temperley-Lieb monoids, planar tangles, presentations.MSC: 16S15, 20M05, 20M20, 16S36, 05E15, 57M99.
Dedicated to the memory of Prof. Vaughan F. R. Jones.
Temperley-Lieb algebras were introduced by their namesakes in [21] to study lattice models in statis-tical mechanics. These algebras have since appeared naturally and extensively in many mathematicalcontexts, and have found particularly strong applications in knot theory; see especially the works ofKauffman and Jones, such as [10–16]. More information may be found in surveys such as [1, 17, 20];the introductions to [3, 22] also contain valuable discussions.One of the most important tools for working with Temperley-Lieb algebras is a well-known pre-sentation by generators and relations (stated in Theorem 2.3 below), which provides an algebraicaxiomatisation of the topological definition in terms of homotopy-classes of planar tangles. As far asthe author is aware, the only complete proof of this presentation is the one given by Borisavljević,Došen and Petrić in [3], where they also discuss the lack of a proof up to that point, even (surprisingly)of the fact that the so-called hooks or diapsides form a generating set; see the tangles denoted e i inFigure 3. The proof given in [3] involves working with a (countably infinite) monoid of tangles in placeof the algebra itself, and is rather ingeneous, albeit quite involved, as the authors themselves note atthe end of their introduction: “Our proof exhibits some difficulties, which we think cannot be evaded” .The main purpose of the current article is to give an alternative proof of the presentation, which webelieve exhibits no such difficulties. Our method also involves replacing the algebra with a monoid, butthis time with a finite one, the so-called Temperley-Lieb monoid , T L n . This monoid is sometimes calledthe Jones monoid in the literature, and denoted J n ; see for example [2, 5, 8, 9], and especially [18] for adiscussion of naming conventions. The main innovation in our proof is in the use of two apparently newsubmonoids L n and R n of T L n ; roughly speaking, these each capture half of the complexity of T L n itself, and we have a natural factorisation T L n = L n R n (Proposition 4.1). We first give presentationsfor L n and R n in Section 3 (Theorem 3.4), and in Section 4 show how to stitch these together to obtaina new presentation for T L n (Theorem 2.1). In Section 5, we show how to rewrite the new presentationto obtain the original one (Theorem 2.2), and in Section 2.3 explain how to convert any presentation forthe monoid into a presentation for the algebra, using general results on twisted semigroup algebras [7].Some aspects of the proof presented here bear similarities to that given in [3], the most obviousbeing the normal forms used; compare our Lemma 4.6 with [3, Lemmas 1 and 4–6]. Although theseforms utilise different generating sets, they both capture natural combinatorial data associated to thestrings of a tangle, as encoded in certain tuples defined in Section 2.4; however, our normal formshave length at most n , while those of [3] grow quadratically in n . As an application of our results,we are also able to establish the (well-known) equivalence of the topological framework of tangles and1he combinatorial approach via set partitions; compare our Corollary 4.2 with [3, Remark 3]. Beyondthese parallels, however, the author believes that the current proof is rather simpler, and hopes it maytherefore be of some benefit. For the duration of the article, we fix a positive integer n , and write n = { , . . . , n } . To avoid trivialitieswe assume that n ≥ .For i ∈ n , we define the points in the plane by P i = ( i, and P ′ i = ( i, . A string is a smoothnon-self-intersecting embedding s of the unit interval [0 , into the rectangle R n = [1 , n ] × [0 , suchthat• s (0) , s (1) ∈ { P , . . . , P n , P ′ , . . . , P ′ n } , and• other than these endpoints, the image of s is contained in the interior of R n .A planar tangle is a collection α = { s , . . . , s n } of n pairwise non-intersecting strings. Such a planartangle is typically identified with the union of the images of its strings, which is a collection of n curves contained in R n . Note that { s (0) , . . . , s n (0) , s (1) , . . . , s n (1) } = { P , . . . , P n , P ′ , . . . , P ′ n } . Someexamples with n = 9 are given in Figure 1.Two planar tangles α = { s , . . . , s n } and β = { t , . . . , t n } are equivalent , written as α ≡ β , if thereis a homotopy through planar tangles from α to β , or more formally if there is a family of continuousmaps F i : [0 , × [0 , → R n , i ∈ n , such that, relabelling the strings if necessary, and writing s ui ( x ) = F i ( u, x ) for i ∈ n and u, x ∈ [0 , ,• γ u = ( s u , . . . , s un ) is a planar tangle for all u ∈ [0 , ,• γ = α and γ = β .We denote by T L n the set of all ≡ -classes of planar tangles.The product αβ of two planar tangles α and β is formed by shifting α one unit in the positive y -direction, attaching it to β , rescaling to the whole object until it lies in the rectangle R n , and finallyremoving any loops contained entirely in the interior of R n . (For later use, let m ( α, β ) be the numberof such loops.) Figure 1 gives an example calculation with n = 9 (and where m ( α, β ) = 1 ). It is clearthat α β ≡ α β whenever α ≡ α and β ≡ β . Thus, we have an induced product on ≡ -classes,which is easily seen to be associative. It follows that T L n is a semigroup under this product, indeed amonoid whose identity is the ≡ -class of the planar tangle id n with n vertical strings. This monoid iscalled the Temperley-Lieb monoid (of degree n ) . In what follows, we typically identify a planar tanglewith its ≡ -class. α = β = = αβ Figure 1: Calculating a product αβ , where α, β ∈ T L .A transversal of a planar tangle α is a string connecting points P i and P ′ j for some i, j ∈ n . Everyother string of α is an upper or lower non-transversal , with obvious meanings. We write rank( α ) for thenumber of transversals, noting that rank( α ) ≡ n (mod 2) . We also define the domain and codomain, dom( α ) = { i ∈ n : P i is the endpoint of a transversal of α } , odom( α ) = { i ∈ n : P ′ i is the endpoint of a transversal of α } . Note that rank( α ) = | dom( α ) | = | codom( α ) | . With α ∈ T L as in Figure 1, we have dom( α ) = { , , } , codom( α ) = { , , } and rank( α ) = 3 . Certain obvious identities will come in handy, including dom( αβ ) ⊆ dom( α ) , codom( αβ ) ⊆ codom( β ) , rank( αβ ) ≤ min(rank( α ) , rank( β )) . Another is that codom( α ) ⊆ dom( β ) ⇒ rank( αβ ) = rank( α ) . These will typically be used withoutexplicit reference.There is a natural involution T L n → T L n : α α † , defined by reflection in the line y = . SeeFigure 2 for an example when n = 9 . Since ( α † ) † = α = αα † α and ( αβ ) † = β † α † for all α, β , it followsthat T L n is a regular ∗ -semigroup in the sense of [19]. We also have identities such as dom( α † ) = codom( α ) , codom( α † ) = dom( α ) , rank( α † ) = rank( α ) . The symmetry/duality afforded by the involution will allow us to simplify many proofs. α = = α † Figure 2: The involution α α † .Now let k be a field, and δ an arbitrary element of k . The Temperley-Lieb algebra
T L n ( k , δ ) is thevector space over k with basis T L n , and with product ⋆ defined on basis elements α, β ∈ T L n (andthen extended by k -linearity) by α ⋆ β = δ m ( α,β ) ( αβ ) , where m ( α, β ) was defined above. So the product in T L n ( k , δ ) of two basis elements is always a scalarmultiple of a basis element, meaning that T L n ( k , δ ) is a so-called twisted semigroup algebra [23]. For a set X , we write X ∗ for the free monoid over X , which consists of all words over X underconcatenation; the empty word will be denoted by . For a set Ω ⊆ X ∗ × X ∗ of pairs of words, wewrite Ω ♯ for the congruence on X ∗ generated by Ω . We say a monoid M has presentation h X : Ω i if M ∼ = X ∗ / Ω ♯ : i.e., if there is a monoid surmorphism X ∗ → M with kernel Ω ♯ . If φ is such asurmorphism, we say M has presentation h X : Ω i via φ . Elements of X and Ω are called generatorsand relations, respectively, and a relation ( u, v ) ∈ Ω will often be denoted as an equation: u = v .If k is a field, then the free (unital) k -algebra over X is k [ X ∗ ] , the semigroup algebra of X ∗ . For Ω ⊆ k [ X ∗ ] × k [ X ∗ ] , we say an algebra A has presentation h X : Ω i if there is an algebra surmorphism φ : k [ X ∗ ] → A with kernel (in the vector space sense) spanned by { uφ − vφ : ( u, v ) ∈ Ω } . The main results of this paper are presentations for the Temperley-Lieb monoid (Theorems 2.1 and 2.2)and algebra (Theorem 2.3). Theorem 2.1 is new, and will be used to prove the well-known Theorems 2.2and 2.3.To state these results, we begin by defining three alphabets: L = { λ , . . . , λ n − } , R = { ρ , . . . , ρ n − } , E = { e , . . . , e n − } . We think of the elements of these sets as abstract letters, but to each ≤ i ≤ n − , we associate threebasic planar tangles, λ i , ρ i and e i , from T L n , as shown in Figure 3.It is easy to see that λ † i = ρ i , ρ † i = λ i , e † i = e i , λ i ρ i = e i , ρ i λ i = e n − for all i .3 i n i n i n Figure 3: The planar tangles λ i (left), ρ i (middle) and e i (right), for ≤ i ≤ n − .We define two morphisms φ : ( L ∪ R ) ∗ → T L n : λ i λ i , ρ i ρ i and ψ : E ∗ → T L n : e i e i , and we extend the over-line notation to words, writing u = uφ for u ∈ ( L ∪ R ) ∗ , and v = vψ for v ∈ E ∗ .Now consider the set Ω consisting of the following relations over L ∪ R , where in each relation thesubscripts range over all meaningful values, subject to any stated constraints: λ i λ n − = λ i (L1) λ i λ j = λ j +2 λ i if i ≤ j ≤ n − (L2) λ in − i +1 λ n − i = λ in − i +1 (L3) ρ n − ρ i = ρ i (R1) ρ j ρ i = ρ i ρ j +2 if i ≤ j ≤ n − (R2) ρ n − i ρ in − i +1 = ρ in − i +1 (R3) λ n − λ j ρ i − if j ≤ i − (RL1) ρ i λ j = λ n − = ρ n − if i − ≤ j ≤ i + 1 (RL2) λ n − λ j − ρ i if j ≥ i + 2 . (RL3)Our main new result is the following, expressed in terms of the above notation: Theorem 2.1.
The Temperley-Lieb monoid
T L n has presentation h L ∪ R : Ω i via φ . Now consider the set Ξ consisting of the following relations over E : e i = e i for all i (E1) e i e j = e j e i if | i − j | > (E2) e i e j e i = e i if | i − j | = 1 . (E3) Theorem 2.2.
The Temperley-Lieb monoid
T L n has presentation h E : Ξ i via ψ . From any presentation for the monoid
T L n , it is easy to deduce a presentation for the algebra T L n ( k , δ ) , for any field k and any non-zero scalar δ ∈ k \ { } , using results of [7, Section 6]. We define Ψ : k [ E ∗ ] → T L n ( k , δ ) to be the k -linear extension of ψ . By [7, Theorem 44], the monoid presentation h E : Ξ i for T L n maybe transformed into an algebra presentation h E : Ξ ′ i for T L n ( k , δ ) by replacing each relation u = v from Ξ by δ m ( v ) u = δ m ( u ) v ; here for a word w = e i · · · e i k ∈ E ∗ , m ( w ) is defined to be the number ofloops created when forming the product e i · · · e i k in T L n . The only relation from Ξ that needs to bemodified in this way is (E1), so let Ξ ′ be the set of relations obtained from Ξ by replacing (E1) with e i = δe i for all i . (E1) ′ Theorem 2.3.
For δ ∈ k \ { } , the Temperley-Lieb algebra T L n ( k , δ ) has presentation h E : Ξ ′ i via Ψ . .4 Partitions and tuples Certain combinatorial data associated to planar tangles will be of use throughout. First, we write n ′ = { ′ , . . . , n ′ } . To each string s of α ∈ T L n , we associate a two-element subset p ( s ) of n ∪ n ′ :• If s joins P i to P j for some i, j ∈ n , then p ( s ) = { i, j } .• If s joins P i to P ′ j for some i, j ∈ n , then p ( s ) = { i, j ′ } .• If s joins P ′ i to P ′ j for some i, j ∈ n , then p ( s ) = { i ′ , j ′ } .We then define p ( α ) = { p ( s ) : s ∈ α } ; this is clearly a (set) partition of n ∪ n ′ , whose blocks all havesize ; cf. [4]. It turns out that the partition p ( α ) provides enough information to uniquely specify α itself, as follows from Corollary 4.2 below.In fact, this partition provides more information than is strictly necessary. Let the left-mostendpoints of the upper and lower non-transversals of α ∈ T L n be P x , . . . , P x k and P ′ y , . . . , P ′ y k , where x > · · · > x k and y > · · · > y k . We then define l ( α ) = ( x , . . . , x k ) and r ( α ) = ( y , . . . , y k ) .By convention, if k = 0 (i.e., if α has no non-transversals), we write l ( α ) = r ( α ) = ∅ . Note that l ( α † ) = r ( α ) and r ( α † ) = l ( α ) .For example, with α, β ∈ T L as in Figure 1, we have l ( α ) = (5 , , , r ( α ) = (7 , , , l ( β ) = (8 , , , , r ( β ) = (8 , , , . As a foreshadowing of things to come, the reader may check that α = λ λ λ ρ ρ ρ (cf. Lemma 4.6).It will be important to know which tuples can occur as l ( α ) or r ( α ) for some α ∈ T L n . With thisin mind, let T n be the set of integer tuples x = ( x , . . . , x k ) such that k ≥ , x > · · · > x k ≥ , x i ≤ n − i + 1 for all i .The third item says that x ≤ n − , x ≤ n − , x ≤ n − , and so on. For such a tuple x , we write | x | = k . Note that ≤ | x | ≤ ⌊ n ⌋ . Lemma 2.4. If α ∈ T L n , then l ( α ) and r ( α ) both belong to T n . Proof.
We just prove the statement for l ( α ) , as that for r ( α ) is dual. Write l ( α ) = ( x , . . . , x k ) . If k = 0 then l ( α ) = ∅ ∈ T n , so for the rest of the proof we assume that k ≥ . By definition, we have x > · · · > x k ≥ , and for each ≤ i ≤ k , α has a string joining P x i to P y i for some x i < y i ≤ n . Itquickly follows that x i = min { x , . . . , x i , y , . . . , y i } . For A ⊆ n with | A | = 2 i , the greatest possiblevalue min( A ) could take is n − i + 1 , so the result follows.Conversely, it follows from Corollary 3.9 below that for any x , y ∈ T n with | x | = | y | , there exists α ∈ T L n with l ( α ) = x and r ( α ) = y (and from Corollary 4.2 that this α is unique). L n and R n In this section and the next we prove Theorem 2.1, which gives a presentation for the Temperley-Lieb monoid
T L n . Our strategy involves two submonoids L n and R n of T L n ; these will be definedshortly, and the main goal of the current section is to give presentations for them (Theorem 3.4).These presentations will then be used in the next section, together with a factorisation T L n = L n R n (Proposition 4.1), to complete the proof of Theorem 2.1. The approach just described is based on thatof [6], which employed similar methods to treat monoids of (planar/order-preserving) partial bijections.We begin by defining the submonoids L n and R n . We say a planar tangle α ∈ T L n is:• right-simple if r ( α ) = ( n − k + 1 , n − k + 3 , . . . , n − , n − for some k ≥ ,• left-simple if l ( α ) = ( n − k + 1 , n − k + 3 , . . . , n − , n − for some k ≥ .5quivalently, α ∈ T L n is right-simple if codom( α ) = { , . . . , r } for some r ≡ n (mod 2) , and thepartition p ( α ) contains the blocks { r + 1 , r + 2 } ′ , { r + 3 , r + 4 } ′ , . . . , { n − , n } ′ . Similar commentsapply to left-simple planar tangles. Note that α is right-simple if and only if α † is left-simple, and viceversa. Also note that λ i is right-simple, and ρ i left-simple, for any i .We define L n (respectively, R n ) to be the set of all right-simple (respectively, left-simple) planartangles. Figure 4 shows elements γ ∈ L and δ ∈ R ; note that γδ is equal to the planar tangle α ∈ T L pictured in Figure 1. γ = δ = Figure 4: Right- and left-simple planar tangles γ ∈ L and δ ∈ R . Lemma 3.1. (i) If α ∈ L n , and if l ( α ) = ( x , . . . , x k ) , then α = λ x · · · λ x k . (ii) If α ∈ R n , and if r ( α ) = ( x , . . . , x k ) , then α = ρ x k · · · ρ x . Proof.
We just prove (i), as (ii) is dual. The proof is by induction on k . If k = 0 , then α has nonon-transversals, so we may write α = ( s , . . . , s n ) , where for some permutation π of n , the string s i joins P i to P ′ iπ for each i ∈ n . If π was not the identity permutation, say with i < j and iπ > jπ , thenthe strings s i and s j would cross, a contradiction. So π is the identity permutation, and by planarity α = id n . The result holds in this case, since the stated product is empty.Now suppose k ≥ . Let s be the string of α with P x as one of its endpoints, and let the otherendpoint of s be P y . We first claim that y = x + 1 . To see this, note first that y ≥ x + 1 by definitionof l ( α ) . Aiming for a contradiction, suppose y ≥ x + 2 , and let t be the string of α with P x +1 as oneof its endpoints. Since x < x + 1 < y , and since s joins P x to P y , it follows by planarity that t isa non-transversal, and is contained entirely in the region bounded by s and the upper border of therectangle R n . In particular, t joins P x +1 to P z for some x + 1 < z < y . But then x + 1 must be oneof the entries in the tuple l ( α ) , and this contradicts the maximality of x .So we have shown that P x is joined to P x +1 by a string of α . Because of this string, it followsthat α = e x α = ( λ x ρ x ) α ; cf. Figure 5. Let β = ρ x α , and note that β contains a string joining P n − to P n , and a string joining P ′ n − k +1 to P ′ n − k +2 ; these strings are coloured red in Figure 5. In fact, bymaximality of x , we have l ( β ) = ( n − , x , . . . , x k ) and r ( β ) = r ( α ) = ( n − k + 1 , n − k + 3 , . . . , n − , n − . (Again, see Figure 5.) Let γ be the planar tangle obtained from β by replacing the two strings justmentioned by two transversals: one joining P n − to P ′ n − k +1 , and the other joining P n to P ′ n − k +2 .Since λ x has a string joining P ′ n − to P ′ n , it follows that λ x β = λ x γ . Moreover, we have l ( γ ) = ( x , . . . , x k ) and r ( γ ) = ( n − k + 3 , . . . , n − , n − . The latter gives γ ∈ L n . By induction, since l ( γ ) has length k − , we have γ = λ x · · · λ x k . It followsthat α = ( λ x ρ x ) α = λ x β = λ x γ = λ x · λ x · · · λ x k , and the proof is complete. Corollary 3.2. (i) If α, β ∈ L n , then α = β ⇔ l ( α ) = l ( β ) . (ii) If α, β ∈ R n , then α = β ⇔ r ( α ) = r ( β ) . Proof.
We just prove (i), as (ii) is dual. The forwards implication is clear, while the converse followsfrom Lemma 3.1(i): if l ( α ) = l ( β ) = ( x , . . . , x k ) , then α = λ x · · · λ x k = β .In what follows, we will write L = { λ , . . . , λ n − } and R = { ρ , . . . , ρ n − } .6 β h n ρ x λ x x n Figure 5: Partitions constructed during the proof of Lemma 3.1. For simplicity, we write h = n − k +1 . Proposition 3.3.
The sets L n and R n are submonoids of T L n , and we have L n = h L i and R n = h R i . Proof.
As usual, it suffices to prove the statement for L n . In light of Lemma 3.1, it remains to showthat h L i ⊆ L n . Since L ⊆ L n , it is enough to show that L n is closed under right multiplication byelements of L . With this in mind, let α ∈ L n , and let ≤ i ≤ n − be arbitrary; we must showthat αλ i ∈ L n . Also let r = rank( α ) . If r = n then α = id n (see the first paragraph of the proof ofLemma 3.1), and so αλ i = λ i ∈ L n . For the rest of the proof, we assume that r < n , and we note that codom( α ) = { , . . . , r } and r ( α ) = ( r + 1 , r + 3 , . . . , n − , n − . Using this, we show in Figure 6 that αλ i ∈ L n in the cases:(a) i < r ,(b) i = r , (c) i > r and i n (mod 2) ,(d) i > r and i ≡ n (mod 2) ,respectively. In fact, in all but case (a), we have αλ i = α . α n (a) λ i α n (b) λ i α n (c) λ i α n (d) λ i Figure 6: Verification that αλ i ∈ L n , as in the proof of Proposition 3.3.Now that we know L n and R n are monoids, we wish to establish presentations. Keeping Proposi-tion 3.3 in mind, we have surmorphisms φ L : L ∗ → L n : λ i λ i and φ R : R ∗ → R n : ρ i ρ i , which are restrictions of the morphism φ : ( L ∪ R ) ∗ → T L n defined in Section 2.3. Let Ω L and Ω R bethe sets of relations (L1)–(L3) and (R1)–(R3), respectively. Our main goal in this section is to provethe following: 7 heorem 3.4. The monoids L n and R n have presentations h L : Ω L i and h R : Ω R i , via φ L and φ R ,respectively. We will just prove Theorem 3.4 for L n , as the result for R n = L † n is dual. Since φ L is a surmorphism,it remains to show that ker( φ L ) = Ω ♯L . For the rest of this section, we write ∼ L = Ω ♯L . Lemma 3.5.
We have ∼ L ⊆ ker( φ L ) . Proof.
This amounts to checking that for each relation ( u, v ) ∈ Ω L we have u = v in L n . For (L1)and (L2), this is easily accomplished diagrammatically; see Figure 7 for the latter. For (L3), one mayfirst show by a simple induction that l ( λ jn − i +1 ) = ( n − i + 1 , n − i + 3 , . . . , n − i + 2 j − for all ≤ j ≤ i . In particular, taking j = i , we have l ( λ in − i +1 ) = ( n − i + 1 , n − i + 3 , . . . , n − , n − . (3.6)Since therefore rank( λ in − i +1 ) = n − i , we have r ( λ in − i +1 ) = ( n − i + 1 , n − i + 3 , . . . , n − , n − as well. Thus, (L3) may be checked diagrammatically, as also shown in Figure 7. λ i i j n λ j λ j +2 i j n λ i λ ik k n λ k − λ ik k n Figure 7: Left: relation (L2). Right: relation (L3), writing k = n − i + 1 .To complete the proof of Theorem 3.4, we must prove the reverse inclusion: ker( φ L ) ⊆ ∼ L . Firstwe need a simple consequence of the relations. Note that (L1) says λ n − is a right identity for each λ j . Since the i = 1 case of (L3) says λ n − λ n − = λ n − , it follows that λ n − is also a right identity: Lemma 3.7. If ≤ i ≤ n − and n − ≤ j ≤ n − , then λ i λ j ∼ L λ i . For a tuple x = ( x , . . . , x k ) ∈ T n (as defined in Section 2.4), we define the words λ x = λ x · · · λ x k ∈ L ∗ and ρ x = ρ x k · · · ρ x ∈ R ∗ . Note that when x = ∅ is the empty tuple, λ ∅ = ρ ∅ = 1 is the empty word. The next lemma refers tothe elements λ x = λ x · · · λ x k ∈ L n . Lemma 3.8. If x ∈ T n , then l ( λ x ) = x . Proof.
This is clear if | x | ≤ . Otherwise, write x = ( x , . . . , x k ) , where k ≥ . Put y = ( x , . . . , x k − ) ,noting that y ∈ T n , and that l ( λ y ) = y by induction. Since λ y ∈ L n has k − non-transversals, wehave codom( λ y ) = { , . . . , n − k + 2 } . Since x k ≤ n − k + 1 , it follows that x k and x k + 1 both belongto codom( λ y ) . Since x k < x k − < · · · < x , each of λ x , . . . , λ x k − contains a transversal joining P x k to P ′ x k ; so too therefore does λ x · · · λ x k − = λ y . It follows that λ x = λ y λ x k contains a non-transversaljoining P x k to P y for some y > x k . Together with the fact that each non-transversal of λ y is containedin λ y λ x k = λ x , it follows that l ( λ x ) = ( x , . . . , x k − , x k ) = x .The following important consequence will be used in the next section:8 orollary 3.9. If x , y ∈ T n and | x | = | y | , then l ( λ x ρ y ) = x and r ( λ x ρ y ) = y . Proof.
Write k = | x | = | y | . Then codom( λ x ) = { , . . . , n − k } = dom( ρ y ) , so rank( λ x ρ y ) = n − k .Thus, λ x ρ y has precisely the same upper (respectively, lower) non-transversals as λ x (respectively, ρ y ),and so l ( λ x ρ y ) = l ( λ x ) and r ( λ x ρ y ) = r ( ρ y ) . The result now follows from Lemma 3.8 and its dual.We now wish to show that every word over L is ∼ L -equivalent to one of the form λ x with x ∈ T n .This is accomplished in Lemma 3.12 below, for whose inductive proof we require some technical lemmasanalysing products of the form λ x λ j , where x ∈ T n and ≤ j ≤ n − . Lemma 3.10. If x ∈ T n with k = | x | ≥ , then λ x λ j ∼ L λ x for any j ≥ n − k . Proof. If j ≥ n − , then the result follows from Lemma 3.7. This includes the k = 1 case, so wenow assume that k ≥ and j ≤ n − , and we proceed by induction on k . Write x = ( x , . . . , x k ) ,and also let y = ( x , . . . , x k − ) . If j ≥ n − k + 1 , then since x k ≤ n − k + 1 ≤ j ≤ n − , we mayapply (L2) and induction to obtain λ x λ j = λ y λ x k λ j ∼ L λ y λ j +2 λ x k ∼ L λ y λ x k = λ x . This leaves onlythe case in which j = n − k . For this we use the j = n − k + 1 case (just proved) and (L3) to obtain λ x λ n − k ∼ L λ x λ kn − k +1 λ n − k ∼ L λ x λ kn − k +1 ∼ L λ x For the next statement, we define min( x ) to be the minimum entry of x ∈ T n if | x | ≥ , and also min( ∅ ) = ∞ . Lemma 3.11. If x ∈ T n with k = | x | ≥ , and if j ≤ n − k − , then λ x λ j ∼ L λ y for some y ∈ T n with | y | = k + 1 and min( y ) = min(min( x ) , j ) . Proof.
We use induction on k . If k = 0 , then λ x λ j = λ j = λ y , where y = ( j ) ∈ T n . Now suppose k ≥ and write x = ( x , . . . , x k ) , noting that min( x ) = x k . If x k > j , then λ x λ j = λ y where y = ( x , . . . , x k , j ) ∈ T n . So for the rest of the proof we will assume that x k ≤ j , and we define u = ( x , . . . , x k − ) , which is empty if k = 1 . Since x k ≤ j ≤ n − k − ≤ n − (as k ≥ ), weapply (L2) and then induction to obtain λ x λ j = λ u λ x k λ j ∼ L λ u λ j +2 λ x k ∼ L λ v λ x k for some v ∈ T n with | v | = k and min( v ) = min(min( u ) , j + 2) .Write v = ( v , . . . , v k ) , noting that v k = min( v ) is either j +2 or min( u ) ; also note that min( u ) = ∞ if k = 1 , or else min( u ) = x k − . In any case, we have v k > x k (since j + 2 ≥ x k + 2 , and since x k − > x k if k ≥ ), and so λ v λ x k = λ y , where y = ( v , . . . , v k , x k ) ∈ T n . (Note that x k ≤ j ≤ n − k − .)Here is the most important technical lemma of this section. Lemma 3.12. If w ∈ L ∗ , then w ∼ L λ x , where x = l ( w ) . Proof.
First note that it suffices to show that w ∼ L λ x for some x ∈ T n , since we will then have l ( w ) = l ( λ x ) = x , by Lemma 3.8. To do this, we use induction on l , the length of w . If l ≤ , then w is already of the form λ x , where | x | ≤ . If l ≥ , then w = uλ j for some u ∈ L ∗ of length l − ,and some ≤ j ≤ n − . By induction, we have u ∼ L λ y for some y ∈ T n with k = | y | ≥ . If j ≥ n − k , then Lemma 3.10 gives w = uλ j ∼ L λ y λ j ∼ L λ y . If j ≤ n − k − , then Lemma 3.11gives w ∼ L λ y λ j ∼ L λ x for some x ∈ T n .We may now tie together the loose ends. Proof of Theorem 3.4.
Recall that we only need to give the proof for L n . For this, it remains toshow that ker( φ L ) ⊆ ∼ L . To do so, let ( u, v ) ∈ ker( φ L ) ; so u, v ∈ L ∗ and u = v . By Lemma 3.12, wehave u ∼ L λ x , where x = l ( u ) . Since v = u , we also have v ∼ L λ x , and so u ∼ L λ x ∼ L v . Remark 3.13.
The proofs of Lemmas 3.10–3.12 give an algorithm for transforming an arbitrary wordover L into a normal form λ x , for x ∈ T n . 9 First presentation for
T L n The previous section gave presentations for the monoids L n and R n (see Theorem 3.4), and we nowuse these to complete the proof of Theorem 2.1. Proposition 4.1. If α ∈ T L n , then α = λ x ρ y , where x = l ( α ) and y = r ( α ) . Consequently, T L n = L n R n = h L ∪ R i . Proof.
In light of Proposition 3.3, it suffices to prove the first statement. Clearly we have α = βγ ,where β ∈ L n and γ ∈ R n are such that β (respectively, γ ) has the same upper (respectively, lower)non-transversals as α ; see Figure 8. Since β has the same upper non-transversals as α , we have l ( β ) = l ( α ) = x = l ( λ x ) , where we used Lemma 3.8 in the last step. It follows from Corollary 3.2(i)that β = λ x . A symmetrical argument shows that γ = ρ y . Thus, α = βγ = λ x ρ y . αβγ Figure 8: Partitions constructed during the proof of Proposition 4.1; note that α = βγ .The next simple consequence refers to the partitions p ( α ) as defined in Section 2.4. The equivalenceof (i) and (ii) was stated without proof in [3, Remark 3]. Corollary 4.2.
For α, β ∈ T L n , the following are equivalent: (i) α = β , (ii) p ( α ) = p ( β ) , (iii) l ( α ) = l ( β ) and r ( α ) = r ( β ) . Proof.
The implications (i) ⇒ (ii) and (ii) ⇒ (iii) are clear, while (iii) ⇒ (i) follows from Proposi-tion 4.1: if l ( α ) = l ( β ) = x and r ( α ) = r ( β ) = y , then α = λ x ρ y = β .It also follows from Proposition 4.1 that the morphism φ : ( L ∪ R ) ∗ → T L n defined in Section 2.3is surjective. To complete the proof of Theorem 2.1, we must show that ker( φ ) = Ω ♯ , where Ω is theset of relations (L1)–(L3), (R1)–(R3) and (RL1)–(RL3). Throughout this section, we write ∼ = Ω ♯ . Lemma 4.3.
We have ∼ ⊆ ker( φ ) . Proof.
This again amounts to a simple diagrammatic check; see Figure 9 for (RL3).We now work towards the reverse inclusion, beginning with an obvious consequence of (RL1)–(RL3):
Lemma 4.4.
For any w ∈ ( L ∪ R ) ∗ , we have w ∼ uv for some u ∈ L ∗ and v ∈ R ∗ . It follows from Lemma 4.4, along with Lemma 3.12 and its dual, that any word over L ∪ R is ∼ -equivalent to λ x ρ y for some x , y ∈ T n . The main step remaining is to show that such a factorisationexists with | x | = | y | , and this is achieved in Lemma 4.6. Roughly speaking, we use the longer of λ x or ρ y to lengthen the shorter, and the next lemma provides the precise mechanism to do this.10 i i j n λ j λ n − i j n λ j − ρ i Figure 9: Relation (RL3).
Lemma 4.5. If x ∈ T n with k = | x | ≥ , then λ x ∼ λ x ρ n − k +1 and ρ x ∼ λ n − k +1 ρ x . Proof.
By the asymmetry in relations (RL1)–(RL3), the two statements are not dual, so both mustbe treated. We proceed by induction on k . Write x = ( x , . . . , x k ) .If k = 1 , then λ x ∼ λ x λ n − ∼ λ x ρ n − and ρ x ∼ ρ n − ρ x ∼ λ n − ρ x , using (L1), (R1) and (RL2).Now suppose k ≥ , and write y = ( x , . . . , x k − ) ∈ T n . Then λ x = λ y λ x k ∼ λ y ρ n − k +3 λ x k ∼ λ y λ n − λ x k ρ n − k +1 ∼ λ y λ x k ρ n − k +1 = λ x ρ n − k +1 , where we first used induction, and then (RL1) and (L1), noting that x k ≤ n − k + 1 . We also have ρ x = ρ x k ρ y ∼ ρ x k λ n − k +3 ρ y by induction ∼ λ n − λ n − k +1 ρ x k ρ y by (RL3) ∼ λ n − k +1 λ n − ρ x k ρ y by (L2), with i = n − k + 1 and j = n − ∼ λ n − k +1 λ n − λ n − ρ x k ρ y by (L1) ∼ λ n − k +1 ρ x k λ n − ρ y by (RL3), with i = x k and j = n − ∼ λ n − k +1 ρ x k ρ n − ρ y by (RL2) ∼ λ n − k +1 ρ x k ρ y = λ n − k +1 ρ x by (R1).Here are the promised normal forms: Lemma 4.6. If w ∈ ( L ∪ R ) ∗ , then w ∼ λ x ρ y , where x = l ( w ) and y = r ( w ) . Proof.
It is enough to show that w ∼ λ x ρ y for some x , y ∈ T n with | x | = | y | . Indeed, then we willhave l ( w ) = l ( λ x ρ y ) = x , by Corollary 3.9, and similarly r ( w ) = y .If w = 1 then w = λ ∅ ρ ∅ , so we assume that w = 1 . By Lemma 4.4 we have w ∼ uv for some u ∈ L ∗ and v ∈ R ∗ . Write k = | l ( u ) | and l = | r ( v ) | , and assume that k ≥ l , the other case beingsymmetrical. Since w = 1 , we have k ≥ . Since u ∈ L n and v ∈ R n have k and l non-transversals,respectively, we have codom( u ) = { , . . . , n − k } ⊆ { , . . . , n − l } = dom( v ) . It follows that rank( w ) = rank( u v ) = rank( u ) = n − k. (4.7)Writing x = l ( u ) and z = r ( v ) , Lemma 3.12 and its dual give u ∼ λ x and v ∼ ρ z . Together with k applications of Lemma 4.5, it follows that w ∼ uv ∼ λ x ρ z ∼ λ x ρ kn − k +1 ρ z . Combining this with (3.6)and (4.7), it follows that rank( ρ kn − k +1 ρ z ) ≤ rank( ρ kn − k +1 ) = n − k = rank( w ) = rank( λ x ρ kn − k +1 ρ z ) ≤ rank( ρ kn − k +1 ρ z ) , so that in fact rank( ρ kn − k +1 ρ z ) = n − k . Thus, ρ kn − k +1 ρ z has k non-transversals and so, writing y = r ( ρ kn − k +1 ρ z ) , we have | y | = k = | x | , and the dual of Lemma 3.12 gives ρ kn − k +1 ρ z ∼ ρ y . Puttingeverything together, we have w ∼ λ x ρ kn − k +1 ρ z ∼ λ x ρ y .As in Remark 3.13, the above lemmas lead to an algorithm to compute the normal form of anarbitrary word over L ∪ R . We may now complete the proof of our main new result. Proof of Theorem 2.1.
It remains to show that ker( φ ) ⊆ ∼ . To do so, let ( u, v ) ∈ ker( φ ) ; so u, v ∈ ( L ∪ R ) ∗ and u = v . By Lemma 4.6, we have u ∼ λ x ρ y , where x = l ( u ) and y = r ( u ) . Since v = u , we also have v ∼ λ x ρ y , and so u ∼ λ x ρ y ∼ v .11 Second presentation for
T L n To prove Theorem 2.2, we need to show that ψ : E ∗ → T L n : e i e i is surjective, and that ker( ψ ) = Ξ ♯ . Proposition 5.1.
We have
T L n = h E i . Consequently, ψ is surjective. Proof.
Since λ i = e i e i +1 · · · e n − and ρ i = e n − · · · e i +1 e i for all i , this follows from Proposition 4.1.For the rest of this section we write ≈ = Ξ ♯ . The next result is proved in the usual way. Lemma 5.2.
We have ≈ ⊆ ker( ψ ) . As ever, proving the reverse inclusion is more of a challenge, though we will be greatly aided bythe fact that we already have the presentation h L ∪ R : Ω i from Theorem 2.1.For ≤ i ≤ n − , we define the following words from E ∗ : b λ i = e i e i +1 · · · e n − and b ρ i = e n − · · · e i +1 e i . We extend this to a morphism ( L ∪ R ) ∗ → E ∗ : w b w , and we note that b w = w for all w ∈ ( L ∪ R ) ∗ ,meaning that b wψ = wφ for all such w . Lemma 5.3.
For any ≤ i ≤ n − , we have b λ i b ρ i ≈ e i . Proof.
We use reverse induction on i . For i = n − , we have b λ n − b ρ n − = e n − e n − ≈ e n − , by (E1).If i ≤ n − , then by induction and (E3), we have b λ i b ρ i = e i b λ i +1 b ρ i +1 e i ≈ e i e i +1 e i ≈ e i .The next result follows immediately: Corollary 5.4.
For any w ∈ E ∗ , we have w ≈ b u for some u ∈ ( L ∪ R ) ∗ . The main technical result we need provides the link between the relations in the two presentations:
Lemma 5.5.
For any relation ( u, v ) from Ω , we have b u ≈ b v . Proof.
For (L1) this follows immediately from (E1). For (L2), if i ≤ j ≤ n − , then b λ j +2 b λ i = e j +2 · · · e n − · e i · · · e j − · e j · e j +1 · · · e n − ≈ e i · · · e j − · e j · e j +2 · · · e n − · e j +1 · · · e n − by (E2) ≈ e i · · · e j − · e j e j +1 e j · e j +2 · · · e n − · e j +1 · · · e n − by (E3) ≈ e i · · · e j − · e j e j +1 · e j +2 · · · e n − · e j e j +1 · · · e n − = b λ i b λ j by (E2).We treat (L3) by induction. For i = 1 , (E3) gives b λ n − b λ n − = e n − e n − e n − ≈ e n − = b λ n − . For i ≥ , b λ in − i +1 b λ n − i = b λ n − i +1 · b λ i − n − i +1 · b λ n − i ≈ b λ i − n − i +3 · b λ n − i +1 · b λ n − i by (L2), proved above = b λ i − n − i +3 · e n − i +1 b λ n − i +2 · b λ n − i ≈ e n − i +1 · b λ i − n − i +3 b λ n − i +2 · b λ n − i by (E2) ≈ e n − i +1 · b λ i − n − i +3 · b λ n − i by induction ≈ b λ i − n − i +3 · e n − i +1 b λ n − i by (E2) = b λ i − n − i +3 · e n − i +1 e n − i e n − i +1 b λ n − i +2 ≈ b λ i − n − i +3 · e n − i +1 b λ n − i +2 by (E3) = b λ i − n − i +3 · b λ n − i +1 ≈ b λ n − i +1 · b λ i − n − i +1 = b λ in − i +1 by (L2) again.We have dealt with relations (L1)–(L3). Relations (R1)–(R3) are dual.12or (RL2), first note that b ρ n − b λ n − = e n − e n − ≈ e n − = b λ n − = b ρ n − , while if i ≤ n − , then b ρ i b λ i = e n − · · · e i +2 e i +1 e i · e i e i +1 e i +2 · · · e n − ≈ e n − · · · e i +2 e i +1 e i · e i +1 e i +2 · · · e n − by (E1) ≈ e n − · · · e i +2 · e i +1 e i +2 · · · e n − by (E3) ≈ e n − · · · e i +2 e i +1 · e i +1 e i +2 · · · e n − by (E1) = b ρ i +1 b λ i +1 ≈ b λ n − by induction.The second line in the previous calculation also shows that b ρ i b λ i ≈ b ρ i b λ i +1 = b ρ i +1 b λ i for i ≤ n − , andthis finishes the proof for (RL2). For (RL1), if j ≤ i − , then b ρ i b λ j = e n − · · · e i · e j · · · e i − e i − · e i − e i · · · e n − ≈ e j · · · e i − e i − · e n − · · · e i · e i − e i · · · e n − by (E2) = e j · · · e i − e i − · b ρ i b λ i − ≈ e j · · · e i − e i − · b λ n − by (RL2), proved above = e j · · · e i − e i − · e n − ≈ e n − · e j · · · e i − e i − by (E2) ≈ e n − · e j · · · e i − b λ i − b ρ i − = b λ n − b λ j b ρ i − by Lemma 5.3.Relation (RL3) is treated in almost identical fashion to (RL1).We now have everything we need to complete the proof of the second main result. Proof of Theorem 2.2.
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