A new basis for the Homflypt skein module of the solid torus

Transcription

1 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) Journal of Pure and Applied Algebra ) Contents lists available at ScienceDirect Journal of Pure and Applied Algebra A new basis for the Homflypt sein module of the solid torus Ioannis Diamantis, Sofia Lambropoulou Department of Mathematics, National Technical University of Athens, Zografou Campus, GR Athens, Greece a r t i c l e i n f o a b s t r a c t Article history: Received 10 December 2014 Received in revised form 13 May 2015 Available online xxxx Communicated by C. Kassel MSC: 57M27; 57M25; 57N10; 20F36; 20C08 In this paper we give a new basis, Λ, for the Homflypt sein module of the solid torus, SST), which topologically is compatible with the handle sliding moves and which was predicted by J.H. Przytyci. The basis Λis different from the basis Λ, discovered independently by Hoste and Kidwell [1] and Turaev [2] with the use of diagrammatic methods, and also different from the basis of Morton and Aiston [3]. For finding the basis Λ we use the generalized Hece algebra of type B, H 1,n, which is generated by looping elements and braiding elements and which is isomorphic to the affine Hece algebra of type A [4]. More precisely, we start with the well-nown basis Λ of SST) and an appropriate linear basis Σ n of the algebra H 1,n. We then convert elements in Λ to sums of elements in Σ n. Then, using conjugation and the stabilization moves, we convert these elements to sums of elements in Λ by managing gaps in the indices, by ordering the exponents of the looping elements and by eliminating braiding tails in the words. Further, we define total orderings on the sets Λ and Λ and, using these orderings, we relate the two sets via a bloc diagonal matrix, where each bloc is an infinite lower triangular matrix with invertible elements in the diagonal. Using this matrix we prove linear independence of the set Λ, thus Λ is a basis for SST). SST) plays an important role in the study of Homflypt sein modules of arbitrary c.c.o. 3-manifolds, since every c.c.o. 3-manifold can be obtained by integral surgery along a framed lin in S 3 with unnotted components. In particular, the new basis of SST) is appropriate for computing the Homflypt sein module of the lens spaces. In this paper we provide some basic algebraic tools for computing sein modules of c.c.o. 3-manifolds via algebraic means Published by Elsevier B.V. 1. Introduction Let M be an oriented 3-manifold, R = Z[u ±1, z ±1 ], L the set of all oriented lins in M up to ambient isotopy in M and let S be the submodule of RL generated by the sein expressions u L + ul zl 0, This research has been co-financed by the European Union European Social Fund ESF) and Gree national funds, MIS , through the Operational Program Education and Lifelong Learning of the National Strategic Reference Framewor NSRF) Research Funding Program: THALES: Reinforcement of the interdisciplinary and/or inter-institutional research and innovation. * Corresponding author. addresses: diamantis@math.ntua.gr I. Diamantis), sofia@math.ntua.gr S. Lambropoulou). URL: S. Lambropoulou) / 2015 Published by Elsevier B.V.

2 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 2 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Fig. 1. The lins L +,L,L 0 locally. Fig. 2. A basic element of SST). where L +, L and L 0 are oriented lins that have identical diagrams, except in one crossing, where they are as depicted in Fig. 1. For convenience we allow the empty not,, and add the relation u u = zt 1, where T 1 denotes the trivial not. Then the Homflypt sein module of M is defined to be: S M) =S M; Z [ u ±1,z ±1],u L + ul zl 0 ) = RL / S. Unlie the Kauffman bracet sein module, the Homflypt sein module of a 3-manifold, also nown as Conway sein module and as third sein module, is very hard to compute see [5] for the case of the product of a surface and the interval). Let ST denote the solid torus. In [2,1] the Homflypt sein module of the solid torus has been computed using diagrammatic methods by means of the following theorem: Theorem 1 Turaev, Kidwell Hoste). The sein module SST) is a free, infinitely generated Z[u ±1, z ±1 ]- module isomorphic to the symmetric tensor algebra SR π 0, where π 0 denotes the conjugacy classes of nontrivial elements of π 1 ST). A basic element of SST) in the context of [2,1], is illustrated in Fig. 2. In the diagrammatic setting of [2] and [1], ST is considered as Annulus Interval. The Homflypt sein module of ST is particularly important, because any closed, connected, oriented c.c.o.) 3-manifold can be obtained by surgery along a framed lin in S 3 with unnotted components. A different basis of SST), nown as Young idempotent basis, is based on the wor of Morton and Aiston [3] and Blanchet [6]. In [4], SST) has been recovered using algebraic means. More precisely, the generalized Hece algebra of type B, H 1,n q), is introduced, which is related to the affine Hece algebra of type A, H n q) [4]. Then, a unique Marov trace is constructed on the algebras H 1,n q) leading to an invariant for lins in ST, the universal analogue of the Homflypt polynomial for ST. This trace gives distinct values on distinct elements of the [2,1]-basis of SST). The lin isotopy in ST, which is taen into account in the definition of the sein module and which corresponds to conjugation and the stabilization moves on the braid level, is captured by the conjugation property and the Marov property of the trace, while the defining relation of the sein module is reflected into the quadratic relation of H 1,n q). In the algebraic language of [4] the basis of SST), described in Theorem 1, is given in open braid form by the set Λ in Eq. 4). Fig. 8 illustrates the basic

3 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 3 Fig. 3. An element of the new basis Λ. element of Fig. 2 in braid notation. Note that in the setting of [4] ST is considered as the complement of the unnot the bold curve in the figure). The looping elements t i H 1,nq) in the monomials of Λ are all conjugates, so they are consistent with the trace property and they enable the definition of the trace via simple inductive rules. In this paper we give a new basis Λfor SST), which was predicted by J.H. Przytyci, using the algebraic methods developed in [4] see Fig. 3). The motivation of this wor is the computation of S Lp, q)) via algebraic means. The new basic set is described in Eq. 1) in open braid form see Fig. 9). The looping elements t i are in the algebras H 1,n q) and they are commuting. For a comparative illustration and for the defining formulas of the t i s and the t i s the reader is referred to Fig. 7 and Eq. 3) respectively. Moreover, the t i s are consistent with the handle sliding move or band move used in the lin isotopy in Lp, q), in the sense that a braid band move can be described naturally with the use of the t i s see for example [7] and references therein). Our main result is the following: Theorem 2. The following set is a Z[q ±1, z ±1 ]-basis for SST): Λ={t t n n, i Z \{0}, i i+1 i, n N}. 1) Our method for proving Theorem 2 is the following: We define total orderings in the sets Λ and Λ, we show that the two ordered sets are related via a lower triangular infinite matrix with invertible elements on the diagonal, and using this matrix, we show that the set Λ is linearly independent. More precisely, two analogous sets, Σ n and Σ n, are given in [4] as linear bases for the algebra H 1,n q). See Theorem 4 in this paper. The set n Σ n includes Λas a proper subset and the set n Σ n includes Λ as a proper subset. The sets Σ n come directly from the wors of S. Arii and K. Koie, and M. Brouè and G. Malle on the cyclotomic Hece algebras of type B. See [4] and references therein. The second set n Σ n includes Λ as a proper subset. The sets Σ n appear naturally in the structure of the braid groups of type B, B 1,n ; however, it is very complicated to show that they are indeed basic sets for the algebras H 1,n q). The sets Σ n play an intrinsic role in the proof of Theorem 2. Indeed, when trying to convert a monomial λ in Λ into a linear combination of elements in Λ we pass by elements of the sets Σ n. This means that in the converted expression of λ we have monomials in the t i s with possible gaps in the indices and possible non-ordered exponents, followed by monomials in the braiding generators g i. So, in order to reach expressions in the set Λ we need: to manage the gaps in the indices of the t i s, to order the exponents of the t i s and to eliminate the braiding tails.

4 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 4 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Fig. 4. A mixed lin in S 3. The paper is organized as follows. In Section 2 we recall the algebraic setting and the results needed from [4]. In Section 3 we define the orderings in the two sets Σ n and Σ n, which include the sets Λ and Λ as subsets, and we prove that these sets are totally ordered. In Section 4 we prove a series of lemmas for converting elements in Λ to elements in the sets Σ n. In Section 5 we convert elements in Σ n to elements in Λusing conjugation and the stabilization moves. Finally, in Section 6 we prove that the sets Λ and Λare related through a lower triangular infinite matrix mentioned above and that the set Λis linearly independent. A computer program converting elements in Λ to elements in Σ n has been developed by K. Karvounis and will be soon available on The algebraic techniques developed here will serve as basis for computing Homflypt sein modules of arbitrary c.c.o. 3-manifolds using the braid approach. The advantage of this approach is that we have an already developed homogeneous theory of braid structures and braid equivalences for lins in c.c.o. 3-manifolds [8,9,7]. In fact, these algebraic techniques are used and developed further in [10] for nots and lins in 3-manifolds represented by the 2-unlin. The authors wish to express their gratitude for some critical comments of the Referee, which led to the improvement of the presentation of the paper. 2. The algebraic settings 2.1. Mixed lins in S 3 We now view ST as the complement of a solid torus in S 3. An oriented lin L in ST can be represented by an oriented mixed lin in S 3 see Fig. 4), that is a lin in S 3 consisting of the unnotted fixed part Î representing the complementary solid torus in S 3 and the moving part L that lins with Î. A mixed lin diagram is a diagram Î L of Î L on the plane of Î, where this plane is equipped with the top-to-bottom direction of I. Consider now an isotopy of an oriented lin L in ST. As the lin moves in ST, its corresponding mixed lin will change in S 3 by a sequence of moves that eep the oriented Î pointwise fixed. This sequence of moves consists in isotopy in the S 3 and the mixed Reidemeister moves. In terms of diagrams we have the following result for isotopy in ST: The mixed lin equivalence in S 3 includes the classical Reidemeister moves and the mixed Reidemeister moves, which involve the fixed and the standard part of the mixed lin, eeping Î pointwise fixed Mixed braids in S 3 By the Alexander theorem for nots in solid torus, a mixed lin diagram Î L of Î L may be turned into a mixed braid I β with isotopic closure see Fig. 5). This is a braid in S 3 where, without loss of generality, its first strand represents Î, the fixed part, and the other strands, β, represent the moving part L. The subbraid β shall be called the moving part of I β.

5 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 5 Fig. 5. The closure of a mixed braid to a mixed lin. Fig. 6. The generators of B 1,n. The sets of braids related to the ST form groups, which are in fact the Artin braid groups of type B, denoted B 1,n, with presentation: B 1,n = t, σ 1,...,σ n σ 1 tσ 1 t = tσ 1 tσ 1 tσ i = σ i t, i > 1, σ i σ i+1 σ i = σ i+1 σ i σ i+1, 1 i n 2 σ i σ j = σ j σ i, i j > 1 where the generators σ i and t are illustrated in Fig. 6. Isotopy in ST is translated on the level of mixed braids by means of the following theorem. Theorem 3. See [11, Theorem 3].) Let L 1, L 2 be two oriented lins in ST and let I β 1, I β 2 be two corresponding mixed braids in S 3. Then L 1 is isotopic to L 2 in ST if and only if I β 1 is equivalent to I β 2 in n=1 B 1,n by the following moves: i) Conjugation: α β αβ, if α, β B 1,n. ii) Stabilization moves: α ασ n ±1 B 1,n+1, if α B 1,n The generalized Iwahori Hece algebra of type B It is well nown that B 1,n is the Artin group of the Coxeter group of type B, which is related to the Hece algebra of type B, H n q, Q) and to the cyclotomic Hece algebras of type B. In [4] it has been established that all these algebras form a tower of B-type algebras and are related to the not theory of ST. The basic one is H n q, Q), a presentation of which is obtained from the presentation of the Artin group B 1,n by adding the quadratic relations g 2 i =q 1)g i + q 2) and the relation t 2 =Q 1) t + Q, where q, Q C\{0} are seen as fixed variables. The middle B-type algebras are the cyclotomic Hece algebras of type B, H n q, d), whose presentations are obtained by the quadratic relation 2) and t d =t u 1 )t u 2 )...t u d ). The topmost Hece-lie algebra in the tower is

6 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 6 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Fig. 7. The elements t i and t i. the generalized Iwahori Hece algebra of type B, H 1,n q), which, as observed by T. tom Diec, is related to the affine Hece algebra of type A, H n q) cf.[4]). The algebra H 1,n q) has the following presentation: g 1 tg 1 t = tg 1 tg 1 tg i = g i t, i > 1 H 1,n q) = t, g 1,...,g n g i g i+1 g i = g i+1 g i g i+1, 1 i n 2. g i g j = g j g i, i j > 1 g 2 i =q)g i + q, i =1,...,n 1 That is: H 1,n q) = Z [ q ±1] B 1,n σ 2 i q 1) σ i q. Note that in H 1,n q) the generator t satisfies no polynomial relation, maing the algebra H 1,n q) infinite dimensional. Also that in [4] the algebra H 1,n q) is denoted as H n q, ). In [12] V.F.R. Jones gives the following linear basis for the Iwahori Hece algebra of type A, H n q): S = { g i1 g i1...g i1 1 )g i2 g i2...g i2 2 )...g ip g ip...g ip p ) }, for 1 i 1 <...<i p n 1. The basis S yields directly an inductive basis for H n q), which is used in the construction of the Ocneanu trace, leading to the Homflypt or 2-variable Jones polynomial. In H 1,n q) we define the elements: t i := g i g i...g 1 tg 1...g i g i and t i := g i g i...g 1 tg 1...g i g i, 3) as illustrated in Fig. 7. In [4] the following result has been proved. Theorem 4. See [4, Proposition 1, Theorem 1].) The following sets form linear bases for H 1,n q): i) Σ n = { i 1 t 2 i 2...t r i r σ, where 1 i 1 <...<i r n 1}, ii) Σ n = {t 1 i 1 t 2 i 2...t r i r σ, where 1 i 1 <...<i r n}, where 1,..., r Z and σ is a basic element in H n q). Remar 1. i) The indices of the t i s in the set Σ n are ordered but are not necessarily consecutive, neither do they need to start from t. ii) A more straightforward proof that the sets Σ n form bases for H 1,n q) can be found in [13].

7 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 7 In [4] the basis Σ n is used for constructing a Marov trace on n=1 H 1,nq). Theorem 5. See [4, Theorem 6].) Given z, s, with Z specified elements in R = Z [ q ±1], there exists a unique linear Marov trace function determined by the rules: tr : H 1,n q) R z,s ), Z n=1 1) trab) = trba) for a, b H 1,n q) 2) tr1) = 1 for all H 1,n q) 3) trag n ) = z tra) for a H 1,n q) 4) trat n )=s tra) for a H 1,n q), Z. Note that the use of the looping elements t i enables the trace tr to be defined by just extending the three rules of the Ocneanu trace on the algebras H n q) [12] by rule 4). Using tr Lambropoulou constructed a universal Homflypt-type invariant for oriented lins in ST. Namely, let L denote the set of oriented lins in ST. Then: Theorem 6. See [4, Definition 1].) The function X : L Rz, s ) X α = [ 1 λq ] n ) e λ tr π α)), λ 1 q) where λ := z+1 q qz, α B 1,n is a word in the σ i s and t i s, α is the closure of α, e is the exponent sum of the σ i s in α, and π the canonical map of B 1,n in H 1,n q), such that t t and σ i g i, is an invariant of oriented lins in ST. The invariant X satisfies a sein relation [4]. Theorems 4, 5 and 6 hold also for the algebras H n q, Q) and H n q, d), giving rise to all possible Homflypt-type invariants for nots in ST. For the case of the Hece algebra of type B, H n q, Q), see also [11] and [14] The basis of SST) in algebraic terms Let us now see how SST) is described in the above algebraic language. We note first that an element α in the basis of SST) described in Theorem 1 when ST is considered as Annulus Interval, can be illustrated equivalently as a mixed lin in S 3 when ST is viewed as the complement of a solid torus in S 3. So we correspond the element α to the minimal mixed braid representation, which has decreasing order of twists around the fixed strand. Fig. 8 illustrates an example of this correspondence. Denoting Λ = {t 0 t 1 1 t t n n, i Z \{0}, i i+1 i, n N}, 4) we have that Λ is a subset of n H 1,n. In particular Λ is a subset of n Σ n. Applying the inductive trace rules to a word w in n Σ n will eventually give rise to linear combinations of monomials in Rz, s ). In particular, for an element of Λ we have: trt 0 t t n n )=s n...s 1 s 0.

8 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 8 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Fig. 8. An element in Λ. Fig. 9. An element of Λ. Further, the elements of Λ are in bijective correspondence with decreasing n-tuples of integers, 0, 1,..., n ), n N, and these are in bijective correspondence with monomials in s 0, s 1,..., s n. See Fig. 8.) Remar 2. The invariant X recovers the Homflypt sein module of ST since it gives different values for different elements of Λ by rule 4) of the trace. 3. An ordering in the sets Λ and Λ In this section we define an ordering relation in the sets Σ n and Σ n, which include Λ and Λ as subsets. Before that, we will need the notion of the index of a word in Λ or in Λ. Definition 1. The index of a word w in Λ or in Λ, denoted indw), is defined to be the highest index of the t i s, resp. of the t i s, in w. Similarly, the index of an element in Σ n or in Σ n is defined in the same way by ignoring possible gaps in the indices of the looping generators and by ignoring the braiding part in H n q). Moreover, the index of a monomial in H n q) is equal to 0. For example, indt 0 t t n n ) = indt u 0...t u n n ) = n. Definition 2. We define the following ordering in the sets Σ n. Let w = t i 1 1 t i t i μ μ and σ = t j 1 λ 1 t j 2 λ 2...t j ν λ ν, where t, λ s Z, for all t, s. Then: a) If μ i=0 i < ν i=0 λ i, then w <σ. b) If μ i=0 i = ν i=0 λ i, then: i) if indw) < indσ), then w <σ, ii) if indw) = indσ), then: α) if i 1 = j 1, i 2 = j 2,..., i s = j s, i s <j s, then w >σ, β) if i t = j t t and μ = λ μ, μ = λ μ,..., i+1 = λ i+1, i < λ i, then w <σ, γ) if i t = j t t and μ = λ μ, μ = λ μ,..., i+1 = λ i+1, i = λ i and i >λ i, then w <σ, δ) if i t = j t t and i = λ i, i, then w = σ.

9 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 9 c) In the general case where w = t i 1 1 t i t i μ μ β 1 and σ = t j 1 λ 1 t j 2 λ 2...t j ν λν β 2, where β 1, β 2 H n q), the ordering is defined in the same way by ignoring the braiding parts β 1, β 2. The same ordering is defined on the set Λ by ignoring the braiding parts. Moreover, the same ordering is defined on the sets Σ n and Λ, where the t i s are replaced by the corresponding t i s. Proposition 1. The set Σ n equipped with the ordering given in Definition 2, is a totally ordered set. Proof. In order to show that the set Σ n is a totally ordered set when equipped with the ordering given in Definition 2, we need to show that the ordering relation is antisymmetric, transitive and total. We only show that the ordering relation is transitive. Antisymmetric property follows similarly. Totality follows from Definition 2 since all possible cases have been considered. Let w, σ, v Σ n such that: w = t i 1 1 t i t i m m β 1, σ = t j 1 λ 1 t j 2 λ 2...t j n λn β 2, v = t φ 1 μ 1 t φ 2 μ 2...t φ p μp β 3, where β 1, β 2, β 3 H n q) and let w <σand σ <v. Since w <σ, one of the following holds: a) Either m i=1 i < n i=1 λ i and since σ <v, we have that n i=1 λ i p i=1 μ i and so m i=1 i < p i=1 μ i. Thus w <v. b) Either m i=1 i = n i=1 λ i and indw) = m <n = indσ). Then, since σ<vwe have that either n i=1 λ i < p i=1 μ i same as in case a)) or n i=1 λ i = p i=1 μ i and indσ) p = indv). Thus, indw) = m < p = indv) and so we conclude that w <v. c) Either m i=1 i = n i=1 λ i, indw) = indσ) and i 1 = j 1,..., i s = j s, i s >j s. Then, since σ <v, we have that either: n i=1 λ i < p i=1 μ i, same as in case a), or n i=1 λ i = p i=1 μ i and indσ) < indv), same as in case b), or indσ) = indv) and j 1 = ϕ 1,..., j p >ϕ p. Then: i) if p = s we have that i s >j s >ϕ s and we conclude that w <v, ii) if p < swe have that i p = j p >ϕ p and thus w <vand if s < pwe have that i s >j s = ϕ s and so w <v. d) Either m i=1 i = n i=1 λ i, indw) = indσ) and n = λ n,..., q < λ q. Then, since σ <v, we have that either: n i=1 λ i < p i=1 μ i, same as in case a), or n i=1 λ i = p i=1 μ i and indσ) < indv), same as in case b), or indσ) = indv) and j 1 = ϕ 1,..., j q >ϕ q, same as in case c), or j n = ϕ n, for all n and μ n = λ n,..., μ c+1 = λ c+1, μ c λ c for some c, then: 1) If μ c > λ c, then: i) If c > qthen c = λ c < μ c and thus w <v. ii) If c < qthen q < λ q = μ q and thus w <v. iii) If c = q then q < λ q < μ q and thus w <v. 2) If μ c = λ c, such that μ c <λ c, then: i) If c > qthen c = λ c = μ c and c = λ c >μ c. Thus w <v. ii) If c q then q < λ q = μ q and thus w <v. e) Either m i=1 i = n i=1 λ i, indw) = indσ) and n = λ n,..., q = λ q, such that q >λ q. Then, since σ <v, we have that either:

10 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 10 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) n i=1 λ i < p i=1 μ i, same as in case a), or n i=1 λ i = p i=1 μ i and indσ) < indv), same as in case b), or indσ) = indv) and j 1 = ϕ 1,..., j q >ϕ q, same as in case c), or j n = ϕ n, for all n and μ n = λ n,..., μ c+1 = λ c+1, μ c λ c for some c, then: 1) If μ c > λ c, then: i) If c > qthen c = λ c < μ c, thus w <v. ii) If c q then q = λ q = μ q and q >λ q = μ q, thus w <v. 2) If μ c = λ c such that λ c >μ c, then: i) If c > qthen c = λ c = μ c and c = λ c >μ c, thus w <v. ii) If c < qthen q = λ q = μ q and q >λ q = μ q, thus w <v. iii) If c = q, then q = λ q = μ q and q >λ q >μ q, thus w <v. So, we conclude that the ordering relation is transitive. Remar 3. Proposition 1 also holds for the sets Σ n, Λ and Λ. Definition 3. We define the subset of level, Λ, of Λ to be the set Λ := {t t m m and similarly, the subset of level of Λ to be m i =, i Z \{0}, i i+1 i} i=0 Λ := {t 0 t t m m m i =, i Z \{0}, i i+1 i}. i=0 Remar 4. Let w Λ be a monomial containing gaps in the indices and u Λ a monomial with consecutive indices such that indw) = indu). Then, it follows from Definition 2 that w <u. Proposition 2. The sets Λ are totally ordered and well-ordered for all. Proof. Since Λ Λ,, Λ inherits the property of being a totally ordered set from Λ. Moreover, t is the minimum element of Λ and so Λ is a well-ordered set. We also introduce the notion of homologous words as follows: Definition 4. We shall say that two words w Λ and w Λare homologous, denoted w w, if w is obtained from w by turning t i into t i for all i. With the above notion the proof of Theorem 2 is based on the following idea: Every element w Λ can be expressed as linear combinations of monomials w i Λwith coefficients in C, such that: i) j such that w j w, ii) w j <w i, for all i j, iii) the coefficient of w j is an invertible element in C.

11 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) From Λ to Σ n In this section we prove a series of lemmas relating elements of the two different basic sets Σ n, Σ n of H 1,n q). In the proofs we underline expressions which are crucial for the next step. Since Λ is a subset of Σ n, all lemmas proved here apply also to Λ and will be used in the context of the bases of SST) Some useful lemmas in H 1,n q) We will need the following results from [4]. The first lemma gives some basic relations of the braiding generators. Lemma 1. See [4, Lemma 1].) For ɛ {±1} the following hold in H 1,n q): i) gi m = q m q m ) m) g i + q m q m 2 + +) m 2 q ) g m i = q m q 1 m +...+) m q ) g i + + q m q 1 m + +) m q +) m) ii) g ɛ i g ±1 g ±1...g j ±1 )=g ±1 g ±1...g j ±1 )g ɛ i+1, for >i j, g ɛ i g ±1 j g ±1 j+1...g ±1 )=g ±1 j g ±1 j+1...g ±1 )g ɛ i, for i>j, where the sign of the ±1 exponent is the same for all generators. iii) g i g i...g j+1 g j g j+1...g i = g j g j+1...g i g i g i...g j+1 g j g i g i...g j+1 g j ɛ g j+1...g i = g j g j+1...g i g ɛ i g i...g j+1 g j iv) g ɛ i...g ɛ n g 2ɛ n g ɛ n...g ɛ i = n i+1 r=0 q ɛ 1) ɛ r q ɛr g ɛ i...g ɛ n r...g ɛ i ), where ɛ r =1if r n i and ɛ n i+1 =0. Similarly, v) g i ɛ...g 2 ɛ g 1 2ɛ g 2 ɛ...g i ɛ = i r=0 qɛ 1) ɛ r q ɛr g i ɛ...g r+2 ɛ g r+1 ɛ g r+2 ɛ...g i ɛ ), where ɛ r =1if r i 1 and ɛ i =0. The next lemma comprises relations between the braiding generators and the looping generator t. Lemma 2. Cf. [4, Lemmas 1, 4, 5].) For ɛ {±1}, i, N and λ Z the following hold in H 1,n q): i) t λ g 1 tg 1 = g 1 tg 1 t λ ii) t ɛ g ɛ 1 t ɛ ɛ g 1 = g ɛ 1 t ɛ g ɛ 1 t ɛ +q ɛ 1)t ɛ g ɛ 1 t ɛ +1 q ɛ )t ɛ g ɛ 1 t ɛ t ɛ g ɛ 1 t ɛ ɛ g 1 = g ɛ 1 t ɛ g ɛ 1 t ɛ +q ɛ 1)t ɛ) g ɛ 1 +1 q ɛ )g ɛ 1 t ɛ) iii) t ɛi g ɛ 1 t ɛ ɛ g 1 = g1t ɛ ɛ g1t ɛ ɛi +q ɛ 1) i j=1 tɛj g1t ɛ ɛ+i j) + +1 q ɛ ) i j=0 tɛ+j) g1t ɛ ɛi j) t ɛi g ɛ 1 t ɛ g ɛ 1 = g ɛ 1 t ɛ g ɛ 1 t ɛi +q ɛ 1) i j=1 tɛ j) g1t ɛ ɛi j) + +1 q ɛ ) i j=1 tɛi j) g1t ɛ ɛ j) The next lemma gives the interactions of the braiding generators and the loopings t i s and t i s.

12 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 12 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Lemma 3. See [4, Lemmas 1 and 2].) The following relations hold in H 1,n q): i) g i t ɛ = t ɛ g i for >i,<i 1 g i t i = qt i g i +q 1)t i g i t i = q t i g i +q 1)t i = t i g i g i t i = qt i g i +q 1)t i ii) g i t i = q t i g i +q 1)t i = t i g i t ng n =q) j=0 qj t j n t j n + q g n t n, if N t ng n =1 q) j=0 qj t j n t j n + q g n t n, if Z N iii) t i t λ j = t λ j t i for i j and, λ Z iv) g i t g i t i ɛ ɛ = t ɛ g i for >i, <i 1 = t iɛ g i +q 1)t iɛ +1 q)t i ɛ v) g i t iɛ = t iɛ g i t i = g i...g 1 t g 1...g i for Z. Using now Lemmas 1, 2 and 3 we prove the following relations, which we will use for converting elements in Λ to elements in Σ n. Note that whenever a generator is overlined, this means that the specific generator is omitted from the word. Lemma 4. The following relations hold in H 1,n q) for N: i) g m+1 t m = q ) t m+1g m+1 + j=1 q j) q 1)t j mt j m+1 see Fig. 10), ii) gm+1 t m = q ) t m+1 g m+1 + j=1 q j) q 1)t j m t j) m+1. Proof. We prove relation i) by induction on. Relation ii) follows similarly. For =1we have that g m+1 t m = t m+1 gm+1, which holds from Lemma 3i). Suppose that the relation holds for 1. Then, for we have: g m+1 t m = g m+1 t m ind. t m = step q 2) t m+1 g m+1 t m j=1 q 2 j) q 1)t j mt j m+1 t m = = q 1 g m+1 t m + q 2 q 1)t m t m j=1 q 2 j) q 1)t j+1 m = q ) t m+1 gm+1 + j=1 q j) q 1)t j mt j m+1. Lemma 5. In H 1,n q) the following relations hold: t j m+1 i) For the expression A = g r g r...g r s ) t the following hold for the different values of N: 1) A = t g r...g r s ) for >ror <r s 2) A = t r g r...gr s ) for = r s 1 3) A = qt r g r...g r s )+q 1)t r g r...g r s ) for = r 4) A = qt r s g r...g r s )+q 1)t r g r...gr s+1) for = r s 5) A = t m g r...g r s )+q 1)t r g r...gm+1) gm...g r s ) for = m {r s +1,...,r 1}.

13 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 13 Fig. 10. Illustrating Lemma 4i) for = 2. Fig. 11. Illustrating Lemma 5ii) for = r s. ii) For the expression A = g r g r...g r s ) t the following hold for the different values of N: 1) A = t g r...g r s ) for >ror <r s 2) A = t r s gr...g r s+1 gr s ) for = r s see Fig. 11) 3) A = t ) m gr g r...g m+1 gm g m...g s for = m {r s +1,...,r} 4) A = q s+1 t r g r...g r s )+q 1) s+1 j=1 qs j+1 t r j g r...g r j+2 g r j...g r s ) for = r s 1. Proof. We only prove relation ii) for = r s 1by induction on s case 4)). All other relations follow from Lemma 3i). For s = 1we have: g r g r t r 2 = g r[qt r g r +q 1)t r 2 ] = qg rt r g r +q 1)g r t r 2 = q[qt r = q 2 t r g r +q 1)t r ]g r +q 1)t r 2 g r g r g r )+q 1) [ qt r g r + q 0 t r 2 g r], and so the relation holds for s = 1. Suppose that the relation holds for s = n. We will show that it holds for s = n +1. Indeed we have: g r...g r n )t r n 2 =g r...g r n )g r n t r n 2 )= g r...g r n ) [ qt r n g r n +q 1)tr n 2] = = qg r...g r n t r n )g r n +q 1)g r...g r n )t r n 2 = q n+2 t r g r...g r n )+ +q) n+1 j=1 qn j+2 t r j g r...g r j+2 g r j...g r n )+ +q)t r n 2 g r...g r n ) = q n+2 tr g r...g r n )+ +q) n+2 j=1 qn+1) j+1 t r j g r...g r j+2 g r j...g r n ). ind. step =

14 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 14 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Before proceeding with the next lemma we introduce the notion of length of w H n q). For convenience we set δ,r := g g...g r+1 g r for >rand by convention we set δ, := g. Definition 5. We define the length of δ,r H n q) to be the number of braiding generators, that is, lδ,r ) := r +1and since every element of the Iwahori Hece algebra of type A can be written as n i=1 δ i,r i so that j < j+1 j, we define the length of an element w H n q) as: n n lw) := l i δ i,r i )= i r i +1. i=1 Note that lg ) = lδ, ) = +1 = 1. Lemma 6. For >rthe following relations hold in H 1,n q): i=1 r t δ,r = q i q 1)δ, i,r t i + q lδ,r) δ,r t r, where δ, i,r := g g...g i+1 g i...g r := g...g i...g r. i=0 Proof. We prove relations by induction on. For =1we have that t 1 g 1 Suppose that the relation holds for 1), then for we have: = q 1)t 1 + qg 1 t, which holds. t δ,r = t g δ,r =q 1)t δ,r + qg t δ,r = =q 1)δ,r t + qg r i=0 q i q 1)δ, i,r t i + + q lδ,r)+1 g δ,r t r = = r i=0 qi q 1)δ, i,r t i + q lδ,r) δ,r t r. Lemma 7. In H 1,n q) the following relations hold: i) For the expression A = g r g r+1...g r+s ) t the following hold for the different values of N: 1) A = t g r...g r+s ) for r + s +1 or <r 2) A = t +1 gr...g g +1 g ) +2...g r+s for r 1 <r+ s 3) A =q) r+s i=r qr+s i t i g r...g i...g r+s )+q s+1 t r g r...g r+s ) for = r + s ii) For the expression A = g r g r+1...g r+s ) t the following hold for the different values of N: 1) A = t g rg r+1...g r+s ) for r + s +1 or <r 2) A = qt +1 g r...g r+s )+q 1) tr g r...g g ) +2...g r+s for r 1 <r+ s 3) A = t r g r...gr+s ) for = r + s Proof. We prove relation i) for r + s = by induction on case 3)). All other relations follow from Lemmas 1 and 3.

15 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 15 Fig. 12. Illustrating Lemma 8i) for i = 2. For =1we have: g 1 t 1 = g 2 1tg 1 = qtg 1 +q 1)t 1. Suppose that the relation holds for = n. Then, for = n +1we have that: ind. step g r...g n+1 t n+1 = qg r...g n t n )g n+1 + q 1)g r...g n )t n+1 = = q [ q 1) n i=r qn i t i g r...g i...g n )+q n r+1 t r g r...g n ) ] g n+1 + +q)t n+1 g r...g n ) = = q 1) n i=r qn i+1 t i g r...g i...g n g n+1 )+q 1)t n+1 g r...g n ) ) + + q n+1 r+1 t r g r...g n g n+1 ) = =q) n+1 i=r qn+1 i t i g r...g i...g n+1 )+q n+1 r+1 t r g r...g n+1 ). Lemma 8. The following relations hold in H 1,n q) for N: i) ii) iii) iv) g1...g i gi 2g ) i...g 1 t = q 1) i =1 qi t g1...g g ) g...g 1 + q i t see Fig. 12) g 1...g i g 2 i g ) i...g 1 t = q 1) i =1 q i ) t g 1...g g ) g...g 1 + q i t g...g2 g 2 1 g 2...g ) t = q 1) i=1 q t i g...g i+2 g ) i+1g i+2...g + q t g...g2 g 2 1 g 2...g t q q 1)g + i=0 t i + t + q +1) q 2 q +1) ]. ) t =...g 1...g + q +i q 1)g...g1 2...g i g i+2...g + [ i=2 q +i q 1) 2 g i...g 2 g 2 1 g 2...g i + Proof. We prove relation i) by induction on i. All other relations follow similarly. For i =1we have: g1t 2 = g 1 g 1 tg 1 g1 = g 1 t 1 g1 =q)t 1 g1 + qt. Suppose that the relation holds for i = n. Then, for i = n +1we have: ) g1...g n gn+1g 2 n...g 1 t = q 1) g1...g n+1 g n...g 1 ) t + + q ) g 1...g n gng 2 n...g 1 t = = q 1)g 1...g n t n+1 gn+1...g 1 + q n =1 qn q 1)t g1...g g...g1 ) + q n+1 t = = q 1)t n+1 g1...g n gn+1 ) n...g 1 + =1 qn+1 q 1)t g1...g g...g1 ) + q n+1 t = = n+1 =1 qn+1 q 1)t g1...g g...g1 ) + q n+1 t.

16 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 16 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 4.2. Converting elements in Λ to elements in Σ n We are now in the position to prove a set of relations converting monomials of t i s to expressions containing the t i s. In [13] we provide lemmas converting monomials of t i s to monomials of t i s in the context of giving a simple proof that the sets Σ n form bases of H 1,n q). Lemma 9. The following relations hold in H 1,n q) for N: i) t 1 = q t 1 + j=1 q j q 1)t j t j 1 g1, ii) t 1 = q + j=1 q j) q 1)t j t +1 j 1 g1. Proof. We prove relation i) by induction on. Relation ii) follows similarly. For =1we have: t 1 = g 1 t g1 = qg1 t g1 + q 1) t g1 = qt 1 + q 1) t g1. Suppose that the relation holds for 1. Then, for we have: t 1 = t 1 ) t ind. 1 = step q t ) 1 t j=1 q j q 1)t j t j ) 1 g1 t 1 = = q t 1 + q t t ) 1 g1 + j=1 q j q 1)t j t j ) 1 t g1 = q t 1 + q q 1)t t ) 1 g1 + + j=1 q j q 1)t j t j ) 1 g1 = = q t 1 + j=1 q j q 1)t j t j 1 g1. Lemma 10. The following relations hold in H 1,n q) for N: t = q t + q 1) i=0 q i t i g g... g i+2 g i+1... g g ). Proof. We prove the relations by induction on. For =1we have: t 1 = g1 t g1 = qg1 t g1 + q 1) t g1 = qt1 + q 1) t g1. Suppose that the relations hold for = n. Then, for = n +1we have that: t n+1 = g n+1 t n gn+1 ind. step = [ = g n+1 q n t n + q 1) n = q n g n+1 t n g = q n[ qt n+1 g n+1 + q 1)t n i=0 qi t i g n...g i+2 g i+1...g n ) ] gn+1 = n+1 + q 1) n i=0 qi g n+1 t i g n...g i+2 g i+1...g n ] g n+1 + q 1) n i=0 qi t i g n+1...g i+2 g i+1...g n+1 ) = = q n+1 t n+1 + qn q 1)t n gn+1 + q 1) n g n+1...g i+2 g i+1...g n+1 ) = = q n+1 t n+1 + q 1) n i=0 qi t i Lemma 11. The following relations hold in H 1,n q) for Z\{0}: i=0 qi t i g n+1... g i+2 g i+1...g n+1 ). g n+1 ) = t m = q m t m + i f i q)t mw i + i g i q)t λ 0 t λ tλ m m u i, where w i, u i H m+1 q), i, m i=0 λ i = and λ i 0, i, if >0 and λ i 0, i, if <0.

17 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 17 Fig. 13. Illustrating Theorem 7. Proof. We prove relations by induction on m. The case m = 1is Lemma 9. Suppose now that the relations hold for m 1. Then, for m we have: t m = g m t m gm ind. = step q m) g m t mgm + i g iq)t λ 0 t λ tλ m 2 m 2 g mt λ m m u igm = q m) q ) t mgm i f iq)g m t mw i g m + L. 4) = j=1 q j) q 1)t j m t j m = q m t m + i f iq)t mw i + i g iq)t λ 0 t λ tλ m i m u i. g m = Using now Lemma 11 we have that every element u Λ can be expressed to linear combinations of elements v i Σ n, where j : v j u. More precisely: Theorem 7. The following relations hold in H 1,n q) for Z: t 0 t t m m = q m n=1 nn t t m m + i f iq) t t m m w i + + j g jq)τ j u j, where w i, u j H m+1 q), i, τ j Σ n, such that τ j <t t m m, j. See Fig. 13.) Proof. We prove relations by induction on m. Le N, then for m = 1we have: t 0 t 1 1 L. 9) = q 1 t j=1 q 1 j) q 1)t 0+j +1 j 1 g1 = = q 1 t q 1 q 1)t 0 1 g j=2 q 1 j) q 1)t 0+j +1 j 1 g1 On the right hand side we obtain a term which is the homologous word of t 0 t 1 1 with scalar q 1 C, the homologous word again followed by g1 2 H 2 q) and with scalar q 1) q 1) C and the terms t 0+j +1 j 1, which are of less order than the homologous word t 0 1, since 1 > 1 +1 j, for all j {2, 3,..., 1 }. So the statement holds for m = 1and 1 N. The case m = 1and 1 Z\N is similar. Suppose now that the relations hold for m 1. Then, for m we have: t 0 t t m m ind. m = step q n=1 nn t 0...t m m t m m + + i f iq) t t m m w i t m + j g jq)τ j u j t m m. Now, since w i, u i H m q), i we have that w i t m m = t m m w i and u i t m m = t m m u i, i. Applying now Lemma 11 to t m we obtain the requested relation. m

18 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 18 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Fig. 14. Conjugating t i by g 1...g i. Example 1. We convert the monomial tt 1t 2 2 Λ to linear combination of elements in Σ n. We have that: and so: t 1 = q t 1 + q 1)t 1 g1 Lemma 9), t 2 2 = q 4 t q 3 q 1)t 1 t 2 g 2 + q 2 q 1)t t 2 g 2g1 g q 2 q 1)t 2 1 g 2 + qq 1) 2 t t1 g 1 g 2 + q 1)t 2 g 2 g1 g 2 Lemma 10), tt 1t 2 2 = q 3 tt 1 t q 4 q 1) tt 1 t 2 2 g1 +1 u + + tt 1 q 1)q 2 q +1) g2 q 1) 2 g 1 g 2 g1 ) g tt 2 q 2 q 1) g2 + qq 1) 3 g2 qq 1) 2 g 2 g + t 1 t 2 qq 1) g 2 g1 g 2 qq 1) 2 g1 ) g t t 1 q 1) g 2 g1 g 2 q q 1) 2 g1 ) g 2 1 g 2 where u = q 1) 2 g1 g 2 q 1) 3 g1 2 g 2 q q 1) 3 g 2 g1 g 2 + q q 1) 3 g2., the homologous word again followed by the braiding generator g1 and terms in Σ n of less order than w, since either their index is less that indw) the terms tt 1, 1and t t 1 ), either they contain gaps in the indices the terms tt2 and t 1 t 2 ). We obtain the homologous word w = tt 1 t From Σ n to Λ ) + In order to prove Theorem 2 we need to show that the set Λ is a spanning set of SST) and also that it is linearly independent. In this section we show that every element in Λ can be expressed in terms of elements in the set Λ. Linear independence of the set Λ is shown in the next section. Before proceeding we need to discuss the following situation. According to Lemma 9, for a word w = t t 1 λ Λ, where, λ N and <λwe have that: w = t t 1 λ = t t 1 λ+1 α 1 + t 2 t 1 λ+2 α t 0 t 1 λ+ α + t t 1 λ++1 α t λ+ α λ, where α i H n q), i. We observe that in this particular case, in the right hand side there are terms which do not belong to the set Λ. These are the terms of the form t q t p 1, where p > qand the term tm 1. So these elements cannot be compared with the highest order term w w. The point now is that these terms are elements in the basis Σ n on the Hece algebra level, but, when we are woring in SST), such elements must be considered up to conjugation by any braiding generator and up to stabilization moves. Topologically, conjugation corresponds to closing the braiding part of a mixed braid. Conjugating t 1 by g1 we obtain tg1 2 view Fig. 14) and similarly conjugating t m 1 by g1 we obtain tg1tg tg1. 2 Then, applying Lemma 3 we obtain the expression m =1 t t m 1 v, where v H n q), for all, that is, we obtain now elements with consecutive indices but not necessarily with ordered exponents.

19 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 19 We shall first deal with elements where the looping generators do not have consecutive indices, and then with elements where the exponents are not in decreasing order. For the expressions that we obtain after appropriate conjugations we shall use the notation = Managing the gaps We will call gaps in monomials of the t i s, gaps occurring in the indices and size of the gap t i i t j j number s i,j = j i N. Lemma 12. For 0, 1... i Z, ɛ = 1or ɛ = and s i,j > 1 the following relation holds in H 1,n q): the t t i i t i i t ɛ j = t t i i t i i t ɛ i+1 g ɛ i+2...gjg ɛ j 2ɛ gj ɛ...gi+2) ɛ. Proof. We have that t ɛ j = g ɛ j...gɛ i+2) t ɛ i+1 g ɛ i+2...g ɛ j) and so: t t i i t i i tɛ j = t t i i t i i gɛ j...gɛ i+2 ) tɛ i+1 gɛ i+2...gɛ j ) = =gj ɛ...gɛ i+2 ) t t i i t i i tɛ i+1 gɛ i+2...gɛ j ) = = t 0...t i i t i i tɛ i+1 gɛ i+2...gɛ j g2ɛ j gɛ j...gɛ i+2 ). In order to pass to a general way for managing gaps in monomials of t i s we first deal with gaps of size one. For this we have the following. Lemma 13. For N, ɛ = 1or ɛ = and α H 1,n q) the following relations hold: t ɛ i α = q ɛu) q ɛ 1)t ɛu u=1 it ɛ u) i αg ɛ i ) + q ɛ) t ɛ ig ɛ i αg ɛ i ). Proof. We prove the relations by induction on. For =1we have t ɛ i α = gɛ i tɛ i gɛ i α = tɛ i gɛ i α gɛ i. Suppose that the assumption holds for 1 > 1. Then for we have: t ɛ i α = t ɛ) i t ɛ i α) tɛ i α = β) = t ɛ) i β = ind. step = 2 u=1 qɛu) q ɛ 1)t ɛu i = 2 u=1 qɛu) q ɛ 1)t ɛu i = 2 u=1 qɛu) q ɛ 1)t ɛu i + q ɛ) t ɛ+1) i gi ɛtɛ i αgɛ i ) = = u=1 qɛu) q ɛ 1)t ɛu i We now introduce the following notation. Notation 1. We set τ i,i+m i,i+m i tɛ u) i tɛ u) i tɛ u) i tɛ u) βgi ɛ) + qɛ 2) t ɛ) i gi ɛβgɛ i ) β = tɛ i = α) t ɛ i αgɛ i ) + qɛ 2) t ɛ) i gi ɛtɛ i αgɛ i ) = αgi ɛ) + qɛ 2) t ɛ) i t ɛ i αgɛ i + αg ɛ i ) + qɛ) t ɛ i gɛ i αgɛ i ). := t i i t i+1 i+1...t i+m i+m, where m N and j 0for all j and { gi g i+1...g j g j if i<j δ i,j := g i g i...g j+1 g j if i>j, { gi g i+1...g g +1...g j g j if i<j δ i,,j := g i g i...g +1 g...g j+1 g j if i>j We also set w i,j an element in H j+1 q) where the minimum index in w is i.

20 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 20 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Using now the notation introduced above, we apply Lemma 13 s i,j -times to 1-gap monomials of the form τ 0,i 0,i t j j and we obtain monomials with no gaps in the indices, followed by words in H n q). Example 2. For s i,j > 1and α H n q) we have: i) ii) iii) τ i 0,i t j α = τ i 0,i t i+1 δ i+2,j αδ j,i+2 τ i 0,i t2 j α = τ i 0,i t2 i+1 δ i+2,j αδ j,i+2 + τ i 0,i t i+1t i+2 β, where [ β = q 1) ] j s=i+2 qj s δ i+3,s δ i+2,s δ s+1,j αδ j,i+2 δ s,i+3 τ i 0,i t3 j α = [ q j i+2)+1] 2 τ i 0,i t3 i+1 δ i+2,j αδ j,i τ i 0,i t2 i+1 t i+2 β + τ i 0,i t i+1t 2 i+2 γ + + τ i 0,i t i+1t i+2 t i+3 μ, where γ = q j i+3)+1 q 1)δ i+3,j δ i+2,s δ s+1,j αδ j,i+2 δ s,i+3, and μ = j j s=i+2 r=s+1 q2j r s q 1) 2 δ i+4,r δ i+2,s δ s+1,r δ r+1,j αδ j,i+2 δ s,i+3 δ r,i+4 + j s s=i+2 r=i+3 q2j r s q 1) 2 δ i+4,r δ i+3,r δ r+1,s δ i+2,s δ s+1,j αδ j,i+2 δ s,i+3. Applying Lemma 13 to the one gap word τ 0,i 0,i t j j, where j Z\{0} and α H n q) we obtain: τ 0,i 0,i t j j α = λ τ 0,i 0,i tλ i+1 i+1...tλ i+ j i+ j α if j <s i,j λ τ 0,i 0,i tλ i+1 i+1...tλ j j β if j s i,j, where α, β H n q), i+ j μ=i+1 λ μ = j, λ μ 0, μ and if λ u =0, then λ v =0, v u. More precisely: Lemma 14. For the 1-gap word A = τ 0,i 0,i t j j α, where α H n q) we have: i) If j <s i,j, then: A = q j ) j i+1) τ 0,i 0,i t j i+1 δ i+2,j αδ j,i j fq)τ 0,i 0,i τ i+1,i+ j i+1,i+ j βαβ. ii) If j s i,j, then: A = q j ) j i+1) τ 0,i 0,i t j i+1 δ i+2,j αδ j,i j fq)τ 0,i 0,i τ i+1,j i+1,j βαβ, where β and β are of the form w i+1,j H j+1 q), j fq, z)τ 0,i 0,i τ i+1,i+ j i+1,i+ j means a sum of elements in Σ n, such that in each one of them, the sum of the exponents of the looping generators t i+1,..., t i+j is equal to j, and such that i+1 < j. Moreover, if μ =0, for some index μ, then s =0for all s > μ. Proof. We prove the relations by induction on j. Let 0 < j <j i. For j =1we have A = [ q 1)] j i+1) τ 0,i 0,i t i+1 δ i+2,j αδ j,i+2 Lemma 12). Suppose that the relation holds for j 1 > 1. Then for j we have: A = τ 0,i 0,i t [ j j t j α) = q j 2 ] j i+1) τ 0,i 0,i t j i+1 δ i+2,j t j αδ j,i+2 ind. step }{{} + i1,i+ j B fq)τ 0,i 0,i τ i 1,i+ j i+1,i+ j βt j β. } {{ } C +

21 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) 21 We now consider B and C separately and apply Lemma 4 to both expressions: B L. 4) = = [ q ] j 2 j i+1) τ 0,i 0,i t j i+1 [ q 1) ] j +i+2 qj t δ i+2, δ +1,j + q j i+2)+1 t i+1 δ i+2,j αδ j,i+2 = [ q ] j 2 j i+1) q 1)τ 0,i 0,i t i+1 j +i+2 qj t δ i+2, δ +1,j αδ j,i [ q ] j j i+1) τ 0,i 0,i t j i+1 δ i+2,jαδ j,i+2. We now do conjugation on the j i + 3))-one gap words that occur and since t β = t i+2 δ i+3, βδ,i+3 we obtain: B = [ q ] j j i+1) τ 0,i 0,i t j i+1 δ i+2,j αδ j,i τ 0,i 0,i t j i+1t i+2 =i+2 fq, z)δ i+3,δ i+2, δ +1,j αδ j,i+2 δ,i+3 = = [ q ] j j i+1) τ 0,i 0,i t j i+1 δ i+2,j αδ j,i+2 + τ i 0,i t i+1t i+2 β 1, where β 1 H j+1 q). Moreover, C = r fq)τ 0,i 0,i τ i+1,i+ j i+1,i+ j βt j β L. 4) and since β = w i+j,j, we have that: β t j = j s=i+ j t s γ s, where γ s H j+1 q) and so: C = v r fq)τ 0,i 0,i τ v i+1,i+ j i+1,i+ j β 2, where β 2 H j+1 q). This concludes the proof. We now pass to the general case of one-gap words. Proposition 3. For the 1-gap word B = τ 0,i 0,i τ j,j+m j,j+m α, where α H nq) we have: B = m s=0 qj+s ) j i+1) τ 0,i 0,i τ j,j+m i+1,i+m m s=0 δ i+m+2 s,j+s) α m s=0 δ j+s,i+m+2 s) + + u r fq)τ 0,i 0,i τ u 1,m i+1,i+m ) α where α H n q), u 1,m = j such that u 1 < j and if u μ =0, then u s =0, s > μ. Proof. The proof follows from Lemma 14. The idea is to apply Lemma 14 on the expression τ 0,i 0,i t j j ρ 1, where ρ 1 = τ j+1,j+m j+1,j+m and obtain the terms τ 0,i 0,i t j i+1 ρ 2 and τ 0,i 0,i τ i+1,i+q i+1,i+q ρ 2 and follow the same procedure until there is no gap in the word. We are now ready to deal with the general case, that is, words with more than one gap in the indices of the generators. Theorem 8. For the φ-gap word: C = τ 0,i 0,i τ i+s 1,i+s 1 +μ 1 i+s 1,i+s 1 +μ 1 τ i+s 2,i+s 2 +μ 2 i+s 2,i+s 2 +μ 2...τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ α, where i Z\{0} for all i, α H n q), s j, μ j N, such that s 1 > 1 and s j >s j + μ j for all j we have: C = φ j=1 φ p=1 α p q i+s j ) s j j j p=1 μ p τ u 0,i+φ+ φ p=1 μp ) 0,i+φ+ φ p=1 μ p + v f vq)τ 0,v 0,v w v, where φ ) p=0 α φ p α

22 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) 22 I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) i) α j ii) α j iii) τ iv) τ u 0,v = μ j λ j =0 δ i+j+1+ j =1 μ λ j, i+s j +μ j λ j, j = {1, 2,..., φ}, = μ j λ j =0 δ i+j+1+ j =1 μ +λ j, i+s j +λ j, j = {1, 2,..., φ}, = τ 0,i 0,i φ j=1 τ i+s j,i+s j +μ j i+j+ j, u φ 0,i+φ+ p=1 μp 0,i+φ+ φ p=1 μ p u φ 0,i+φ+ 0,v <τ p=1 μp 0,i+φ+ φ p=1 μ p, for all v, p=1 μ p,i+j+ j p=1 μ p v) w v of the form w i+2,i+sφ +μ φ H i+sφ +μ φ +1q), for all v, vi) the scalars f v q) are expressions of q C for all v. Proof. We prove the relations by induction on the number of gaps. For the 1-gap word τ 0,i 0,i τ i+s,i+s+μ i+s,i+s+μ α, where α H n q), we have: A = [ μ λ=0 ) ] q i+s+λ s τ 0,i 0,i τ i+s,i+s+μ i+1,i+1+μ μ μ λ=0 δ i+2+μ+λ,i+s+λ + v f vq) τ u 0,v 0,v w v, λ=0 δ i+2+μ λ,i+s+μ λ α which holds from Proposition 3. Suppose that the relation holds for φ 1)-gap words. Then for a φ-gap word we have: τ 0,i 0,i τ i+s 1,i+s 1 +μ 1 i+s 1,i+s 1 +μ 1 τ i+s 2,i+s 2 +μ 2 i+s 2,i+s 2 +μ 2...τ i+s φ,i+s φ +μ φ φ j=1 q i+s j ) s j j j =1 μ τ u 0,i+φ+ φ =1 μ i+s φ,i+s φ +μ φ ) 0,i+φ+ φ =1 μ s φ >s φ +μ φ v f vq) τ u 0,v 0,v w τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ = φ j=1 q ) s i+s j j j j =1 μ u φ 0,i+φ+ τ =1 μ 0,i+φ+ φ =1 μ v f vq) τ u 0,v 0,v τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ w q ) s i+s j j j j =1 μ μφ p=0 φ j=1 τ i+s φ,i+s φ +μ φ i+φ+ φ =1 μ,i+φ+ φ φ =1 μ +μ [ φ μ λ=0 q i+s+λ ) s ] τ 0,i v f vq) τ u 0,v 0,v w v. Prop. 3) τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ α = ind. step φ 2 =0 α φ τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ α φ =1 α + τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ φ 2 =0 α φ α φ =1 α + = q ) s φ φ φ =1 i+s φ +p μ u φ 0,i+φ+ τ =1 μ 0,i+φ+ φ =1 μ =0 α φ α φ =1 α + v f vq) τ u 0,v 0,v τ i+s φ,i+s φ +μ φ i+s φ,i+s φ +μ φ w 0,i τ i+s,i+s+μ i+1,i+1+μ μ λ=0 δ i+2+μ λ,i+s+μ λ α μ λ=0 δ i+2+μ+λ,i+s+λ + All results are best demonstrated in the following example on a word with two gaps. Example 3. For the 2-gap word t 0 1 t 3t 2 5t 6 Σ n we have: t 0 1 t 3t 2 5t 6 = t 0 1 g 3t 2 g 3 t 2 5t 6 = g 3 t 0 1 t 2t 2 5t 6 g 3 = t 0 1 t 2t 2 5t 6 g2 3 = = t 0 1 t 2t 5 t 5 t 6 g2 3 = t 0 1 t 2g 5 g 4 t 3 g 4 g 5 t 5 t 6 g2 3 = = g 5 g 4 t 0 1 t 2t 3 g 4 g 5 t 5 t 6 g2 3 = t 0 1 t 2t 3 g 4 g 5 t 5 t 6 g2 3g 5 g 4 = = t 0 1 t 2t 3 [ q 2 t 3 g 4 g 5 + qq 1)t 4 g 5 + q 1)t 5 g 4 ] t 6 g2 3g 5 g 4 = = q 2 t 0 1 t 2t 2 3g 4 g 5 t 6 g2 3g 5 g 4 + qq 1)t 0 1 t 2t 3 t 4 g 5 t 6 g2 3g 5 g 4 + +q 1)t 0 1 t 2t 3 t 5 g 4 t 6 g2 3g 5 g 4 = q 2 t 0 1 t 2t 2 3t 6 g 4g 5 g 2 3g 5 g 4 + +q 1)t 0 1 t 2t 3 t 5 t 6 g 4g 2 3g 5 g 4 + qq 1)t 0 1 t 2t 3 t 4 t 6 g 5g 2 3g 5 g 4 = = q 2 t 0 1 t 2t 2 3g 6 g 5 t 4 g 5 g 6 g 4g 5 g 2 3g 5 g 4 + Prop. 3) =

23 JID:JPAA AID:5294 /FLA [m3l; v1.159; Prn:3/09/2015; 7:22] P ) I. Diamantis, S. Lambropoulou / Journal of Pure and Applied Algebra ) Ordering the exponents + qq 1)t 0 1 t 2t 3 t 4 g 6 t 5 g 6 g 5g 2 3g 5 g 4 +q 1)t 0 1 t 2t 3 g 5 t 4 g 5 t 6 g 4 g3g 2 5 g 4 )= q 2 g6 g 5 t 0 1 t 2t 2 3t 4 g 5 g 6 g 4g 5 g3g 2 5 g qq 1)g 6 t 0 1 t 2t 3 t 4 t 5 g 6 g 5g 2 3g 5 g 4 + +q 1)g 5 t 0 1 t 2t 3 t 4 t 6 g 5g 4 g 2 3g 5 g 4 = = q 2 t 0 1 t 2t 2 3t 4 g 5 g 6 g 4g 5 g 2 3g 5 g 4 g 6 g qq 1)t 0 1 t 2t 3 t 4 t 5 g 6 g 5g 2 3g 5 g 4 g 6 + q 1)t 0 1 t 2t 3 t 4 t 6 g 5 g 4 g 2 3g 5 g 4 g 5 ) = q 2 t 0 1 t 2t 2 3t 4 g 5 g 6 g 4g 5 g 2 3g 5 g 4 g 6 g qq 1)t 0 1 t 2t 3 t 4 t 5 g 6 g 5g 2 3g 5 g 4 g 6 + +q 1)t 0 1 t 2t 3 t 4 g 6 t 5 g 6 g 5g 4 g 2 3g 5 g 4 g 5 = = q 2 t 0 1 t 2t 2 3t 4 g 5 g 6 g 4g 5 g 2 3g 5 g 4 g 6 g qq 1)t 0 1 t 2t 3 t 4 t 5 g 6 g 5g 2 3g 5 g 4 g 6 + +q 1)t 0 1 t 2t 3 t 4 t 5 g 6 g 5g 4 g 2 3g 5 g 4 g 5 g 6. We now deal with elements in Σ n, where the looping generators have consecutive indices but their exponents are not in decreasing order. More precisely, we will show that these elements can be expressed as sums of elements in the n H nq)-module Λ, namely, as sums of elements in Λ followed by a braiding tail. We will need the following lemma. Lemma 15. The following relations hold in H 1,n q) for λ N: t i t +λ i+1 = j t u j i t v j i+1 w j, where u j + v j =2 + λ, u j v j and w j H n q), j. Proof. We have that t i t+λ i+1 = t i t i+1 tλ i+1 L. 13 = q λ g i+1 t λ i g i+1 = t i t i+1 + λ 2 j=0 qj q 1)t j+1 i = q λ t i t i+1 g i+1t λ i g i+1 + λ 2 j=0 qj q 1)t +j+1 i ) t λ j i+1 = t +λ j i+1. We obtained the term t i t i+1 g i+1t λ i g i+1, terms where the exponent of t i is greater than the exponent of t i+1 and terms of the form t p 1 i tp 2 i+1, where <p 1 >p 2 <+ λ. We apply Lemma 13 on the terms of the last form and repeat the same procedure until there are only elements of the form t u 1 i t u 2 i+1, u 1 >u 2 left in each sum. Note that each time Lemma 13 is performed, a term of the form t m 1 i t m 1 i+1 g i+1t m 2 i g i+1 appears. For these elements we have: t m 1 i t m 1 i+1 g i+1t m 2 L. 3 i g i+1 = t m 1 i q 1) ) m 1 j=0 q j t j i tm 1 j i+1 + q m 1 g i+1 t m 1 i t m 2 i g i+1 = q 1) m 1 j=0 q j t m 1+m 2 +j i t m 1 j i+1 g i+1 + q m 1 t m 1 i g i+1 t m 1+m 2 i g i+1.