ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES STEEN RYOM-HANSEN

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1 1 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES STEEN RYOM-HANSEN arxiv: v2 [math.rt] 26 Jun 2008 Abstract. We consider the algebra E n (u) introduced by F. Aicardi and J. Juyumaya as an abstraction of the Yokonuma-Hecke algebra. We construct a tensor space representation for E n (u) and show that this is faithful. We use it to give a basis of E n (u) and to classify its irreducible representations. 1. Introduction We initiate in this paper a systematic study of the representation theory of an algebra E n (u) defined by F. Aicardi and J. Juyumaya. The origen of E n (u) is in the Yokonuma-Hecke algebra Y n (u) defined as the endomorphism ring of ind G U 1, where G is a a Chevalley group over F q and U a maximal unipotent subgroup. Y n (u) is a unipotent Hecke algebra in the sense of [T] and it has a presentation on generators T i, i = 1,..., n 1 and f i, i = 1,..., n where the f i generate a free abelian group and the T i satisfy the usual braid relation of type A, but where the quadratic relation takes the form T 2 i = 1 + (u 1)e i (1 + T i ) for e i a complicated idempotent involving the f i and f i+1. The algebra E n (u) of the paper is obtained through an abstraction process where e i is declared a new generator E i. It was introduced by F. Aicardi and J. Juyumaya in [AJ]. They show that E n (u) is finite dimensional and that it has connections to knot theory via the Vasiliev algebra. They construct a diagram calculus where the T i are represented by braids in the usual sense and the E i by ties. Using results from [CHWX], they also show that E n (u) can be Yang- Baxterized in the sense of V. Jones, [Jo]. 1 Supported in part by Programa Reticulados y Ecuaciones and by FONDECYT grant

2 2 STEEN RYOM-HANSEN In this paper we initiate a systematic study of the representation theory of E n (u), obtaining a complete classification of its simple modules for the parameter u generic. In [AJ], this was achieved only for n = 2, 3. A key feature for this classification is the construction of a tensor space V n for E n (u) which turns out to be faithful. One can view the combinatorial construction of V n as parallel to the construction of the tensor space for the Ariki-Koike algebra in [ATY] see also [RH] where the [ATY] tensor space is shown to induce a tensor space for the blob algebra. The proof of the faithfulness of V n goes hand in hand with the calculation of the dimension of E n (u), which turns out to be B n n! where B n is the Bell number. Given the tensor space, the classification of irreducible modules follows the principles laid out in [Ja]. We give the organization of paper. Section 2 contains the definition of the algebra E n (u). In section 3 we start out by giving the construction of the tensor space V n. We then construct a certain subset G E n (u) and show that it generates E n (u). Finally we show that it maps to a linearly independent set in EndV n, thereby obtaining the faithfulness and the dimension of E n (u). In section 4 we recall the basic representation theory of the symmetric group and the Iwahori-Hecke algebra, and use the previous sections to construct simple modules of E n (u) as pullbacks of the simple modules of these. In section 5 we show that E n (u) is selfdual by constructing a nondegenerate invariant form on it. This involves the Moebius function for the usual partial order on set partition of {1, 2,..., n}. Finally, in section 6 we give the classification of the simple modules of E n (u), to a large extent following the approach of James, [Ja]. Thus, we especially introduce a parametrizing set Λ for the irreducible modules, analogues of the permutations modules and prove James s submodule theorem in the setup. The simple modules, the Specht modules, turn out to be a combination of the Specht modules for the Hecke algebra and for the symmetric group and hence E n (u) can be seen as a combination of these two.

3 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 3 It is a great pleasure to thank J. Juyumaya for telling me about E n (u) and for many useful conversations. 2. Definition of E n (u) In this section we introduce the algebra E n (u), the main object of our work. Let A be the principal ideal domain C[u, u 1 ] where u is an unspecified variable. We first define the algebra E A n (u) as the associative unital A-algebra on the generators T 1,..., T n 1 and E 1,..., E n 1 and relations (E1) T i T j = T j T i if i j > 1 (E2) E i E j = E j E i i, j (E3) E i T j = T j E i if i j > 1 (E4) E = E i (E5) E i T i = T i E i (E6) T i T j T i = T j T i T j if i j = 1 (E7) E j T i T j = T i T j E i if i j = 1 (E8) E i E j T j = E i T j E i = T j E i E j if i j = 1 (E9) T = 1 + (u 1)E i (1 + T i ) It follows from (E9) that T i is invertible with inverse T 1 i = T i + (u 1 1)E i (1 + T i ) so the presentation of E n (u) is not efficient, since the generators E i for i 2 can be expressed in terms of E 1. However, for the sake of readability, we prefer the presentation as it stands. We now define E n (u) as E n (u) := E A n (u) A C(u) where C(u) is considered as an A-module through inclusion. This algebra is our main object of study. It was introduced by Aicardi and Juyumaya, in [AJ], although the relation (E9) varies slightly from theirs since we have changed T i to T i. They show, among other things, that it is finite dimensional.

4 4 STEEN RYOM-HANSEN From E A n (u) we can consider the specialization to a fixed value u 0 of u which we denote E n (u 0 ). However, we shall in this paper only need the case u 0 = 1, corresponding to E n (1) = E A n (u) A C where C is made into a A-module by taking u to 1. There is a homomorphism ι : CS n E n (1), which can be shown to be injective, using the results of the next sections. 3. The tensor space In the rest of the paper we write k = C(u) (but still allow the letter k as an index, this should not cause confusion). Let V be the vector space V = span k { v j i i, j = 1,..., n}. We consider the tensor product V 2 and define E End k (V 2 ) by the rules { E(v j 1 v j 2 v j 1 ) = v j 2 if j 1 = j 2 0 otherwise Furthermore we define T End k (V 2 ) by the rules v j 2 v j 1 if j 1 j 2 T(v j 1 v j 2 ) = u v j 1 v j 2 if j 1 = j 2, = v j 2 v j 1 i i if j 1 = j 2, < u v j 2 v j 1 + (u 1) v j 1 v j 2 if j 1 = j 2, > We extend these operators to operators E i, T i acting in the tensor space V n by letting E, T act in the factors (i, i+1). In other words, E i acts as a projection in the factors at the positions (i, i + 1) with equal upper index, whereas T i acts as a transposition if the upper indices are different and as a Jimbo matrix for the action of the Iwahori-Hecke algebra in tensor space if the upper indices are equal, see [Ji]. Theorem 1. With the above definitions V n becomes a module for the algebra E n (u).

5 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 5 Proof. We must show that the operators satisfy the defining relations (E1),..., (E9). Here the relations (E1),..., (E5) are almost trivially satisfied, since E i acts as a projection. To prove the braid relation (E6) we may assume that n = 3 and must evaluate both sides of (E6) on the basis vectors v j 1 v j 2 v j 3 i 3 of V 3. The case where j 1, j 2, j 3 are distinct corresponds to the symmetric group case and (E6) certainly holds. Another easy case is j 1 = j 2 = j 3, where (E6) holds by Jimbo s classical result, [Ji]. We are then left with the case j 1 = j 2 j 3 and its permutations. In order to simplify notation, we omit the upper indices of the factors of the equal j s and replace the third j by a prime, e.g. v j 1 v j 2 v j 3 i 3 is written v i1 v i2 v i 3 and so on. We may assume that the lower indices of the unprimed factors are 1 or 2 since the action of T just depends on the order. Furthermore we may assume that the lower index of the primed factor is always 1 since T always acts as a transposition between a primed and an unprimed factor. This gives 12 cases. On the other hand, the cases where the two unprimed factors have equal lower indices are easy, since both sides of (E6) act through u σ 13, where σ 13 is the permutation of the first and third factor of the tensor product. So we are left with the following 6 cases v 1 v 2 v 1 v 1 v 1 v 2 v 1 v 1 v 2 v 2 v 1 v 1 v 2 v 1 v 1 v 1 v 2 v 1 Both sides of (E6) act through σ 13 on the first three of these subcases whereas the last three subcases involve each one Hecke- Jimbo action. For instance T 1 T 2 T 1 (v 2 v 1 v 1) = u v 1 v 1 v 2 + (u 1) v 1 v 2 v 1 which is the same as acting with T 2 T 1 T 2. The other subcases are similar. Let us now verify that (E7) holds for our operators. We may once again assume that n = 3 and must check (E7) on all basis elements v j 1 v j 2 v j 3 i 3. Once again, the cases of j 1, j 2, j 3 all distinct or all

6 6 STEEN RYOM-HANSEN equal are easy. We then need only consider j 1 = j 2 j 3 and its permutations and can once again use the prime/unprime notation as in the verification of (E6). Let us first verify that E 1 T 2 T 1 = T 2 T 1 E 2. We first observe that E 2 acts as the identity on exactly those basis vectors that are of the form v v i2 v i3. Hence T 2 T 1 E 2 (v v i2 v i3 ) = v i2 v i3 v = E 1 T 2 T 1 (v 1 v i2 v i3 ) The missing basis vectors are of the form v i1 v v i3 or v i1 v i2 v i 3 and are hence killed by E 2 and therefore T 2 T 1 E 2. But one easily checks that they are also killed by E 1 T 2 T 1. The relation E 2 T 1 T 2 = T 1 T 2 E 1 is verified similarly. Let us then check the relation (E8). Once again we take n = 3 and consider the action of E 1 E 2 T 2, E 1 T 2 E 1 and T 2 E 1 E 2 in the basis vector v j 1 v j 2 v j 3 i 3. If the j 1, j 2, j 3 are distinct, the action of the three operators is zero, and if j 1 = j 2 = j 3 they all act as T 2. Hence we may once again assume that exactly two of the j s are equal. But it is easy to check that each of the three operators acts as zero on all vectors of the form v v i2 v i3, v i1 v v i3 and v i1 v i2 v i 3. and so we have proved that E 1 E 2 T 2 = E 1 T 2 E 1 = T 2 E 1 E 2. Similarly one proves that E 2 E 1 T 1 = E 2 T 1 E 2 = T 1 E 2 E 1. Finally we check the relation (E9), which by (E5) can be transferred into T = 1 + (u 1)(1 + T i )E i It can be checked taking n = 2. We consider vectors of the form v j 1 v j 2. If j 1 j 2 E i acts as zero and we are done. And if j 1 = j 2, the relation reduces to the usual Hecke algebra square. The theorem is proved. Since the above proof is only a matter of checking relations, it also works over E A n (u) and hence we get Remark 1. There is a module structure of E A n (u) on V n.

7 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 7 Our next goal is to prove that V n is a faithful representation of E n (u). Our strategy for this will be to construct a subset G of E A n (u) that generates E A n (u) as a A-module and maps to a linearly independent subset of End A (V n ) under the representation. We will then also have determined the dimension of E n (u). Let us start out by stating the following useful lemma. Lemma 1. The following formulas hold in E n (u) and E A n (u). (a) (b) (c) = Ti 1 E j T i if i j] = 1 T j E i = E j Ti 1 T j if i j] = 1 = T i E j Ti 1 if i j] = 1 T j E i T 1 j T 1 i T j E i T 1 j Proof. The formula (a) is just a reformulation of (E7) whereas the formula (b) follows from T 1 i = T i + (u 1 1)E i (1 + T i ) combined with (E7) and (E8). Formula (c) is a variation of (b). For 1 i < j n we define E ij by E i if j = i + 1, and otherwise E ij := T i T i+1...t j 2 E j 1 Tj Ti+1 T i 1 We shall from now on use the notation n := {1, 2,..., n}. For any subset I n with I 2 we extend the definition of E ij to E I := (i,j) I,i<j We now aim at showing that this product is independent of the order in which it is taken. Let us denote by s i the transposition (i, i + 1). Thus s i acts on n and hence also on the subsets of n. Write E {j,k} for E min{j,k},max{j,k}. Lemma 2. We have for all i, j, k that (a) (b) T i E jk T 1 i T 1 i E ij = E si {j,k} E jk T i = E si {j,k}

8 8 STEEN RYOM-HANSEN Proof. Let us prove (a). We first consider the case where i is not any of the numbers j 1, j, k 1 or k. In that case we must show that T i and E j,k commute. For i < j 1 and i > k this is clear since T i then commutes with all of the factors of E j,k. And for j < i < k 1 one can commute T i through E j,k using (E6) and (E3). For i = j 1 the formula follows directly from the definition of E j,k. For i = k we get that T i commutes with all the T l factors of E j,k and hence the formulas reduce to showing that T k E k 1 T 1 k = T k 1 E k T 1 k 1 which is true by formula (c) of lemma 1. For i = k 1 the formula follows from the definitions and (E7). Finally, we consider the case i = j. To deal with this case, we first rewrite E j,k, using formula (c) of lemma 1 repeatedly starting with the innermost term, in the form E j,k = T k 1 T k 2...T j+1 E j Tj Tk 2 T k 1 1 (1) The formula of the lemma now follows from relation (E7). Formula (b) is proved the same way. With this preparation we obtain the commutativity of the factors involved in E I. We have that Lemma 3. The E ij are commuting idempotents of E n (u) and E A n (u). Proof. The E ij are obviously idempotents in E n (u) and E A n (u) so we just have to prove that they commute. Thus, given E ij and E kl we show by induction on (j i) + (l k) that they commute with each other. The induction starts for (j i)+(l k) = 2, in which case E ij = E i and E kl = E k, that commute by (E2). Suppose now (j i) + (l k) > 2 and that E ij, E kl is not a pair of the form E s 1,s+2, E s,s+1 for any s. One checks now there is an r such that E sr {i,j}, E sr {k,l} is covered by the induction hypothesis.

9 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 9 But then, using (a) from the previous lemma together with the induction hypothesis, we find that E ij E kl = T 1 r T 1 r E sr {i,j}t r Tr 1 E sr {k,l}t r = Tr 1 E sr {i,j}e sr {k,l}t r = E sr {k,l}e sr {i,j}t r = T 1 E sr {k,l}t r T 1 r r E sr {i,j}t r as needed. Finally, if our pair is of the form E s 1,s+2, E s,s+1 we use (E8) to finish the proof the lemma. E s 1,s+2 E s,s+1 = T s 1 T s E s+1 Ts 1 E s,s+1 E s 1,s+2 T 1 s 1 E s = E s T s 1 T s E s+1 Ts 1 Ts 1 1 = We have now proved that the product involved in E I is independent of the order taken. We then go on to show that many of the factors of this product can be left out: Lemma 4. Let I n with I 2 and set i 0 := mini. Then E I = (i 0,j) I E i0 j Proof. It is enough to show the lemma for I of cardinality three. Moreover, by a direct calculation using the definition of E kl one sees that this case can be reduced to I = {1, 2, i}. Set now E 1 := E 1 T 1 T 2...T i 1 E i T 1 i 1 E 2 := T 2 T 3...T i 1 E i T 1 i T2 T T3 T2 1 Then the left hand side of the lemma is E 1 E 2 while the right hand side is E 1, so we must show that E 1 E 2 = E 1. But using formula (a) of lemma 1 repeatedly this identity reduces to E 1 T 1 E 2 T1 1 E 2 = E 1 T 1 E 2 T1 1 which is true by relations (E5) and (E8). We need a generalization of the lemma. Let R be a symmetric subset of n n. Write i R j if (i, j) R and write for the equivalence relation induced by i R j (that is i i for all i and i j if there is a chain i =,,..., i k = j such that i s R i s+1 for all s). Assume furthermore that has precisely one block (by

10 10 STEEN RYOM-HANSEN which we mean equivalence class) C(R) n of cardinality larger than 2. Lemma 5. In the situation described above the following formula is valid E i,j = E C(R) (i,j) R Proof. It is enough to show the following equations for i < j < k E ij E ik = E ij E jk = E ik E jk = E ij E jk E ik since one can then gradually add all the elements to C(R). The equation E ij E ik = E ij E jk E ik was shown in the previous lemma so we only need show that E ik E jk = E ij E jk E ik and E ij E jk = E ij E jk E ik. We consider the involution inv of En A (u) given by the formulas inv(t i ) = T n i Using equation (1) we find that inv(e i ) = E n i inv(e ij ) = E n j,n i But then E ik E jk = E ij E jk E ik follows from E ij E ik = E ij E jk E ik. We then show that E ij E jk = E ij E jk E ik. By the above, it can be reduced to showing the identity E ij E jk = E ij E ik Using the definition of the E ij it can be reduced to the case i = 1, j = 2, i.e. E 1 E 2k = E 1 E 1k. Using formula (a) of lemma 1 it becomes the valid identity E 1 E 2 = E 1 T 1 E 2 T1 1, We are now in position to construct the subset G of En A (u). For w = s i1 s i2...s in S n in reduced form, we write as usual T w = T i1 T i2...t in. Moreover, we shall need the notation I for the family of disjoint subsets A = {I 1, I 2,..., I k } of n all satisfying I j 2. For A I we write E A := j E I j. We finally define G := { E A T w A I, w S n } (2) Now from (a) of lemma 2 we obtain a formula involving the E A.

11 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 11 Corollary 1. For A = {I 1, I 2,..., I k } I we define s i A componentwise, i.e. s i A := {s i I 1, s i I 2,..., s i I k }. Then we have T i E A T 1 i = E si A With the theory developed so far we can prove the following theorem. Theorem 2. The set G generates En A (u) over A. Proof. Consider a word w = X i1 X i2 X ik in the generators T i and E i, i.e. X ij = T ij or X ij = E ij for all j. Using the corollary we can move all the E i to the front position, at each step changing the index set by its image under some reflection, and are finally left with a word in the T i s, which is possibly not reduced. If it is not so, it is equivalent under the braid relations (E6) to a word with two consecutive T i s, see [H] chapter 8. Expanding the Ti 2 gives rise to a linear combination of 1, E i and T i E i, where the E i s can be commuted to the front position the same way as before. Continuing this way we eventually reach a word in reduced form, that is a linear combination of elements of the form (i,j) R, x S n E ij T x. Let be the equivalence relation on n induced by R and let A := {I 1, I 2,..., I k } be its blocks. Then A I. Using lemma 5 we conclude the proof of the theorem. Any element A = {I 1,..., I k } I defines canonically a set partition of n, whose blocks are the I i along with the one point blocks {a} for a I i. We shall denote this set partition by A as well. For B I we write A B if A is less than B in the usual partial order on set partitions. This means that each block of the set partition B is a union of certain blocks of the set partition A. From lemma 5 we have Corollary 2. Assume A, B I and let C I be minimal with respect to A C and B C. Then E A E B = E C. With this piece of notation we are in position to prove our main result:

12 12 STEEN RYOM-HANSEN Theorem 3. The set G is a basis of E n (u). Proof. Since E n (u) = E A n (u) A C(u) it is enough to show that G is linearly independent over A. Assume that there exists a linear dependence g G λ g g = 0 with 0 λ g A. It maps to a linear dependence g G λ g ψ(g) = 0 in End A (V n ) where ψ : E A n (u) End A(V n ) is the representation homomorphism. But End A (V n ) is free over A, hence also torsionfree and we may assume that not all λ g are divisible by u 1. From this we obtain a nontrivial linear dependence g G λ g(1) ψ(g) = 0 in End A (V n ). To arrive at the desired contradiction we specialize ψ(g) at u = 1. But for u = 1, the action of T i in V n is just permutation of the factors (i, i+1). Hence E kl acts as a projection in the space of equal upper indices in the kl th factors of V n. In formulas E kl (v j 1... v j k i k... v j l i l... v j n { in ) = v j 1... v j k i k... v j l i l... v j n in if j k = j l 0 otherwise Hence, for a family A = {I 1, I 2..., I s } I we get that E A acts as the projection π A on the space of equal upper indices in factors corresponding to each of the I k. In formulas E A (v j 1... v j r i r... v j s i s... v j n { in ) = 0 if there exist r, s, k such that r, s Ik and j r j s v j 1... v j r i r... v j s i s... v j n in otherwise Let us now consider a linear dependence: λ w,a T w π A = 0 (3) w S n, A I with λ w,a C. Take A 0 such that λ w,a0 0 for some w S n and A 0 is minimal with respect to this condition; minimality refers to the partial order introduced above. Write A 0 = {I 1, I 2,..., I s }, including the one-point blocks. Take a basis vector v A 0 = v j 1... v j k i k... v j l i l... v j n in

13 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 13 such that j k = j l if and only if k, l belong to the same I i, we get on evaluation in (3), using the minimality of A 0 that w S n λ w,a0 T w v A0 = 0 We now furthermore take v A 0 such that its lower i-indices are all distinct. But then {T w v A 0, w S n} is a linearly independent set and we conclude that λ w,a0 = 0 for all w, which contradicts the choice of I 0. This shows that the set {T w π A w S n, A I} is linearly independent. To get the linear independence of {π A T w w S n, A I} we apply corollary (1). Corollary 3. We have dim E n (u) = n!b n, where B n is the Bell number, defined as the number of set partitions of n. For example dim E 2 (u) = 4, dim E 3 (u) = 30, etc. The appearance of set partitions in the above, notably corollary 2, might indicate a connection with the partition algebra A n (k) introduced by P. Martin [M], see also [HR], but we could not find it. The relation (E9) makes it especially awkward to relate E n (u) to classical objects of combinatorial representation theory. It reveals the origen of E n (u) in the Yokonuma-Hecke algebra. Corollary 4. The tensor space V n is a faithful E n (u)-module. Proof. We proved that G is a basis of E n (u) that maps to a linearly independent set in End k (V n ). 4. Representation theory, first steps We initiate in this section the representation theory of E n (u). We construct a family of irreducible representations of E n (u) as pullbacks of representations of the symmetric group and of the Hecke algebra.

14 14 STEEN RYOM-HANSEN Let I E n (u) be the two-sided ideal generated by E i for all i (actually E 1 is enough to generate I). Let furthermore J E n (u) be the two-sided ideal generated by the E i 1 for all i (once again E 1 1 is enough to generate J). We write S n for the symmetric group on n letters and H n (u) for the Hecke algebra of type A n 1 generated over k by T 1,..., T n 1 on the relations T i T j = T j T i if i j > 1 and T i T i±1 T i = T i±1 T i T i±1, (T i u)(t i + 1) = 0 Lemma 6. a) There is an isomorphism ϕ : ks n E n (u)/i b) There is an isomorphism ψ : H n (u) E n (u)/j Proof. We first prove a). In E n (u)/i we have Ti 2 = 1 and hence we obtain a surjection ϕ : ks n E n (u)/i. Consider once again the vector space V = span k {v j i i, j = 1,..., n} and its tensor space V n as a representation of E n (u). We consider the subspace M V n given by M = span k { v j 1... v j n in the upper indices are all distinct } It is easy to check from the rules of the action of E n (u) that M is a submodule of V n. Since the E i act as zero in M we get an induced homomorphism ρ : E n (u)/i End k (M), where ρ(t i ) is the switching of the i th and i+1 th factors of the tensor product. But then the image of ρ ϕ has dimension n! and we conclude that ϕ indeed is an isomorphism. In order to prove b) we basically proceed the same way: In the quotient E n (u)/j we have T = 1 + (u 1)(1 + T i ) which implies the existence of a surjection ψ : H n (u) E n (u)/j. To show that ψ is injective we this time consider the submodule N = span k { v j 1... v j n in the upper indices are all equal to 1} All E i act as 1 in N and so we get a induced map ρ : E n (u)/j End k (N). The composition ρ ψ is the regular representation of H n (u) and hence dim Im(ρ ψ) = n! which proves that also ψ is an isomorphism.

15 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 15 We now recall the well known basic representation theory of ks n and of H n (u). Let λ = (λ 1, λ 2,..., λ k ) be an integer partition of n and let Y (λ) be its Young diagram. Let t λ (resp. t λ ) be the λ-tableau in which the numbers {1, 2,..., n} are filled in by rows (resp. columns). Denote by R(λ) (resp. C(λ)) the row (resp. column) stabilizer of t λ. Define now r λ = w, c λ = ( 1) l(w) w, s λ = c λ r λ w R(λ) w C(λ) Then s λ is the Young symmetrizer and S(λ) = ks n s λ is the Specht module associated with λ. Since Chark = 0 the Specht modules are simple and classify the simple modules of ks n. To give the Specht modules for H n (u), we use Gyoja s Hecke algebra analogue of the Young symmetrizer, [G], [Mu]. In our setup it looks as follows: For X S n, define ι(x) = T w, ǫ(x) = w X w X( u) l(w) T w If for example X = S n, we have T w ι(s n ) = u l(w) ι(s n ), T w ǫ(s n ) = ( 1) l(w) ǫ(s n ) for all T w. We now define x λ = ι(r(λ)), y λ = ǫ(r(λ)) Let w λ S n be the element such that w λ t λ = t λ. Then the Hecke algebra analogue of the Young symmetrizer is e λ = T w 1 y λ T w λ λ x λ = c λ (u) r λ (u) where c λ (u) := T w 1 y λ T w λ λ and r λ (u) := x λ (u). The permutation module and the Specht module associated with λ are defined as M(λ) := H n (u)x λ and S(λ) = H n (u)e λ. Since u is generic, S(λ) is irreducible. For future reference, we recall the following result, see eg. [DJ], [Mu]. Lemma 7. Suppose that c λ M(µ) 0. Then µ λ.

16 16 STEEN RYOM-HANSEN Here refers to the dominance order on partitions of n, defined by λ = (λ 1, λ 2,...) µ = (µ 1, µ 2,...) iff λ 1 + λ λ i µ 1 + µ µ i for all i. The dominance order is only a partial order, but we shall embed it into the total order < on partitions of n, defined by λ = (λ 1, λ 2,...) < µ = (µ 1, µ 2,...) iff λ 1 +λ λ i µ 1 +ν µ i for some i and λ 1 +λ λ j = µ 1 +µ µ j for j < i. We extend < to a total order on all partitions by declaring λ < µ if λ < µ. Corresponding to Schur s lemma we have that e λ H n (u)e µ = δ λ,µ k e λ and s λ S n s µ = δ λ,µ k s λ (4) where δ λ,µ is the Kronecker delta. This gives a way of determining the partition λ that corresponds to a given simple module L for ks n (or H n (u)): it is the unique λ such that s λ L 0 or (e λ L 0). Now, if S(λ) is a Specht module for either ks n or H n (u) we use ϕ or ψ to obtain a simple module for E n (u), which is also denoted S(λ). On the other hand, these two series of simple modules do not exhaust all the simple modules for E n (u) as we shall see in the next sections. 5. E n (u) as a E n (u)-module In this section we return to E n (u). We show that it is selfdual as a left module over E n (u) itself. As a consequence of this we get that all simple modules occur as left ideals in E n (u). Denote by ω : E n (u) E n (u) the k-linear antiautomorphism given by ω(t i ) = T i and ω(e i ) = E i. To check that ω exists we must verify that ω leaves the defining relations (E1),..., (E9) invariant. This is obvious for all of them, except possibly for (E7) where it follows by interchanging i and j. We now make the linear dual E n (u) of E n (u) into a left E n (u)- module using ω: (xf)(y) := f(ω(x)y) for x, y E n (u), f E n (u)

17 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 17 We need to consider the linear map ǫ : E n (u) k, x coeff En (x) where coeff En (x) is the coefficient of E n = E {1,2,...,n} when x E n (u) is written in the basis elements T w E A of G, see (2). With this we may construct a bilinear form, on E n (u) by x, y = ǫ(ω(x)y) for x, y E n (u) And then we finally obtain a homomorphism ϕ by the rule ϕ : E n (u) E n (u) : x (y x, y ) Theorem 4. With the above definitions, we get that ϕ is an isomorphism of left E n (u)-modules. Proof. One first checks that the bilinear form satisfies xy, z = y, ω(x)z for all x, y, z E n (u) which amounts to saying that ϕ is E n (u)-linear. Since E n (u) is finite dimensional, it is now enough to show that, is non-degenerate. For this we first observe that our construction of, is valid over A as well and hence also defines a bilinear form, A on E A n (u). It is enough to show that, A is nondegenerate. Any a E A n (u) can be written a = (u 1) N a where a = g G λ g g and where λ g (1) is not constantly 0. Then, letting π : E A n (u) E n (1) be the specialization map we have π(a ) 0 since it was shown in the proof of theorem 3 that G is a basis of E A n (1). Let us denote by, 1 the bilinear form on E n (1) constructed similarly to,. Then we have that π(a), π(b) 1 = a, b A A C for all a, b E A n (u) since π is multiplicative and satisfies π(ω(a) = ω(π(a)). We are now reduced to proving that, 1 is nondegenerate. Let us therefore consider an arbitrary a = w,i λ w,ie I T w E n (1), where λ w,i C.

18 18 STEEN RYOM-HANSEN Let I 0 I be minimal subject to the condition that a w,i0 0 for some w. Take now w 0 S n with λ w0,i 0 0 and define b = E I0 (1 E I ) T w0 I 0 I We claim that b, a 1 0. Indeed, since u = 1 we have ω(b)a = Tw 1 0 (1 E I )E I0 a I 0 I Since I 0 was chosen minimal, there can be no cancellation of the coefficient of E I0 T w0 in E I0 a which hence is λ w0,i 0. All E I appearing in the expansion of E I0 a with respect to the basis E I T w satisfy I 0 I, and are therefore killed by I 0 I (1 E I). By this we get T 1 w 0 I 0 I (1 E I )E I0 a = λ w0,i 0 T 1 w 0 (1 E I )E I0 T w0 I 0 I The coefficient of E n in this expression is by corollary (1) equal to the coefficient of E n in λ w0,i 0 (1 E I )E I0 I 0 I On the other hand, the coefficient of E n in I 0 I (1 E I)E I0 is given by the Moebius function associated with the partial order on I. It is equal to ( 1) k 1 k!, where k is the number of blocks of I 0. Summing up we find that b, a 1 = ( 1) k 1 λ w0,i 0 k! 0 which proves the theorem. 6. Classification of the irreducible representations In this section we give the classification of the irreducible representations of E n (u). For M an left E n (u)-module we make its linear dual M into a left E n (u)-module using the antiautomorphism ω. If M is a simple E n (u)-module then any m M \ {0} defines a surjection E n (u) M, x xm forx E n (u)

19 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 19 By duality and by the last section, we then get an injection of M into E n (u). On the other hand, the canonical isomorphism M M is E n (u)-linear because ω 2 = Id and so we conclude that all simple E n (u)-modules appear as left ideals in E n (u). Let now I be a simple left ideal of E n (u) and let x 0 I \{0}. Since the tensor space V n is a faithful E n (u)-module, we find a v V n such that x 0 v 0. But then the E n (u)-linear map I V n, x xv for x I is nonzero, and therefore injective. We conclude that all simple E n (u)-modules appear as submodules of V n. Consider now a simple submodule M of V n. Take A 0 n maximal such that E A0 M 0. By section 3, in the two extreme situations A 0 = or A 0 = n we can give a precise description of M, since in those cases M is a module for ks n or for H n (u). Hence, the irreducible representations are parametrized by partitions of n as we have already mentioned in section 3. Let us denote by P the set of { (λ s ), (m s ), (µ s ) s = 1,..., l } where λ s is a partition, m s an integer and µ s a partition of m s such that s m s λ s = n and such that (λ s ) is ordered increasingly with respect to the total order < on partitions. For any integer partition λ = (λ 1, λ 2,...) we define the vector v λ V λ as v λ = v tλ (1,1) v tλ (1,3) v tλ (1,2)... where t λ (i, j) are the coordinates of the λ-tableaux t λ introduced in section 3. Thus the first λ 1 indices are all 1, the next λ 2 indices are 2 and so on. Suppose now Λ P. We associate to it the vector v Λ V n in the following way v Λ := v 1 λ 1 v2 λ 1... vm 1+1 λ 2 v m 1+2 λ 2... where v 1 λ 1 means that all factors of v λ 1 come with un upper index 1 and so on.

20 20 STEEN RYOM-HANSEN The element Λ P gives naturally rise to an A I, the first m 1 blocks being of size λ 1, the next m 2 blocks of size λ 2 and so on. The numbers 1,..., n are filled in increasingly. Note that it is possible that λ 1 = λ 2 making the first m 1 + m 2 blocks of equal size and so on. Now the product of symmetric groups S m := S m1 S m2... S ml acts blockwise in the factors of V n, i.e. S m1 permutes the first m 1 blocks of A and so on. Take for example n = 6, l = 1 and Λ = (λ, 2, µ) where λ = (2, 1) and µ = (1, 1). Then A = {(1, 2, 3), (4, 5, 6)} and S m is the group of order two that permutes the two blocks, thus generated by σ = (1, 4)(2, 5)(3,6). If we write σ i = (i, i + 1) we have σ = σ 3 σ 4 σ 5 σ 2 σ 3 σ 4 σ 1 σ 2 σ 3 S m and hence T σ = T 3 T 4 T 5 T 2 T 3 T 4 T 1 T 2 T 3 E n (u) The action of σ and T σ on v Λ is the same since the action of the various T i involves different upper indices. This remark generalizes to all elements σ S m and so we may regard the row and column (anti)-symmetrizers r µ i and c µ i as elements of E n (u). By corollary 1, E A commutes with r µ i and c µ i. We now define the permutation module as M(Λ) := E n (u)(r µ 1 r µ 2... r µ s)v Λ := E n (u)w Λ where w Λ := (r µ 1 r µ 2... r µ s)v Λ and define furthermore e Λ := (c µ 1 c µ 2... c µ s)(c λ 1(u) c λ 1(u)... c λ 2(u)...)E A where c λ i(u) is as in section 4 and where c λ 1(u) occurs m 1 times, c λ 2(u) occurs m 2 times etc. We define the Specht module as S(Λ) := E n (u)e Λ w Λ M(Λ) The main result of this section is now Theorem 5. S(Λ) is a simple module for E n (u). The simple E n (u)- modules are classified by S(Λ) for Λ P.

21 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 21 Proof. The first step is to show that e Λ M(Λ) = k e Λ w Λ. Let us for this take x E n (u) and consider the element E A xw Λ M(Λ). Since w Λ is a linear combination of yv Λ where y S m, we break the analysis up by firstly investigating the conditions on x for E A xyv Λ to be nonzero, when y S m. We can write x as a linear combination of elements E B T w from our basis G and hence E A x as a linear combination of E A E B T w. But by corollary 2, E A E B is equal to a E C for C with A C. Since E C M(Λ) = 0 for A C it is enough to take B such that E B E A = E A into account. Thus, E A x can be assumed to be a linear combination of elements of the form E A T w, where T w permutes the blocks of A of equal cardinality, since by corollary 1, otherwise E A T w acts as zero. Let thus S m S n be the subgroup consisting of the permutations of the blocks of A of equal cardinality. Note that S m S m, the inclusion being strict in general. As in the case of S m, the elements of S m can be seen as elements of E n (u), by the map z T z. In this notation, if E A xyv Λ is nonzero it is a linear combination of elements of the form T z (T w 1 1 vλ 1 T 1 w1 2 v2 λ1...) (5) where T w 1 1 H λ1 (u), T w 2 1 H λ1 (u) (here m 1 2) etc and where T z S m. However, we need to show that actually T z S m and therefore consider the action on (c λ 1(u) c λ 1(u)... c λ 2(u)...) on (5). Let from this λ 1, λ 2,..., λ t be the partitions with λ i = λ 1. Since the λ i are ordered increasingly, we get by lemma 7 that the product is nonzero only if each factor c λ k(u) of c λ 1(u) c λ 1(u)... c λ t(u) acts in a T w a k v a λ k factor of (5), i.e. a factor with the same λ k appearing as index. This argument extends to all of (c λ 1(u)... c λ 2(u)...) and so T z S m as claimed. Let us now return to E A xw Λ. If it is nonzero, at least one of its terms E A xyv Λ must be nonzero and hence, using the above, we get that x S m.

22 22 STEEN RYOM-HANSEN We can now commute E A ahead to w Λ, where it cancels. And then we find that e Λ xw Λ is equal to C(c µ 1... c µ s)(c λ 1(u)...)x(r λ 1(u)...)(r µ 1... r µ s)w Λ where C k accounts for the insertion of the two tensor factors in front of w Λ. But using (4) twice, we get that this is equal to ke Λ w Λ. For x = 1, the constant involved is nonzero. We have proved that e Λ M(Λ) = ke Λ w Λ. Since S(Λ) M(Λ) we of course have e Λ S(Λ) ke Λ w Λ. The equality comes from the fact that the Young symmetrizers are idempotents up to nonzero scalars. We now proceed to prove that S(Λ) is a simple module for E n (u). We do it by setting up of version of James s submodule theorem, [Ja]. Assume therefore N S(Λ) is a submodule. If e Λ N 0, we have by the above that e Λ N is a scalar multiple of w Λ and so N = S(Λ). In order to treat the other case e Λ N = 0, we define a bilinear form on V n by declaring the natural basis orthonormal (just as in the submodule theorem): v j 1... v j n in, v j 1 i... v j n 1 i n = δ i=i,j=j where i = (,,..., i n ) etc. It has the following invariance property xv, w = v, ω(x)w where ω is as in section 4. We have that for all x E n (u), v, w V n ω(c λ ) = c λ, ω(r λ ) = r λ, ω(c λ (u)) = c λ (u), ω(r λ (u)) = r λ (u) where we used that ω is an antiautomorphism to show for instance that ω(t w 1 y λ T w λ λ ) = T w 1 y λ T w λ λ. Since the group S m permutes blocks corresponding to equal λ i, we have that the factors of e Λ all commute, and so ω(e Λ ) = e Λ We may now calculate as follows 0 = e Λ N, M(Λ) = N, e Λ M(Λ) N, e Λ w Λ

23 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 23 which implies that N, S(Λ) = 0 or N S(Λ). But since u is generic, S(Λ) S(Λ) = 0. This a contradiction unless N = 0. We have proved that S(Λ) is simple. Take Λ = ((λ s ), (m s ), (µ s )) P and assume S(Λ) = S(Λ). The element A I associated with S(Λ) is maximal with respect to having blocks of consecutive numbers such that E A S(Λ) 0. Hence, if A I is the element associated with S(Λ), we have that A = A. Then (λ s ) and (λ s ) must be partitions of the same numbers, corresponding to the block sizes of A (and A). They are both ordered increasingly, and c λ s and c λ s both act nontrivially in E A S(Λ) = E A S(Λ). But then we must have (λ s ) = (λ s ) in view of (7). Similarly, we get (µ s ) = (µ s ) and have proved the claim. It remains to be shown that any simple module L is of the form S(Λ) for some Λ P. We saw in the remarks preceding the theorem, that it can be assumed that L V n. Choose A I maximal with respect to having blocks of consecutive numbers and such that E A L 0. Then E A L is a module for the tensor product of Hecke algebras H A (u) = H I1 (u) H I2 (u)... where A = {I 1, I 2... }. Choose for each I i a c λ i H Ii such that c λ 1 c λ 2... acts nontrivially in E A L. The I i can be ordered to make λ i increasing. Choose at last s µ 1 s µ 2..., acting nontrivially in (c λ 1 c λ 2...)E A L. The data collected gives rise to a Λ with S(Λ) = E n (u)e Λ v Λ L. But since L is simple, there must be equality. We have finally proved all statements of the theorem. Let us work out some low dimensional cases. For n = 2 we have the following possibilities for Λ: (λ 1, m 1, µ 1 ) = (, 1, ), (λ 1, m 1, µ 1 ) = (, 1, ) (λ 1, m 1, µ 1 ) = (, 2, ), (λ 1, m 1, µ 1 ) = (, 2, ) They all give rise to irreducible representations of dimension one. The first two are the one dimensional representations of H 2 (u). By

24 24 STEEN RYOM-HANSEN our construction the third is given by v 1 1 v v 2 1 v 1 1 and the last by v 1 1 v 2 1 v 2 1 v 1 1. They correspond to the trivial and sign representation of ks 2. The square sum of the dimensions is 4, which is also the dimension of E 2 (u). For n = 3 we first write down the multiplicity free possibilities of Λ, i.e. those having m s = 1 and so µ s = for all s. They are (λ 1 ) = ( ), (λ 1 ) = ( ), (λ 1 ) = ( ) ( λ 1, λ 2 ) = (, ), ( λ 1, λ 2 ) = (, ) The first three of these are the Specht modules for H 3 (u), their dimensions are respectively 1,2 and 1. The fourth is given by the vector v 1 1 v 1 1 v 2 1 and the last by the vector (v 1 1 v 1 2 v 1 2 v 1 1) v 2 1, according to our construction. In both cases, one gets dimension three. Allowing multiplicities, we have the following possibilities: (λ 1, m 1, µ 1 ) = (, 3, ), (λ 1, m 1, µ 1 ) = (, 3, ) and (λ 1, m 1, µ 1 ) = (, 3, ). We get the Specht modules of ks 3 of dimensions 1,2 and 1. The square sum of all the dimensions is 30, in accordance with the dimension of E 3 (u). We have thus proved that E n (u) is semisimple for n = 2 and n = 3. The classification of the simple modules for n = 2 and n = 3 has also been done in [AJ] with a different method. To finish off, we present the following problems, that we hope to address in a future work. Problem 1. Show that E n (u) is semisimple for all n. Problem 2. Calculate the dimensions of the Specht modules S(Λ).

25 ON THE REPRESENTATION THEORY OF AN ALGEBRA OF BRAIDS AND TIES 25 Problem 3. Calculate End En (u)(v n ). Describe a Schur-Weyl duality. Problem 4. Calculate the center of E n (u). Problem 5. Analyse the case where u is a root of unity and characteristic p. References [AJ] F. Aicardi, J. Juyumaya, An algebra involving braids and ties, ictp preprint IC , [ATY] S. Ariki, T. Terasoma, H. Yamada, Schur-Weyl reciprocity for the Hecke algebra of Z/rZ S n, J. Algebra 178 (1995), [CHWX] Y. Cheng, M.L. Ge, Y.S. Wu, K. Xue, Yang-Baxterization of braid group representations, Comm. Math. Phys., 136, [DJ] R. Dipper, G. D. James, Representation of Hecke Algebras of general linear groups, Proc. London Math. Soc. 54 No. 3, 1987, [HR] T. Halverson, A. Ram, Partition algebras, European J. of Combinatorics, 26 (2005) [H] J.E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge studies in advanced mathematics 29. [G] A. Gyoja, A q-analogue of Young symmetrizer, Osaka J. of Math. 23 (1986), [Ja] G. D. James, The representation theory of the symmetric groups, Lecture Notes in Math., vol. 682, Springer-Verlag, Berlin and New York, [Jo] V.F.R Jones, Comm. Math. Phys., 125, 459, 1989 [Ji] M. Jimbo, A q-analog of U(gl(N +1)), Hecke algebra and the Yang Baxter equation, Lett. Math. Phys. 11, (1986), [J1] J. Juyumaya, A new algebra arising from the representation theory of finite groups, Rev. Math. Phys., 11, [J2] J. Juyumaya, S. Kennan, Braid relations in the Yokonuma-Hecke algebra, J. Algebra, 239, [M] P.P. Martin, Temperley-Lieb Algebras for non-planar statistical mechanics - the partition algebra construction, J. Knot Theory Ramifications, 3, 1994, [Mu] G.E. Murphy, On the Representation Theory of the Symmetric group and Associated Hecke Algebras, J. Algebra, 152, , (1992). [RH] S. Ryom-Hansen, The Ariki-Terasoma-Yamada tensor space and the blob algebra, arxiv RT/ [T] N. Thiem, Unipotent Hecke algebras of Gl n (F q ), J. Algebra 294 (2005), Instituto de Matemática y Física, Universidad de Talca, Chile, steen@instmat.utalca.cl