The hook fusion procedure James Grime Department of Mathematics, University of York, York, YO10 5DD, UK

[email protected]

arXiv:math/0502521v2 [math.RT] 8 May 2007 Submitted: Feb 25, 2005; Accepted: Apr 25, 2005; Published: June 25, 2005 Mathematics Subject Classifications: 05E10, 20C30 Abstract We derive a new expression for the diagonal matrix elements of irreducible representations of the symmetric group. We obtain this new expression using Cherednik’s fusion procedure. However, instead of splitting Young diagrams into their rows or columns, we consider their principal hooks. This minimises the number of auxiliary parameters needed in the fusion procedure. 1 Introduction In this article we present a new expression for the diagonal matrix elements of irreducible representations of the symmetric group. We will obtain this new expression using Cherednik’s fusion procedure. This method originates from the work of Jucys [J], and has already been used by Nazarov and Tarasov [N1, NT]. However our approach differs by minimizing the number of auxiliary parameters needed in the fusion procedure. This is done by considering hooks of Young diagrams, rather than their rows or columns as in [N1, NT]. Irreducible representations of the group Sn are parameterized by partitions of n, [CR]. The Young diagram of a partition λ is the set of boxes (i, j) ∈ Z2 such that 1 6 j 6 λi . In drawing such diagrams we let the first coordinate i increase as one goes downwards, and the second coordinate j increase from left to right. For example the partition λ = (3, 3, 2) gives the diagram the electronic journal of combinatorics 12 (2005), #R00 1 If (i, j) is a box in the diagram of λ, then the (i, j)-hook is the set of boxes in λ {(i, j ′ ) : j ′ > j} ∪ {(i′ , j) : i′ > i}, We call the (i, i)-hook the ith principal hook. A standard tableau, Λ, is a filling of the diagram λ in which the entries are the numbers 1 to n, each occurring once. The Young Symmetrizer of a standard tableau Λ of shape λ generates an irreducible CSn -module, [CR], denoted Vλ . In 1986 Ivan Cherednik proposed another description of the Young symmetrizer, [C]. The following version of Cherednik’s description can be found in [NT]. There is a basis for the space Vλ called the Young basis, with its vectors labeled by the standard tableaux of shape λ, [JK], [OV]. Now, for any standard tableau Λ consider the diagonal matrix element, FΛ , of the representation Vλ corresponding to the Young basis vector vΛ , defined by, X FΛ = (vΛ , gvΛ )g ∈ CSn . g∈Sn Multiplying FΛ on the left and right by certain invertible elements returns the Young sym- metrizer. Hence FΛ may also be used to generate the irreducible module Vλ , [JK], [N3, section 2]. Cherednik’s fusion procedure can be used to find an alternative expression for FΛ . Consider a standard tableaux Λ of shape λ. For each p = 1, . . . , n, put cp = j − i if the box (i, j) ∈ λ is filled with the number p in Λ. The difference j − i is the content of box (i, j). For any two distinct numbers p, q ∈ {1, . . . , n}, let (pq) be the transposition in the symmetric group Sn . Consider the rational function of two complex variables u, v, with values in the group ring CSn : (pq) fpq (u, v) = 1 − . (1) u−v Now introduce n complex variables z1 , . . . , zn . Consider the set of pairs (p, q) with 1 6 p < q 6 n. Ordering the pairs lexicographically we form the product → Y FΛ (z1 , . . . , zn ) = fpq (zp + cp , zq + cq ). (2) (p,q) the electronic journal of combinatorics 12 (2005), #R00 2 If p and q sit on the same diagonal in the tableau Λ, then fpq (zp + cp , zq + cq ) has a pole at zp = zq . Let RΛ be the vector subspace in Cn consisting of all tuples (z1 , . . . , zn ) such that zp = zq whenever the numbers p and q appear in the same row of the tableau Λ. By direct calculation we can show the following the identities are true; fpq (u, v)fpr (u, w)fqr (v, w) = fqr (v, w)fpr (u, w)fpq (u, v) (3) for all pairwise distinct indices p, q, r, and fpq (u, v)fst (z, w) = fst (z, w)fpq (u, v) (4) for all pairwise distinct p, q, s, t. We call (3) and (4) the Yang-Baxter relations. Using these relations we obtain reduced expres- sions for FΛ (z1 , . . . , zn ) different from the right hand side of (2). For more details on different expressions for FΛ (z1 , . . . , zn ) see [GP]. Using (3) and (4) we may reorder the product FΛ (z1 , . . . , zn ) such that each singularity is contained in an expression known to be regular at z1 = z2 = . . . = zn , [N2]. It is by this method that it was shown that the restriction of the rational function FΛ (z1 , . . . , zn ) to the subspace RΛ is regular at z1 = z2 = . . . = zn . Furthermore, by showing divisibility on the left and on the right by certain elements of CSn it was shown that the value of FΛ (z1 , . . . , zn ) at z1 = z1 = . . . = zn is the diagonal matrix element FΛ , [CR]. Thus we have the following theorem, 1.1 Restriction to RΛ of the rational function FΛ (z1 , . . . , zn ) is regular at z1 = z2 = . . . = zn . The value of this restriction at z1 = z2 = . . . = zn coincides with the element FΛ ∈ CSn . Similarly, we may form another expression for FΛ by considering the subspace in Cn consisting of all tuples (z1 , . . . , zn ) such that zp = zq whenever the numbers p and q appear in the same column of the tableau Λ [NT]. In this article we present a new expression for the diagonal matrix elements which minimizes the number of auxiliary parameters needed in the fusion procedure. We do this by considering hooks of standard tableaux rather than their rows or columns. the electronic journal of combinatorics 12 (2005), #R00 3 Let HΛ be the vector subspace in Cn consisting of all tuples (z1 , . . . , zn ) such that zp = zq whenever the numbers p and q appear in the same principal hook of the tableau Λ. We will prove the following theorem. Theorem 1.2 Restriction to HΛ of the rational function FΛ (z1 , . . . , zn ) is regular at z1 = z2 = . . . = zn . The value of this restriction at z1 = z2 = . . . = zn coincides with the element FΛ ∈ CSn . In particular, this hook fusion procedure can be used to form irreducible representations of Sn corresponding to Young diagrams of hook shape using only one auxiliary parameter, z. By taking this parameter to be zero we find that no parameters are needed for diagrams of hook shape. Therefore if ν is a partition of hook shape we have → Y Fν = Fν (z) = fpq (cp , cq ), (5) (p,q) with the pairs (p, q) in the product ordered lexicographically. To motivate the study of modules corresponding to partitions of hook shape first let us consider the Jacobi-Trudi identities [M, Chapter I3]. We can think of the following identities as dual to the Jacobi-Trudi identities. If λ = (λ1 , . . . , λk ) such that n = λ1 + . . . + λk then we have the following decomposition of the induced representation of the tensor product of modules corresponding to the rows of λ; ⊗ V(λ2 ) ⊗ · · · ⊗ V(λk ) ∼ M IndSSnλ ×Sλ2 ×···×Sλk V(λ1 ) = (Vµ )⊕Kµλ , 1 µ where the sum is over all partitions of n. Note that V(λi ) is the trivial representation of Sλi . The coefficients Kµλ are non-negative integers known as Kostka numbers, [M]. Importantly, we have Kλλ = 1. On the subspace RΛ , if zi − zj ∈ / Z when i and j are in different rows of Λ then the above induced module may be realised as the left ideal in CSn generated by FΛ (z1 , . . . , zn ). The irreducible representation Vλ appears in the decomposition of this induced module with coefficient 1, and is the ideal of CSn generated by FΛ (z1 , . . . , zn ) when z1 = z2 = . . . = zn . The fusion procedure of theorem 1.1 provides a way of singling out this irreducible component. the electronic journal of combinatorics 12 (2005), #R00 4 Similarly we have the equivalent identity for columns, ∼ M IndSSn′ ×S V ′ ⊗ V(1λ′2 ) ⊗ · · · ⊗ V λ′ = (Vµ )⊕Kµ′ λ′ , λ1 λ′2 ×···×Sλ ′ (1λ1 ) (1 l ) l µ where l is the number of columns of λ. In this case V ′ is the alternating representation of (1λi ) Sλ′i . This induced module is isomorphic to the left ideal of CSn generated by FΛ (z1 , . . . , zn ) considered on the subspace RΛ′ , with zi − zj ∈ / Z when i, j are in different columns of Λ. Again the irreducible representation Vλ appears in the decomposition of this induced module with coefficient 1, and is the ideal of CSn generated by FΛ (z1 , . . . , zn ) when z1 = z2 = . . . = zn . There is another expression known as the Giambelli identity [G]. Unlike the Jacobi-Trudi iden- tities, this identity involves splitting λ into its principal hooks, rather than its rows or columns. A combinatorial proof of the Giambelli identity can be found in [ER]. Divide a Young diagram λ into boxes with positive and non-positive content. We may illustrate this on the Young diagram by drawing ’steps’ above the main diagonal. Denote the boxes above the steps by α(λ) and the rest by β(λ). For example, the following figure illustrates λ, α(λ) and β(λ) for λ = (3, 3, 2). λ α(λ) β(λ) Figure 1: The Young diagram (3, 3, 2) divided into boxes with positive content and non- positive content If we denote the rows of α(λ) by α1 > α2 > . . . > αd > 0 and the columns of β(λ) by β1 > β2 > . . . > βd > 0, then we have the following alternative notation for λ; λ = (α|β), where α = (α1 , . . . , αd ) and β = (β1 , . . . , βd ). Here d denotes the length of the side of the Durfee square of shape λ, which is the set of boxes corresponding to the largest square that fits inside λ, and is equal to the number of principal hooks in λ. In our example d = 2 and λ = (2, 1|3, 2). the electronic journal of combinatorics 12 (2005), #R00 5 We may consider the following identity as a dual of the Giambelli identity. V(α1 |β1 ) ⊗ V(α2 |β2) ⊗ · · · ⊗ V(αd |βd ) ∼ M IndSSnh ×Sh2 ×···×Shd = (Vµ )⊕Dµλ , 1 µ where hi is the length of the ith principal hook, and the sum is over all partitions of n. This is a decomposition of the induced representation of the tensor product of modules of hook shape. Further these hooks are the principal hooks of λ. The coefficients, Dµλ , are non-negative integers, and in particular Dλλ = 1. On the subspace HΛ , if zi − zj ∈ / Z when i and j are in different principal hooks of Λ then the above induced module may be realised as the left ideal in CSn generated by FΛ (z1 , . . . , zn ). The irreducible representation Vλ appears in the decomposition of this induced module with coefficient 1, and is the ideal of CSn generated by FΛ (z1 , . . . , zn ) when z1 = z2 = . . . = zn . Hence, in this way, our hook fusion procedure relates to the Giambelli identity in the same way that Cherednik’s original fusion procedure relates to the Jacobi-Trudi identity. Namely, it provides a way of singling out the irreducible component Vλ from the above induced module. The fusion procedure was originally developed in the study of affine Hecke algebras, [C]. Our results may be regarded as an application of the representation theory of these algebras, [OV]. Acknowledgements and thanks go to Maxim Nazarov for his supervision, and for introducing me to this subject. I would also like to thank EPSRC for funding my research and my anonymous referees for their comments. 2 Fusion Procedure for a Young Diagram The diagonal matrix element FΛ determines the irreducible module Vλ of Sn , up to isomorphism, for any tableau Λ of shape λ. Therefore in the sequel we will only use one particular tableau, the hook tableau. In which case we may denote the diagonal matrix element FΛ by Fλ , and the space HΛ by Hλ . We fill a diagram λ by hooks to form a tableau Λ in the following way: For the first principal hook we fill the column with entries 1, 2, . . . , r and then fill the row with entries r + 1, r + 2, . . . , s. We then fill the column of the second principal hook with s + 1, s + 2, . . . , t and fill the the electronic journal of combinatorics 12 (2005), #R00 6 row with t + 1, t + 2, . . . , x. Continuing in this way we form the hook tableau. Example 2.1. On the left is the hook tableau of the diagram λ = (3, 3, 2), and on the right the same diagram with the content of each box. 1 4 5 0 1 2 2 6 8 -1 0 1 3 7 -2 -1 Therefore the sequence (c1 , c2 , . . . , c8 ) is given by (0, −1, −2, 1, 2, 0, −1, 1). Consider (2) as a rational function of the variables z1 , . . . , zn with values in CSn . The factor fpq (zp + cp , zq + cq ) has a pole at zp = zq if and only if the numbers p and q stand on the same diagonal of the tableau Λ. We then call the pair (p, q) a singularity. And we call the corresponding term fpq (zp + cp , zq + cq ) a singularity term, or simply a singularity. Let p and q be in the same hook of Λ. If p and q are next to one another in the column of 1 the hook then, on Hλ , fpq (zp + cp , zq + cq ) = 1 − (pq). And so 2 fpq (zp + cp , zq + cq ) is an idempotent. Similarly, if p and q are next to one another in the same row of the hook then fpq (zp + cp , zq + cq ) = 1 + (pq), and 21 fpq (zp + cp , zq + cq ) will be an idempotent. Also, for distinct p, q, we have 1 fpq (u, v)fqp (v, u) = 1 − . (6) (u − v)2 Therefore, if the contents cp and cq differ by a number greater than one, then the factor fpq (zp + (cp −cq )2 cp , zq + cq ) is invertible in CSn when zp = zq with inverse f (c , c ). (cp −cq )2 −1 qp q p The presence of singularity terms in the product Fλ (z1 , . . . , zn ) mean this product may or may not be regular on the vector subspace of Hλ consisting of all tuples (z1 , . . . , zn ) such that z1 = z2 = . . . = zn . Using the following lemma, we will be able to show that Fλ (z1 , . . . , zn ) is in fact regular on this subspace. Lemma 2.2 The restriction of fpq (u, v)fpr (u, w)fqr (v, w) to the set (u, v, w) such that u = v ±1 is regular at u = w. the electronic journal of combinatorics 12 (2005), #R00 7 Proof. Under the condition u = v ± 1, the product fpq (u, v)fpr (u, w)fqr (v, w) can be written as (pr) + (qr) (1 ∓ (pq)) · 1 − . v−w And this rational function of v, w is clearly regular at w = v ± 1. Notice that the three term product, or triple, in the statement of the lemma can be written in reverse order due to (3). In particular, if the middle term is a singularity and the other two terms are an appropriate idempotent and triple term, then the triple is regular at z1 = z2 = . . . = zn . We may now prove the first statement of Theorem 1.2. Proposition 2.3 The restriction of the rational function Fλ (z1 , . . . , zn ) to the subspace Hλ is regular at z1 = z2 = . . . = zn . Proof. We will prove the statement by reordering the factors of the product Fλ (z1 , . . . , zn ), using relations (3) and (4), in such a way that each singularity is part of a triple which is regular at z1 = z2 = . . . = zn , and hence the whole of Fλ (z1 , . . . , zn ) will be manifestly regular. Define gpq to be the following;   f (z + c , z + c ) if p < q pq p p q q gpq =  1 if p > q Now, let us divide the diagram λ into two parts, consisting of those boxes with positive contents and those with non-positive contents as in Figure 1. Consider the entries of the ith column of the hook tableau Λ of shape λ that lie below the steps. If u1 , u2 , . . . , uk are the entries of the ith column below the steps, we define n Y Ci = gu1 ,q gu2 ,q . . . guk ,q . (7) q=1 Now consider the entries of the ith row of Λ that lie above the steps. If v1 , v2 , . . . , vl are the entries of the ith row above the steps, we define n Y Ri = gv1 ,q gv2 ,q . . . gvl ,q . (8) q=1 the electronic journal of combinatorics 12 (2005), #R00 8 Our choice of the hook tableau was such that the following is true; if d is the number of principal hooks of λ then by relations (3) and (4) we may reorder the factors of Fλ (z1 , . . . , zn ) such that d Y Fλ (z1 , . . . , zn ) = Ci Ri . i=1 Now, each singularity (p, q) has its corresponding term fpq (zp + cp , zq + cq ) contain in some product Ci or Ri . This singularity term will be on the immediate left of the term fp+1,q (zp+1 + cp+1 , zq + cq ). Also, this ordering has been chosen such that the product of factors to the left of any such singularity in Ci or Ri is divisible on the right by fp,p+1 (zp + cp , zp+1 + cp+1 ). Therefore we can replace the pair fpq (zp + cp , zq + cq )fp+1,q (zp + cp , zq + cq ) in the product by the triple 21 fp,p+1 (zp + cp , zp+1 + cp+1 )fpq (zp + cp , zq + cq )fp+1,q (zp+1 + cp+1 , zq + cq ), where 1 2 fp,p+1 (zp + cp , zp+1 + cp+1 ) is an idempotent. Divisibility on the right by fp,p+1(zp + cp , zp+1 + cp+1 ) means the addition of the idempotent has no effect on the value of the product Ci or Ri . By Lemma 2.2, the above triples are regular at z1 = z2 = . . . = zn , and therefore, so too are the products Ci and Ri , for all 1 6 i 6 d. Moreover, this means Fλ (z1 , . . . , zn ) is regular at z1 = z2 = . . . = zn . Example 2.4. As an example consider the hook tableau of the Young diagram λ = (3, 3, 2), as shown in Example 2.1. In the original lexicographic ordering the product Fλ (z1 , . . . , zn ) is written as follows, for sim- plicity we will write fpq in place of the term fpq (zp + cp , zq + cq ). Fλ (z1 , . . . , zn ) = f12 f13 f14 f15 f16 f17 f18 f23 f24 f25 f26 f27 f28 f34 f35 f36 f37 f38 f45 f46 f47 f48 f56 f57 f58 f67 f68 f78 We may now reorder this product into the form below using relations (3) and (4) as described in the above proposition. The terms bracketed are the singularity terms with their appropriate triple terms. Fλ (z1 , . . . , zn ) = f12 f13 f23 f14 f24 f34 f15 f25 f35 (f16 f26 )f36 f17 (f27 f37 )f18 f28 f38 ·f45 f46 f56 f47 f57 (f48 f58 ) · f67 f68 f78 And so, for each singularity fpq , we can replace fpq fp+1,q in the product by the triple 12 fp,p+1fpq fp+1,q , where 21 fp,p+1 is an idempotent, without changing the value of Fλ (z1 , . . . , zn ). Fλ (z1 , . . . , zn ) = f12 f13 f23 f14 f24 f34 f15 f25 f35 ( 12 f12 f16 f26 )f36 f17 ( 12 f23 f27 f37 )f18 f28 f38 ·f45 f46 f56 f47 f57 ( 12 f45 f48 f58 ) · f67 f68 f78 the electronic journal of combinatorics 12 (2005), #R00 9 And since each of these triples are regular at z1 = z2 = . . . = zn then so too is the whole of Fλ (z1 , . . . , zn ). Therefore, due to the above proposition an element Fλ ∈ CSn can now be defined as the value of Fλ (z1 , . . . , zn ) at z1 = z2 = . . . = zn . We proceed by showing this Fλ is indeed the diagonal matrix element. To this end we will need the following propositions. Note that for n = 1, we have Fλ = 1. For any n > 1, we have the following fact. Proposition 2.5 The coefficient of Fλ ∈ CSn at the unit element of Sn is 1. Proof. For each r = 1, . . . , n−1 let sr ∈ Sn be the adjacent transposition (r, r +1). Let w0 ∈ Sn be the element of maximal length. Multiply the ordered product (2) by the element w0 on the right. Using the reduced decomposition → Y w0 = sq−p (9) (p,q) we get the product → Y 1 sq−p − . zp − zq + cp − cq (p,q) For the derivation of this formula see [GP, (2.4)]. This formula expands as a sum of the elements s ∈ Sn with coefficients from the field of rational functions of z1 , . . . , zn valued in C. Since the decomposition (9) is reduced, the coefficient at w0 is 1. By the definition of Fλ , this implies that the coefficient of Fλ w0 ∈ CSn at w0 is also 1. In particular this shows that Fλ 6= 0 for any nonempty diagram λ. Let us now denote by ϕ the involutive antiautomorphism of the group ring CSn defined by ϕ(g) = g−1 for every g ∈ Sn . Proposition 2.6 The element Fλ ∈ CSn is ϕ-invariant. Proof. Any element of the group ring CSn of the form fpq (u, v) is ϕ-invariant. Applying the antiautomorphism ϕ to an element of the form (2) just reverses the ordering of the factors the electronic journal of combinatorics 12 (2005), #R00 10 corresponding to the pairs (p,q). However, the initial ordering can then be restored using relations (3) and (4), for more detail see [GP]. Therefore, any value of the function Fλ (z1 , . . . , zn ) is ϕ-invariant. Therefore, so too is the element Fλ . Proposition 2.7 Let x be last entry in the kth row of the hook tableau of shape λ. Denote by σx the embedding CSn−x → CSn defined by σx : (pq) 7→ (p + x, q + x) for all distinct p, q = 1, . . . , n − x. If λ = (α1 , α2 , . . . , αd |β1 , β2 , . . . , βd ) and µ = (αk+1 , αk+2 , . . . , αd |βk+1 , βk+2 , . . . , βd ), then Fλ = P · σx (Fµ ), for some element P ∈ CSn . Proof. Here the shape µ is obtained by removing the first k principal hooks of λ. By the ordering described in Proposition 2.3, k Y Fλ (z1 , . . . , zn ) = Ci Ri · σx (Fµ (zx+1 , . . . , zn )), i=1 where Ci and Ri are defined by (7) and (8). Since all products Ci and Ri are regular at z1 = z2 = . . . = zn , Proposition 2.3 then gives us the required statement. In any given ordering of Fλ (z1 , . . . , zn ), we want a singularity term to be placed next to an appropriate triple term such that we may then form a regular triple. In that case we will say these two terms are ’tied’. We now complete the proof of Theorem 1.2. If u, v stand next to each other in the same row, or same column, of Λ the following results show that Fλ is divisible on the left (and on the right) by 1 − (uv), or 1 + (uv) respectively, or divisible by these terms preceded (followed) by some invertible elements of CSn . Hence Fλ is the diagonal matrix element. However, proving the divisibilities described requires some pairs to be ’untied’, in which case we must form a new ordering. This is the content of the following proofs. Some explicit examples will then given in Example 2.10 below. Proposition 2.8 Suppose the numbers u < v stand next to each other in the same column of the hook tableau Λ of shape λ. First, let s be the last entry in the row containing u. If cv < 0 the electronic journal of combinatorics 12 (2005), #R00 11 then the element Fλ ∈ CSn is divisible on the left and on the right by fu,v (cu , cv ) = 1 − (uv). If cv > 0 then the element Fλ ∈ CSn is divisible on the left by the product   Y← Y→  fij (ci , cj ) i=u,...,s j=s+1,...,v Proof. By Proposition 2.6, the divisibility of Fλ by the element 1 − (uv) on the left is equivalent to the divisibility by the same element on the right. Let us prove divisibility on the left. By Proposition 2.7, if σx (Fµ ) is divisible on the right by fuv (cu , cv ), or fuv (cu , cv ) followed by some invertible terms, then so too will Fλ . If σx (Fµ ) is divisible on the left by fuv (cu , cv ), or fuv (cu , cv ) preceded by some invertible terms, then, by using Proposition 2.6 twice, first on σx (Fµ ) then on the product Fλ = P · σx (Fµ ), so too will Fλ . Hence we only need to prove the statement for (u, v) such that u is in the first row or first column of Λ. Let r be the last entry in the first column of Λ, s the last entry in the first row of Λ, and t the last entry in the second column of Λ, as shown in Figure 2. 1 r + 1r + 2 ... u ... ... s 2 s + 1t + 1 ... v ... .. .. . . .. t . .. . r Figure 2: The first two principal hooks of the hook tableau Λ We now continue this proof by considering three cases and showing the appropriate divisibility in each. (i) If cv < 0 (i.e. u and v are in the first column of Λ) then v = u + 1 and Fλ (z1 , . . . , zn ) can be written as Fλ (z1 , . . . , zn ) = fuv (zu + cu , zv + cv ) · F . the electronic journal of combinatorics 12 (2005), #R00 12 Starting with Fλ (z1 , . . . , zn ) written in the ordering described in Proposition 2.3 and simply moving the term fu,v (zu + cu , zv + cv ) to the left results in all the singularity terms in the product F remaining tied to the same triple terms as originally described in that ordering. Therefore we may still form regular triples for each singularity in F , and hence F is regular at z1 = z2 = . . . = zn . So by considering this expression for Fλ (z1 , . . . , zn ) at z1 = z2 = . . . = zn we see that Fλ will be divisible on the left/right by fuv (cu , cv ) = (1 − (uv)). (ii) If cv = 0 then v = s + 1, and Fλ (z1 , . . . , zn ) can be written as ← Y Fλ (z1 , . . . , zn ) = fi,s+1 (zi + ci , zs+1 + cs+1 ) · F ′ . i=u,...,s Again, starting with the ordering described in Proposition 2.3, this results in all the singularity terms in the product F ′ remaining tied to the same triple terms as originally described in that ordering. Hence F ′ is regular at z1 = z2 = . . . = zn . And so Fλ is divisible on the left by ← Y fi,s+1 (ci , cs+1 ). i=u,...,s (iii) If cv > 0 (i.e. v is above the steps) then fuv (zu + cu , zv + cv ) is tied to the singularity fu−1,v (zu−1 + cu−1 , zv + cv ) as a triple term. To show divisibility by fuv (zu + cu , zv + cv ) in this case we need an alternative expression for Fλ (z1 , . . . , zn ) that is regular when z1 = z2 = . . . = zn . Define a permutation τ as follows,   → Y ← Y τ=  (ij) i=u,...,s j=s+1,...,v   1 2 ... u − 1 u u + 1 ... ... ... ... v − 1 v v + 1 ... n =  1 2 ... u − 1 s + 1 s + 2 ... v − 1 v u u + 1 ... s − 1 s v + 1 ... n From the definition of C1 in (7) we now define C1′ = τ C1 , where τ acts on the indices of the product C1 such that n r ! Y Y τ C1 = gi,τ ·j . j=1 i=1 For the rest of this proof we will simply write fij instead of fij (zi + ci , zj + cj ). Define R1′ as,        Y← Y→ Y→ Y→ Y← → Y R1′ =  fij  ·  fij  ·  fr+1,j   fr+1,j  i=r+2,...,u−1 j=s+1,...,v i=s+1,...,t−1 j=i+1,...,t j=s+1,...,t j=t+1,...,v the electronic journal of combinatorics 12 (2005), #R00 13     → Y → Y → Y → Y ×  fij  ·  fij  . i=r+1,...,s−1 j=i+1,...,s j=v+1,...,n i=r+1,...,s Finally, define C2′ as,   → Y → Y C2′ =  fij  . j=t+1,...,n i=s+1,...,t Then,   ← Y → Y d Y Fλ (z1 , . . . , zn ) =  fij  · C1′ R1′ C2′ R2 · Ci Ri , i=u,...,s j=s+1,...,v i=3 where d is the number of principal hooks of λ. The product C1′ R1′ C2′ R2 · Q Ci Ri is regular at z1 = z2 = . . . zn since, as before, for any singularity (p, q) the terms fpq fp+1,q can be replaced by the triple 12 fp,p+1 fpq fp+1,q – except in the expression R1′ where the terms fpl fpq are replaced by 12 fpl fpq flq , where l is the entry to the immediate left of q. Note that l = q − 1 when cq > 1 and l = s + 1 when cq = 1. And so by letting z1 = z2 = . . . = zn we see that Fλ is divisible on the left by   Y← Y→  fij (ci , cj ) . i=u,...,s j=s+1,...,v Proposition 2.9 Suppose the numbers u < v stand next to each other in the same row of the hook tableau Λ of shape λ. Let r be the last entry in the column containing u. If cu > 0 then the element Fλ ∈ CSn is divisible on the left and on the right by fu,v (cu , cv ) = 1 + (uv). If cu 6 0 then the element Fλ ∈ CSn is divisible on the left by the product   Y← Y→  fij (ci , cj ) i=u,...,r j=r+1,...,v We omit the proof of this proposition as it is very similar to that of Proposition 2.8. This completes our proof of Theorem 1.2. Let us now consider an example that allows us to see how the product Fλ (z1 , . . . , zn ) is broken down in the proof of Proposition 2.8. the electronic journal of combinatorics 12 (2005), #R00 14 Example 2.10. We again consider the hook tableau of the Young diagram λ = (3, 3, 2), as shown in Example 2.1. We begin with the product Fλ (z1 , . . . , zn ) in the ordering described in Proposition 2.3. For simplicity we again write fpq in place of the term fpq (zp + cp , zq + cq ). We have also marked out the singularities in this expansion along with their triple terms, but no idempotents have yet been added which would form regular triples. Fλ (z1 , . . . , zn ) = f12 f13 f23 f14 f24 f34 f15 f25 f35 (f16 f26 )f36 f17 (f27 f37 )f18 f28 f38 (10) ·f45 f46 f56 f47 f57 (f48 f58 ) · f67 f68 f78 Let u = 4 and v = 6 in Proposition 2.8. Then by that proposition we may arrange the above product as follows. Notice since cv = 0 all singularity-triple term pairs remain the same. Fλ (z1 , . . . , zn ) = f56 f46 · f12 f13 f23 (f16 f26 )f36 f14 f24 f34 f15 f25 f35 f17 (f27 f37 ) f18 f28 f38 · f45 f47 f57 (f48 f58 ) · f67 f68 f78 We may now add the appropriate idempotents so that all singularities remain in regular triples. And so by considering the product at z1 = z2 = . . . = zn we have that Fλ is divisible on the left by (1 − (46)), preceded only by invertible terms, as desired. Now let u = 5 and v = 8. As described by Proposition 2.8 (iii) we may arrange (10) in the following way. Singularities in R1 have been marked out with their alternative triple terms, while all other singularity-triple term pairs remain the same. Fλ (z1 , . . . , zn ) = f56 f57 f58 · f12 f13 f23 f14 f24 f34 (f16 f26 )f36 f17 (f27 f37 )f18 f28 f38 f15 f25 f35 · f67 f47 (f46 f48 )f45 · f68 f78 In moving f58 to the left it is untied from the singularity f48 . So we must form new triples which are regular at z1 = z2 = . . . = zn . Therefore, by considering the product at z1 = z2 = . . . = zn , we have that Fλ is divisible on the left by (1 − (58)), again preceded only by invertible terms, as desired. the electronic journal of combinatorics 12 (2005), #R00 15 References [C] I. V. Cherednik, On special bases of irreducible finite-dimensional representations of the degenerate affine Hecke algebra, Funct. Anal. Appl. 20 (1986) 87-89. [CR] C. Curtis and I. Reiner, Representation theory of finite groups and associative algebras, Interscience Publishers, 1962. [ER] Ö. N. Eǧecioǧlu and J.B. Remmel, A combinatorial proof of the Giambelli Identity for Schur functions, Adv. in Math. 70 (1988), no. 1, 59-86. [G] G. Z. Giambelli, Alcuna propreita delle funzioni simmetriche charatteristiche, Atti Accad. Sci. Torino Cl. Sci. Fis. Mat. Natur. 38 (1903), 323-344. [GP] V. Guizi and P. Papi, A combinatorial approach to the fusion process for the symmetric group, European Journal of Combinatorics, 19 (1998), 835-845. [J] A. Jucys, On the Young operators of the symmetric group, Lietuvos Fizikos Rinkinys. 6 (1966), 163-180. [JK] G. James and A. Kerber, The representation theory of the symmetric group, Encyclopedia of mathematics and its applications 16 (1981). [M] I. G. Macdonald, Symmetric functions and Hall polynomials, Oxford Univ. Press, Lon- don/New York, 1979. [N1] M. Nazarov, Young’s symmetrizers for projective representations of the symmetric group, Adv. Math. 127 (1997), no. 2, 190–257. [N2] M. Nazarov, Yangians and Capelli Identities, Amer. Math. Soc Translations 181 (1998), 139-163. [N3] M. Nazarov, Representations of twisted Yangians associated with skew Young diagrams, Selecta Math. 10 (2004), 71-129. [NT] M. Nazarov and V. Tarasov, On irreducibility of tensor products of Yangian modules as- sociated with skew Young diagrams., Duke Math Jour. 112 (2002), 343-378. [OV] A. Okounkov and A. Vershik, A new approach to representation theory of symmetric groups, Selecta Math, 2 (1996), no. 4, 581–605. the electronic journal of combinatorics 12 (2005), #R00 16