YOUNG SEMINORMAL REPRESENTATION MURPHY ELEMENTS AND CONTENT EVALUATIONS

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1 A. Garsia Lecture Notes in Algebraic Combintorics March 7, YOUNG SEMINORMAL REPRESENTATION MURPHY ELEMENTS AND CONTENT EVALUATIONS ABSTRACT These are notes resulting from a course in Algebraic Combinatorics given at UCSD in winter The lectures where based on writings of A. Young [8], R. M. Thrall [7] and D. E. Rutherford [6], A. Jucy [3] and G, Murphy [5]. The material covers basic identities leading to the construction of Young s seminormal units. Murphy elements are used first as an aid to the construction of the entries in the seminormal matrices corresponding to the simple reflections then used to express characters and conjugacy classes. As a by-product we obtain a new way of constructing certain polynomials introduced in a recent paper by A. Goupil et Al. [2] and by Diaconis and Greene in an earlier unpublished manuscript [1]. These polynomials yield some truly remarkable formulas for the irreducible characters of the symmetric groups. More precisely, it is shown in [1] and [2] that for each partition γ k we can construct a symmetric polynomial W γ which evaluated at the contents of a partition λ n k yield the central character value ω λ γ,1 n k. The polynomials W γ are remarkably simple when expressed in terms of the power basis. Moreover, their power basis expansion are of the form W γ = ρ cγ ρ(n) p ρ with coefficients c γ ρ(n) polynomials in n. The approach followed in [2] is quite intricate and somewhat indirect. In [1] Diaconis and Greene follow a purely combinatorial approach and derive an algorithm for constructing the polynomials W γ.in these notes we obtain explicit closed form expressions for W γ. Our approach is based on a method introduced by Macdonald [4] in an exercise where an explicit formula is obtained for the central character value ω λ k,1 n k. Table of Contents 1.Young idempotents. In this section we introduce the Young last letter order and establish some of the basic properties of Young idempotents under this order. 2.Young s seminormal units. We construct here the Young seminormal units and prove their orthogonality. We also construct the seminormal matrix units and compute the characters of the corresponding representations. We terminate by proving seminormality. 3.The Murphy elements. In this section we introduce the Murphy elements and prove their commutativity. We derive the action of the class C 2 of transposition on a seminormal tableau unit. We thus obtain in a purely combinatorial way that the central character χ λ 21 n 2 C 2 /n λ / is n(λ ) n(λ). The basic result here is that the seminormal units are simultaneous eigenfunctions of the Murphy elements. We show that the eigenvalue of m k on the tableau unit e(t ) is given by the content of the cell of k in T.Wealso obtain in a canonical way a polynomial P T (m 2,...,m n ) which gives e(t ).

2 A. Garsia Lecture Notes in Algebraic Combintorics March 7, The seminormal matrices. In this section we derive the entries in Young s seminormal matrices from the action of Murphy elements on the seminormal matrix units. 5.Murphy elements and conjugacy classes. We start by giving two different proofs that symmetric polynomials evaluated at the Murphy elements yield class functions. This done our main goal is to obtain explicit formulas for the symmetric polynomials that yield the Characters and the Classes. Tablesof these polynomials are included in a few special cases. 1. The Young idempotents In this section we recall some basic properties of the Young idempotents and set notation. Most of the material covered in this section is presented in full detail in the lecture notes on the Young s Natural Representation.References to these notes will be indicated by the symbol [YNR]. We shall use the French convention of depicting Ferrers diagrams as left justified rows of lattice cells of lengths decreasing from bottom to top. Recall that a tableau T of shape λ n is a filling of the cells of the Ferrers diagram of λ with the letters 1, 2,...,n.Ifthe filling is row and column increasing then T is said to be standard. The shape of T will be simply denoted λ(t ). Given a tableau T,welet R(T ) and C(T ) respectively denote the Row Group and Column Group oft.asin[ynr] we set P (T ) = α, N(T ) = sign(β) β, 1.1 α R(T ) β C(T ) and E(T ) = P (T )N(T ). 1.2 Note that if T has n cells and σ S n then σt denotes the tableau obtained by replacing in T the index i by σ i for i =1,...,n.Itiseasily seen that we have R(σT) = σr(t ) σ 1, C(σT) = σc(t ) σ 1 thus P (σt) = σp(t ) σ 1, N(σT) = σn(t ) σ In particular if T 1 and T 2 are tableaux of the same shape and σ T1,T 2 is the permutation that sends T 2 into T 1 we have the identities σ T1,T 2 P (T 2 ) = P (T 1 )σ T1,T 2, σ T1,T 2 N(T 2 ) = N(T 1 )σ T1,T 2, σ T1,T 2 E(T 2 ) = E(T 1 )σ T1,T We have the following basic fact

3 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Proposition 1.1 If T 1 and T 2 are two tableaux with n cells, then we have N(T 2 )P (T 1 ) 0 or P (T 1 )N(T 2 ) only if In particular λ(t 2 ) λ(t 1 ) (in dominance) 1.5 λ(t 2 ) <λ(t 1 ) = N(T 2 )P (T 1 )=P (T 1 )N(T 2 )=0 Construct the diagram T 1 T 2 by placing in the lattice cell (i, j) the intersection of row i of T 1 with column j of T 2. Note that if one of these intersections contains two elements r and s then R(T 1 ) and C(T 2 ) would have the transposition (r, s) in common, but then we would have, for instance N(T 2 )P (T 1 ) = N(T 2 )(r, s)p (T 1 ) = N(T 2 )P (T 1 ) in plain contraddiction with 1.4. Thus 1.4 forces the cells of T 1 T 2 to contain at most one element. Note that if we lower these elements down their columns by eliminating the empty cells we will obtain a tableau T 3 of shape λ(t 2 ), thus the number of cells in the first i rows of T 3 is given by λ 1 (T 2 )+λ 2 (T 2 )+ + λ i (T 2 ). But all the elements of the first i rows of T 1 are in the first i rows of T 1 T 2 and a fortiori must all be in the first i rows of T 3. This gives the inequality λ 1 (T 2 )+λ 2 (T 2 )+ + λ i (T 2 ) λ 1 (T 1 )+λ 2 (T 1 )+ + λ i (T 1 ). Since this must hold true for all i we necessarily have 1.5 as asserted. It will be convenient to denote the group algebra of S n by A[S n ]. This given, Proposition 1.1 implies the following fundamental fact Theorem 1.1 If the tableaux T 1,T 2 have both n cells, then for any element f A[S n ] we have In particular P (T 1 )fn(t 2 ) 0 or N(T 2 )fp(t 1 ) 0 = λ(t 2 ) λ(t 1 ) 1.6 λ(t 1 ) λ(t 2 ) = E(T 1 )fe(t 2 ) = 0 (for all f A[S n ]) Note that we may write (using 1.3) P (T 1 )fn(t 2 ) = σ S n f(σ) P (T 1 )σn(t 2 ) = σ S n f(σ) P (T 1 )N(σT 2 )σ 1,

4 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Thus the non vanishing of the left hand side forces the non vanishing at least one term in the sum in the right hand side. But then 1.6 follows from 1.5. Of course the same will hold true if N(T 2 )fp(t 1 ) 0. But now note that P (T 1 )N(T 1 )fp(t 2 )N(T 2 ) forces N(T 1 )fp(t 2 ) 0yielding λ(t 1 ) λ(t 2 ). On the other hand, ( again by 1.6), 1.8 itself forces λ(t 2 ) λ(t 1 ). Thus only the equality λ(t 1 )=λ(t 2 ) is compatible with the non vanishing of E(T 1 )fe(t 2 ). This proves 1.7 and completes the proof. We have the following fundamental identity whose proof may be found in [YNR]. Proposition 1.2 (Von Neuman Sandwich Lemma) Forany element f A(S n ) and any tableau T P (T ) fn(t ) = c T (f) P (T )N(T ), 1.9 with c T (f) = fn(t )P (T ), where denotes the identity permutation. This result has the following important corollary. Theorem 1.2 Forany tableau T, the group algebra element E(T ) is idempotent. More precisely there is a non-vanishing constant h(t ) depending only on the shape of T such that E(T )E(T ) = h(t ) E(T ) 1.10 Using 1.9 with f = N(T )P (T ), the identity in 1.9 gives E(T )E(T ) = h(t )E(T ). with h(t ) = N(T )P (T )N(T )P (T ) Note that if h(t ) were to vanish then E(T ) would be nilpotent. Now a cute argument shows that that the coefficient of the identity of any nilpotent element of a group algebra necessarily vanishes. Now this immediately leads to a contraddiction since we can easily see that E(T ) = sign(β) αβ = 1. α R(T ) β C(T )

5 A. Garsia Lecture Notes in Algebraic Combintorics March 7, In fact the identity αβ = holds only when α = β =. Weshall show later that if T has shape λ then h ( T )= n! n λ where n λ denotes the number of standard tableaux of shape λ. But for the moment it will suffice to show that h(t ) depends only on λ(t ). Now this follows immediately from the identities h(t )=N(T)P(T)N(T)P(T) = 1 σn(t )P (T )N(T )P (T ) σ 1 n! σ S n = 1 N(σT)P (σt)n(σt)p(σt) n! = 1 n! σ S n λ(t )=λ(t ) N(T )P (T )N(T )P (T ) This given, here and after we will use the symbol todenote h(t ) for any tableau T of shape λ. In the construction of Young s Seminormal Representation we will make systematic use of Young s last letter order. Given two standard tableaux T 1,T 2 of the same shape we shall say that k is the last letter of disagreement between T 1 and T 2 if the letters k +1,k +2,...,n occupy exatly the same positions in T 1 and T 2, but k does not. This given we shall say that T 1 precedes T 2 in the Young last letter order and write if and only if the last letter of disagrement is higher in T 1 than in T 2. T 1 < LL T If T is a standard tableau we shall denote by T (k) the tableau obtained by removing from T the letters larger than k together with their cells. Note that if λ(t 1 )=λ(t 2 ) and k +1is the last letter of disagreement between T 1 and T 2 then λ ( T 1 (k +1) ) = λ ( T 2 (k + 1)) and 1.11 holds true if and only if λ(t 1 (k)) >λ(t 2 (k)) (in dominance) 1.12 For f A(S n ) let us set f = f(σ) σ σ S n We shall refer to this operation as flipping f There are interesting identities connected with the operation of passing from T to T (k), they may be stated as follows Proposition 1.3 Forany tableau T of shape λ n and k n we have two elements p k (T ) and n k (T ) such that a) p k (T )P (T (k)) = P (T ) = P (T (k)) p k (T ), 1.13 b) n k (T )N(T (k)) = N(T ) = N(T (k)) n k (T ).

6 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Note that the second equality in 1.13 a) follows by flipping both sides of the first equality. The same holds true for 1.13 b). So in each case we need only show the first equality. Moreover since we can pass from T to T (k) by successive removals of the largest letter, we can see that we need only show our equalities in the case k = n 1. Tothis end let us suppose that the row of T that contains the letter n consists of the letters a 1,a 2,...,a r,n. In this case denoting by (a 1,a 2,...,a r,n) and (a 1,a 2,...,a r ) the formal sums of all the permutations of S n that leave fixed the elements of the complements of {a 1,a 2,...,a r,n} and {a 1,a 2,...,a r } respectively, we have, by left coset decomposition (a 1,a 2,...,a r,n) = ( +(a 1,n)+(a 2,n)+ +(a r,n) ) (a 1,a 2,...,a r ). Multiplying this identity by the contributions to P (T ) coming from the other rows of T yields P (T ) = ( +(a 1,n)+(a 2,n)+ +(a r,n) ) P (T (n 1)) which is precisely the first equality in 1.13 a) for k = n 1. For 1.13 b) we can proceed in a similar way. Indeed let us suppose that the column of T that contains the letter n consists of the letters b 1,b 2,...,b s,n. Denoting by (b 1,b 2,...,b s,n) and (b 1,b 2,...,a s ) the formal sums of all the signed permutations of S n that leave fixed the elements of the complements of {b 1,b 2,...,b s,n} and {b 1,b 2,...,b s } respectively, we have, by left coset decomposition (b 1,b 2,...,b s,n) = ( (b 1,n) (b 2,n) (b s,n) ) (b 1,b 2,...,b r ). Multiplying this identity by the contributions to N(T ) coming from the other columns of T yields N(T ) = ( (b 1,n) (b 2,n) (b r,n) ) N(T (n 1)) which is precisely the first equality in 1.13 b) for k = n 1. This completes our proof. The following result will play a crucial role here. Theorem 1.3 For two standard tableau T 1 and T 2 of the same shape we have N(T 1 )P (T 2 ) 0 = T 1 < LL T

7 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Let k +1be the last letter of disagrement between T 1 and T 2. Using Proposition 1.3 we derive the factorization N(T 1 )P (T 2 ) = n k (T 1 ) N(T 1 (k))p (T 2 (k)) p k (T 2 ) Thus N(T 1 )P (T 2 ) 0 = N(T 1 (k))p (T 2 (k)) 0 and Proposition 1.1 gives λ((t 1 (k)) >λ(t 2 (k)) However we have seen that this holds true if and only if This completes our argument. T 1 < LL T Young s seminormal units We have shown in the Representation Theory notes (here and after referred to by the symbol [RT]) that the group algebra A(G) of a finite group G has a basis { {e λ ij } n } λ i,j=1 2.1 λ Λ with the property that 0 if λ µ, e λ ije µ rs = 0 if λ = µ,but j r 2.2 e µ is if λ = µ and j = r. Moreover, we have also derived the identities 0 if i j, e λ ij = (with = G /n λ ) 2.3 1/ if i = j. From these identities it follows that we have the expansions f = λ Λ n λ i,j=1 fe λ ji e λ ij, (for all f A(G)). 2.4 This basis allows us to construct a complete set of representatives of irreducible representations of G. These are simply given by the collection {A λ } λ Λ obtained by setting A λ (σ) = a λ ij(σ)) n λ i,j=1 2.5 with a λ ij(σ) = e λ ji σ 1 ( for all σ G) 2.6

8 A. Garsia Lecture Notes in Algebraic Combintorics March 7, In fact, from 2.4 and 2.6 we derive the expansion σ = λ Λ and the relations in 2.3 immediately yield that n λ i,j=1 a λ ij(σ) e λ ij. 2.7 σe µ r,s = n µ e µ i,s aµ i,r (σ) 2.8 or in matrix form σ e µ i,s = e µ 1 i n µ i,s A µ (σ) i n µ Now we see from 2.6 that the representation A λ is orthogonal (that is a ij (σ 1 )=a ji (σ))ifand only if we have e λ ji(σ) = e λ ij(σ 1 ) ( for all σ G) 2.10 and this, in compact form may be simply be rewritten as e λ ij = e λ ji 2.11 When Young set himself the task of finding a complete set of irreducible orthogonal representations of S n, his point of departure (see [8] QSA VI) was the construction of units e λ ij satisfying conditions 2.2 and 2.3 together with the additional condition e λ ij = dλ i d λ j e λ ji These constructs have come to be referred to as Young s seminormal units. It is easily seen that setting fij λ d λ j = d λ e λ ij 2.13 i the orthogonality condition in 2.11 will be satisfied by the fij λ and the desired orthogonal representations can then be readily obtained. Surprisingly Young s seminormal units can be written down with a minimal amount of additional notation from the material which is already in our possession. To begin, we let T λ 1,T λ 2,...,T λ n λ denote the standard tableaux of shape λ in Young s last letter order. Moreover the permutation that sends T λ j onto T λ i will be denoted σ λ ij.insymbols σ λ ij T λ j = T λ i 2.14

9 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Finally, for a tableau T with n cells, it will be convenient, to use the abreviations T (n 1) = T, T(n 2) = T, T(n 3) = T This given, Young s seminormal units are simply given by the formulas e λ ij = e(t λ i ) P (T λ i ) σλ ij N(T λ j ) e(t λ j ), 2.15 where for a given standard tableau T of shape λ the group algebra element e(t ) is recursively defined by setting if T = [1], e(t ) = 2.16 P (T )N(T ) e(t ) e(t ) otherwise. The remainder of this section is dedicated to proving that these units satisfy the identities in 2.2, 2.3 and But before we can carry this out we need to derive a few identities. Our basic tool is provided by the following Proposition 2.1 For any standard tableau T we have a) E(T )e(t )E(T ) = h(t ) E(T ) b) E(T )e(t )E(T ) = h(t ) E(T ) c) e(t )E(T )e(t ) = h(t ) e(t ) 2.17 where we should set h(t )= when λ(t )=λ. Lemma 2.1 To prove this we need two auxiliary identities which are of independent interest. a) e(t )P (T )N(T ) = 1, b) e(t )P (T )N(T ) 2.18 = 1. We proceed by induction on the size of T.Soassume that 2.18 a) is true for any tableau with n 1 cells and let T be a tableau with n cells. Note that the induction hypothesis then implies that e(t )P (T )N(T ) = Now let the elements of T that are in the same row and column as n be given as in the proof of Proposition 1.3. We may then write σp(t )N(T ) = P (T ) ( +(a 1,n)+ +(a r,n) ) ( (b 1,n) (b s,n) ) N(T ) σ Note that for any pair i, j, the product of the two cycles (a i,n)(b j,n) is the 3-cycle (a i,n,b j ) which is not in S n 1, and since all of the permutations in P (T ) or N(T ) are in S n 1, the only terms in 2.20

10 A. Garsia Lecture Notes in Algebraic Combintorics March 7, that will yield permutations in S n 1 are those produced by the identity term. Thus we necessarily have σp(t )N(T ) = P (T )N(T ) σ 1 = σp(t )N(T ). (for all σ S n 1 ). Multiplying this relation by e(t ) σ and summing over σ S n 1 we get e(t ) P (T )N(T ) = e(t ) P (T )N(T ). and 2.19 gives e(t ) P (T )N(T ) = 1, proving 2.18 b). Now 2.16 gives h(t )e(t )P (T )N(T ) = e(t )P (T )N(T )e(t )P (T )N(T ) (by Prop. 1.2) = e(t ) (N(T )e(t )P (T )N(T )P (T ) ) P (T )N(T ) (by 2.18 b) = (e(t )P (T )N(T )P (T )N(T ) ) (by 1.10) = h(t ) e(t )P (T )N(T ) (by 2.18 b) = h(t ). This proves e(t )P (T )N(T ) = 1. and completes the induction. of Proposition 2.1 To begin we have This proves 2.17 a). Next we have This proves 2.17 b). E(T )e(t )E(T ) = P (T )N(T )e(t )P (T )N(T ) (byprop. 1.2) = N(T )e(t )P (T )N(T )P (T ) P (T )N(T ) = e(t )P (T )N(T )P (T )N(T ) P (T )N(T ) (by1.10) = h(t )e(t )P (T )N(T ) P (T )N(T ) (by2.18 b)) = h(t ) P (T )N(T ) = h(t )E(T ). E(T )e(t )E(T ) = P (T )N(T )e(t )P (T )N(T ) (byprop. 1.2) = N(T )e(t )P (T )N(T )P (T ) P (T )N(T ) = e(t )P (T )N(T )P (T )N(T ) P (T )N(T ) (by1.10) = h(t )e(t )P (T )N(T ) P (T )N(T ) (by2.18 a)) = h(t ) P (T )N(T )=h(t)e(t).

11 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Finally from the definition in 2.16 we get This proves 2.17 c). e(t )E(T )e(t ) = e(t ) E(T ) e(t )E(T )e(t ) h(t ) (by2.17 a)) = e(t )E(T )e(t ) (by2.16) = e(t )E(T )e(t ) E(T ) h(t ) e(t ) (by2.17 a)) = e(t )E(T )e(t ) (by2.16) = h(t ) e(t ). We are now ready to establish the basic properties of Young s seminormal units. We begin with a result which is an immediate consequence of the definition. Theorem 2.1 For any standard tableau T the group algebra element e(t ) is idempotent. For T = [1] this is true by definition, we can thus proceed by induction on the number of cells of T. Now we have e(t )e(t ) = e(t ) E(T ) E(T ) e(t ) e(t ) h(t ) h(t ) e(t ) (by induction) = e(t ) E(T ) E(T ) e(t ) h(t ) h(t ) e(t ) (by 2.17 a)) = e(t ) E(T ) h(t ) e(t ) = e(t ). Q.E.D. These idempotents are orthogonal, more precisely we have Theorem 2.2 If T 1 and T 2 are standard tableaux with n cells then e(t 1 ) e(t 2 ) = 0 if λ(t 1 ) λ(t 2 ) 0 if λ(t 1 )=λ(t 2 ) but T 1 T e(t 1 ) if T 1 = T 2 Since by definition e(t 1 ) e(t 2 ) = e(t 1 ) E(T 1) h(t 1 ) e(t 1) e(t 2 ) E(T 2) h(t 2 ) e(t 2), 2.21 the first equality is an immediate consequence of 1.7. The last equality is a restatement of Theorem 2.1. We are left to prove the second equality. Since for n =1, 2 there is nothing to prove, we shall

12 A. Garsia Lecture Notes in Algebraic Combintorics March 7, proceed by induction on n and assume its validity for n 1. This given we have the following sequence of implications e(t 1 ) e(t 2 ) 0 (by 2.21) e(t 1 ) e(t 2 ) 0 (by induction) T 1 = T 2 (when λ(t 1 )=λ(t 2 )) T 1 = T 2 This proves our assertion. Theorem 2.3 Young s seminormal units statisfy the identities e λ ije µ rs = 0 if λ µ, 0 if λ = µ,but j r e µ is if λ = µ and j = r From the definition in 2.15 we derive that a) e λ ij = e(t λ i ) σ λ ij E(T λ j ) e(t λ j ), b) e µ rs = e(t µ r ) E(T µ r ) h µ σ µ rs e(t µ s ) Thus the first equality in 2.1 is an immediate consequence of 1.7. In the case µ = λ, 2.33 a) and b) give e λ ij e λ rs = e(t λ i ) σij λ E(Tj λ) e(t λ j )e(t λ r ) E(T r µ ) σ h rse(t λ λ s ), 2.24 λ Now if r j, Theorem 2.2 gives e(t λ j )e(t λ r )=0,proving the second case of Finally for λ = µ and j = r, 2.24 becomes e λ ij e λ js = e(t λ i ) σij λ E(Tj λ) e(t λ j ) e(t λ j ) E(T µ j ) σjs λ e(t λ s ), (by Theorem 2.1) = e(t λ i ) σij λ E(Tj λ) e(t λ j ) E(T µ j ) σjs λ e(t λ s ), ( ) λ by 2.17 a) = e(t i ) σij λ E(Tj λ) σjs λ e(t λ s ), (by 1.4) = e(t λ i ) P (T i λ) σλ is N(T s λ ) e(t λ s )=e λ is This proves the last equality in 2.22 and completes our argument. Note that we also have

13 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Theorem 2.4 The definition in 2.15 gives e λ ij e λ ij = 0 if i j, 1/ if i = j. = e(t λ i ) P (T i λ) σλ ij N(T j λ) e(t λ j ) = e(t λ j ) e(t λ i ) P (T λ i ) σλ ij N(T λ j ) 2.26 and the first case of 2.26 follows from Theorem 2.2. But for i = j this becomes e λ ii = e(t λ i ) e(t λ i ) P (T i λ) N(T i λ) (by Theorem 2.1) = 1 e(t λ i ) P (Ti λ ) N(T λ i ) (by 2.18 b)) = 1 This completes our proof. From the identities in 2.22 and formula 2.9, it follows that, for λ n, the character of the representation resulting from the action of S n on a linear span L [ e λ 1s,e λ 2s,...,e λ ] n λ s is the same as the character of the action of S n on the left ideal A(S n )e(t λ s ). From this observation we derive the following important fact. Proposition 2.2 The character of the action of S n on L [ ] e λ 1s,e λ 2s,...,e λ n λ s depends only on λ and it is given by the formula χ λ = T St(λ) P (T )N(T ) where T St(λ) istoindicate that the sum is over all tableaux of shape λ. It is shown in the [RT] notes that if e is an idempotent of the group algebra A(G) then the character χ e of the left multiplication action of G on the ideal A(G)e is given by the formula χ e = σ e σ 1 σ G

14 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Taking T = T λ s, from this formula we obtain that χ λ = σe(t ) σ 1 = σe(t ) σ G σ S n (Circular rearrangements are OK) = σe(t )e(t ) σ S n and 2.18 b) gives (by 1.3) (by Theorem 2.1) = (by Von Neuman lemma) = σ S n σe(t ) = σ S n σ P (T )N(T ) P (T )N(T ) P (T )N(T ) N(T ) e(t ) P (T ) (e(t )P (T )N(T ) ) e(t ) σ 1 σ 1 σ 1 σ S n σ σ 1 N(T )P (T ) σ 1. χ λ = N(σT)P (σt). σ S n This proves the Theorem since σt describes all tableaux of shape λ, asσ varies in S n. Remark 2.1 It is easily derived from the identities in 2.22 that the set { } {e λ ij }n λ ij=1 is independent. λ n Furthermore, we proved in [YNR] that the numbers of standard tableaux n λ satisfy the identity n 2 λ = n!. λ n Thus { } {e λ ij }n λ ij=1 is an independent subset of A(S λ n n) of cardinality equal to the dimension of A(S n ). Soitmust be a basis. From this it is easy to derive that the expansion of any element f A(S n ) in terms of the Young s seminormal units is given by formula 2.4. We terminate this section with two important applications of this formula. We begin with the following beautiful identity of Alfred Young. Theorem 2.5 For any standard tableau T e(t ) = S χ(s = T ) e(s) 2.27 where the sum is over all standard tableaux S yielding T upon removal of n. Using formula 2.4 for f = e(t ) and Young s units we get, (if T has n cells) e(t ) = λ n n λ i,j=1 e(t ) e λ ji e λ ij 2.28

15 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Now we have e(t ) e λ ji = e(t )e(t λ j ) P (T j λ)σλ ji N(T i λ) e(t λ i ) hλ = e(t λ i )e(t )e(t λ j ) P (T j λ)σλ ji N(T i λ),. hλ and we immediately derive from Theorem 2.2 that this term doesn t vanish only if 2.29 T λ i = T = T λ j In particular this forces i = j, and 2.29 becomes e(t ) e λ ii = e(t λ i )e(t )e(t λ i ) P (T i λ)n(t i λ) Using 2.30 and 2.31 reduces 2.28 to (by Theorem 2.1) = e(t λ i ) P (T j λ)n(t i λ) (by 2.18 b)) = e(t ) = λ n λ Since e λ ii = e(t λ i ) P (T i λ)n(t i λ) e(t λ i ) = e(ti λ ) formula 2.32 is only another way of writing χ ( T λ i = T ) e λ ii A immediate corollary of Theorem 2.5 is the identification of the constant. Proposition 2.3 = n! n λ 2.33 Equating coefficients of the identity on both sides of 2.27, and using 2.26 we get 1 h(t ) = S χ(s = T ) 1 h(s). Assuming that T has n cells, and that λ(t )=µ we may rewrite this identity in the form 1 = h µ λ χ(µ λ) 1, 2.34 where the symbol µ λ is to express that λ is obtained by adding a cell to µ. Multiplying both sides of 2.33 by n! and setting k µ = (n 1)! h µ, k λ = n!

16 A. Garsia Lecture Notes in Algebraic Combintorics March 7, converts 2.34 into the identity nk µ = λ χ(µ λ) k λ. which is precisely the recursion statisfied by the number of standard tableaux. From this it easily follows that we must have k λ = n λ λ as desired. The seminormality of the Young units is based on the following remarkable fact. Theorem 2.6 For any standard tableaux T we have e(t ) = e(t ) Since 2.35 is obviously true for T = [1] we can proceed by induction on the number of cells of T. This given we have N(T )P (T ) e(t ) = e(t ) e(t ) h(t ) Using 2.27 this may be rewritten as e(t ) = N(T )P (T ) e(t 1 ) e(t 2 ) 2.36 h(t ) T 1=T T 2=T Since N(T )P (T ) e(t 1 ) e(t 2 ) = e(t 1 ) E(T 1) h(t ) h(t 1 ) e(t N(T )P (T ) 1) e(t 2 ) E(T 2) h(t ) h(t 2 ) e(t 2) from Theorem 1.1 we derive that T 1, T 2 and T must all have the same shape. But then the conditions T 1 = T = T 2 force them all to be the same, converting 2.36 to e(t ) = e(t ) E(T ) N(T )P (T ) e(t ) e(t ) E(T ) h(t ) h(t ) h(t ) e(t ) = P (T )N(T ) N(T )P (T ) P (T )N(T ) e(t ) e(t ) e(t ) e(t ) h(t ) h(t ) h(t ) (by 1.9 used twice) = ab e(t )P (T )N(T ) P (T )N(T )e(t ) h(t ) 3 (by 1.10) = ab ab e(t )P (T )N(T )e(t ) = h(t ) 2 h(t ) e(t ) 2.37 where a = N(T )e(t )N(T )P (T ) and b = e(t )P (T )N(T )P (T ). To avoid computing these constants we simply observe that e(t ) and e(t ) must have the same coefficient of the identity, and since this coefficient must be 1/h(T ) by 2.26, the identity in 2.37 forces ab = h(t )

17 A. Garsia Lecture Notes in Algebraic Combintorics March 7, and proves We are finally ready to prove seminormality. Theorem 2.7 For each partition λ we have constants d λ 1,d λ 2,...,d λ n λ 2.38 giving e λ ij = dλ i d λ j e λ ji (for all 1 i, j n λ ) Since 2.39 may also be written as e λ ij = dλ i /dλ 1 d λ j /dλ 1 e λ ji, there is no loss in assuming that d λ 1 = 1. This given, we need only show that for some constants d j we have e 1j = 1 d j e j1 (for 1 j n λ ), 2.40 where to lighten our notation we shall for a moment omit the superscript λ. Infact, 2.40 implies and then 2.22 gives e i1 = d i e 1i (for 1 i n λ ), e ij = (e i1 e 1j )=( e 1j )( e i1 )= d i d j e j1 e 1i = d i d j e ji, proving So let us then prove To this end we use the expansion formula 2.4 and get e 1j = n λ n λ r=1 s=1 (( e 1j )e sr ) e rs 2.41 However, note that Theorem 2.6 gives ( e 1j )e sr = e(t j ) N(T j)p (T j ) σ j1 e(t 1 )e(t s ) N(T s)p (T s ) σ sr e(t r ) = N(T j)p (T j ) σ j1 e(t 1 )e(t s ) N(T s)p (T s ) σ sr e(t r )e(t j )

18 A. Garsia Lecture Notes in Algebraic Combintorics March 7, and the non vanishing of the two products e(t 1 )e(t s ) and e(t r )e(t j ) forces s =1and r = j reducing 2.41 to e 1j = (( e 1j )e 1j ) e j1 and this proves 2.40 with 1 = ( e 1j )e 1j. d j This completes the proof and this section. The actual nature of the constants in 2.38 will come to the surface in the next section. 3. The Murphy elements Aremarkable set of group algebra elements was shown by Murphy [5] to play an elegant role in the study of representations of S n. It develops that these elements considerably simplify manipulations with Young seminormal units. Their definition is quite simple. We set m k =(1,k)+(2,k)+(3,k)+ +(k 1,k) (for k =2, 3,...,n) 3.1 These elements generate a commutative subalgebra of A(S n ).Infact we have the following basic relations. Theorem 3.1 Letting s h denote the transposition (h, h +1) a) m h m k = m k m h (for 2 h k n), b) s h m k s h = m k (for h k, k 1), c) s k m k s k = m k+1 s k. d) s k 1 m k s k 1 = m k 1 + s k. 3.2 Note first that in A(S 3 ) we have (1, 2) ( (1, 2)+(1, 3)+(2, 3) ) = ( (1, 2)+(1, 3)+(2, 3) ) (1, 2) and this immediately implies m 2 m 3 = m 3 m 2. This relation will clearly remain valid in A(S n ) for all n 3. Sowemay proceed by induction and suppose that m 2,m 3,...,m n 1

19 A. Garsia Lecture Notes in Algebraic Combintorics March 7, have been shown to commute in A(S n 1 ). Since they will necessarily commute also in A(S n ). and C 2 = m 2 + m m n 3.3 is a conjugacy class, we deduce that for all 2 k n 1 we have m 2 m k + + m n 1 m k + m k m n = m k (m 2 + m m n ) =(m 2 + m m n ) m k = m 2 m k + + m n 1 m k + m n m k This yields m k m n = m n m k (for all 2 k n 1) and proves 3.2 a). Now note that 3.2 b) is trivial when h>k. On the other hand, when h<k, conjugation of m k by s h only interchanges the terms (h, k) and (h +1,k) in m k. Thus 3.2 b) must hold true for all h k precisely as asserted. Finally, we see that we also have ) s k ((1,k)+(2,k)+ +(k 1,k) s k = ((1,k+1)+(2,k+1)+ +(k 1,k+1) m k+1 (k, k+1). This proves 3.2 c). The proof is now complete since 3.2 d) immediately follows from 3.2 c) upon replacing k by k 1. What is truly remarkable is that the Murphy elements have the Young seminormal units e λ ij as their common eigenvectors. This is an immediate consequence of the following identity. Theorem 3.2 Forevery standard tableaux T of shape λ n we have where for a partition λ =(λ 1,λ 2,...,λ k ) we set C 2 e(t ) = ( n(λ ) n(λ) ) e(t ), 3.4 n(λ) = k (i 1) λ i 3.5 and λ denotes the conjugate of λ Since C 2 commutes with every element of A(S n ) we derive that and Von Neuman s lemma then yields C 2 e(t ) = e(t ) P (T )C 2N(T ) e(t ), 3.6 P (T )C 2 N(T ) = C 2 N(T )P (T ) E(T ).

20 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Substituting this in 3.6 we get C 2 e(t ) = = (C 2 N(T )P (T ) ) P (T )N(T ) e(t ) e(t ) (C 2 N(T )P (T ) ) e(t ). It remains to prove C 2 N(T )P (T ) = n(λ ) n(λ). 3.7 Now the definition in 1.1 gives C 2 N(T )P (T ) = 1 i<j n sign(β)χ ( βα =(i, j) ). 3.8 α R(T ) β C(T ) However, it should be apparent that unless the indices i, j are in the same row or column of T it is impossible to interchange their positions in T by a row permutation followed by a column permutation without messing up the positions of the other entries in T. Thus the only way we can have the equality βα =(i, j) is β =(i, j) or α =(i, j). This reduces the evaluation of the right hand side of 3.8 to counting transpositions in C(T ) and R(T ). Now if λ =(λ 1,λ 2,...,λ k ) and λ =(λ 1,λ 2,...,λ h) then the number of transpositions in C(T ) and R(T ) are respectively given by h ( ) λ i 2 and k ( ) λi. 2 This reduces 3.8 to C 2 N(T )P (T ) = k ( ) λi 2 h ( ) λ i, 2 and it easily seen that this is only another way of writing the equality in 3.7. Our proof of 3.5 is thus complete. Remark 3.1 We should mention that the identity in 3.7 implies a classical identity (see [4]) satisfied by the characters of S n.tosee this note that since conjugation by an elemnt σ S n does not change the coefficient of the identity, formula 3.7 may also be rewritten as n! But then Proposition 2.1 gives σ S n C 2 σ N(T )P (T ) σ 1 = n(λ ) n(λ). n! C 2 χ λ = n(λ ) n(λ).

21 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Since n!/ = n λ we finally obtain that C 2 n λ χ λ 21 n 2 = n(λ ) n(λ), 3.9 where χ λ 21 n 2 gives the value of χ λ at the conjugacy class C 2 and C 2 gives the cardinality of C 2. It is customary to call the difference j i the content ofthe lattice cell (i, j). For instance in the figure below we display the Ferrers diagram of the partition (5, 3, 3) with its cells filled by their contents For a given tableau T with n cells and an integer 2 k n let us denote by c T (k) the content of the cell that contains k in T. This given it is easy to see that for any tableau T of shape λ we necessarily have n c T (k) = n(λ ) n(λ) k=2 This simple observation causes Theorem 3.2 to have the following truly remarkable corollary. Theorem 3.3 Forevery standard tableau T with n cells we have for 2 k n 3.10 a) m k e(t ) = c T (k) e(t ), b) e(t ) m k = c T (k) e(t ) Note that since by Theorem 2.6 e(t ) is self flipping and m k is trivially so, 3.12 b) can be obtained by flipping both sides of 3.12 a). So we need only prove the latter. From the definition in 2.16 it follows that 2 e( 1 ) = (1, 2) and e( 1 2 ) = +(1, 2). Thus we see that 2 2 m 2 e( 1 ) = e( 1 ) and m 2 e( 1 2 ) = e( 1 2 ). This verifies 3.12 a) for n =2.Sowemay proceed by induction and suppose that 3.12 a) has been verified up to n 1. Inparticular if T is any standard tableau with n cells we will necessarily have m k e(t )=c T (k)e(t )=c T (k)e(t ) (for 2 k n 1) Thus P (T )N(T ) m k e(t )=m k e(t ) e(t ) h(t ) P (T )N(T ) (by 3.13) = c T (k) e(t ) e(t )=c T (k) e(t ). h(t ) 3.14

22 A. Garsia Lecture Notes in Algebraic Combintorics March 7, This proves 3.12 for for 2 k n 1. But for k = n we may use 3.4 with n(λ ) n(λ) given by 3.11 and get (m 2 + m m n ) e(t ) = (c T (2) + c T (3) + + c T (n)) e(t ). Subtracting the identities in 3.14 for 2 k n 1 yields completing the induction and the proof. m n e(t ) = c T (n) e(t ). A beautiful consequence of this result is a completely explicit formula for Young s seminormal units. To present this development we need a few preliminary observations. To begin note that if P (x 2,x 3,...,x n ) is any polynomial in its arguments and T is any standard tableau, then from Theorem 3.3 it follows that P (m 2,m 3,...,m n ) e(t ) = P (c T (2),c T (3),...,c T (n)) e(t ) now we may construct P (x 2,x 3,...,x n ) in such manner that P (c T (2),c T (3),...,c T (n)) =1while at the same time the operator P (m 2,m 3,...,m n ) kills all the other seminormal idempotents. To give a precise and efficient construction of such a polynomial we need notation. Let us recall that the addable cells apartition µ of n 1 are the cells we may add to the Ferrers diagram of µ to obtain of the Ferrers diagram of a partition of n. The collection of contents of the addable cells of µ will be simply denoted AC µ. For instance it is easily seen from figure 3.10 that AC 533 = { 3, 2, 5}. This given we have Theorem 3.4 Forany standard tableau T we recursively define a polynomial P T (x 2,x 3,...,x n ) by setting Then P T (x 2,x 3,...,x n ) = P T (x 2,x 3,...,x n 1 ) P T (m 2,m 3,...,m n ) e(s) = c AC λ(t ) c c T (n) x n c c T (n) c, ( with P[1] =1 ) if S is standard and S T, e(t ) if S = T For T = [1] there is nothing to prove. We can thus proceed by induction and assume 3.17 to be valid for all for tableaux with n 1 cells. Note that the induction hypothesis immediately implies that the expression P T (m 2,m 3,...,m n 1 ) e(s) 3.17

23 A. Garsia Lecture Notes in Algebraic Combintorics March 7, fails to vanish only if and only if S = T and in that case we have P T (m 2,m 3,...,m n 1 ) e(t ) = 1 This also proves the first case of 3.17 when S T.Soweare left with the case S = T. Now from from 3.16 it follows that for S = T ( ) x k c P T (m 2,m 3,...,m n ) e(s) = e(s) c T (n) c c AC λ(t ) c c T (n) ( by 3.12 a) ) = ( c AC λ(t ) c c T (n) ) c S (n) c e(s) c T (n) c This may also be written as P T (m 2,m 3,...,m n ) e(s) = ( c AC µ c c T (n) ) c S (n) c e(s) 3.18 c T (n) c where for convenience we have set µ = λ(s) =λ(t ) Note further that the equality S = T forces c S (n) AC µ.sofor the right hand side of 3.18 not to vanish we must have c S (n) =c T (n). But this holds true if and only if S = T.Inthis case we have c AC λ(t ) c c T (n) c S (n) c c T (n) c = c AC λ(t ) c c T (n) c T (n) c c T (n) c = 1 and 3.18 reduces to P T (m 2,m 3,...,m n ) e(t ) = e(t ), completing the proof of the Theorem. We can now prove the following remarkable fact Theorem 3.5 For any standard tableau T we have e(t ) = P T (m 2,m 3,...,m n ) Assume that T = T µ r. 3.20

24 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Then P T (m 2,m 3,...,m n )e λ ji = P T (m 2,m 3,...,m n )e λ jje λ ji e λ ii Now, by 3.20 and Theorem 3.4, we have 3 e( )=(m 2 + 1)(m 3 +1)/6, e( )=(m )(m 3 2)/ = P T (m 2,m 3,...,m n )e ( T λ j )e λ ji e ( T λ i ) = e ( T λ j )e λ ji e ( T λ i ) P T (m 2,m 3,...,m n ) 3.21 P T (m 2,m 3,...,m n )e ( T λ j ) 0 = λ = µ and j = r 3.22 Similarly, in view of 3.12 b) we also derive that e ( T λ i ) P T (m 2,m 3,...,m n ) 0 = λ = µ and i = r 3.23 Using 3.21, 3.22 and 3.23 the expansion in 2.4 with f = P T (m 2,m 3,...,m n ) reduces to P T (m 2,m 3,...,m n ) = h µ P T (m 2,m 3,...,m n )e ( T r µ ) e ( T r µ ) (by 3.20) = h µ P T (m 2,m 3,...,m n )e ( T ) e(t ) (by Theorem 3.4) = h µ e ( T ) e(t ) (by 2.26) = e(t ) completing the proof of the Theorem. It might be good at this point to exhibit a few instances of the identity in For the tableaux with three cells Theorem 3.4 gives We should also note that we have e( ) = (m 2 1)(m 3 1)/6, e( ) = (m 2 1)(m 3 +2)/ e( ) = (m 2 1)(m 3 + 2)(m 4 2)(m 4 + 2)(m 5 2)/96. These expressions are easily derived if we take the view that the successive linear factors must be selected to kick each additional label into the position that it occupies in the target tableau. 4. The seminormal matrices In this section we shall work out explicit formulas for the seminormal matrices corresponding to simple reflections for any given partition µ. Since we shall keep µ fixed throughout our derivation, to simplify the displays we will sometimes omit the superscript µ. Soweassume that T 1,T 2,...,T nµ 4.1 are the standard tableaux of shape µ in the Young Last Letter Order. We pick a simple transposition s k =(k, k +1)(for 1 k n 1) and proceed to construct all the entries of the seminormal matrix A µ (s k ). Suppose first that for a pair 1 r<s n µ we have s k T r = T s. 4.2

25 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Note that since T r and T s only differ in the positions of k and k +1, the last letter of disagreement between T r and T s is necessarily k +1. Moreover, since r<s T r < LL T s it follows that k +1 is higher in T r than in T s. Clearly k and k +1cannot be in the same row or column of T r for then 4.2 would force T s not to be standard. Furthermore two successive integers can never be in the same diagonal of any standard tableau. In conclusion we see that these two tableaux can only be as indicated in the figure below k+1 T r = k T s = k+1 This given our first step is to compute the image by s k of the seminormal idempotent e rr = e(t r ).Tothis end we use the expansion formula in 2.4 and start by writing k 4.3 s k e(t r ) = λ n λ i,j=1 ( sk e(t r ) e λ ) ji e λ ij. 4.4 However, since by our conventions e(t r )=e µ rr a use of Theorem 2.3 quickly reduces 4.4 to n µ s k e(t r ) = h µ ( sk e(t r ) e µ ) ri e µ ir n µ ( = h µ sk e µ ) ri e µ ir n µ = h µ e µ ri (s k) e µ ir. 4.5 Now it develops that this expansion can be reduced drammatically further by use of Murphy elements. To this end set n c 2 (m h c) Π = 4.6 (c Tr (h) c) h=2 h k,k+1 c=c 1 c c Tr (h) where c 1 and c 2 respectively denote the minimum and the maximum of the contents of the cells of the diagram of µ. Note that since T r and T s only differ in the positions of k and k +1we will have c Tr (h) =c Ts (h) (for all h k, k +1). 4.7 It follows from 4.6 and 3.12 a) that Π e(t i ) = 0 if i r, s, e(t r ) if i = r, e(t s ) if i = s. 4.8

26 A. Garsia Lecture Notes in Algebraic Combintorics March 7, In fact, the last two cases of 4.8 are immediate consequences of 4.6 and 4.7. On the other hand we see from 4.6 that Π e(t i ) 0 = c Ti (h) =c Tr (h) h k, k +1. Since standard tableaux increase along diagonals these equalities force T i to be identical with T r except for the positions of k and k +1.Inother words the non-vanishing of Πe(T i ) forces T i = T r or T i = T s. This proves the first case of 4.8. Note next that, since there is no occurrence of m k and m k+1 in the right hand side of 4.6, it follows from Theorem 3.1 that Π s k = s k Π 4.9 This given, we get This may be rewritten as s k e(t r ) = s k Π e(t r ) (by 4.9) = Πs k e(t r ) n µ (by 4.5) = h µ e µ ri (s k)πe µ ir n µ (by 2.22) = h µ e µ ri (s k)πe µ ii eµ ir (by 4.9) = h µ e µ rr(s k ) e µ rr + h µ e µ rs(s k ) e µ sr. where a and b are constants we shall soon determine. s k e µ rr = ae µ rr + be µ sr, 4.10 To begin we multiply both sides of 4.10 by the Murphy element m k and get (using 3.2 c)) and 3.12 a) gives (s k m k+1 1) e µ rr = am k e µ rr + bm k e µ sr, c Tr (k +1)s k e µ rr e µ rr = ac Tr (k) e µ rr + bc Tr (k +1)e µ sr, Subtracting from this 4.10 multiplied by c Tr (k +1)yields e µ rr = a (c Tr (k) c Tr (k + 1)) e µ rr. Thus 1 a = c Tr (k +1) c Tr (k) Our next step is to multiply 4.10 by s k. This gives e µ rr = as k e µ rr + bs k e µ sr. (by 4.10) = a(ae µ rr + be µ sr ) + bs k e µ sr. = a 2 e µ rr + ab e µ sr + bs k e µ sr. 4.12

27 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Note that c Tr (k+1) c Tr (k) =1forces k and k+1to be in the same row of T r and c Tr (k+1) c Tr (k) = 1 forces k and k +1to be in the same column. Since these two alternatives have been excluded we can t have a 2 =1. But then 4.12 shows that we can t have b =0. This allows us to extract s k e µ sr out of 4.12 and obtain s k e µ sr = (1 a2 ) b e µ rr ae µ sr To determine b we need a further consequence of Theorem 1.3 which is quite interesting in its own right. This may be stated as follows. Proposition 4.1 For two standard tableaux T 1 T 2 of the same shape we have a) E(T 2 ) e(t 1 ) 0 = T 2 < LL T 1, b) e(t 1 ) E(T 2 ) 0 = T 1 < LL T Let k +1be the last letter of disagreement between T 1 and T 2. In view of the recursive definition of the seminormal unit e(t 2 ) given in 2.16 we may write e(t 1 ) in the form e(t 1 ) = e ( T 1 (k 1) ) P ( T 1 (k) ) R where R 1 is a residual factor of no concern. In the same vein, using 1.13 b) we may write Using 4.13 and 4.14 we see that E(T 2 ) = P (T 2 ) n k (T 2 )N ( T 2 (k) ) E(T 2 ) e(t 1 ) 0 = N ( T 2 (k) ) e ( T 1 (k 1) ) P ( T 1 (k) ) 0. Thus from Proposition 1.1 it follows that E(T 2 ) e(t 1 ) 0 = λ ( T 2 (k) ) λ ( T 1 (k) ) (in dominance) and 4.12 a) necessarily follows from the definition of Young s Last Letter Order. The proof of 4.12 b) is entirely analogous. As a corollary of 4.12 we can now immediatly derive the following surprising result. Proposition 4.1 The constant b appearing in 4.10 and 4.13 is plainly and simply equal to 1. Note that since e µ sr = e µ sse µ sre µ rr we may write h µ e µ sr = e(t s ) E(T ( ) s) e(t s ) e(t s )E(T s )σ µ h sre(t r ) e(t r ) E(T r) e(t r ) µ h µ = e(t s ) E(T ( ) s) e(t s )E(T s )σ µ E(Tr ) h sre(t r ) e(t r ) µ h µ = e(t s )E(T s )σ sre(t µ r )

28 A. Garsia Lecture Notes in Algebraic Combintorics March 7, and since by assumption we have s k T r = T s we can set σ sr µ = s k in this last identity and obtain h µ e µ sr = e(t s )E(T s )s k e(t r ) (by 4.10) = e(t s )E(T s ) ( ae µ rr + be µ ) sr = ae(t s )E(T s )e(t r ) + be(t s )E(T s )e µ sr ( = ae(t s )E(T s )e(t r ) + be(t s )E(T s ) e(t s ) E(T ) s) s k e(t r ) h µ (Using 2.17 c)) = ae(t s )E(T s )e(t r ) + be(t s )E(T s )s k e(t r ) = ae(t s )E(T s )e(t r ) + bh µ e µ sr Now note that we cannot have E(T s )e(t r ) 0 for otherwise our Proposition 4.1 would give s<rwhich contraddicts our original assumptions. Thus 4.15 necessarily reduces to h µ e µ sr = bh µ e µ sr which forces b = 1 and completes our proof. To continue our construction of the seminormal matrix corresponding to the simple transposition s k we need to compute the image by s k of the idempotent e(t r ) when k and k +1are in the same row or column. Our point of departure, also in these cases is formula 4.5. Moreover, since now the tableau s k T r is no longer standard, from 4.6 we derive that Π e(t i ) = 0 if i r, e(t r ) if i = r. This given, multiplying both sides of 4.5 by Π we derive that 4.16 n µ s k e(t r ) = s k Π e(t r ) = Πs k e(t r ) = h µ e µ ri (s k)πe µ ir n µ = h µ e µ ri (s k)πe µ ii eµ ir 4.17 (by 4.16) = h µ e µ rr(s k )e µ rr = ae(t r ). with a = h µ e µ rr(s k ). 4.18

29 A. Garsia Lecture Notes in Algebraic Combintorics March 7, To determine a we multiply both sides of 4.18 by the Murphy element m k and use 3.2 c) to get ac Tr (k) e µ rr = m k s k e µ rr = s k m k+1 e(t r ) e(t r ) = c Tr (k +1)s k e(t r ) e(t r ) (by 4.18) = c Tr (k +1)ae(T r ) e(t r ), and this gives a = 1 c Tr (k +1) c Tr (k) = { 1 if k and k +1are in the same row of Tr, 1 if k and k +1are in the same column To state our final result in a compact form we need one further observation concerning the case when k and k +1are not in the same row or column of T r. Then if k +1is in cell (i 1,j 1 ) and k is in cell (i 2,j 2 ) we may write c Tr (k) c Tr (k +1) = (j 2 i 2 ) (j 1 i 1 ) = j 2 j 1 + i 1 i 2 When k +1is in a higher cell than k in T r,asitwas assumed at the beginning of this section, then the quantity π k (T r ) = c Tr (k) c Tr (k +1) = 1/a 4.20 gives the taxi-cab distance between k and k +1. That is the length of any path that joins k +1 to k by EAST and SOUTH steps. See figure below where we have drawn such a path by a dashed line. j 1 j 2 T r = k+1 i 1 k i 2

30 A. Garsia Lecture Notes in Algebraic Combintorics March 7, Theorem 4.2 Let This given, the identities proved in this section yield the following fundamental result T µ 1,Tµ 2,...,Tµ n µ be the standard tableaux of shape µ n in Young s Last Letter Order, then for any pair 1 r, i n µ and 1 k<nwe have e µ ru if k and k +1 are in the same row of T µ r e µ ru if k and k +1 are in the same column of T r µ s k e µ ru = 1 π k (T r µ ) eµ ru + e µ su if T s µ = s k T r µ and k +1 is higher than k in T r µ 1 ( 1 ) π k (T r µ ) eµ ru + 1 πk 2(T r µ e µ su if T s µ = s k T r µ and k is higher than k +1 in T r µ ) 4.21 Given 4.19, the first two cases are simply a restatement of 4.17 right multiplied by e µ ru. Note next that if we combine 4.10, 4.11, 4.20 and Proposition 4.1 we obtain s k e µ rr = 1 π k (T r µ ) eµ rr + e µ sr 4.22 This gives the third case after right multiplication by e µ ru. Similarly, 4.13 gives s k e µ sr = ( 1 1 ) πk 2(T r µ e µ rr + ) 1 π k (T µ r ) eµ sr 4.23 Since π k (T µ r )=π k (T µ s ), the fourth case of 4.21 is obtained by an interchange of r and s followed by right multiplication by e µ su. This completes the proof. From 4.21 we obtain a rather simple recipe for constructing the seminormal representation matrix A µ (s k ). More precisely we have Theorem 4.3 If A µ (s k ) is the matrix yielding the action of the simple transposition s k = (k, k +1) on the basis e µ 1,u,eµ 2,u,...,eµ n µ,u, that is then A µ (s k )= a µ ij (s k) nµ i,j=1 with s k e µ 1,u,eµ 2,u,...,eµ n µ,u = e µ 1,u,eµ 2,u,...,eµ n µ,u A µ (s k ) 4.24 a rr (s k ) = 1 π k (T µ r ) 1 π k (T µ r ) if k +1 is higher than k in T µ r, if k is higher than k +1 in T µ r, 4.25

31 A. Garsia Lecture Notes in Algebraic Combintorics March 7, where π k (T r µ ) is the taxi-cab distance between k and k +1 in T r µ. Moreover for i j we have 0 if s k T µ i T µ j, a ij (s k ) = 1 1 π 2 k (T µ i ) if s k T µ i = T µ j and i<j, 1 if s k T µ i = T µ j and i>j The equation in 4.23 simply says that for each 1 j n µ we have n µ s k e µ ju = e µ iu aµ ij (s k) Thus the identities in 2.25 and 2.26 are obtained by equating coefficients of the units e µ iu on both sides of We are now in a position to obtain explicit expressions for the factors d µ i Our starting point is the following auxiliary identity. occurring in Proposition 4.2 Let T µ s = s k T µ r with r<sthen d µ s /d µ r = ( 1 1/π 2 k(t r ) ) Equating coefficients of the identity in 4.23 and using 2.26 gives s k e µ sr = ( 1 1 ) 1 πk 2(T r) h µ On the other hand we have s k e µ sr = (s k e µ sr) = ( e µ sr) s k (by2.39) = dµ s d µ r = dµ s d µ r (by4.22 right multiplied by e µ rs) = dµ s d µ r e µ rs s k s k e µ rs ( 1/πk (T µ r ) e µ rs + e µ ss) = dµ s d µ r 1 h µ and 4.27 follows by combining 4.28 and 4.29.

32 A. Garsia Lecture Notes in Algebraic Combintorics March 7, To complete the construction of these factors we need one further result. Proposition 4.3 For any λ and any 1 s n λ we can join T λ 1 to T λ s by achain of tableaux with the following two basic properties T λ 1 = T λ i 1 T λ i 2 T λ i m 1 T λ i m = T λ s 4.30 (a) Ti λ r 1 < LL Ti λ r for all 1 r m 1, 4.31 (b) Ti λ r = s kr Ti λ r 1 with s kr =(k r,k r +1). The assertion is trivial for λ 2. So, by induction, let us assume we have proved the result for any µ n 1. Let then λ n and 1 s n λ. To construct the chain in 4.30 we need some notation. To begin let c s be the lattice cell that contains n in Ts λ and let c 1 be the lattice cell that contains n in T1 λ. We should note that c 1 is necessarily be the highest corner of the diagram of λ. Now set λ(t λ s )=µ. Using the induction hypothesis we construct a chain for T λ s : T µ 1 = T µ j 1 T µ j 2 T µ j m 1 T µ j m = T λ s This given, let Ti λ r, for 1 r m, bethe tableau obtained by adding n to T µ j r in the cell c s. If it happens that c 1 = c s then the tableaux Ti λ r will be satisfy 4.30 and 4.31 and we are done. If c s is a lower corner of the diagram of λ then the chain T λ i 1 T λ i 2 T λ i m 1 T λ i m = T λ s will satisfy 4.31 a) and b). Now we only need to construct a chain that joins T1 λ to Ti λ 1. The crucial observation is that when c s c 1 then c 1 is both the highest corner of the diagrams of µ and λ. Thus in T µ 1 the label n 1 is necessarily in c 1.Inparticular Ti λ 1 has n in c s and n 1 in c 1. This given, the tableau s n 1 Ti λ 1 will have n in c 1 and n 1 in c s. Thus s n 1 T λ i 1 < LL T λ i 1. This reduces us to the previous case since now n is the highest corner of λ. Thus the constrution of the desired chain can now be carried out by prepending the chain in 4.33 with the chain that joins T λ 1 to s n 1 T λ i 1. This completes the induction and the proof. Theorem 4.4 If Combining the last two propositions we obtain T λ 1 = T λ i 1 T λ i 2 T λ i m 1 T λ i m = T λ s is any chain satisfying 4.31 a) and b) then d λ s = m 1 r=1 ( 1 π 2 kr (T µ i r ) ) 4.34