Hook-lengths and pairs of compositions

arXiv:math/0410466v1 [math.CO] 21 Oct 2004 Hook-lengths and Pairs of Compositions Charles F. Dunkl Abstract. The nonsymmetric Jack polynomials are defined to be the simultaneous eigenfunctions of a parametrized commuting set of first-order differentialdifference operators. These polynomials form a basis for the homogeneous polynomials and they are labeled by compositions, just like the monomials. The coefficients of the polynomials when expanded in the standard monomial basis are rational functions of the parameter. The poles are determined by certain hook-length products. Another way of locating the poles depends on possible degeneracies of the eigenvalues under the defining set of operators, when the parameter takes on certain negative rational values. This property can be described in an elementary geometric way. Here is an example: consider the two compositions (2,7,8,2,0,0) and (5,1,2,5,3,3), then the respective ranks (permutations of the index set {1,2,...,6} sorting the compositions) are (3,2,1,4,5,6) and (1,6,5,2,3,4), and the two vectors of differences (between the compositions and the ranks, respectively) are (-3,6,6,-3,-3,-3) and (2,-4,-4,2,2,2), which are parallel, with ratio -3/2. It is this parallelism property which is associated with the degeneracy of eigenvalues. For a given composition and associated negative rational number there is an algorithm for constructing another composition with the parallelism property and which is comparable to it in a certain partial order on compositions, derived from the dominance order. This paper presents the background on the polynomials and hook-lengths, and establishes the properties of the algorithm. There is a discussion of some open problems. 1. Introduction A composition is an element of NN 0 (where N0 := {0, 1, 2, 3, . . .}); a typical composition is α = (α1 , . . . , αN ) and the components αi are called the parts of α. Compositions have the obvious application of labeling the monomial basis of polynomials in the variables x1 , . . . , xN and they also serve as labels for the nonsymmetric Jack polynomials (a set of homogeneous polynomials which are simultaneous eigenfunctions of a certain parametrized and commuting set {Ui : 1 ≤ i ≤ N } of difference-differential operators). In this context the ranks of the parts of a composition become significant. The ranks are based on sorting on magnitude and index so that the largest part has rank 1; if the maximum value is repeated then the one with least index has rank 1. This is made precise in the following (the cardinality of a set E is denoted by #E): 2000 Mathematics Subject Classification. Primary 05E10, Secondary 05E35, 33C52. Key words and phrases. nonsymmetric Jack polynomials, compositions. During the preparation of this article the author was partially supported by NSF grant DMS 0100539. 1 2 CHARLES F. DUNKL Definition 1. For α ∈ NN 0 and 1 ≤ i ≤ N let r (α, i) := # {j : αj > αi } + # {j : 1 ≤ j ≤ i, αj = αi } be the rank function. A consequence of the definition is that r (α, i) < r (α, j) is equivalent to αi > αj , or αi = αj and i < j. For any α the function i 7→ r (α, i) is one-to-one on {1, 2, . . . , N }. A partition is a composition satisfying αi ≥ αi+1 for all i, equivalently, r (α, i) = i for all i. For a fixed α ∈ NN 0 the values {r (α, i) : 1 ≤ i ≤ N } ′ are independent of trailing zeros, that is, if α′ ∈ NM 0 , αi = αi for 1 ≤ i ≤ N and ′ ′ αi = 0 for N < i ≤ M then r (α, i) = r (α , i) for 1 ≤ i ≤ N , and r (α′ , i) = i for N < i ≤ M . A formal parameter κ appears in the construction of nonsymmetric Jack polynomials; their coefficients are in Q (κ), a transcendental extension of Q. The relevant information in a composition label is encoded as the function i 7→ αi − κr (α, i). We will be concerned with situations where a pair (α, β) of compositions have the property that αi − κr (α, i) = βi − κr (β, i) for all i, when κ is specialized to some negative rational number. This is equivalent to the condition that (r (β, i) − r (α, i)) κ + αi − βi is a rational multiple of mκ + n for some fixed N N m, n > 0 (or that the vectors [αi − βi ]i=1 and [r (α, i) − r (β, i)]i=1 are parallel). For our application an additional condition is imposed on the pair (α, β) which is stated in terms of a partial order on compositions. Let SN denote the symmetric group on N objects, considered as the permutation group of {1, 2, . . . , N }. The action of SN on compositions is defined by (wα)i = αw−1 (i) , 1 ≤ i ≤ N . PN Definition 2. For a composition α ∈ NN 0 let |α| := i=1 αi and let ℓ (α) := max {j : αj > 0} be the length of α. + + Definition 3. For α ∈ NN 0 let α denote the unique partition such that α = N wα for some w ∈ SN . For α, β ∈ N0 the partial order α ≻ β (α dominates β) P P means that α 6= β and ji=1 αi ≥ ji=1 βi for 1 ≤ j ≤ N ; and α ⊲ β means that |α| = |β| and either α+ ≻ β + or α+ = β + and α ≻ β. For a given α ∈ NN 0 let w be the inverse function of i 7→ r (α, i) then r (α, w (j)) = j for 1 ≤ j ≤ N and α = wα+ .
n Definition 4. A pair (α, β) of compositions is a − m -critical pair (where m, n ≥ 1) if α ⊲ β and mκ + n divides (r (β, i) − r (α, i)) κ + αi − βi (in Q [κ]) for each i. By elementary arguments we show why only negative numbers appear in the critical pairs, and we also find a bound on ℓ (β). A simple example shows that m = 0 is possible: let α = (3, 0) and β = (2, 1), then both α and β have ranks (1, 2) . Proposition 1. Suppose α, β ∈ NN 0 , α ⊲ β and there are integers m, n such that ((r (β, i) − r (α, i)) κ + αi − βi ) / (mκ + n) ∈ Q for 1 ≤ i ≤ N , then mn ≥ 0 and n 6= 0. Proof. The case n = 0 is impossible since that would imply αi − βi = 0 for all i, that is, α = β. So we assume n ≥ 1 and then show m ≥ 0. Let w be the inverse function of i 7→ r (α, i) (so that r (α, w (i)) = i). By definition either α+ ≻ β + or α+ = β + and α ≻ β. Suppose that α+ ≻ β + and let k ≥ 1 have the property that βw(j) = αw(j) and r (β, w (j)) = j for 1 ≤ j < k and at least one of βw(k) 6= αw(k) and r (β, w (k)) > k = r (α, w (k)) holds. Define l by r (β, l) = k, then by the HOOK-LENGTHS 3 definition of the dominance order ≻ we have that αw(k) ≥ βl . Also βl ≥ βw(k) because r (β, w (k)) ≥ k. The case αw(k) = βw(k) and r (β, k) > k (thus n = 0) is impossible hence αw(k) > βw(k) . If r (β, k) = k then m = 0 or else r (β, k) > k and m > 0. Now suppose α+ = β + and α ≻ β, and let k ≥ 1 have the property that βj = αj for 1 ≤ j < k and αk > βk (the existence of k follows from the definition of α ≻ β). Since β is a permutation of α we have that r (a, j) = r (β, j) for 1 ≤ j < k and r (α, k) < r (β, k). This implies m > 0. 
n -critical pair, for some m, n ≥ Proposition 2. Suppose that (α, β) is a − m 1, then ℓ (β) ≤ ℓ (α) + |α|. Proof. First we show that if i > ℓ (α) and βi = 0 then βj = 0 for all j > i. By hypothesis (mκ + n) divides (r (β, i) − r (α, i)) κ + (αi − βi ) = (r (β, i) − i) κ, hence r (β, i) = i. This implies that 0 ≤ βj ≤ βi = 0 for all j > i. Thus if ℓ (β) > ℓ (α) then βi ≥ 1 for ℓ (α) < i ≤ ℓ (β). Since |β| = |α| this shows that ℓ (β) − ℓ (α) ≤ |α|.  2. Nonsymmetric Jack polynomials and hook-lengths QN αi α For α ∈ NN 0 the corresponding monomial is x := i=1 xi and the degree α of x is |α|. For 1 ≤ i, j ≤ N and i 6= j the transposition of i and j is denoted by (i, j) (that is, the permutation w with w (i) = j, w (j) = i and w (k) = k for k 6= i, j). The action of SN on coordinates is defined by (xw)i = xw(i) and is extended to polynomials by (wp) (x) := p (xw) with the effect that w (xα ) = xwα . The operators Ui for 1 ≤ i ≤ N are defined by Ui p (x) = N X ∂ (xi p (x)) + κ ∂xi j=1,j6=i i−1 X xi p (x) − xj p (x (i, j)) p (x (i, j)) , −κ xi − xj j=1 where p is a polynomial (∈ Q (κ) [x1 , x2 , . . . , xN ]). Then (see [2, pp.291-2] for details) Ui Uj = Uj Ui for 1 ≤ i, j ≤ N and there is a crucial triangularity (in the sense of matrices) property: Ui xα = ξi (α) xα + qα,i (x) where qα,i (x) is a sum of terms of the form ±κxβ with certain β ⊳ α and ξi (α) = (N − r (α, i)) κ + αi + 1 = (αi − r (α, i) κ) + (N κ + 1) , for 1 ≤ i ≤ N and α ∈ NN 0 . The existence of the nonsymmetric  Jack polynomials follows from a theorem of elementary linear algebra. Suppose A(i) : 1 ≤ i ≤ N is a collection of pairwise commuting lower triangular M × M matrices over a field. If for each pair (i, j) , 1 ≤ (k) (k) i < j ≤ M there is at least one matrix A(k) such that Aii 6= Ajj , then there exists a unique set of M linearly independent simultaneous (column) eigenvectors  T for A(i) with each eigenvector of the form (0, . . . , 0, 1, . . .) . Equivalently, there is a unique lower triangular unipotent matrix V such that V −1 A(i) V is diagonal for each i. Now apply this result to the action of {U i : 1 ≤ i ≤ N } on the spaces of homogeneous polynomials with the standard basis xα : α ∈ NN 0 , |α| = k , ordered and α = 6 β implies that ξi (α) 6= ξi (β) by ⊲, for k ∈ N0 . It is clear that α, β ∈ NN 0 for any i with αi 6= βi andPgeneric κ. Thus for each α ∈ NN 0 there is a unique polynomial ζα (x) = xα + β⊳α Aβα xβ with coefficients Aβα ∈ Q (κ) such that 4 CHARLES F. DUNKL Ui ζα = ξi (α) ζα for 1 ≤ i ≤ N . The coefficients, as rational functions of κ, can have poles only at certain negative rational numbers, which in turn are linked to the critical pairs (α, β). This is a sketch of the argument (for a detailed proof see [1] ): by the P triangularity property there are coefficients Bβα ∈ Q (κ) such that xα = ζα + β⊳α Bβα ζβ for each α ∈ NN 0 . Now apply the following operator to this equation and obtain: Y PN vi (Ui − ξi (β)) i=1 xα = ζα ; PN v (ξ (α) − ξ (β)) i i i i=1 β⊳α where {v1 , . . . , vN } are formal commuting variables (extending the field Q (κ)). Since the variables {vi } do not appear in ζα the denominators of the coefficients Bβα must come from reducible factors of the form N X vi (ξi (α) − ξi (β)) = N X vi ((r (β, i) − r (α, i)) κ + αi − βi ) i=1 i=1 = N X ci vi i=1 ! (mκ + n)
with ci ∈ Q and m, n ≥ 1. This is exactly the property that (α, β) is a − nn -critical pair. By combinatorial means Knop and Sahi [3] found a polynomial h (α, κ + 1) in κ (the definition will come later) such that h (α, κ + 1) ζα is a polynomial with coefficients in N0 [κ] (polynomials in κ with nonnegative integer coefficients). The expression h (α, κ + 1) is a product of linear factors, called hook-lengths, associated to the Ferrers diagram of the composition α. Suppose α ∈ NN 0 and ℓ (α) = m; the (modified for compositions) Ferrers diagram of α is the set {(i, j) : 1 ≤ i ≤ m, 0 ≤ j ≤ αi } . For each node (i, j) with 1 ≤ j ≤ αi there are two special subsets of the Ferrers diagram, the arm {(i, l) : j < l ≤ αi } and the leg {(l, j) : l > i, j ≤ αl ≤ αi } ∪ {(l, j − 1) : l < i, j − 1 ≤ αl < αi }. The node itself, the arm and the leg make up the hook. The cardinality of the leg is called the leg-length, formalized by the following: Definition 5. For α ∈ NN 0 , 1 ≤ i ≤ ℓ (α) and 1 ≤ j ≤ αi the leg-length is L (α; i, j) := # {l : l > i, j ≤ αl ≤ αi } + # {l : l < i, j ≤ αl + 1 ≤ αi } . For t ∈ Q (κ) the hook-length and the hook-length product for α are given by h (α, t; i, j) = (αi − j + t + κL (α; i, j)) h (α, t) = ℓ(α) αi YY h (α, t; i, j) , i=1 j=1 Note that the indices {i : αi = 0} are omitted in the product h (α, t). (In the present paper t almost always has the value κ + 1, but t = 1 and t = κ do occur in some formulae for Jack polynomials.) Knop and Sahi [3] showed that all coefficients of h (α, κ + 1) ζαx are in N0 [κ] and the coefficient of xk+1 xk+2 . . . xk+l in ζαx is l!κl /h (α, κ + 1) for k = ℓ (α) and l = |α|. HOOK-LENGTHS 5 We conclude from the above results that for any node (i, j) , 1 ≤ j ≤ αi with h (α, κ + 1; i, j) = (L (α; i, j) + 1) κ + αi + 1 − j there must exist at least one β such that (α, β) is − (αi + 1 − j) / (L (α; i, j) + 1)-critical. The main purpose of this paper is to construct such a composition β by algorithmic means. First we assume that the node is in the largest part, that is, r (α, i) = 1. The modification for other parts is trivial - one merely ignores all larger parts αk (with r (α, k) < r (α, i)). This will be explained in detail later. We illustrate how the algorithm works on a partition α, thereby avoiding some technical complexity. Choose n ≤ α1 and suppose L (α; 1, α1 + 1 − n) = m − 1 (thus h (α, κ + 1; 1, α1 + 1 − n) = mκ + n). Then αm > α1 − n ≥ αm+1 . Define a sequence by ξmk+i = αi − nk for 1 ≤ i ≤ m and k ≥ 0. Then ξj ≥ ξj+1 for all j; indeed if j = mk + i with i < m then ξj − ξj+1 = (αi − nk) − (αi+1 − nk) ≥ 0 and if j = mk then ξmk − ξmk+1 = (αm − n (k − 1)) − (α1 − nk) = αm − (α1 − n) > 0. Also αm+1 ≤ α1 − n = ξm+1 . Since the values {ξj } are eventually negative there exists a unique T such that αm+s ≤ ξm+s+1 for 0 < s < T and αm+T > ξm+T +1 (or T = 1 when αm+1 > ξm+2 ). Set t = ((T − 1) mod m) + 1 and k = (T − t) /m (thus T = mk + t and 1 ≤ t ≤ m). Define β ∈ NN 0 by  ξ = αi − (k + 1) n, 1 ≤ i ≤ t    (k+1)m+i ξkm+i = αi − kn, t<i≤m βi = α + n, m +1≤i≤m+T  i   αi , m + T < i. In this context, an upper bound on N is not needed; that is αi is defined for all i ≥ 1 and αi = 0 for  However one can show that l (β) ≤ max (l (α) , m + T0 ) Pmi > ℓ (α). where T0 := i=1 αni (⌊r⌋ is the largest integer ≤ r) and it suffices to take N as large as this bound. Then αm+T +n = βm+T > βt+1 ≥ . . . ≥ βm > β1 ≥ . . . ≥ βt ≥ βm+T +1 = αm+T +1 and r (β, m + i) = i for 1 ≤ i ≤ T , r (β, i) = T + m − t + i = m (k + 1) + i for 1 ≤ i ≤ t, r (β, i) = mk + i for t < i ≤ m and r (β, i) = i for i > m + T . The proof of these facts is a special case of the general result. 3. The Construction of Critical Pairs The main difficulty in extending the method from partitions to compositions is to deal with tied values. Recall that for αi = αj we have r (α, i) < r (α, j) if and only if i < j. The definition of leg-length L is more subtle for compositions. The introduction of small deformations in the values makes it possible to use essentially the same method as for partitions. Loosely speaking we use an infinitesimal quantity υ which satisfies 0 < iυ < 1 for all i ≥ 1, but of course the inequality will only be needed for all i ≤ N for some N ≥ max (ℓ (α) , ℓ (β)). In the sequel we let N = ℓ (α) + |α| which suffices by Proposition 2 and we let υ = N1+1 . Definition 6. For α ∈ NN e ∈ QN be given by α ei := αi −iυ for 1 ≤ i ≤ N . 0 let α Proposition 3. For any α ∈ NN 0 the following hold: (i) if i 6= j then α ei − α ej ∈ / Z, in particular, α ei 6= α ej , (ii) r (α, i) = r (e α, i) for 1 ≤ i ≤ N , (iii) there is a unique permutation w of {1, . . . , N } such that r (α, w (i)) = i. Proof. To show part (i) suppose αi > αj then α ei − α ej = αi − αj − (i − j) υ ≥ 1 − (i − j) υ > 0, or suppose αi = αj and i < j then α ei − α ej = (j − i) υ > 0. In 6 CHARLES F. DUNKL both cases, α ei − α ej ∈ / Z. This shows that r (e α, i) = # {j : α ej > α ei } + 1 (extending the definition of the rank function to elements of QN ). Now r (α, i) = # {j : j > i, αj > αi } + # {j : j < i, αj ≥ αi } + 1 = # {j : j > i, α ej > α ei } + # {j : j < i, α ej > α ei } + 1 = r (e α, i) for 1 ≤ i ≤ N . The fact that the sorting permutation w is unique follows trivially from part (i).  As noted before, part (iii) implies that α+ i = αw(i) , also that αi = αj and i < j implies w (i) < w (j) (consider this as the formal proof that i 7→ r (α, i) is one-to-one). Here is the formula for leg-length in terms of α e. Proposition 4. For α ∈ NN 0 and for 1 ≤ i ≤ ℓ (α) , 1 ≤ j ≤ αi the leg-length L (α; i, j) = # {l : j − iυ − 1 < α el < α ei } . Proof. For l < i the inequalities are j − iυ − 1 < αl − lυ and αl − lυ < αi − iυ, equivalent to αl + 1 − j > − (i − l) υ > −1 (hence αl + 1 − j ≥ 0) and αi − αl > (i − l) υ > 0 (hence αi − αl ≥ 1), respectively. For l > i the inequalities αl + 1 − j > (l − i) υ and αi − αl > − (l − i) υ are equivalent to αl + 1 − j ≥ 1 and αi − αl ≥ 0 respectively.  We begin the construction for a given α and the hook-length for a node in the largest part. We use the permutation w described in
part (iii) of Proposition 3. Suppose 1 ≤ n ≤ αw(1)
and L α; w (1) , αw(1) + 1 − n = m − 1. We construct β n -critical pair. so that (α, β) is a − m There is another characterization of m: it has the property that α ew(m) > α ew(1) − n > α ew(j) for j > m; note  m − 1 = # l : αw(1) + 1 − n − w (1) υ − 1 < α el < α ew(1)  =# l:α ew(1) − n < α el < α ew(1) . This shows that the m largest parts of α, excluding α ew(1) , are in the latter interval; the equation α el = α ew(1) − n is impossible by part (i) of Proposition 3. Algorithm 1. Define a sequence ξmk+i = α ew(i) −nk for 1 ≤ i ≤ m and k ≥ 0. The sequence is strictly decreasing and ξj+1 < α ew(j) for large enough j . Thus there is a unique T such that α ew(m+s) < ξm+s+1 for 1 ≤ s < T and α ew(m+T ) > ξm+T +1 −t (the case of equality is ruled out). Then let t = ((T − 1) mod m) + 1 and k = Tm N (so that T = mk + t and 1 ≤ t ≤ m) and define β ∈ N0 by  α − (k + 1) n, 1 ≤ i ≤ t    w(i) αw(i) − kn, t<i≤m βw(i) = α + n, m+1≤i≤m+T    w(i) αw(i) , m + T < i. Proof. The decreasing property has

two cases. If 1 ≤ i < m then ξmk+i − ξmk+i+1 = α e − nk − α e − nk > 0. If i = m
then ξmk+m − ξmk+m+1 = w(i) w(i+1)

α ew(m) − nk − α ew(1) − n (k + 1) = α ew(m) − α ew(1) − n > 0. Since α ew(i) −e αw(j) ∈ / Z for i < j it is impossible for α e = ξ when s ≥ 1. m+s+1 w(m+s)  Pm  Let T0 = i=1 αw(i) /n . We claim ξm+T0 +1 < −1. Set αw(i) = nqi + ri with qi ≥ 0 and 0 ≤ ri ≤ n − 1 for 1 ≤ i ≤ m. Then q1 ≥ 1 since n ≤ αw(1) and HOOK-LENGTHS 7 q1 ≥ qi ≥ q1 − 1. This follows from the inequalities αw(1) ≥ αw(i) ≥ αw(i) − n, that is, nq1 + r1 ≥ nqi + ri ≥ (n − 1) q1 + r1 , r1 − ri ≥ n (qi − q1 ) ≥ r1 − ri − n, but 1 − n ≤ r1 − ri ≤ n − 1 so 1 − 2n ≤ n (qi − q1 ) ≤ n − 1. Furthermore qi ≥ qi+1 for 1 ≤ i < m since αw(i) ≥ αw(i+1) . Thus there exists k with 1 < l ≤ m so that qi = q1 for 1 ≤ i ≤ l and qi = q1 − 1 for l < i ≤ m. Write T0 + m + 1 = mk + i with k ≥ 0 and 1 ≤ i ≤ m . Then mk + i = (lq1 + (m − l) (q1 − 1)) + m + 1 = mq1 + l + 1 and so i = (l + 1) mod m. If l < m then k = q1 , i = l + 1 and ξT0 +m+1 = αw(l+1) − nq1 − w (l + 1) υ = rl+1 − n − w (l + 1) υ < −1. If l = m then k = q1 + 1, i = 1 and ξT0 +m+1 = αw(1) − (n + 1) q1 − w (1) υ = r1 − n− w (1) υ < −1. Finally T0 ≤ |α| n and m + T0 + 1 ≤ ℓ (α) + |α| + 1; also w (j) = j for j > ℓ (α) and thus α ew(j) > −1 > ξj+1 for all sufficiently large j ≤ N .  The following is the main result. The notations α, β, ξ, w, m, n, T, t continue with the definitions given above. The proof is broken up in several lemmas. It is possible that m = 1, in which case one makes the obvious modifications in the following statements. Theorem 1. For α ∈ NN 0 the composition β produced by the algorithm has the property that mκ + n divides (r (β, w (i)) − i) κ + αw(i) − βw(i) for all i, and α ⊲ β,
n that is, (α, β) is a − m -critical pair. Lemma 1. The following inequalities hold: (i) α ew(m+T +1) < ξT +m < ξT +1 < α ew(m+T ) + n (omit ξT +m if m = 1) e e (ii) for t < m, βw(m+T ) > βw(t+1) > . . . > βew(m) > βew(1) > . . . > βew(t) > βew(m+T +1) (iii) for t = m, βew(m+T ) > βew(1) > . . . > βew(m) > βew(m+T +1) . Proof. By construction α ew(m+T ) + n > ξm+T +1 + n = (ξT +1 − n) + n > ξT +m ,by the decreasing property of {ξj }. If T > 1 then ξT +m > α ew(m+T −1) > α ew(m+T +1) . If T = 1 then α ew(m+1) + n > ξ2 > ξm+1 > α ew(m+1) > α ew(m+2) . This proves part (i). As before T = km + t. By construction, βew(i) = ξmk+m+i for 1 ≤ i ≤ t, = ξmk+i for t < i ≤ m, = α ew(i) + n for m + 1 ≤ i ≤ m + T and =α ew(i) for i > m + T . Thus the inequality in part (i) shows βew(m+T +1) < βew(t) < βew(t+1) < βew(m+T ) .  Lemma 2. The following rank values hold: (i) r (β, w (m + i)) = i for 1 ≤ i ≤ T, (ii) r (β, w (t + i)) = mk + t + i for 1 ≤ i ≤ m − t, (iii) r (β, w (i)) = m (k + 1) + i for 1 ≤ i ≤ t, (iv) r (β, w (i)) = i for m + T + 1 ≤ i ≤ N . Proof. By parts (ii) and (iii) of the previous lemma, in decreasing order the m largest values of βe are βew(m+1) , . . . , βew(m+T ) , the next m − t values are βew(t+1) , . . . , βew(m) , the next t values are βew(1) , . . . , βew(t) , and the remaining are βew(m+T +1) , . . .. Note that r (β, w (t + i)) = T + i = mk + t + i in part (ii) and r (β, w (i)) = T + (m − t) + i = m (k + 1) + i.  8 CHARLES F. DUNKL Lemma 3. The composition β satisfies the condition that mκ + n divides (r (β, w (i)) − i) κ + αw(i) − βw(i) for all i ≤ N . Proof. Let γi = (r (β, w (i)) − i) κ + αw(i)
− βw(i) . For 1 ≤ i ≤ t, γi (m (k + 1) + i − i) κ + αw(i) − αw(i) − (k + 1) n = (k + 1) (mκ + n). For t < i
m, γi = mkκ + αw(i) − αw(i) − kn
= k (mκ + n). For m + 1 ≤ i ≤ m + T , γi ((i − m) − i) κ + αw(i) − αw(i) + n = − (mκ + n), and γi = 0 for i > m + T . = ≤ =  We must show that α ⊲ β to complete the proof of the theorem. For 1 ≤ i ≤ N let ε (i) ∈ NN 0 denote the standard basis element, that is, ε (i)j = δij . The idea is to describe the construction as a sequence of compositions, each of which is produced by adding n (ε (w (m + i)) − ε (w ((i − 1) mod m + 1))) to the previous one, for i = 1, 2, . . . T . The effect of the argument ((i − 1) mod m) + 1 is to cycle through the values w (1) , . . . , w (m). Ps Lemma 4. Let β (s) := α − n i=1 (ε (w ((i − 1) mod m + 1)) − ε (w (m + i))) for 0 ≤ s ≤ T . Then β (0) = α, β (T ) = β and β (s) ⊲ β (s+1) for 0 ≤ s < T . Proof. We use Lemma 8.2.3 from [2, p.289]. This states that if λ is a partition such that 1 ≤ n < λi − λj (for some i < j) then λ ≻ (λ − n (ε (i) − ε (j)))+ . As a consequence we have that if γ ∈ NN 0 and 1 ≤ n < γi − γj for some i, j + then γ + ≻ (γ − n (ε (i) − ε (j))) , that is, γ ⊲ (γ − n (ε (i) − ε (j))). Suppose (s) s = ml + i < m + T with 0 ≤ i < m and l ≥ 0. Then βw(j) = αw(j) + n for (s) (s) m + 1 ≤ j ≤ m + s, βw(j) = αw(j) − n (l + 1) for 1 ≤ j ≤ i, βw(j) = αw(j) − nl for i + 1 ≤ j ≤ m and βw(m+s+1) = αw(m+s+1) . By definition, β (s+1) = β (s) − n (ε (w (i + 1)) − ε (w (m + s + 1))). Let δs denote the difference between the two affected values, that is, (s) (s) δs = βw(i+1) − βw(m+s+1) = αw(i+1) − nl − αw(m+s+1) = ξml+i+1 − α ew(m+s+1) + (w (s + 1) − w (m + s + 1)) υ. By the construction of the Algorithm ξm+s+1 > α ew(m+s) for s < T and α ew(m+s) > α ew(m+s+1) . Thus δs − n = ξm+s+1 − α ew(m+s+1) + (w (s + 1) − w (m + s + 1)) υ > (w (s + 1) − w (m + s + 1)) υ > −1. If δs − n ≥ 1 then by the above argument we (s) (s+1) (s) (s+1) have β (s) ⊲ β (s+1) . If δs − n = 0 then βw(m+s+1) = βw(i+1) , βw(i+1) = βw(m+s+1)
+
+ and w (s + 1) < w (m + s + 1); thus β (s+1) = β (s) and β (s) ≻ β (s+1) (for any composition γ, if i < j and γi > γj then γ ≻ (i, j) γ), that is, β (s) ⊲ β (s+1) .  This completes the proof of the theorem when the hook length mκ+ n is associated with the largest part (either αw(1) > αw(2) or αw(1) = αw(i) and i 6= 1 implies
w (1) < w (i)). Suppose for some l ≥ 1 that L α; w (l + 1) , αw(l+1) + 1 − n = m − 1. Then the algorithm is applied with the arguments of w and the ranks all shifted by l; and the largest l parts are not changed. Thus βw(i) = αw(i) for 1 ≤ i ≤ l and  α − (k + 1) n, 1 ≤ i ≤ t    w(l+i) αw(l+i) − kn, t<i≤m βw(l+i) = α + n, m+1≤i≤m+T    w(l+i) αw(l+i) , m + T < i, HOOK-LENGTHS 9 using the same notations k, T, t as above. 4. Examples and Discussion The first example is a partition type: α = (9, 8, 8, 7, 4, 3, 3, 2, 2) , n = 3, m = 13 4, T = 9 and β = (0, 2, 2, 1, 7, 6, 6, 5, 5, 3, 3, 3, 3). The ranks of β are (r (β, i))i=1 = (13, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9). To illustrate the situation where the node of the hook is not in the largest row, consider the hook at (2, 5) in α = (9, 8, 8, 5, 4, 4), then L (α; 2, 5) = 3, n = 3 and β = (9, 4, 4, 5, 8, 8) with ranks (1, 5, 6, 4, 2, 3). The next example is a composition α = (0, 3, 5, 6, 6, 4, 1) (ranks (7, 5, 3, 1, 2, 4, 6)) with n = 3; then m = L (α; 4, 4) + 1 = 5 and β = (3, 0, 2, 0, 0, 1, 4, 3, 3, 3, 3, 3) (ranks (2, 10, 8, 11, 12, 9, 1, 3, 4, 5, 6, 7)). There is an analogous situation for critical pairs when β is the given composition; in this case the expansion of nonsymmetric Jack polynomials in terms of the p-basis
(see [2, p.298]) suggests that the linear factors (mκ + n) of h (β, 1) lead to n −m -critical pairs (α, β). However we consider the uniqueness problem described below as more important for applications, and will not further investigate h (β, 1). A natural question occurs: for a given α and hook-length mκ + n does one step of the algorithm produce all possible solutions for β? If this were true then there would be a uniqueness result of the form: suppose the multiplicity of the linear factor (mκ + n) in the hook-length
product h (α, κ + 1) is one, then there n -critical pair. However, this may fail if is a unique β so that (α, β) is a − m m, n are not relatively prime: for α = (6, 3, 1, 1) , n = 6, m = 4 the algorithm produces β (1) = (0,
3, 1, 1, 6) ;the multiplicity of (2κ + 3) is one in both h (α, κ + 1) and h β (1) , κ + 1 . Apply the algorithm to β (1) with n = 3, m = 2 to obtain
β (2) = (0, 3, 4, 1, 3) and h β (2) , κ + 1 does not have (2κ + 3) as a factor. We conjecture there is uniqueness if m, n are relatively prime (and the multiplicity is one); a particular case of this was established in [1]. This is crucial because it is n when the number of variables is less used to show that ζα has no pole at κ = − m than ℓ (β). here is an example: α = (7, 6, 6, 4, 4) with n = 2, so that m = 3, then the unique β = (1, 0, 0, 6, 6, 2, 2, 2, 2, 2, 2, 2); the application is to ζα on R10 and the uniqueness of β implies that κ = − 32 is not a pole of ζα (for less than 12 variables). However the uniqueness result uses specific properties of a class of partitions and m, n are relatively prime. There is a weak uniqueness result concerning the sign  changes in the sequence α ew(m+i) − ξm+i+1 : i = 1, 2, . . . . Recall that T is chosen so that α ew(m+s) − ξm+s+1 < 0 for 1 ≤ s < T and α ew(m+T ) − ξm+T +1 > 0. Proposition 5. Suppose for some s > T that α ew(m+s) − ξm+s+1 > 0 and
α ew(m+s+1) − ξm+s+2 < 0, then the hook-length at the node w (i) , αw(i) + 1 − nl is l (mκ + n) where m + s + 1 = ml + i and 1 ≤ i ≤ m. Proof. By hypothesis α ew(ml+i+1) >α ew(i) − nl > ξm+s+2 > α ew(ml+i) . This
implies L α; w (i) , αw(i) + 1 − nl = ml − 1.  The Proposition implies that if there is only one hook-length divisible by mκ+n in the rows w (i) for 1 ≤ i ≤ m then α ew(m+i) > ξm+i+1 for all i ≥ T ; so there is only one sign-change. It appears that there can be a considerably larger number of solutions than the multiplicity. Here is an example: α = (9, 7, 6, 5, 2), the multiplicity of (2κ + 3) in h (α, κ + 1) is 4. The relevant hook lengths are 2κ + 3 at nodes (1, 7) and 10 CHARLES F. DUNKL (3, 4), and 4κ + 6 at nodes (1, 4) and (2, 2). The algorithm produces (6, 7, 9, 5, 2) , (9, 7, 0, 2, 5, 3, 3) , (3, 7, 6, 5, 8) , (9, 1, 0, 5, 2, 6, 6) respectively for these nodes. But one can continue the process: for example the multiplicity of (2κ + 3) in h ((6, 7, 9, 5, 2) , κ + 1) is 3, and the algorithm produces three more solutions for β, one being (6, 7, 3, 5, 8). One could speculate that
there is a lattice of solutions, n ordered by ⊲. Finally, one can ask if there are − m -critical pairs without a corresponding hook-length mκ+n. It seems doubtful, but we will leave this unanswered. References [1] C. Dunkl, Singular polynomials for the symmetric groups, Int. Math. Research Not. 2004 (2004), #67, 3607-3635, arXiv:math.RT/0403277. [2] C. Dunkl and Y. Xu, Orthogonal Polynomials of Several Variables, Encycl. of Math. and its Applications 81, Cambridge University Press, Cambridge, 2001. [3] F. Knop and S. Sahi, A recursion and a combinatorial formula for Jack polynomials. Invent. Math. 128 (1997), 9–22. Department of Mathematics, P.O. Box 400137, University of Virginia, Charlottesville, VA 22904-4137 U.S. E-mail address: [email protected] URL: http://www.people.virginia.edu/~cfd5z