arXiv:1204.5117v4 [math-ph] 24 Feb 2013
CLUSTERING PROPERTIES OF RECTANGULAR MACDONALD POLYNOMIALS CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE Abstract. The clustering properties of Jack polynomials are relevant in the theoretical study of the fractional Hall states. In this context, some factorization properties have been conjectured for the (q, t)-deformed problem involving Macdonald polynomials (which are also the quantum eigenfunctions of a familly of commuting difference operators with signifiance in the relativistic Ruijsenaars-Schneider model). The present paper is devoted to the proof of this formula. To this aim we use four families of Jack/Macdonald polynomials: symmetric homogeneous, nonsymmetric homogeneous, shifted symmetric and shifted nonsymmetric.
1. Introduction The symmetric (homogeneous) Jack polynomials are relevant in the study of the quantum many-body wave functions. In particular, fractional quantum Hall states of particles in the lowest Landau levels are described by such polynomials [3, 4, 5, 6]. Pioneered by Laughlin [24], the theoretical study of the fractional quantum Hall states use multi-variable polynomials [17] describing the full many-body state of the interacting electrons on a plane or on a sphere (In the case of the sphere, the polynomials appear after stereographic projection). The special polynomials that are relevant in this context are not general solutions of the true eigenvalue problem involving the Coulomb interaction but they are constructed to be adiabatically related to the true eigenstates. The most famous example Q is the Laughlin wave function which is the cube of the Vandermonde determinant i 0. Let γ = [λ[1], . . . , λ[m − 1], λ[m] − 1, 0N −m ]. If Mγ (0) = P (γ) then Mλ (0) = P (λ). Proof. Let β = [λ [m] − 1, λ [1] , . . . , λ [m − 1] , 0, . . .] α = [λ [1] , . . . , λ [m − 1] , 0, . . . 0, λ [m]] . We evaluate Mβ (0) /Mγ (0) , Mλ (0) /Mα (0) and use Mα (0) = −Mβ (0) to prove Mλ (0) /Mγ (0) = P (λ) /P (γ). The formula (34) is used repeatedly in the following calculations. First we have Mλ (0) 1 − q λ[m] tN −m+1 = . Mα (0) 1 − q λ[m] t Since λ [m − 1] ≥ λ [m] it follows that γ [i] > γ [m] for 1 ≤ i < m. Then m−1 Y 1 − q λ[i]−λ[m]+1 tm−i Mβ (0) = . Mγ (0) 1 − q λ[i]−λ[m]+1 tm−i+1 i=1
qtN ; q, t λ h (γ, qt) λ[m] N +1−m Next . In this ratio the = 1 − q t and we evaluate q,t N h qt ; q, t γ q,t (λ, qt) only cells that have a changed factor are those in column λ [m] and in row m. So
RECTANGULAR MACDONALD POLYNOMIALS
13
bλ (i, λ [m]) = m − i and bγ (i, λ [m]) = m − i − 1 for 1 ≤ i < m. The cells in {(1, λ [m]) , · · · (m − 1, λ [m])} contribute m−1 Y i=1
1 − q λ[i]−λ[m]+1 tm−i 1 − q λ[i]−λ[m]+1 tm−i+1
to the ratio and the cells in row m contribute Qλ[m]−1 1 − tq λ[m]−i 1 i=1 = . Qλ[m] λ[m] λ[m]+1−i ) 1 − tq (1 − tq i=1
Thus
m−1 qtN ; q, t λ hq,t (γ, qt) 1 − tN −m+1 q λ[m] Y 1 − q λ[i]−λ[m]+1 tm−i = . λ[i]−λ[m]+1 tm−i+1 (qtN ; q, t)γ hq,t (λ, qt) 1 − tq λ[m] 1 − q i=1
This agrees with −Mλ (0) /Mγ (0) and proves the lemma.
End of the proof of Proposition 5. Any vector v can be obtained from 0N by adding 1 to a null part or to a minimal nonzero part. We will denote this operation by + →. For instance, +
+
+
+
[0, 0, 0] →[0, 1, 0] →[0, 2, 0] →[0, 3, 0] →[1, 3, 0]. Let v →+ v ′ . Suppose v ′+ = [λ1 , . . . , λm , 0, . . . , 0] then v + = [λ1 , . . . , λm−1 , λm −1, 0, . . . , 0]. Hence, if Mv+ (0, . . . , 0) = P (v + ) then by lemma 6 Mv′+ (0) = P (v ′+ ). And using repeatedly eq. (36), we obtain Mv (0) = P (v) implies Mv′ (0) = P (v ′ ). Since M0N (0) = P (0N ) the result is shown by a straightforward induction. As a direct consequence Corollary 7. Ev (h0i) = (⋆)
(tN q; q, t)v+ . hq,t (v, qt)
Lascoux [21] gave an equivalent expression using infinite vectors v ∞ = [v[1], . . . , v[N], v[1] + 1, . . . , v[N] + 1, . . . ] and hvi∞ = [hvi[1], . . . , hvi[N], qhvi[1], . . . , qhvi[N], . . . ]. Note hvi∞ 6= hv ∞ i. Lascoux showed (37)
P (v) = (−1)
|v|
N Y i=1
Y
j>i v ∞ [i]>v ∞ [j]
∞
hvi [i] t hvi ∞ [j] − 1 hvi∞ [i] hvi∞ [j]
−1
.
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CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
4.2. Principal specialization of Pλ . From eq. (31) and (19) one has (38)
Pλ (h0i) = (⋆)MSλ (0) = (⋆)
φt (SN ) Mλ− (0). φt (Sλ )
So one has to determine the value of the constant term Mλ− (0). The t-multinomial coefficient is defined as follows: for any λ ∈ NN 0 let mλ (i) = partition Q # {j : λ [j] = i} (defined for 0 ≤ i ≤ λ [1]) and let mNλ = (t, t)N / λ[1] i=0 (t, t)mλ (i) t Qn i (note (t, t)n = i=1 (1 − t )). In fact, the Poincar´e series for λ, denoted φ (Sλ), equals Qλ[1] N i=0 (t, t)mλ (i) /(1 − t) where Sλ is the stabilizer subgroup of λ in SN (the symmetric group on N letters), and φ (SN ) = (t, t)N /(1 − t)N . For application of proposition 5, we find a convenient formula for Mλ− (0) for partitions λ. Proposition 8. For a partition λ ∈ NN 0 tN ; q, t λ φ (Sλ ) Mλ− (0) = (−1) . hq,t (λ, t) φ (SN ) N |λ| t q; q, t λ so we need to evaluate the Proof. From Proposition 5 Mλ− (0) = (−1) − h q,t λ , qt tN q; q, t λ hq,t λ− , qt and ratios N . It is easy to see that hq,t (λ, t) t ; q, t |λ|
λ
N tN q; q, t λ Y 1 − tN +1−i q λ[i] = . N +1−i 1 − t tN ; q, t λ i=1
If the multiplicity of a particular λ [i] is 1 (in λ) then the leg-length of cell (N + 1 − i, j + 1) in λ− is the same as that of the cell (i, j) in λ for 1 ≤ j < λ [i]. To account for multiplicities let c be the inverse of rλ− , that is rλ− [ci ] = i for 1 ≤ i ≤ N. Then λ− [ci ] = λ [i] (for example suppose λ [1] > λ [2] = λ [3] > λ [4], then c1 = N, c2 = N − 2, c3 = N − 1). As before bλ− (ci , j + 1) = bλ (i, j) , 1 ≤ j < λ [i] .
!
1 − tbλ (i,j)+1 q λ[i]−j which is the hq,t λ− , qt is the product of same as the factor at the cell (i, j) in hq,t (λ, t). Thus hq,t (λ, t) the factors at the cells (ci , 1) in hq,t (λ− , qt) divided by the product of the factors at the cells (i, λ [i]) in hq,t (λ, t), for 1 ≤ i ≤ ℓ(λ), since there are no cells for the zero parts. The factor at (ci , 1) is 1 − tN +1−i q λ[i] . Suppose λ [i] has multiplicity k (that is, The factor in hq,t (λ− , qt) at the cell (ci , j + 1) is
λ [i − 1] > λ [i] = . . . = λ [i + k − 1] > λ [i + k]) then bλ (i + l − 1, λ [i]) = k − l for Q 1 ≤ l ≤ k, and the cells (i, λ [i]) , . . . , (i + k − 1, λ [i]) contribute kl=1 1 − tl = (t, t)k .
RECTANGULAR MACDONALD POLYNOMIALS
Thus
15
Qℓ(λ) N +1−i λ[i] 1 − t q hq,t λ− , qt = i=1Qλ[1] . hq,t (λ, t) (t, t) i=1 m (i) λ
Combining the ratios we find that Qℓ(λ) tN ; q, t λ 1 − tN +1−i q λ[i] tN q; q, t λ YN 1 − tN +1−i i=1 = Qλ[1] i=1 1 − tN +1−i q λ[i] hq,t λ− , qt hq,t (λ, t) i=1 (t, t)mλ (i) tN q; q, t λ (t, t) Yλ[1] = YN N − hq,t λ , qt (t, t)mλ (i) 1 − tN +1−i i=1 i=ℓ(λ)+1 N t q; q, t λ N = , hq,t λ− , qt mλ t since mλ (0) = N − ℓ(λ).
As a direct consequence of Proposition 8 and equality (38), we obtain Corollary 9.
tN ; q, t λ . Pλ (h0i) = (⋆) hq,t (λ, t) 5. Subrectangular Macdonald polynomials
5.1. The denominator of sub-rectangular Macdonald polynomials. The following lemma shows that if λ ⊆ [mk , 0N −k ] with 2k ≤ N then Pλ does not have a pole at m−1 N−k+1 q m−1 tN −k+1 = 1 and q d t d 6= 1 where d > 1 is a common factor of m − 1 and N − k + 1: Lemma 10. Let λ ⊆ [mk , 0N −k ] be a partition with 2k ≤ N. Then (1 − q m−1 tN −k+1 ) is not a factor of any the denominators of coefficients of xv in Pλ . Proof. It suffices to prove that the integral version Jλ does not vanish when 1 − m−1 N−k+1 q m−1 tN −k+1 = 0 and 1 − q d t d 6= 0 where d > 1 divides m − 1 and N − k + 1. In other words, we have to show that cλ (q, t) does not have 1 − q m−1 tN −k+1 as a factor. ′′ ′ Suppose λ = [mk , λk′+1 , . . . , λk′′ , 0N −k ] with k ′ ≤ k ′′ ≤ k and λk′′ > 0 . Then the ′ only factors with the relevant power of q that occur in cλ are (1 − q λi −j tλj −i+1 ) for λi = m (so i = 1..k ′ ) and j = 1 (so λ′j = k ′′ ). Hence, for such a pair (i, j) one has λ′j − i + 1 ≤ k ≤ N − k. This proves the result.
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CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
k 5.2. The (q, t)-binomial mλ . Lassalle [23] gave an explicit formula for the (q, t) k binomial mλ for λ ⊆ mk : k Y ti−1 − q j−1tk (1 − q m−j+1 ti−1 ) m i−k = t (39) λ (i,j)∈λ 1 − q Aλ (i,j) t1+bλ (i,j) 1 − q 1+Aλ (i,j) tbλ (i,j) Using (q, t)-hook products, the formula reads −m k k t ; q, t (q ; q, t)λ m ′ λ = (−1)|λ| t3n(λ)−(k−1)|λ| q m|λ|−n(λ ) (40) . λ hq,t (λ, q) hq,t (λ, t) We deduce
Lemma 11. Let λ ( mk be a strict subrectangular partition with 2k ≤ N. The denommk inator of the reduced fraction λ has no factor 1 − q m−1 tN −k+1 .
Proof. We examine the factor of the denominator of equality (39): Y 1+Aλ (i,j) bλ (i,j) Aλ (i,j)t1+bλ(i,j) t Dλ := 1 − q . 1 − q (i,j)∈λ
′
′
Since λ is a partition, there are only terms of the type 1 − q ∗ tλj −i+1 and 1 − q ∗ tλj −i in Dλ . When λ ( mk , we have λ′j − i + 1 < λ′j + 1 ≤ k + 1. And so k ≤ N − k implies λ′j − i + 1 < N − k + 1. This proves the result.
Note
(41)
mk mk
=
mk 0
= 1.
5.3. Principal specialization subrectangular partitions. We consider certain k for N −k with 2k ≤ N. In particular, examine the ocpoles of Mλ− (0) for λ ⊆ m, 0 m−1 N −k+1 N currences of 1 − q t in t ; q, t λ and hq,t (λ, t). Since the maximum leg k N −k length of any cell in λ ⊂ m , 0 is k − 1 only factors of the form 1 − ta q b with 1 ≤ a ≤ k < N − k + 1 and 0 ≤ b ≤ m − 1 can appear in hq,t (λ, t). Further λ[i]
k Y Y 1 − tN −i+1 q j−1 . t ; q, t λ = N
i=1 j=1
m−1 N −k+1 The factor 1 − q t appears in this product exactly when λ [k] = m, that is, k N −k λ = m ,0 . Thus Mλ− (0) has a zeroat q m−1 tN −k+1 = 1 for λ = mk , 0N −k and no zeros or poles there when λ ( mk , 0N −k . We summarize this in the following lemma
RECTANGULAR MACDONALD POLYNOMIALS
17
Lemma 12. Let λ ⊆ [mk , 0N −k ] with 2k ≤ N then the reduced fractions Mλ− (0) and Pλ (h0i) have 1 − q m−1 tN −k+1 as a factor if and only if λ = [mk , 0N −k ]. Equivalently, if (1 − q m−1 tN −k+1 ) = 0 and for any d > 1 dividing m − 1 and N − k + 1, m−1 N−k+1 (1 − q d t d ) 6= 0 then Pmk (h0i) = 0. Pλ (h0i) Note we can write another proof using eq. (37) (see appendix B). 6. Rectangular Macdonald polynomials 6.1. Clustering properties of MSmk . In this section, we will use finite alphabets with different sizes. Notations 13. For simplicity, when we use a partition λ ∈ NN , we mean that the underlying alphabet is {x1 , . . . , xN } and the symmetric shifted monic Macdonald poly(N ) nomial will be denoted by MSλ = MSλ (x1 , . . . , xN ). In the same way, for v ∈ NN the (N ) nonsymmetric shifted monic Macdonald polynomial will be denoted Mv . The length of v ∈ NN is ℓ (v) := max {i : v [i] > 0}. For k ≥ ℓ(v) we will denote also v (k) := [v1 , . . . , vk ]. The following proposition shows that if v ∈ NN with ℓ(v) ≤ k then the Macdonald (N ) (k) polynomials Mv and Mv(k) are clearly related. Proposition 14. The following assertions hold: (1) Suppose v ∈ NN 0 satisfies ℓ (v) ≤ k for some k < N, then (k) Mv x1 tN −k , . . . , xk tN −k , tN −k−1 , . . . , t, 1 = t(N −k)|v| Mv(k) .
(2) Suppose a partition λ ∈ NN 0 satisfies ℓ (λ) ≤ k for some k < N, then (k) (N ) x1 tN −k , . . . , xk tN −k , tN −k−1 , . . . , t, 1 = t(N −k)|v| MSλ (x) . MSλ
Proof. We prove first (1): Both sides have the same coefficient of xv and satisfy the same vanishing conditions. More precisely, the both sides vanish for x = hui, u ∈ Nk0 , |u| ≤ |v| , u 6= v, and hui [j] = q u[j] tk−ru [i] , for 1 ≤ j ≤ k; and the multiplicative coefficient is obtained considering the coefficient of the dominant term in the left hand side and the right hand side. To prove (2) remark the left-hand side is symmetric in (x1 , . . . , xk ). Hence an obvious modification of the above argument applies.
Example 15. For instance consider v = [2, 0]. The polynomial M[2,0] (x1 , x2 ) vanishes for [x1 , x2 ] = h00i = [t, 1], [x1 , x2 ] = h10i = [qt, 1], [x1 , x2 ] = h01i = [1, qt], [x1 , x2 ] = h11i = [qt, q], [x1 , x2 ] = h02i = [1, q 2 t] whilst M[20] (h20i) 6= 0 (recall that h20i = [q 2 t, 1]). Hence, the polynomial P (x1 ) = M[2,0] (x1 t, 1) vanishes for [x1 ] = [1] = h0i and [x1 ] = [q] = h1i whilst P (h2i) = M[2,0] (q 2 t, 1) 6= 0. This agrees with the definition of M[2] (x1 ) up to a multiplicative factor.
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CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
For almost rectangular partitions vm,k := [mN −k , (m + 1)k ] we prove that the corresponding Macdonald polynomials nicely factorize: Proposition 16. For m ≥ 0 and 0 ≤ k < N we have ) Mv(N mk
= (−1)
N m+k
q
m(k+(m−1)N/2)
N −k Y
xi , q
−1
i=1
m
N Y
xi , q −1
i=N −k+1
m+1
.
Proof. Recall u.Φ = [u [2] , . . . , u [N − 1] , u [1] + 1], thus vm,k .Φ = vm,k+1 for 0 ≤ k < N and vn,N −1 .Φ = vm+1,0 . The claim is obvious for v0,0 . Suppose the claim is true for some (m, k), then ) Mv(N τ m,k
=q
m(k+(m−1)N/2)
(−1)
N m+k
NY −k−1
xi , q
−1
i=1
m
N −1 Y
xi , q −1
i=N −k (N )
m+1
q −1 xN , q −1
m
.
Multiply both sides by (xN − 1) to get an expression for Mvm,k+1 (or vm+1,0 ); observe (xN − 1) (q −1 xN , q −1 )m = − (xN , q −1 )m+1 . This completes the inductive proof.
As a straightforward consequence, setting k = 0 in the previous proposition, Mvn,0 is symmetric in x1 , . . . , xN and then we find: Q (k) Corollary 17. For n ≥ 1 MSvn,0 = q kn(n−1)/2 (−1)kn ki=1 (xi , q −1 )n ; note vn,0 = nk .
Let X = {x1 , . . . , xN } and Y = {y1 , . . . , yk } be two alphabets. We will denote by Q Qk R(X, Y) = N j=1 (xi − yj ) the resultant of X with respect to Y. i=1 Let also y DN,k,m (x1 , . . . , xk ) := R([x1 , . . . , xk ], [ytN −k q 0 , . . . , ytN −k q m−1 ]).
We have:
Proposition 18. Let λ = mk , 0N −k for some 1 ≤ k < N and m ≥ 1, and y 6= 0 then x1 xk N −k−1 (N ) y ,..., ,t , . . . , 1 = y −km DN,k,m (x1 , . . . , xk ). MSλ y y Proof. By the previous propositions k Y x1 xi xk N −k−1 (N ) km k(m−1) k(N −k)m m −1 MSλ (−1) , ,..., ,t ,...,1 = t q2 ,q N −k y y yt m i=1
and
xi , q −1 N −k yt
m
=
m−1 Y
1−
j=0
m
= (−1) q
xi ytN −k q j
−m(m−1)/2 −m(N −k) −m
t
y
m−1 Y j=0
for 1 ≤ i ≤ k.
xi − ytN −k q j ,
RECTANGULAR MACDONALD POLYNOMIALS
19
6.2. Homogeneity of MSmk at 1 − q m−1 tN −k+1 = 0. Let us describe first the solution set Z of the conditions q m−1 tN −k+1 = 1 and q (m−1)/a t(N −k+1)/a 6= 1 for any common divisor a of m − 1 and N − k + 1 with a > 1. Let d := gcd (m − 1, N − k + 1) , m0 = m−1 , n0 = N −k+1 . d d Recall the Euler φ-function, φ (d) = #Kd where Kd := {k : 1 ≤ k < d, gcd (k, d) = 1} . We claim that Z has φ (d) disjoint connected components as a subset of C2 . Let z = re2πiθ ∈ C with r > 0 and θ ∈ R, and set q = z n0 , t = ωz −m0 , ω = e2iπψ , then q m0 tn0 = ω n0 and it is required that ω n0 d = 1 and (ω n0 )j 6= 1 for 1 ≤ j < d. (if d = 1 the latter restriction becomes void). For determining connected components it suffices to let r = 1 and project onto the 2-torus realized as the unit square {(α, β) : 0 ≤ α, β ≤ 1} with identifications (α + 1, β) ≡ (α, β) and (α, β + 1) ≡ (α, β), with the mapping (α, β) 7−→ e2πiα , e2πiβ . The pre-image of (q, t) is {(n0 θ, −m0 θ + ψ) : θ ∈ R} .
We require ψn0 d ∈ Z and ψn0 j ∈ / Z for 1 ≤ j < d, that is ψ = nk0 d for some k and kj ∈ / Z (when d = 1 the value k = 0 is permissible). This is equivalent d e to gcd (k, d) = 1. By the identification assume 0 ≤ ψ < 1. So there are #K e := {k : 1 ≤ k ≤ n0 d, gcd (k, d) = 1}, and #K e = n0 φ (d). solutions for ψ wheren K o e let Zk = For k ∈ K n0 θ, −m0 θ + k : 0 ≤ θ < 1 . Then Zk meets Zk′ where n0 d
n0
k ′ = (k − m0 d) mod (n0 d) (the first coordinate is periodic for θ 7→ θ+ n10 ; and where 0 ≤ a mod b < b for a ∈ Z, b ≥ 1). This is continued to show that Zk is connected to Zk′ with k ′ = (k − m0 dj) mod (n0 d) , 1 ≤ j < n0 . Consider the set e ′ := {(k − m0 dj) mod (n0 d) , 0 ≤ j < n0 , k ∈ Kd } . K
e If k, k ′ ∈ Kd and k 6= k ′ then (k − m0 dj) 6= We claim this set coincides with K. (k ′ − m0 dj ′ ) mod (n0 d) for any j, j ′ (else k = k ′ mod d); suppose 0 ≤ j < j ′ < n0 then (k − m0 dj) 6= (k − m0 dj ′ ) mod (n0 d) or else m0 d (j − j ′ ) = n0 dl for some l, which is e′ ⊂ K e and #K e ′ = #K e and hence K e ′ = K. e impossible for 1 ≤ j − j ′ < n0 . Thus K S Thus Z is the disjoint union of the connected sets {Zk+jd : 0 ≤ j < n0 } × {r : r > 0} for k ∈ Kd . Example 19. (1) First examine the solution of q a tb = 1 where gcd(a, b) = 1 on the example a = 4 and b = 7. Topologically we can collapse the problem to the torus [q = exp(2iπα), t = exp(2iπβ)] , 0 ≤ α, β ≤ 1. On the unit square this
20
CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
becomes 7α + 4β = 0 mod 1, or β = − 47 α, interpreted periodically, period 1 see fig.1.
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
Figure 1. Solution of q a tb = 1 for a = 4 and b = 7. (2) When gcd(a, b) > 1, for instance a = 8 and b = 12, there are several connected solutions. In the example given in fig.2, there are two solutions 2α + 3β = 41 and 2α + 3β = 43 . Proposition 20. Consider q = z a and t = ωz b a specialization of the parameters such m−1 N−k+1 that 1 − q m−1 tN −k+1 = 0 and 1 − q d t d 6= 0 if d > 1 divides m − 1 and N − k + 1. Then, MSmk = Pmk Proof. From Lemma 4, one has MS mk = Pmk
X τµ mk Pmk (h0i) Pµ . + µ q−1 ,t−1 Pµ (h0i) τmk k µ(m
Pmk (h0i) Pµ (h0i)
= 0 for µ ( mk . Furthermore, from Lemmas 10 and 11 there k (otherwise the denominators of are no poles at q = z a and t = ωz b in Pµ and mµ
From Lemma 12,
q −1 ,t−1
the two polynomials have (1 − q m−1 tN −k+1 ) as a factor which contradicts the lemmas). The result follows.
RECTANGULAR MACDONALD POLYNOMIALS
21
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
t
Figure 2. Solution of q a tb = 1 for a = 12 and b = 8. 7. Proof of a conjecture of Forrester 7.1. Clustering properties of Pmk . In the study of fractional quantum Hall states, Bernevig and Haldane [4] identified certain clustering conditions on rectangular Jack polynomials. These can be interpreted as factorization properties under certain specialization of the variables. In this context, Baratta and Forrester conjectured [2] a more general identity involving a rectangular Macdonald polynomial: (42)
1
Pmk (y, yq α , . . . , yq
N−k−1 α
y , xN −k , . . . , xN ) = DN,k,m (x1 , . . . , xk ),
−k+1 where N ≥ 2k, α = − Nm−1 and gcd(N − k + 1, m − 1) = 1. Propositions 20 and 18 allow us to write a more general formula.
Theorem 21. Let y ∈ N. Consider q = z a and t = ωz b a specialization of the m−1 N−k+1 parameters such that 1 − q m−1 tN −k+1 = 0 and 1 − q d t d 6= 0 if d > 1 divides m − 1 and N − k + 1. Then, y Pmk (x1 , . . . , xk , ytN −k−1, . . . , yt, y) = DN,k,m (x1 , . . . , xk ).
Proof. Propositions 20 and 18 shows the formula for y 6= 0. Since the coefficients of xv in the left hand side and right hand side are polynomial in y, the result remains true for y = 0.
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CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
Example 22. Let us illustrate this result with few examples: (1) Set q = t−3 then y y P[2,2,0,0] (x1 , x2 , yt, y) = (x1 − )(x2 − )(x1 − yt2 )(x2 − yt2) t t −1 (2) Let t = −q , one has: y P[3,0] (x1 , y) = (x1 + )(x1 + y)(x1 + qy). q whilst the result does not hold for t = q −1 : ! 2 1 P[3,0] (x1 , y) = x21 − 1 + yx1 + y 2 (x1 + y). q 2πi (3) Let ω = exp 3 and t = ωq −1. We have
P[4,4,0,0] (x1 , x2 , ty, y) = (x1 − ω 2 yq −2)(x2 − ω 2 yq −2)(x1 − ω 2yq −1 )(x2 − ω 2 yq −1)× ×(x1 − ω 2 y)(x2 − ω 2 y)(x1 − ω 2yq)(x2 − ω 2 yq), whilst the result does not hold for t = q −1 , since there is a pole in P[4,4,0,0] : Numer(P[4,4,0,0 (x1 , x2 , ty, y))|t=q−1 = 3
y4 (1 − q 2 )(1 − q 3 )2 (1 − q 4 )x21 x22 , q6
where Numer denotes the numerator of the reduced fraction. ( N−k+1 ) (α) 7.2. The Jack polynomial Pmk 1−m . The Jack polynomial Pλ is usually recovered from the Macdonald polynomial Pλ (x1 , . . . , xN ; q, t) by setting q = tα and taking the limit t → 1. So as a consequence of Theorem 21 we obtain Corollary 23. Let N, k ≥ 1 and m ≥ 2 verifying N ≥ 2k and gcd(N −k+1, m−1) = 1. Then we have ! ×N −k k N−k+1 Y z }| { ( 1−m ) x1 , . . . , xk , y, . . . , y = Pmk (xi − y)m. i=1
See Kakei et al. [14] for related results on shifted Jack polynomials.
7.3. Clustering properties of nonsymmetric Macdonald polynomials. As before we use (∗) to indicate a term q a tb with a, b ∈ Z when its value can be ignored in the context of analyzing zeros and poles. We restate a result of Lascoux et al.[22, Corollary u N 4.3 from Sahi’s binomial formula]. For u ∈ NN 0 let Bu := v ∈ N0 : v 6= 0, v 6= u . As stated before v ∈ Bu implies v + ⊂ u+ and |v| < |u|, but there are important additional consequences as we will show. Theorem 24. For u ∈ NN 0 Mu (x) = Eu (x) +
X
v∈Bu
(∗)
hui v
q −1 ,t−1
Eu (h0i) Ev (x) . Ev (h0i)
RECTANGULAR MACDONALD POLYNOMIALS
23
We need to use this formula for u = mk , 0N −k with 2k ≤ N and (q, t) in a neighborhood Ω of the set of solutions of q m tN −k+1 = 1 and q a tb 6= 1 for 1 ≤ a ≤ m and 0 ≤ b ≤ N − k + 1, except for a = m and b = N − k + 1. (We note here that if q, t ∈ C and neither q nor t are roots of unity then q m tN −k+1 = 1 and q a tb = 1 imply that a = nd m, b = nd (N − k + 1) where d = gcd (m, N − k + 1) and n ∈ Z; roughly the equations imply q am tbm = 1 = q am ta(N −k+1) , etc.) From Knop [16, Prop. 5.3] the coefficients of hq,t (v, qt) Mv (x) are in Z [q, q −1 , t, t−1 ] for any v ∈ NN 0 (the same multiplier works for Ev (x)). Furthermore Mv (hvi) = (∗) hq,t (v, q) (from Lascoux et al.[22, (tN q,q,t) p.8]), and Ev (h0i) = (∗) hq,t (v,qt)v+ by Proposition 5. Thus we consider hq,t (v, qt) and hq,t (v, q) for v ∈ Bu . We showed that v + ⊂ u and |v| < |u| implies tN q, q, t v+ 6= 0 for (q, t) ∈ Ω. Proposition 25. Suppose u ∈ NN 0 and ℓ (u) ≤ k < N then v ∈ Bu implies ℓ (v) ≤ k.
Proof. We use the “extra-vanishing” theorem of Knop [16, Thm. 4.5] applied to v with Mv (hui) 6= 0. This implies that there is a permutation w ∈ SN such that for any i with 1 ≤ i ≤ N the inequality i ≥ w (i) implies v [i] ≤ u [w (i)] otherwise v [i] < u [w (i)]. By hypothesis u [j] = 0 for any j > k hence w −1 (j) ≥ j (since v [w −1 (j)] < 0 is impossible). Thus w −1 (j) ≥ j for each j > k and by downward induction w (j) = j for k < j ≤ N. By applying Knop’s condition it follows that v [j] ≤ u [w (j)] = u [j] = 0 for j > k. Corollary 26. If u = mk , 0N −k and v ∈ Bu then hq,t (v, qt) 6= 0 and hq,t (v, q) 6= 0 for (q, t) ∈ Ω.
Proof. The factors in the hook products are of the form 1 − q a tb with 1 ≤ a ≤ m and 0 ≤ b ≤ k < N − k + 1 (the leg-lengths are bounded by k − 1).
This shows that
h
mk v
i
has no poles for v ∈ Bmk and (q, t) ∈ Ω. The last step is
q −1 ,t−1 m N −k+1
to show Emk (h0i) = 0 for q t
= 1 and (q, t) ∈ Ω, and indeed
k Y m Y tN q; q, t mk = 1 − q j tN −i+1 ; i=1 j=1
the factor with i = k, j = m is 1 − q m tN −k+1 . Proposition 27. Suppose u = mk , 0N −k , 2k ≤ N, q m tN −k+1 = 1 and q m/a t(N −k+1)/a 6= 1 for a > 1 being a common divisor of m and N − k + 1, then N −k+1
Eu x1 , . . . , xk , yt
Mu (x) = Eu (x) , y . . . , yt, y = DN,k,m (x1 , . . . , xk ) .
Proof. The proof is essentially the same as that of Proposition 21.
24
CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
We find that the polynomial Eu (x) is singular for certain values of (q, t) exactly when it coincides with Mu (x), and in that case there is a factorization result of clustering type. 7.4. Highest weight and singular Macdonald polynomials. In the casePof special parameter values when MSλ = Pλ by Proposition 2 it follows that MSλ N i=1 Ξi = PN PN PN Pλ i=1 Ξi = Pλ i=1 ξi and thus Pλ i=1 Di = 0. For rectangular partitions, this matches with a result of one of the authors (J.-G.L.) with Th. Jolicœur [18] which involves the kernel of the operator N Y N X txi − xj ∂ L+ := xi − xj ∂q xi i=1 j=1 j6=i
with ∂ f (x1 , . . . , xN ) − f (x1 , . . . , xi−1 , qxi , xi+1 , . . . xN ) f (x1 , . . . , xN ) = . ∂q xi xi − qxi Note from a result of Baker and Forrester [1, eq. (5.9) p12], the operator L+ satisfies (1 − q)L+ =
N X
Di .
i=1
Theorem 28. (Jolicœur-Luque)[18] If N ≥ 2k then the Macdonald polynomial k (x1 , . . . , xN ; q, t) belongs to the kernel of k−1−N Pmm−1 where d = gcd(m − 1, N − k − 1) and L+ for the specialization (q, t) = z d , ωz d n o ω = exp 2iπ(1+dn) with n ∈ Z. m−1 We observe similar phenomena for nonsymmetric Macdonald polynomials: Note Eu , Mu have the same eigenvalues under the action of ξi , Ξi respectively. If for certain parameters (q, t) Mu = Eu then Eu Di = 0 for 1 ≤ i ≤ N (since Mu Ξi = Eu Ξi = Eu ξi + Eu Di ). According to Proposition 27: Suppose u = mk , 0N −k 2k ≤ N and q m tN −k+1 = 1 and q m/a t(N −k+1)/a 6= 1 for a > 1 being a common divisor of m and N − k + 1, then Emk 0N−k Di = 0 for each 1 ≤ i ≤ N. 8. Conclusion and perspectives One of the authors (J.-G.L.) with Thierry Jolicoeur[18] investigated other families of symmetric Macdonald polynomials belonging to the kernel of L+ for certain specializations of the parameters. These polynomials are indexed by staircase partitions and numerical evidence shows that they have nice factorization properties. A staircase ℓ+1 partition λ = [((β + 1)s + r)k , (βs + r)ℓ , . . . , (s + r)ℓ , 0 s−1 ] is defined by five integer ℓ+1 r + ℓ(β + 1) + k ∈ N. The parameters β, s, r, k and ℓ such that k ≤ ℓ and N = s−1 s−1
l+1
corresponding specialization is (t, q) = (z g , z − g ω) where g = gcd(ℓ + 1, s − 1) and n o ω = exp 2iπ(1+dg) . The simplest polynomials (r = k = 0) can be factorized for any s−1
RECTANGULAR MACDONALD POLYNOMIALS
25
variables x1 , . . . , xN as a (q, t)-discriminant. When g = 1, this is a consequence of a theorem proved by one of the authors (J.-G.L) with A. Boussicault [8, Theorem 3.2]. For instance, P420 (x1 , x2 , x3 ; q = t−2 , t) = (−)t (x1 −tx2 )(x1 −tx3 )(x2 −tx1 )(x2 −tx3 )(x3 −tx1 )(x3 −tx1 ), where (−)t denotes a factor depending only on t. But it seems that a factorization holds also for the other cases (for special values of the variables and the parameters). Let us give a few examples: P630 (x1 , x2 , x3 ; q = −t−1 , t) = (−)t (x1 + x2 )(x1 + x3 )(x2 + x3 )(x1 − tx2 )(x1 − tx3 ) (x2 − tx1 )(x2 − tx3 )(x3 − tx1 )(x3 − tx1 ), P53000 (x1 , x2 , y, yt, yt2; q = t−2 , t) = (−)t (x1 − yt3 )(x1 − yt)(x1 − yt−1 ) (x2 − yt3)(x2 − yt)(x2 − yt−1), P42200 (x1 , y1 , y1t, y2 , y2 t; q = t−3 , t) = (−)t (x1 − y1 t2 )(x1 − y2 t2 )(x1 − y1 t)(x1 − y2 t) (y1 − ty2 )(y1 − t2 y2 )(y2 − ty1 )(y2 − t2 y1 ), P6400000 (x1 , y1 , y1 t, y1 t2 , y2 , y2 t, y2 t2 ; q = t−2 , t) = (−)t (x1 − y1 t3 )(x1 − y1 t3 )(x1 − y1 t−1 ) (x1 − y2 t3 )(x1 − y2 t3 )(x1 − y2 t−1 )(y1 − ty2 )(y1 − t3 y2 )(y2 − ty1 )(y2 − t3 y1 ), P7507 (x1 , y1 , . . . , y1 t4 , y2, . . . , y2 t2 ; q = t2 , t) = (−)t (y1 − x1 t3 )(y1 − y2 t3 )(y1 − x1 t5 ) (y1 − y2 t5 )(y2 − x1 t3 )(y2 − y1 t3 )P420 (x1 , y1 , y2; q = t−2 , t), . . . A correct formula (and of course a proof) remains to be found. Note there are also analogous formulas for singular nonsymmetric Macdonald polynomials indexed by staircase partitions. Consider the following examples: E210 (x1 , x2 , x3 ; q = z −2 , t = z) = (−)t (tx2 − x1 )(tx3 − x1 )(tx3 − x2 ) E630 (x1 , x2 , x3 ; q = z −2 , t = z 3 ) = (−)z (x2 z − x3 ) (−zx3 + x2 ) (x2 − z 3 x3 ) (x1 z − x3 ) (−zx3 + x1 ) (x1 − z 3 x3 ) (x1 z − x2 ) (x1 − x2 z) (x1 − x2 z 3 ) , E420 (x1 , x2 , x3 ; q = −t, t) = (−)t (x2 + x3 )(−tx3 + x2 )(x3 + x1 )(−tx3 + x1 )(x1 + x2 )(x1 − x2 t), E221100 (x1 , x2 , y1 , ty1 , y2, ty2 ; q = t−3 , t) = (−)t (y1 − y2 t2 )(y1 − y2 t)(x2 − y2 t2 ) (x2 − t2 y1 )(x1 − y2 t2 )(x1 − t2 y1 ), E442200 (x1 , y1 , y1 z 2 , y2 , y2z 2 ; q = z −3 , t = z 2 ) = (−)z (y1 − y2 z 4 )(y1 − y2 z)(y1 z − y2 ) (y1 − y2 z 2 )(x1 − y2 z 4 )(x1 − y2 z)(x1 − y1 z 4 )(x1 − y1 z). There are also equations involving permutations of (mk , 0N −k ). For instance: E0022 (t, 1, x3 , x4 ; q = t−3 , t) = (−)t (tx4 − y)(tx3 − y)x4 x3 . In appendix C we show a factorization formula for permutations of mk 0N −k . The connections with singular properties remain to be investigated. In particular, if Eu is singular then it has the same spectral vector for {Ξi } as Mu . If the spectral vector is nondegenerate (multiplicity = 1) then Mu = Eu .
26
CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
Appendix A. About notations For parameters a, b, c, we consider the operator Gia,b,c acting on polynomials (or Laurent polynomials) by bxi − cxi+1 Gia,b,c = a + (1 − si ) , 1 ≤ i < N. xi − xi+1 The braid relations Gi Gi+1 Gi = Gi+1 Gi Gi+1 hold for two families of solutions: (a, b, c) = (t1 , −t1 , t2 ) and (a, b, c) = (t1 , t2 , −t1 ). In both cases the quadratic relation Gia,b,c − t1 Gia,b,c − t2 = 0
holds. The inverses are
Git1 ,−t1 ,t2 Git1 ,t2 ,−t1
−1
−1
= Gi1/t1 ,1/t2 ,−1/t1 , = Gi1/t1 ,−1/t1 ,1/t2 , .
Some pertinent evaluations are 1Gia,b,c = a, xi Git1 ,−t1 ,t2 = −t2 xi+1 , xi+1 Git1 ,t2 ,−t1 = −t2 xi . In our applications we require the quadratic relation Gia,b,c − t Gia,b,c + 1 = 0, and the evaluation 1Gia,b,c = t, because 1 is (trivially) symmetric, thus a = t, b = −1. The version of Ti used in one of our previous papers [9] is Git,−t,−1 , but Lascoux et al.[22] −1 uses Git,−1,−t which equals Gi1/t,−1/t,−1 (that is, Ti (1/t)−1 from our notations[9]). With parameters 1t , 1q in relation to our notations [9] the following hold for the Lascoux et al. [22] versions (and Ti = Git,−1,−t , Ti−1 = G1/t,−1/t,−1 ):
(1) ξi = t1−i Ti−1 . . . T1 τ TN−1−1 . . . Ti−1 (Cherednik operator) (2) pτ (x) = p (xN /q, x1 , . . . , xN −1 ) ; (3) DN = (1 − ξN ) x−1 N ; a simple check for the coefficient of ξN : apply the operator to the constant polynomial 1, since 1Ti = t we have 1ξi = ti−N and it is necessary that 1DN = 0. (4) Di = tTi−1 Di+1 Ti−1 (5) Ξi = t1−i Ti−1 . . . T1 τ 1 − x1N TN−1−1 . . . Ti−1 + x1i ; similarly check the constants 1 Ti−1 = x1i . by verifying 1Ξi = ti−N ; note xi+1 Appendix B. Alternative proof of Lemma 12
′′
′
Let λ ⊂ [mk , 0N −k ] with 2k ≤ N. We have λ = [mk , λk′ +1 , . . . , λk′′ , 0N −k ] with ′ k ≤ k ′′ ≤ k and λk′ +1 < m. The vector hλ− i∞ with λ ( mk is [tN −k
′′ −1
′
, . . . , t, 1, ∗, . . . , ∗, q m tN −1 , . . . , q m tN −k , qtN −k
′′ −1
, . . . , qt, q, . . . ]
So we have to enumerate the pairs (i, j) with i = 1 . . . k ′ and j = 1 . . . N − k ′′ verifying m N −i N −k hλ− i∞ , hλ− i∞ N +j = qt N −k ′ +i = q t
′′ −j
and
hλ− i∞ N−k′ +i hλ− i∞ N+j
= q m−1 tN −k+1 . Then, we
have to enumerate the pairs (i, j) with i = 1 . . . k ′ , j = 1 . . . N − k ′′ such that N − i − (N − k ′′ − j) = N − k + 1. Equivalently, j = N − k − k ′′ + 1 + i. Hence, when i ranges 1 . . . k ′ , j ranges N − k − k ′′ + 2 . . . N − k + k ′ − k ′′ + 1. But since k ≤ N − k and k ′′ ≤ k
RECTANGULAR MACDONALD POLYNOMIALS
27
we have 2 ≤ k − k ′′ + 2 ≤ N − k − k ′′ + 2. If λ ( [mk , 0N −k ] then k ′ < k and we have N − k + k ′ − k ′′ ≤ N − k ′′ . To resume: when λ ( [mk , 0N −k ], j ranges N − k − k ′′ + 2 . . . N − k + k ′ − k ′ + 1 ⊂ 1 . . . N − k ′′ . So there are N − k + k ′ − k ′′ + 1 − (N − k − k ′′ + 2) + 1 = k ′ pairs (i, j) satisfying the property. It follows that the maximal power of (1 − q m−1 tN −k+1 ) in the denominator of equality (37) is k ′ . But if λ = [mk , 0N −k ] then j ranges N − 2k + 2 . . . N − k (k = k ′ = k ′′ ); so the maximal power is k ′ − 1 = k − 1. Similarly, we study the numerator: we have to enumerate the pair (i, j) with i = 1 . . . k ′ , j = 1 . . . N − k ′′ such that N − i − (N − k ′′ − j) + 1 = N − k + 1. Equivalently, j = N −k−k ′′ +i. Hence, when i ranges 1 . . . k ′ , j ranges N −k−k ′′ +1 . . . N −k+k ′ −k ′′ . Hence, when i ranges 1 . . . k ′ , j ranges N − k − k ′′ + 1 . . . N − k + k ′ − k ′′ . Again N − k − k ′′ + 1 . . . N − k + k ′ − k ′′ ⊂ 1 . . . N − k ′′ . So there are N − k + k ′ − k ′′ − (N − k − k ′′ + 1) + 1 = k ′ pairs (i, j) satisfying the property. As a consequence the maximal power of (1 − q m−1 tN −k+1 ) in the numerator of (37) equals k ′ . If λ ( [mk , 0N −k ], the numerator and the denominator simplify and we have no factor of the form (1 − q m−1 tN −k+1 ) in (37). On the other hand, when λ = [mk , 0N −k ], after simplification it remains a factor (1 − q m−1 tN −k+1 ) in (37). This completes the proof. Appendix C. Expressions for Mu where u is a permutation of [mk 0N −k ] Fix m ≥ 1 and k with 1 ≤ k ≤ N. As before we neglect scalar multipliers (∗) of the form q a tb , and then adjust the formulae using the monic properties. Define ΩN,k := β ∈ Nk : 1 ≤ β [1] < β [2] < . . . < β [k] ≤ N
For β ∈ ΩN,k define u (β) ∈ NN by u (β) [β [i]] = m for 1 ≤ i ≤ k otherwise u (β) [j] = 0. Let σ (β) denote the set {β[i] : 1 ≤ i ≤ k}. For 1 ≤ i ≤ k define εi by εi [j] = δij .
Remark 29. Note if i = β[j] − 1 6∈ σ(β) for some j then u(β).si = u(β − εj ). For instance, if N = 7 and β = [1, 4, 5, 7] then u(β) = [m, 0, 0, m, m, 0, m] and u(β)s6 = [m, 0, 0, m, m, m, 0] = u([1, 4, 5, 6]) = u(β − ε4 ). Definition 30. For 1 ≤ i ≤ N let χβ (i) = N − i − # {l : β [l] > i}, (if i = β [l] then χβ (i) = N − i − k + l). Also let x(β) be the point given by xi , i ∈ σ (β) (β) . xi = χβ (i) t ,i ∈ / σ (β) (β)
The meaning of χβ is simply that χβ (i) = N − ru(β) [i] and xi = hu (β)i [i] for i∈ / σ (β). The following is one of the main results of this section. The proof consists of several lemmas. Theorem 31. For β ∈ ΩN,k (β)
Mu(β) x
=
k Y j=1
N +j−k−β[j]
xβ[j] − t
Y m−1 i=1
xβ[j] − tN −k q i
28
CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
Lemma 32. Suppose v ∈ NN , v [i] < v [i + 1] for some i < N and y is a point such that Mv (y) = 0, then yi − tyi+1 Mv (y.si ) . Mv.si (y) = yi − yi+1 Proof. This follows from Definition 1 of Ti .
Lemma 33. Suppose β ∈ ΩN,k and β [j] − 1 ∈ / σ (β) for some j. Set i = β [j] − 1 so that u (β) [i] = 0, u (β) [i + 1] = m and u (β) .si = u (β − εj ) (see Remark 29). Then Mu(β) x(β) .si = 0.
Proof. Let v ∈ NN satisfy v [l] ≥ 1 for l ∈ σ (β − εj ) and v [l] = 0 for l ∈ / σ (β − εj ) (in particular v [i] ≥ 1 and v [i + 1] = 0). By Knop’s “extra-vanishing” theorem [16, Thm. 4.5] Mu(β) (hvi) = 0. To prove by contradiction suppose Mu(β) (hvi) 6= 0 then there is a permutation w ∈ SN such that for any i with 1 ≤ i ≤ N either i ≥ w (i) and u (β) [i] ≤ v [w (i)] or i < w (i) and u (β) [i] < v [w (i)]. Let S1 = {1, . . . , N} \σ (β) and S2 = {1, . . . , N} \σ (β − εj ). The sets S1 and S2 agree with the exception i ∈ S1 \S2 and i + 1 ∈ S2 \S1 . By construction w (l) ∈ S2 implies l ∈ S1 and l ≥ w (l). Thus w maps S1 onto S2 and by induction (using l ≥ w (l)) we see that w (l) = l for each l ∈ S1 with l < i. But then w (i) ≥ i + 1, which is impossible. Consider the vector hvi; by construction v[l] N −r [l] v q t , l ∈ σ (β − εj ) hvi [l] = . χβ (l) t ,l ∈ / σ (β − εj ) Note 1 ≤ rv [l] ≤ k for l ∈ σ (β − εj ), and hvi = x(β) .si where each xl is of the form q a tb with a ≥ 1 and N − k + 1 ≤ b ≤ N. Now Mu x(β) .si is a polynomial in k variables of degree mk which vanishes at infinitely many points of this form; by the uniqueness of the vanishing property of shifted Macdonald polynomials Mu x(β) .si = 0 for all x(β) . Lemma 34. Suppose β = [N − k + 1, . . . , N] then x(β) = tN −k+1 , . . . , t, 1, xN −k+1, . . . , xN and (β)
Mu(β) x
=
N Y
(xi − 1)
m−1 Y j=1
i=N −k+1
xi − tN −k q j .
i h k N −k . By the results of Section 6.1 Proof. Let v = (m − 1) , 0 N −k−1
Mv x1 , . . . , xk , t
,...,1 =
k m−2 Y Y i=1 j=0
xi − tN −k q j ;
RECTANGULAR MACDONALD POLYNOMIALS
29
if m = 1 the product equals 1. From v.Φk = u (β) it follows that Mu(β) (x) = (∗) Mv (xN −k+1 /q, . . . , xN /q, x1 , . . . , xN −k )
N Y
(xi − 1) .
i=N −k+1
Set xi = tN −k−i for 1 ≤ i ≤ N − k, then
QN Mu(β) (x) = (∗) Mv xN −k+1 /q, . . . , xN /q, tN −k−1 , . . . , 1 i=N −k+1 (xi − 1) Qm−2 QN = (∗) i=N −k+1 (xi − 1) j=0 xi /q − tN −k q j .
This proves the Lemma.
For β ∈ ΩN,k define Fβ (x) :=
k Y j=1
xβ[j] − tN +j−k−β[j] .
The special cases are β = [N − k + 1, . . . , N], Fβ = Q [1, 2, . . . , k], Fβ = ki=1 xi − tN −k .
QN
i=N −k+1
(xi − 1) and β =
Proof. (of Theorem 31): We use induction for steps of the form u (β) → u (β) .si applied to u (β) with u (β) [i] = 0 and u (β) [i + 1] = m. The previous Lemma provides the starting point. Suppose for some j, i that β [j] = i + 1 and i ∈ / σ (β), then with the (β) (β ′ ) ′ identification xi = xi+1 we have x .si = x , where β = β − εj . Then ′ xi − tN +j−k−i (β) F x , Fβ ′ x(β ) = β xi − tN +j−k−i−1 because only the factor involving xi changes. Suppose that the claimed formula holds for β, that is, Y m−1 (β) (β) Mu(β) x = Fβ x xβ[i] − tN −k q j . (β)
By Lemma 33 Mu(β) x
′
j=1
.si = 0 and by Lemma 32 ! (β ′ ) (β ′ ) ′ x − tx i i+1 Mu(β).si x(β ) = Mu(β) x(β) , (β ′ ) (β ′ ) xi − xi+1 (β ′ )
(β)
because x(β ) .si = x(β) . Also xi+1 = xi = tχβ (i) where χβ (i) = N −i−# {l : β [l] > i} = N − i − (k − j + 1). Thus ′ xi − tN +j−i−k (β ) Mu(β).si x = Mu(β) x(β) N +j−i−k−1 xi − t ′ Fβ ′ x(β ) = F x(β) Mu(β) x(β) , ) β(
and this proves the formula for β − εj .
30
CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
Example 35. Let β = [2, 5, 6, 9] ∈ Ω10,4 , then u (β) = [0m00mm00m0],
Mu(β)
x(β) = (t5 , x2 , t4 , t3 , x5 , x6 , t2 , t, x9 , 1) , x(β) = (x2 − t5 ) (x5 − t3 ) (x6 − t3 ) (x9 − t) Q Q 6 i × j∈{2,5,6,9} m−1 i=1 (xj − t q ) .
The factorization k N −k result for rectangular Eu can be adapted to this situation. Start with 2k ≤ N. with u = m , 0 For the rest of this section assume q m tN −k+1 = 1 and no relation q a tb = 1 with a < m or b < N − k + 1 holds (for details see Section 6.2). Then by Proposition 27 Mu (x) = Eu (x). We claim this equality can be extended to u (β) provided that u (β) is a reverse lattice permutation; this means that any substring [u (β) [j] , u (β) [j + 1] , . . . , u (β) [N]] contains at least as many 0’s as m’s [34, p.313]. This condition is equivalent to (43)
β [j] ≤ N − 2k + 2j − 1, 1 ≤ j ≤ k.
As above suppose β [j] = i + 1, β [j − 1] < i and v = u (β) , v.si = u (β − εj ). The inverse relation is (1 − ζ)2 1−t Mv = , Mv.si Ti − ζ (tζ − 1) (ζ − t) 1−ζ where ζ = q m tk+1+i−2j . The same transformation takes Ev.si to Ev . We use induction. Suppose Mu(β−εj ) = Eu(β−εj ) when q m tN −k+1 = 1 and both polynomials are defined (that is, the coefficients have no poles at q m tN −k+1 = 1). Then the same properties hold for Mu(β) and Eu(β) if ζ 6= t−1 , 1, t. Set q m = t−(N −k+1) then ζ = ta with a = 2k + i − 2j − N. But i = β [j] − 1 thus by condition (43) we have a ≤ 2k − 2j − N + (N − 2k + 2j − 1) − 1 = −2. The induction starts with β = [1, 2 . . . , k]. We have shown Proposition 36. Suppose 2k ≤ N, β ∈ ΩN,k , β satisfies (43), (and q m = t−(N −k+1) ) then Eu(β) (x) = Mu(β) (x) . (β)
For the specialization result, let z (β) be defined by zi χβ (i) yt if i ∈ / σ (β), for variables xi and y.
(β)
= xi if i ∈ σ (β), and zi
=
Proposition 37. Suppose 2k ≤ N, β ∈ ΩN,k , β satisfies (43), (and q m = t−(N −k+1) ), then k Y Y m−1 (β) N +j−k−β[j] Eu(β) z = xβ[j] − t y xβ[j] − tN −k q i y . j=1
i=1
RECTANGULAR MACDONALD POLYNOMIALS
Proof. With the notation of Definition 30 Eu(β) yx(β) = y mk Eu(β) x(β) = y mk Mu(β) x(β) = y
mk
k Y
N +j−k−β[j]
xβ[j] − t
j=1
Y m−1 i=1
31
xβ[j] − tN −k q i .
Replace xi by xi /y to finish the proof.
Example 38. Let β = [2, 5, 6, 9] ∈ Ω10,4 , q m t7 = 1 then Eu(β)
z (β) = (yt5 , x2 , yt4 , yt3, x5 , x6 , yt2 , yt, x9 , y) , z (β) = (x2 − yt5 ) (x5 − yt3 ) (x6 − yt3 ) (x9 − yt) Q Q 6 i × j∈{2,5,6,9} m−1 i=1 (xj − t q y)
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CHARLES F. DUNKL AND JEAN-GABRIEL LUQUE
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[email protected]., URL: http://people.virginia.edu/∼cfd5z/. Jean-Gabriel Luque: Universit´e de Rouen, Laboratoire d’Informatique , du Traitement de l’Information et des Syst`emes (LITIS), Avenue de lUniversit´e - BP 8 76801 Saint-´etienne-du-Rouvray Cedex, FR., Email:
[email protected]
