Séminaire Lotharingien de Combinatoire 61A (2010), Article B61Ai

GENERALIZATION OF SCOTT’S PERMANENT IDENTITY

Alain Lascoux

IGM, Université de Paris-Est 77454 Marne-la-Vallée CEDEX 2 Alain.Lascoux@univ-mlv.fr

                 http://phalanstere.univ-mlv.fr/∼al

dédié à la mémoire de Pierre Leroux

Abstract. Let x = {x_{1}, …, x_{r}}, y = {y_{1}, …, y_{n}}, z = {z_{1}, …, z_{n}} be three sets of indeterminates. We give the value of the determinant

| ∏_{x∈x}(xy − z)^{−1}|_{y∈y,z∈z}

when specializing y and z to the set of roots of y^{n} − 1 and z^{n} − ξ^{n}, respectively.

In the case where r = 2 and x = {1, 1} the determinant |(y − z)^{−2}|_{y∈y,z∈z} factorizes into the determinant of the Cauchy matrix [(y − z)^{−1}] and its permanent (see [1], respectively [11, vol. 2, pp. 173–175]). Scott [13, 10] found the value of this permanent when specializing y to the roots of y^{n} − 1 and z to the roots of z^{n} +1. Han [4] described more generally the case where z is the set of roots of z^{n} + az^{k} + b instead of z^{n} + 1.

Instead of restricting to r = 2 and specializing x, we shall consider the determinant

| ∏_{x∈x}(xy − z)^{−1}|_{y∈y,z∈z} ,

and obtain in Theorem 2 its value when specializing y and z. The remarkable feature is that this value is a product of sums of monomial functions in x without multiplicities, thus extending the factorized expressions of [13, 4].

We first need a few generalities about symmetric functions [7].

Research partially supported by the ANR project MARS (BLAN06-2 134516).

Given two sets of indeterminates x, z (we say alphabets), the complete functions S_{n}(x − z) are the coefficients of the generating function

∑_{n} γ^{n}S_{n}(x − z) = ∏_{z∈z}(1 − γz) ∏_{x∈x}(1 − γx)^{−1} .

For any r, any λ ∈ ℤ^{r}, S_{λ}(x − z) = det(S_{λ_{i}+j−i}(x − z)).

In the case where z = 0, and x of cardinality r, these functions can be obtained by symmetrization over the symmetric group 𝔖_{r}. Let π_{ω} be the following operator on functions in x:

f → fπ_{ω} := ∑_{σ∈𝔖_{r}} (f ∏_{1≤i<j≤r}(1 − x_{j}/x_{i})^{−1})^{σ} .

Then, when λ ≥ [1 − r, …, −1, 0] (i.e., λ_{1} ≥ 1 − r, …, λ_{r} ≥ 0), the monomial x^{λ} = x_{1}^{λ_{1}} ⋯ x_{r}^{λ_{r}} is sent to S_{λ}(x) under π_{ω}. When λ is a partition (i.e., λ_{1} ≥ ⋯ ≥ λ_{r} ≥ 0), S_{λ}(x) is the usual Schur function of index λ.

Let n be a positive integer, ξ an indeterminate, and z be the set of roots of z^{n} − ξ^{n}. Equivalently, e_{i}(z) = 0 for 1 ≤ i ≤ n − 1, e_{n}(z) = (−1)^{n−1}ξ^{n}. For any integer j, any x, one has

S_{j}(x − z) = S_{j}(x) − ξ^{n}S_{j−n}(x) ,

and, more generally, from the determinantal expression for Schur functions, for any λ ∈ ℕ^{r},

S_{λ}(x − z) = ∑_{u∈{0,n}^{r}} S_{λ−u}(x)(−1)^{|u|/n}ξ^{|u|} ,

where |u| denotes the sum of the coordinates of u.

In particular, when λ = \underbrace{n − 1, …, n − 1}_{r−1}, p (which we shall also denote as

λ = (n−1)^{r}p), then the terms with negative last component −(n−p) vanish, and the set {λ − u : u ∈ {0, n}^{r}} to consider is {[v, p] : v ∈ {n − 1, −1}^{r−1}}. Reordering indices, putting q = p − r + 1, one can rewrite the sum as

S_{(n−1)^{r−1}p}(x − z) = (−1)^{r−1} ∑_{u∈{0,n}^{r−1}} S_{q,u}(x)(−1)^{(|λ|−|u|−q)/n}ξ^{|λ|−|u|−q} . (1)

Since x^{v}π_{ω} = S_{v}(x), for any v ≥ [1 − r, …, −1, 0], one can rewrite (1) as a symmetrization of monomial functions in x − x_{1} = {x_{2}, …, x_{r}} :

S_{(n−1)^{r−1}p}(x − z) = (−1)^{r−1} ∑_{j=0}^{r−1} x_{1}^{q}m_{n^{j}}(x − x_{1})(−1)^{r−1−j}ξ^{|λ|−jn−q} π_{ω} . (2)

From the identity

m_{n^{j}}(x − x_{1}) = m_{n^{j}}(x) − x_{1}^{n}m_{n^{j−1}}(x) + x_{1}^{2n}m_{n^{j−2}}(x) + ⋯ + (−x_{1}^{n})^{j} ,

one sees that x_{1}^{q}m_{n^{j}}(x − x_{1}) π_{ω} is equal to

S_{q}(x)m_{n^{j}}(x) − S_{q+n}(x)m_{n^{j−1}}(x) + ⋯ + (−1)^{j}S_{q+jn}(x) .  (3)

On the other hand, S_{(n−1)^{r−1}p}(x − z) belongs to the linear span of Schur functions indexed by partitions μ such that μ_{1} ≤ n − 1. This implies that S_{(n−1)^{r−1}p}(x − z) belongs to the linear span of monomial functions indexed by the same set of partitions. Consequently, one can restrict the sum in (3) to the term (−1)^{j}S_{q+jn}(x).

In summary, one has the following expression for the specialization of the Schur function that we are considering.

Proposition 1. Let x be an alphabet of cardinality r, z be the set of roots z^{n} − ξ^{n} = 0, p ≤ n − 1, N = (n − 1)(r − 1). Then

S_{(n−1)^{r−1}p}(x − z) = ∑_{μ} m_{μ}(x) ξ^{N+p−|μ|} ,  (4)

where the sum is over all partitions μ ∈ ℕ^{r}, μ_{1} ≤ n−1, |μ| ≡ p−r+1 mod n.

For example, for n = 4, r = 2, one has

S_{30}(x − z) = m_{3}(x) + m_{21}(x), S_{31}(x − z) = m_{31}(x) + m_{22}(x) + ξ^{4} ,

S_{32}(x − z) = m_{32}(x) + ξ^{4}m_{1}(x), S_{33}(x − z) = m_{33}(x) + ξ^{4}(m_{2}(x) + m_{11}(x)) .

Let

D(x, y, z) = | ∏_{x∈x}(xy − z)^{−1}|_{y∈y,z∈z} .

In the case r = 2, this determinant has been obtained by Izergin and Korepin [5] as the partition function of the Heisenberg XXZ-antiferromagnetic model (modulo some normalization factor). Gaudin [3] had previously described the partition function of some other model as the determinant |(x − y)^{−1}(x − y + γ)^{−1}| for some parameter γ.

The Izergin–Korepin determinant is used in the enumeration of alternating sign matrices [2]. In that case, one first specializes x = {e^{2iπ/3}, e^{4iπ/3}}. Kuperberg [6] and Okada [12] evaluate more general partition functions corresponding to similar determinants or Pfaffians, and to other roots of unity (see also [8, Th. 7.2]).

We shall take another point of view, keep x generic, but specialize instead y and z. In [9, Formula 4], it is shown that the function

G(x, y, z) = \frac{D(x, y, z)}{Δ(z)} ∏_{x∈x} ∏_{y∈y} ∏_{z∈z}(xy − z)

is equal to the determinant of the matrix

[S_{(n−1)^{r−1}j}(y_{i}x − z)]_{j=0...n−1, i=1...n} ,  (5)

where Δ(z) = ∏_{1≤i<j≤n}(z_{i} − z_{j}).

For any k ∈ ℕ, let φ_{k} be the sum of all monomial functions m_{μ}(x) of degree k, with μ_{1} ≤ n − 1 (notice that φ_{k} = 0 when k > (n − 1)r). From (4), one sees that S_{(n−1)^{r−1}j}(y_{i}x − z) specializes, when z is the set of roots of z^{n} − ξ^{n}, to

S_{(n−1)^{r−1}j}(y_{i}x − z)

= y_{i}^{N+j}φ_{N+j} + ξ^{n}y_{i}^{N+j−n}φ_{N+j−n} + ξ^{2n}y_{i}^{N+j−2n}φ_{N+j−2n} + ⋯, (6)

where, as before, N = (n − 1)(r − 1). Further specializing y to the roots of y^{n} − 1, one sees that the matrix (5) factorizes into the product of the matrix [y_{i}^{(N+j)}], where the symbol (N + j) stands for the remainder of division of N + j by n, and the diagonal matrix

diag((φ_{N}+ξ^{n}φ_{N−n}+ξ^{2n}φ_{N−2n}+⋯), (φ_{N+1}+ξ^{n}φ_{N+1−n}+ξ^{2n}φ_{N+1−2n}+⋯),

⋯, (φ_{N+n−1} + ξ^{n}φ_{N+n−1−n} + ξ^{2n}φ_{N+1−n} + ⋯)) .

For example, for n = 3, r = 3,

S_{220}(y_{i}x − z) = y_{i}^{4}φ_{4} + y_{i}φ_{1}ξ^{3}, S_{221}(y_{i}x − z) = y_{i}^{5}φ_{5} + y_{i}^{2}φ_{2}ξ^{3},

S_{222}(y_{i}x − z) = y_{i}^{6}φ_{6} + y_{i}^{3}φ_{3}ξ^{3} + ξ^{6} ,

and the matrix factorizes into

[\begin{matrix}y_{1} & y_{1}^{2} & 1 \\ y_{2} & y_{2}^{2} & 1 \\ y_{3} & y_{3}^{2} & 1 \end{matrix}] [\begin{matrix}φ_{4} + φ_{1}ξ^{3} & 0 & 0 \\ 0 & φ_{5} + φ_{2}ξ^{3} & 0 \\ 0 & 0 & φ_{6} + φ_{3}ξ^{3} + ξ^{6} \end{matrix}] .

Taking into account that ∏_{x,y,z}(xy − z) specializes to ∏_{y,x}(x^{n}y^{n} − ξ^{n}) = ∏_{x∈x}(x^{n} − ξ^{n})^{n}, and that the determinant of powers of the y ∈ y is a permutation of the Vandermonde in y, one obtains the following theorem.

Theorem 2. Let n, r be two positive integers, N = (n − 1)(r − 1). Let x be an alphabet of cardinality r, y be the set of roots of y^{n} − 1, z be the set of roots of z^{n} − ξ^{n}. Then

Δ(y)Δ(z) |∏_{k=1}^{r}(x_{k}y_{i} − z_{j})^{−1}|_{i,j=1...n}

= \frac{(−1)^{(n−1)(n/2+r−1)}}{∏_{x∈x}(x^{n} − ξ^{n})^{n}} ∏_{i=0}^{n−1} (∑_{j=0}^{∞} φ_{N+i−nj} ξ^{nj}) .

For x = {1, 1}, this theorem is due to Han[4]. In that case, φ_{i} = i + 1 and φ_{n−1+i} = n − i for i = 0, …, n − 1, and the product appearing in the theorem is

n (n − 1 + ξ^{n}) (n − 2 + 2ξ^{n}) ⋯ (1 + (n − 1)ξ^{n}) .

For r = 5, n = 3, as a further example, the theorem yields the expression

∏_{k=1}^{5}(x_{k}^{3} − ξ^{3})^{−3} (φ_{8} + φ_{5}ξ^{3} + φ_{2}ξ^{6}) (φ_{9} + φ_{6}ξ^{3} + φ_{3}ξ^{6} + ξ^{9})

(φ_{10} + φ_{7}ξ^{3} + φ_{4}ξ^{6} + φ_{1}ξ^{9}) ,

which specializes, for x = {1, 1, 1, 1, 1}, to

(1 − ξ^{3})^{−15}(15 + 51ξ^{3} + 15ξ^{6})(5 + 45ξ^{3} + 30ξ^{6} + ξ^{9})(1 + 30ξ^{3} + 45ξ^{6} + 5ξ^{9}) .

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