The higher spin generalization of the 6-vertex model with domain wall boundary conditions and Macdonald polynomials

Transcription

1 J Algebr Comb (2015) 41: DOI /s The higher spin generalization of the 6-vertex model with domain wall boundary conditions and Macdonald polynomials Tiago Fonseca Ferenc Balogh Received: 18 October 2012 / Accepted: 18 August 2014 / Published online: 9 September 2014 Springer Science+Business Media New York 2014 Abstract The determinantal form of the partition function of the 6-vertex model with domain wall boundary conditions was given by Izergin. It is known that for a special value of the crossing parameter the partition function reduces to a Schur polynomial. Caradoc, Foda and Kitanine computed the partition function of the higher spin generalization of the 6-vertex model. In the present work, it is shown that for a special value of the crossing parameter, referred to as the combinatorial point, the partition function reduces to a Macdonald polynomial. Keywords Quantum integrable systems Combinatorics Mathematical physics Symmetric polynomials 6 Vertex model 1 Introduction The 6-vertex model with domain wall boundary conditions was introduced by Korepin in [11], where the partition function was shown to satisfy certain recursion relations. T. Fonseca Centre de Recherches Mathématiques, Université de Montréal, Montréal, QC, Canada tiago.dinis.da.fonseca@sapo.pt Present Address: T. Fonseca LAPTh and CNRS, 9 chemin de Bellevue, BP 110, Annecy-le-Vieux Cedex, France F. Balogh Centre de Recherches Mathématiques, Concordia University, Montréal, QC, Canada Present Address: F. Balogh (B) SISSA, via Bonomea, 265, Trieste, Italia fbalogh@sissa.it

2 844 J Algebr Comb (2015) 41: In [10], Izergin solved Korepin s recursion relations, expressing the partition function in a determinantal form. An interesting feature of the 6-vertex model is that it has multiple combinatorial interpretations since its configurations are in bijection with several combinatorial objects, such as alternating sign matrices, fully packed loops and states of a square ice model [2,4,13,14,17,18]. For example, Kuperberg [13] used the 6-vertex model to compute the number of alternating sign matrices. The 6-vertex model is an integrable model, meaning that it possesses an R-matrix which satisfies the Yang Baxter equation. It is this algebraic structure that allows the partition function to be calculated explicitly. To each edge of the 6-vertex model, a 1/2 spin is associated, since the corresponding R-matrix is intimately connected with the representation theory of the algebra sl 2.It is natural to search for generalizations of this model, where the representation of the underlying algebra or even the algebra itself is replaced (for example, by sl r ). The idea is to construct an R-matrix for the model that satisfies the Yang Baxter equation (see, e.g. [20]). If we restrict ourselves to studying general irreducible representations of sl 2, there is a systematic way of constructing the R-matrix, referred to as fusion [12,21]. The idea comes from the simple fact that in the representation theory of sl 2 we can build the spin l/2 representation through the fusion of l spin 1/2 representations. This was achieved, for the 6-vertex model with domain wall boundary conditions, in the work of Caradoc, Foda and Kitanine [3], which serves as the basis of our paper. The partition function of the 6-vertex model, as defined in Sect. 2, is a multivariate polynomial which is symmetric in two separate sets of variables (known as spectral parameters). It also depends on an extra parameter, normally denoted by q, and is referred to as the crossing parameter. If we set q = exp(2πi/3), known as the combinatorial point, the partition function becomes symmetric in the two sets of variables as a whole, and it simplifies to a Schur polynomial corresponding to a staircase partition (see [19,23]). The main goal of the present work is to prove an analogous result for the higher spin generalization of the six-vertex model with domain wall boundary conditions: by setting q = exp(2πi/(2l + 1)), the partition function reduces to a Macdonald polynomial corresponding to a staircase partition. As a consequence, the partition function is a symmetric polynomial in the full set of spectral parameters. 1.1 Outline of the paper The first two sections are introductory: Sect. 2 gives the definition of the 6-vertex model with domain wall boundary conditions and the explicit determinantal form of the partition function. In Sect. 3, the higher spin generalization of the 6-vertex model and the analogous representation of its partition function are presented. The following two sections contain the original results of the paper: Sect. 4 presents an alternative representation of the partition function in terms of determinants of scalar products of rational functions, and this reformulation is used to prove some basic but important properties of the partition function, valid for all values of q.

3 J Algebr Comb (2015) 41: In Sect. 5, we prove that the partition function satisfies the wheel condition when q = exp(2πi/(2l + 1)), and, using the results of [6], we show that there is a welldefined unique Macdonald polynomial satisfying the same vanishing constraints. It is shown that, up to a multiplicative constant, there is a unique polynomial with the prescribed degrees and symmetries that satisfies the wheel condition, and therefore, the partition function and the above mentioned uniquely fixed Macdonald polynomial coincide, up to an explicit multiplicative constant. The proof of the uniqueness lemma and the calculation of the proportionality constant are left to the appendices. 2 Review of the 6-vertex model In this section, we give a brief description of the 6-vertex model, on a square grid, with domain wall boundary conditions, following the construction presented in [8]. 2.1 Definition of the model Take a square grid of size n n, in which each edge is given an orientation (an arrow), such that at each vertex there are two incoming and two outgoing arrows, which gives six possibilities. Alternatively, the arrows can be represented by signs according to the rule that arrows pointing right or upward correspond to plus signs and arrows pointing downward or left correspond to minus signs. Impose the domain wall boundary conditions prescribing that the arrows at the top and bottom boundaries are outgoing, and the ones at the left and right boundaries are incoming. See, e.g. Fig. 1. To each vertex configuration we assign a weight a(x, y) = qx q 1 y w(x, y) = b(x, y) = x y c(x, y) = (q q 1 ) xy, (1) Fig. 1 A6 6 configuration of the 6-vertex model in terms of arrows (left)orsigns(right)

4 846 J Algebr Comb (2015) 41: a(x,y) b(x,y) c(x,y) Fig. 2 Weights of vertex configurations according to Fig. 2. The parameter q is called the crossing parameter of the model, while the parameters x and y, called spectral parameters, depend on the row and the column of the vertex, respectively. Let x ={x 1,...,x n } and y ={y 1,...,y n } be the horizontal and vertical spectral parameters, respectively. The weight of a configuration is defined by the product of the weights of the vertices. The partition function is defined as the sum of the weights over all possible configurations: Z n (x, y) := n configurations i, j=1 w ij (x i, qy j ), (2) where we have chosen to rescale the vertical spectral parameters by a factor of q.the partition function is renormalized in the following way: ) Z n (x, y) = ( 1) (n 2) q n 2 /2 (q q 1 ) n ( n i=1 x 1/2 i y 1/2 i Z n (x, y). (3) The function Z n (x, y) is a homogeneous polynomial of total degree n(n 1) and of degree n 1 in each variable x i or y i. 2.2 Integrability Let V be the standard representation of sl 2 spanned by the eigenvectors + and of S z. The key ingredient to the exact solvability of the 6-vertex model is the R-matrix representing an endomorphism a R(x, y) := 0 b c 0 0 c b 0 (4) a R : V V V V ɛ 1 ɛ 2 R ɛ 3ɛ 4 ɛ 1 ɛ 2 ɛ 3 ɛ 4, (5) where ɛ i {+, }. In what follows, we consider the vector space n k=1 V k, where each V k is a labelled copy of V. We use the abbreviated notation

5 J Algebr Comb (2015) 41: ɛ 1 ɛ 2 ɛ n := ɛ 1 ɛ 2 ɛ n, ɛ i =± (i = 1, 2,...,n) (6) for a canonical basis in the tensor product representation. The matrix R ij stands for the map that acts as R on V i V j and as the identity elsewhere. The action of the matrix R can be interpreted as completing an allowed sign configuration at a vertex with prescribed left and bottom signs, as shown below: The R-matrix satisfies the Yang Baxter equation: R 23 (y 2, y 3 )R 13 (y 1, y 3 )R 12 (y 1, y 2 ) = R 12 (y 1, y 2 )R 13 (y 1, y 3 )R 23 (y 2, y 3 ), (7) and the inversion equation: R 21 (y, x)r 12 (x, y) = (qy q 1 x)(qx q 1 y) Id. (8) The transfer matrix of the model is defined as T (x, y) = 0 R 0n (x, qy n )...R 02 (x, qy 2 )R 01 (x, qy 1 ) + 0, (9) where the matrix R 0i acts on the tensor product of the i th space and the so-called auxiliary space V 0. In terms of the transfer matrix, the partition function is given by Z n (x, y) = ++ + T (x 1, y)t (x 2, y)...t (x n, y). (10) Using the Yang Baxter equation, the renormalized partition function Z n (x, y), defined in Eq. (3), is shown to be the so-called Izergin Korepin determinant [10,11] Z n (x, y) = i, j (x i qy j )(x i q 1 y j ) (x) ( y) where (x) = 1 i< j n (x i x j ). ( det 1 i, j n 1 (x i qy j )(x i q 1 y j ) ), (11) 2.3 Combinatorial point When q = exp(2πi/3), the Izergin Korepin determinant (11) dramatically simplifies and becomes a Schur polynomial [19,23]: Z n (x, y) = s δn (x, y), (12) where δ n = (n 1, n 1, n 2,...,2, 1, 1, 0, 0). It follows that the partition function Z n (x, y), at the combinatorial point, is a fully symmetric polynomial in the 2n variables {x, y}.

6 848 J Algebr Comb (2015) 41: Fusion and the higher spin generalization of the 6-vertex model We introduce the higher spin generalization of the 6-vertex model considered in this paper. The corresponding R-matrix is constructed using fusion techniques for the representations of sl 2, as briefly explained below (see [21]). 3.1 Fusion and the generalized R-matrix The representation Sym l V is the irreducible component of V l spanned by the vectors l; l m := 1 ( S ) m m! }{{} m = 0, 1,...,l (13) l (see [9]). The R-matrix (5) can be used to build an endomorphism of the vector space with spectral parameters V l V l = V 1... V l V l+1... V 2l, (14) {x, q 2 x,...,q 2l 2 x, y, q 2 y,...,q 2l 2 y} (15) associated to the 2l factors of the tensor product decomposition (14). Definition 3.1 The operator is defined by R (l) (x, y) : V l V l V l V l (16) R (l) (x, y) := R 1,2l (x, q 2l 2 y)r 1,2l 1 (x, q 2l 4 y) R 1,l+1 (x, y) R 2,2l (q 2 x, q 2l 2 y)r 2,2l 1 (q 2 x, q 2l 4 y) R 2,l+1 (q 2 x, y). R l,2l (q 2l 2 x, q 2l 2 y)r l,2l 1 (q 2l 2 x, q 2l 4 y) R l,l+1 (q 2l 2 x, y). (17) Given that R satisfies the Yang Baxter equation Eq. (7) and the inversion relation (8), it can be shown that so does R (l). Proposition 3.2 The matrix R (l) (x, y) satisfies the Yang Baxter equation R (l) 23 (y 2, y 3 )R (l) 13 (y 1, y 3 )R (l) 12 (y 1, y 2 ) = R (l) 12 (y 1, y 2 )R (l) 13 (y 1, y 3 )R (l) 23 (y 2, y 3 ) (18) and the inversion equation R (l) (y, x)r (l) (x, y) Id. (19)

7 J Algebr Comb (2015) 41: Note that (18) and (19) hold with general spectral parameters; the special choice (15) allows the operator R (l) to be restricted to the subspace Sym l V Sym l V,asshown below. Proposition 3.3 A state v V l belongs to Sym l V if and only if for all 1 i l 1. R i,i+1 (q 2i 2 x, q 2i x) v =0 (20) Proof It is enough to show that the operator R i,i+1 (q 2i 2 x, q 2i x) annihilates a state v if and only if v is invariant under exchanging positions i and i + 1, since the R-matrix is a local operator. The R-matrix for the special choice of spectral parameters (q 2i 2 x, q 2i x) reads as R i,i+1 (q 2i 2 x, q 2i x) = (q 2 1)q 2i 2 x , (21) whose kernel is exactly Sym 2 V V V. Lemma 3.4 The operator R (l) leaves Sym l V Sym l V invariant and therefore the map is well-defined. R (l) (x, y) : Sym l V Sym l V Sym l V Sym l V (22) Proof Let v w Sym l V Sym l V. Note that the commutation relation ( R i,i+1 (q 2i 2 x, q 2i x) = R (l) (x, y) ) R (l) (x, y) v w ) ( R i,i+1 (q 2i 2 x, q 2i x) v w (23) holds for 1 i l 1, as a consequence of the Yang Baxter equation Eq. (7) applied to (17). Since v belongs to Sym l V, the r.h.s. of (23) vanishes for 1 i l 1. Therefore, by Proposition 3.3, the vector R (l) (x, y) v w is symmetric in its first l factors. A similar argument shows that R (l) (x, y) v w is also symmetric in its last l factors, and hence R (l) (x, y) v w Sym l V Sym l V. 3.2 Higher spin model Take a n n grid as in the 6-vertex model above, and to each edge associate an integer 0 α l, which labels the corresponding state l; α Sym l V. We assume

8 850 J Algebr Comb (2015) 41: x 1 x 2 x 3 x 4 x 5 x 6 l l l l l l l l l l l l y 1 y 2 y 3 y 4 y 5 y 6 Fig. 3 Boundary conditions and spectral parameters for the 6 6grid generalized domain wall boundary conditions, that is, l is assigned to the edges on the left and top boundaries, and 0 is fixed along the edges of the bottom and right boundaries. As in the 6-vertex model, 2n spectral parameters {x, y} are associated to the vertical and horizontal lines of the grid (see Fig. 3). Analogously to the 6-vertex model, the following conservation condition is imposed on a vertex configuration: such that α + β = γ + η. Using the standard notation R (l) (l) γ,η (x i, y j ) l; α l; β = R α,β (x i, y j ) l; γ l; η, (24) the weight of a vertex configuration is given by (l) γ,η w i, j (x i, y j ) = R α,β (x i, qy j ). (25) This choice guarantees the integrability of the model. 3.3 Partition function The partition function Z n,l (x, y) of the spin l model is defined exactly as in (2), where now the weights w ij are those given in (25). The fusion process, defining R (l) from R, allows us to express the partition function of the spin l model in terms of the original 6-vertex model partition function as where Z n,l (x, y) = Z ln ( x, ȳ), (26) x ={x 1, q 2 x 1,...,q 2l 2 x 1,...,x n, q 2 x n,...,q 2l 2 x n }, (27)

9 J Algebr Comb (2015) 41: and ȳ is defined similarly. As in (3), the function ˆ Z n,l (x, y) = n (x i y i ) l/2 Z n,l (x, y) (28) i is a homogeneous polynomial in the variables x i and y i. Moreover, Zˆ n,l (x, y) is divisible by the product n i, j l 2 l 1 (q 2k x i q 2p+1 y j ), (29) p=0 k=0 as a consequence of the Izergin Korepin formula (11) evaluated at x and ȳ. The reduced partition function is defined as Z n,l (x, y) := ni, lp=0 l 1 j=1 k=0 (q2k x i q 2p 1 y j ) det A l (x, y), (30) ( x) ( ȳ) where A l (x, y) is the ln ln matrix given by with l l blocks of the form [ A l (x, y) := A l (x, y) := [ A l (x α, y β ) ] n α,β=1 (31) 1 (q 2 j x q 2i 1 y)(q 2 j x q 2i+1 y) ] l 1 Note that we recover the Izergin Korepin determinant when l = 1. The original partition function can be written as i, j=0. (32) Z n,l (x, y) = Const. n n (x i y i ) l/2 i i, j l 2 l 1 (q 2k x i q 2p+1 y j )Z n,l (x, y), (33) p=0 k=0 where the constant can be determined explicitly. 4 Alternative representation of the partition function For rational functions u(z) and v(z) their residue pairing is defined as u(z), v(z) = Res u(z)v(z), (34) z=

10 852 J Algebr Comb (2015) 41: in terms of which the Izergin Korepin formula (11) can be presented as where ( ) 1 1 Z n (x, y) = det p i (x; z),, (35) (x) ( y) 1 i, j n z y j p i (x; z) = n (z qx k )(z q 1 x k ), i = 1,...,n. (36) k=1 k =i In what follows, we use the following alternative representation of the partition function: Proposition 4.1 The standard 6-vertex model partition function Z n (x, y) can be written as where and Z n (x, y) = ( z j 1 ) det r i (x; z),, (37) 1 i, j n w(y; z) w(y; z) = n (z y k ), (38) k=1 r i (x; z) = w(x; qz)(q 1 z) i 1 w(x; q 1 z)(qz) i 1 (q q 1 )z, i = 1,...,n. (39) Proof First observe that if ( y) = 0 then { } { 1 z j 1 } span = span, (40) 1 j n z y j 1 j n w(y; z) and the change of basis is given explicitly as z i 1 w(y; z) = where M is the n n matrix M ij ( y) = n 1 M ij ( y), i = 1,...,n, (41) z y j j=1 y i 1 j k = j (y, i, j = 1,...n. (42) j y k )

11 J Algebr Comb (2015) 41: Similarly, if (x) = 0 then { span p j (x; z) } { = span r j (x; z) }, (43) 1 j n 1 j n and the change of basis can be written, using again (42), as r i (x; z) = n M ij (x)p j (x, z). (44) j=1 To conclude the proof, it is enough to note that det(m(x)) = ( 1)(n 2) (x). (45) Lemma 4.2 The higher spin partition function is given by where Z n,l (x, y) = ( z j 1 ) det r i (x; z),, (46) 1 i, j nl w(ȳ; z) r i (x; z) = q n(l 1) w(x; qz)(q 1 z) i 1 w(x; q 2l+1 z)(qz) i 1 (q q 1 )z, (47) for 1 i nl. Proof Define the polynomial Observe that π(x; z) = q l 1 n(l 1)2 k=1 w(x; q 2k+1 z). (48) and hence w( x; qz) = q n(l 1) π(x; z)w(x; qz) (49) w( x; q 1 z) = q n(l 1) π(x; z)w(x; q 2l+1 z), (50) r i ( x; z) = π(x; z) r i (x; z), i = 1,...,nl. (51)

12 854 J Algebr Comb (2015) 41: The identity (46) follows by factoring out the product n l 1 π(x; q 2k y j ) (52) j=1 k=0 of the l.h.s. of (37). Proposition 4.3 The partition function Z n,l (x, y) satisfies the following properties: (i) Z n,l (x, y) is a homogeneous polynomial in the set of variables {x, y}, (ii) Z n,l (x, y) is symmetric in the variables x and in the variables y, (iii) Z n,l (x, y) = Z n,l ( y, x), (iv) Z n,l (x, y) has total degree at most ln(n 1), (v) Z n,l (x, y) has degree at most l(n 1) in each variable x i or y i. Proof The exchange symmetry (iii) follows easily from the representation (30). The determinantal representation (46) shows that Z n (x, y) is a symmetric homogeneous polynomial in x, and therefore in y. The total degree of the partition function can be read off from (30). To compute the degree of Z n,l (x, y) in the variable y 1, note that (46) can be slightly modified as Z n,l (x, y) = ( 1)(nl 2 ) ( ) 1 det r i (x; z),, (53) ( ȳ) 1 i, j nl z ȳ j where only the first l columns of the determinant depend on y 1. The degree of the polynomial r i (x; z) in z is equal to n+i 2, and therefore, the ith row in the first nl l block of the matrix in (53) consists of polynomials of degree n + i 2 in the variable y 1. This means that the highest possible exponent of y 1 appearing in the determinant (53) is equal to l k=1 (n + nl k 1) = nl(l + 1) 2 1 l(l + 3). To conclude (v), it is enough to recall that the degree of ( ȳ) in the variable y 1 is 2 1 l(l 1) + (n 1)l2. Remark 4.4 In general, the partition function Z n,l (x, y) is not fully symmetric in the set of 2n variables {x, y}. 5 The combinatorial point In this section we present our main result: at the combinatorial point, q = exp(2πi/(2l + 1)), the partition function Z n,l (x, y) is a certain Macdonald polynomial. 5.1 The main result Let lδ n be the staircase partition with n steps 2 l, that is lδ n = (l(n 1), l(n 1),...,l,l,0, 0). Notice that the total degree and the degree in each variable x i or y i

13 J Algebr Comb (2015) 41: of the partition function Z n,l (x, y) are equal to, respectively, lδ n =ln(n 1) and the first part of lδ n.let ρ l = e 2πi/(2l+1), (54) and see [16] for the definition of the Macdonald polynomials P λ (x; q, t). Theorem 5.1 At the combinatorial point q = ρ l, the partition function Z n,l (x, y) is a Macdonald polynomial up to a multiplicative constant, more precisely Z n,l (x, y) = γ n,l P lδn (x, y; ρ 2 l,ρ l). (55) The proportionality constant γ n,l will be given explicitly in Proposition Remark 5.2 The coefficients u λμ (q, t) in the expansion of the Macdonald polynomial P λ (z; q, t) = μ λ u λμ (q, t)m μ (z) (56) are rational functions of q and t, and therefore, the specialization q = ρ 2 l and t = ρ l has to be done carefully. This issue is addressed in Sect Theorem 5.1 has the following important consequence: Corollary 5.3 At the combinatorial point q = ρ l, the partition function Z n,l (x, y) is a fully symmetric polynomial. In order to prove Theorem 5.1, we show below that the partition function Z n,l (x, y) satisfies a set of constraints, known as the wheel condition, see Theorem 5.9. The wheel condition together with the properties described in Proposition 4.3 is sufficient to characterize the partition function up to a multiplicative constant, see Lemma Using a result of Feigin et al., we prove that there is a unique Macdonald polynomial of total degree ln(n 1) which also satisfies the wheel condition, see Proposition Therefore the uniqueness result of Lemma 5.11 implies Theorem The wheel condition Let z ={z 1,...,z 2n }. Definition 5.4 [Wheel condition] Let q and t be such that q r 1 t k+1 = 1, for some non-negative integers k and r. A function f (z) is said to obey the (r, k) q,t -wheel condition if f (z) vanishes whenever z iα+1 z iα = tq s α for any s α N such that and for any choice of 1 i 1 < i 2 <...<i k+1 2n. k s α r 1, (57) α=1

14 856 J Algebr Comb (2015) 41: Lemma 5.5 Let n 3. At the combinatorial point q = ρ l, the partition function Z n,l (x, y) vanishes whenever for any s 1,s 2 N such that s 1 + s 2 l 1. Proof Let S = span {r i ( x; z)}. The polynomial 1 i n ( n l 1 a( x; z) = (z q 2k x i ) x 3 = q 1+2s 2 x 2 = q 2+2s 1+2s 2 x 1, (58) i=4 k=0 s 1 +s 2 1 of degree l(n 2) 1, is such that )( s1 1 ) (z q 2k x 1 ) k=0 l 2 (z q 1+2k x 1 ) (z q 2+2k x 1 ), (59) k=s 1 k=s 1 +s 2 w( x; qz)a( x; q 1 z) w( x, q 1 z)a( x; qz) (q q 1 )z = 0. (60) Thus dim S <ln. This implies that the matrix appearing in Eq. (46) is singular. Lemma 5.6 Let n 2. At the combinatorial point q = ρ l, the partition function Z n,l (x, y) vanishes whenever for any s 1,s 2 N such that s 1 + s 2 l 1. y 1 = q 1+2s 2 x 2 = q 2+2s 1+2s 2 x 1, (61) Proof Notice that ȳ l s1 s 2 + j = q 2 j 1 x 1 for 0 j s 1, because y 1 = q 2+2s 1+2s 2 x 1. Let S = span { r i (x; z)} and let 1 i n for 1 i ln s 1. Then the polynomials s 1 1 a i (x; z) = z i 1 (z q 2 j x), (62) j=0 r i ( x; z) = w(x; qz)a i (x 1 ; q 1 z) w(x; q 2l+1 z)a i (x 1 ; qz) (q q 1 )z belonging to S, vanish at z = q 2 j 1 x 1 for 0 j s 1., (63)

15 J Algebr Comb (2015) 41: Let A ij = 1 r i (x; z), z ȳ j, then the partition function is given by Z n,l (x, y) = ( 1)(ln 2 ) ( ) det Aij. (64) ( ȳ) 1 i, j nl Let S = span {r i (x; z)} be a subspace of S, and let A ij = r i (x; z), 1 z ȳ j.the 1 i ln s 1 entries A ij vanish when l s 1 s 2 j l s 2,soA is of rank at most ln s 1 1. Therefore A is of rank at most ln 1. Lemma 5.7 Let n 2. At the combinatorial point q = ρ l, the partition function Z n,l (x, y) vanishes whenever for any s 1,s 2 N such that s 1 + s 2 l 1. y 2 = q 1+2s 2 y 1 = q 2+2s 1+2s 2 x 1, (65) Proof This follows from Lemma 5.6, using the symmetries of the partition function. Lemma 5.8 Let n 3. At the combinatorial point q = ρ l, the partition function Z n,l (x, y) vanishes whenever for any s 1,s 2 N such that s 1 + s 2 l 1. y 3 = q 1+2s 2 y 2 = q 2+2s 1+2s 2 y 1, (66) Proof Recall that Z n,l (x, y) = Z n,l ( y, x), and hence this follows from Lemma 5.5. By summarizing the above, we obtain the following crucial property of the partition function at the combinatorial point: Theorem 5.9 At the combinatorial point q = ρ l, the partition function Z n,l (x, y) satisfies the (l, 2) ρ 2 l,ρ l -wheel condition. Definition 5.10 Let x ={x 1,...,x n } and y ={y 1,...,y n }. The vector space V n is defined as the space of polynomials p(x, y) such that (i) p(x, y) is a homogeneous polynomial in the set of variables {x, y}, (ii) p(x, y) is symmetric in the variables x, and also in the variables y, (iii) p(x, y) = p( y, x), (iv) p(x, y) has total degree at most ln(n 1), (v) p(x, y) has degree at most l(n 1) in each variable x i or y i, (vi) p(x, y) satisfies the (l, 2) ρ 2 l,ρ l -wheel condition. Lemma 5.11 The vector space V n is at most one-dimensional. The proof of the above lemma is given in Appendix 1.

16 858 J Algebr Comb (2015) 41: The wheel condition and Macdonald polynomials Definition 5.12 [Admissible partitions] A partition λ = (λ 1,λ 2,...) is said to be (r, k)-admissible if and only if λ i λ i+k r for all i. According to the result of Feigin et al [6], the space of symmetric polynomials satisfying the wheel condition is spanned by Macdonald polynomials. More precisely Theorem 5.13 ([6]). Let q and t be two generic scalars such that q r 1 t k+1 = 1, and let V denote the space of symmetric polynomials in z satisfying the (r, k) q,t -wheel condition. Let M be the space spanned by the Macdonald polynomials P λ (z; q, t) indexed by (r, k)-admissible partitions. Then V = M. In the case of relevance for our purposes, this simplifies to Proposition 5.14 Let q and t be two generic scalars such that q l 1 t 3 = 1, and let p(z) be a homogeneous symmetric polynomial of total degree ln(n 1) satisfying the (l, 2) q,t -wheel condition. Then p(z) P lδn (z; q, t). (67) Proof The partition λ = lδ n is the only (l, 2)-admissible partition such that λ = ln(n 1). Because we do not use generic q and t, extreme care is required before we apply the theorem of Feigin et al. Let m = gcd(3,l 1). The locus of the equation q l 1 t 3 = 1 splits into m branches: q (l 1)/m t 3/m = ω, where ω is a m-th root of unity. We consider the branch corresponding to ω = exp(2πi/m), parametrized by a variable u: q(u) = u 3 t(u) = u (l 1) for l 0, 2 (mod 3) q(u) = u t(u) = u l 1 3 e 2πi 3 for l 1 (mod3). (68) Let u 0 be such that q(u 0 ) = ρ 2 l and t(u 0) = ρ l. Explicitly u 0 = e 3(2l+1) 4πi e 2πi 3 for l 0 (mod3) u 0 = e 2l+1 4πi for l 1 (mod3) u 0 = e 3(2l+1) 4πi e 2πi 3 for l 2 (mod 3). (69) In [6], it was shown that the coefficients u λμ (q(u), t(u)) in the expansion (56) are well-defined rational functions in u. When u approaches u 0, the coefficients u lδn,μ(q(u), t(u)) behave like (u u 0 ) n μ for some power n μ.letn = min μ {n μ }, which is non-negative because n lδn = 0. Then the renormalized polynomial P lδn (z; q(u), t(u)) = (u u 0 ) N P lδn (z; q(u), t(u)) (70) is a regular function when u approaches u 0.

17 J Algebr Comb (2015) 41: Notice that P lδn (z; q(u), t(u)) is a homogeneous symmetric polynomial of total degree ln(n 1) satisfying the wheel condition when u = u 0. The limit lim u u0 P lδn (z; q(u), t(u)) is well-defined, and it is simply given by lim u u 0 P lδn (z; q(u), t(u)) = P lδn (z; ρ 2 l,ρ l). (71) Lemma 5.15 The polynomial P lδn (z; ρ 2 l,ρ l) satisfies the (l, 2) ρ 2 l,ρ l -wheel condition. Proof This lemma follows from the regularity of approaches u 0. P lδn (z; q(u), t(u)) when u 5.4 The proof of the main theorem Let [l] q!=[l] q [l 1] q...[1] q, where [a] q = qa 1 q 1. (72) Proposition 5.16 The constant γ n,l, defined in Theorem 5.1, is given by γ n,l = ( ( 1) (l 2) ρ 2( l 2)(2n 3) l [l] ρ 2 l!) n. (73) In order to prove this proposition, in Appendix 2 we compute the coefficient of ni=1 (x i y i ) l(i 1) in the partition function. Since this coefficient is non-zero, the Macdonald polynomial P lδn (x, y; ρ 2 l,ρ l) is well-defined and P lδn (x, y; ρ 2 l,ρ l) = P lδn (x, y; ρ 2 l,ρ l). (74) Proof (Proof of Theorem 5.1) Note that the definition of V n is such that Z n,l (x, y) V n, and P lδn (x, y; ρl 2,ρ l) V n. But by Lemma 5.11, V n is at most one-dimensional, and therefore, Z n,l (x, y) = γ n,l P lδn (x, y; ρl 2,ρ l), with coefficient γ n,l computed in Proposition Final remarks 6.1 Combinatorial interpretation A canonical higher spin generalization of an alternating sign matrix (ASM), as defined by Behrend and Knight [1], is a square matrix with entries in { l,..., 1, 0, 1,...,l}, such that sum of all entries in each row or column is l, and the partial sum of the first or last r entries in each row or column is non-negative for each r. The bijection between

18 860 J Algebr Comb (2015) 41: such matrices and the configurations of the model considered in this paper can be established similarly to the case l = 1: reading each column of the grid from the bottom to the top, if the value of the edge goes from β to η, the corresponding entry of the higher spin ASM is η β. Alternatively, the conservation condition α +β = γ +η guarantees that the same result is obtained by reading each row from right to left. Unfortunately, the weights considered in this paper seem very unnatural in the combinatorial setting. By setting all variables of the partition function Z n (x, y) to be equal to 1 at the combinatorial point q = exp(2πi/3), we get the famous sequence 1, 1, 2, 7, 42, 429, 7436, ,... (sequence A in OEIS) (75) that counts several objects, including alternating sign matrices and totally symmetric self-complementary plane partitions. We have seen that at the combinatorial point q = exp((2l + 1)/2πi), the higher spin partition function Z n,l (x, y) is a Macdonald polynomial. Therefore, it would be interesting to find some combinatorial interpretation of the values obtained for l>1 in the homogeneous limit when x i = y i = 1for all i. 6.2 The wheel condition The wheel condition for the higher spin partition function is deduced using the special determinantal form Z n,l (x, y) that involves rational function entries with a specific dependence on the crossing parameter q. This motivates a more systematical study of the connection between the wheel condition and more general determinants of a similar type. See, for example [15], for a different generalization of the Izergin Korepin determinant that satisfies a wheel condition. There are other integrable statistical models that are of interest from this point of view, such as, for example, the so n model in [5] and the eight-vertex model in [22,25]. 6.3 Symmetry It is very intriguing that we start with a function that exhibits an S n S n symmetry but not S 2n symmetry in general. It is natural to ask what the special features of this determinant are that imply the extra symmetry at q = exp((2l + 1)/2πi). It would be expected that the symmetry comes from the physics of the model, that is, there exists some mechanism (like the Yang Baxter equation) which, for this very special value of q, allows us to exchange a row with a column. 6.4 Relation to KP τ-functions The alternative representation (46) of the partition function has a Grassmannian manifold interpretation, and this point of view is intimately connected with the fact that the 6-vertex model with domain wall boundary conditions can be seen as a τ-function of the KP hierarchy, where the spectral parameters x and y play the role of Miwa variables associated to the commuting flows of the hierarchy. This was first observed

19 J Algebr Comb (2015) 41: by Foda et al. [7], based on a fermionic vacuum expectation value representation of the partition function. In his survey paper [24], Takasaki extends this and other related results by showing how partition functions of certain 2D solvable models and scalar products of Bethe vectors from integrable spin chain models can also be written as KP τ-functions. Acknowledgments The authors thank the Centre de Recherches Mathématiques in Montréal where most of the research presented here was carried out. The work of T. F. was partially supported by ANR project DIADEMS (Developing an Integrable Approach to Dynamical and Elliptic Models). The work of F. B. was partially supported by the FP7 IRSES project RIMMP (Random and Integrable models in Mathematical Physics), the ERC project FroM-PDE (Frobenius Manifolds and Hamiltonian Partial Differential Equations) and the MIUR Research Project Geometric and analytic theory of Hamiltonian systems in finite and infinite dimensions. The authors are grateful to the referees for their constructive remarks to improve the structure and the readability of the paper. Appendix 1: Uniqueness The goal of this section is to prove Lemma That is, if there exists a non-zero polynomial p(x, y) V n, then it is unique up to a multiplicative constant. Let z ={z 1,...,z 2n }={x, y} and q = ρ l. We generalize the (l, 2) q 2,q-wheel condition: Definition 6.1 (r-wheel condition). Let r be such that 3r <l. A polynomial p(z) is said to obey the r-wheel condition if it vanishes whenever z k = q 1+2r+2s 2 z j = q 2+4r+2s 1+2s 2 z i, (76) for any s 1, s 2 N such that s 1 +s 2 l 1 3r and any choice of 1 i < j < k 2n. The following lemma holds: Lemma 6.2 Let a(x, y) V n be such that a(x, y) y j =qx i = 0 for a given i and j. Then a(x, y) = 0. Proof The result is trivial when n = 1. Assume n > 1. By symmetry, a(x, y) i, j (y j qx i )(x i qy j ), therefore, there exists a polynomial b(x, y) such that a(x, y) = (y j qx i )(x i qy j ) b(x, y). (77) i, j The polynomial b(x, y) has total degree at most (l 2)n(n 1) 2n and degree at most (l 2)(n 1) 2 in each variable x i or y i. When l 2, b(x, y) = 0 and the lemma is proved. By the wheel condition, the following holds: b(x, y) x j =qx i = k =i, j s=1 ( ) l l 1 (x k q 2s x i ) (y k q 2s x i ) b (x, y). (78) k s=2

20 862 J Algebr Comb (2015) 41: The polynomial b (x, y) has degree at most 2n in x i, and therefore it vanishes identically. Thus, a(x, y) = w 1 (x, y)a 1 (x, y), (79) for some polynomial a 1 (x, y) obeying the 1-wheel condition, and w α (z) = i = j(z i q α z j ). (80) The polynomial a 1 (z) has total degree at most (l 4)n(n 1) 2n, and it has degree at most (l 4)(n 1) 2 in each variable x i or y i. By the 1-wheel condition, the following holds: a 1 (z) z j =q 3 z i = l 4 (z k q 6+2s z i ) a 1 (z). (81) k =i, j s=0 The polynomial a 1 (z) has degree at most 2(n 1) 4inz i, and therefore it vanishes identically. Thus, a 1 (z) = w 3 (z)a 2 (z). (82) This procedure can be iterated. In the r-th step, we define a r (z) by a r 1 (z) = w 2r 1 (z)a r (z). (83) The polynomial a r (z) obeys the r-wheel condition, has total degree at most (l 4r)n(n 1) 2rn and has degree at most (l 4r)(n 1) 2r in each variable z i. The r-wheel condition implies a r (z) z j =q 1+2r z i = k =i, j l 1 3r s=0 (z k q 2+4r+2s z i ) a r (z). (84) The polynomial a r (z) viewed as a function of z i has degree at most 2r(n 1) 4r, and therefore, it vanishes identically. This closes the iteration step. We should stop the iteration at r = min{r N such that 3r l}. The polynomial a r (z) has negative total degree, and therefore, it vanishes identically. Proof (Proof of Lemma 5.11) Use induction on n. Letp n (x, y) V n. The lemma holds for n = 1. By the wheel condition the following holds: p n (x, y) y j =qx i = k =i l (x k q 2s x i ) k = j s=1 l (y k q 2s x i ) ˆp n (x, y). s=1

21 J Algebr Comb (2015) 41: The polynomial ˆp n (x, y) does not depend either on x i or on y j. It can then be checked that ˆp n (z) V n 1, which by the induction hypothesis is one-dimensional. For any non-zero polynomial p n (x, y) V n, there is a constant α such that ( pn (x, y) α p n (x, y)) y = 0. (85) j =qx i By application of Lemma 6.2 the proof is complete. Appendix 2: The coefficient γ n,l Proposition 6.3 The recursion Z n,l (x, y) xn =y n =0 = ( 1)(l 2) q 4( l 2)(n 1) ( l 2) [l]q 2! ( n 1 k=1 holds, where ˆx = (x 1,...,x n 1 ) and ŷ = (y 1,...,y n 1 ). x l k yl k Proof The determinant representation (46) ofz n,l (x, y) implies that ) Z n 1,l ( ˆx, ŷ) (86) Z n,l (x, y) xn =y n =0 = ( det r i (x; z) xn =0, 1 i, j nl z j 1 w( ŷ; z)z l ). (87) Observe that r l+i (x; z) xn =0 = zl+1 r i ( ˆx; z) i 1. (88) Note also, assuming y i = 0for1 i n 1, that { z j 1 } ( { } 1 l { z j 1 span = span 1 j nl w( ŷ; z)z l zl j+1 j=1 w(y; z) } (n 1)l j=1 ), (89) with the change of basis z i 1 w( ŷ; z)z l = l M i, j ( ŷ) z j=1 1 l j+1 + (n 1)l j=1 M i,l+ j ( ŷ) z j 1 w( ŷ; z), (90) where the matrix M( ŷ) defined by (90) is upper triangular, and its determinant is equal to 1 det(m( ŷ)) = w( ŷ; 0) q (n 1)l2(l 1) = ( 1)(n 1)l l n 1 k=1 yl2 k. (91)

22 864 J Algebr Comb (2015) 41: A simple calculation gives that r i (x; z) xn =0, 1 zl j+1 { 0 ifi >l j + 1 = ( 1) n 1 q n(l 1)+i 2l+1 [l i + 1] n 1 q 2 k=1 x k if i = l j + 1. (92) By combining the above, we obtain Z n,l (x, y) xn = det(m( ŷ)) =y n =0 det A short calculation gives n det and we also have 1 i, j l j=1 1 i, j l n j=1 ( det z l+1 r i ( ˆx; z), 1 i, j (n 1)l r i (x; z) xn =0, 1 zl j+1 r i (x; z) xn =0, 1 zl j+1 z j 1 ). (93) w( ŷ; z) ( n 1 ) l = ( 1) (l 2)+(n 1)l 3l(l 1) n(l 1)l q 2 [l] q 2! x k, (94) ( det z l+1 r i ( ˆx; z), 1 i, j (n 1)l ( n 1 = q (l 1)l(l+1)(n 1) k=1 z j 1 ) k=1 w( ŷ; z) ) Z n 1,l ( ˆx, ŷ), (95) y l(l+1) k from which the recursion formula (86) follows. Proof (Proof of Prop. 5.16) Note that and hence r i (x; z) = q i l [l i + 1] q 2z i 1 + lower order terms in z, (96) ( Z 1,l (x, y) = q (l 2) [l]q 2! det z i 1 z j 1 ) nl, w(ȳ; z) i, j=1 = ( 1) (l 2) q ( l 2) [l]q 2!. (97)

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