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44(2008), 213–257

A φ

1,3

-Filtration of the Virasoro Minimal Series M (p, p

) with 1 < p

/p < 2

Dedicated to Professor Heisuke Hironaka on the occasion of his seventy-seventh birthday

By

B.Feigin, E.Feigin∗∗, M.Jimbo∗∗∗, T.Miwaand Y.Takeyama

Abstract

The filtration of the Virasoro minimal series representationsMr,s(p,p) induced by the (1,3)-primary fieldφ1,3(z) is studied. For 1 < p/p < 2, a conjectural basis of Mr,s(p,p) compatible with the filtration is given by using monomial vectors in terms of the Fourier coefficients of φ1,3(z). In support of this conjecture, we give two results. First, we establish the equality of the character of the conjectural basis vectors with the character of the whole representation space. Second, for the unitary series (p=p+ 1), we establish for eachmthe equality between the character of the degree m monomial basis and the character of the degree m component in the associated graded module gr(Mr,s(p,p+1)) with respect to the filtration defined byφ1,3(z).

Communicated by M. Kashiwara. Received March 10, 2006. Revised February 9, 2007.

2000 Mathematics Subject Classification(s): 81R10.

Landau Institute for Theoretical Physics, Chernogolovka, 142432, Russia and Indepen- dent University of Moscow, Russia, Moscow, 119002, Bol’shoi Vlas’evski per., 11.

e-mail: feigin@mccme.ru

∗∗Tamm Theory Division, Lebedev Physics Institute, Russia, Moscow, 119991, Leninski pr., 53 and Independent University of Moscow, Russia, Moscow, 119002, Bol’shoi Vlas’evski per., 11.

e-mail: evgfeig@mccme.ru

∗∗∗Graduate School of Mathematical Sciences, The University of Tokyo, Tokyo 153-8914, Japan.

e-mail: jimbomic@ms.u-tokyo.ac.jp

Department of Mathematics, Graduate School of Science, Kyoto University, Kyoto 606- 8502, Japan.

e-mail: tetsuji@math.kyoto-u.ac.jp

Institute of Mathematics, Graduate School of Pure and Applied Sciences, University of Tsukuba, Tsukuba, Ibaraki 305-8571, Japan.

e-mail: takeyama@math.tsukuba.ac.jp

c 2008 Research Institute for Mathematical Sciences, Kyoto University. All rights reserved.

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§1. Introduction

In this paper we study a filtration on the Virasoro minimal modules by the φ1,3 primary field. We first state the problem in a general scheme. Let S =α(z)}α∈A be a set of vertex operators acting on a graded vector space V. In the actual setting, the representation space V is a direct sum V =

β∈BV(β)such that the index setAis a subset ofB, the grading is of the form V(β)=

i∈∆(β)+Z≥0Vi(β), and the action of the vertex operator φα(z) :V(β)→V(γ)

is decomposed as

n∈Z+∆(γ)−∆(β)φ(γ,β)α,−nzn−∆(α), where φ(γ,β)α,−n :Vi(β)→Vi+n(γ). Now for some fixed vectorv0 ∈V0), one can define a sequence of sub- spacesE0(V)⊂E1(V)⊂E2(V)⊂ · · · ⊂V by setting

Em(V) = spanα01)

1,−n1φα12)

2,−n2. . . φαk−1k)

k,−nk v0| (1.1)

αi∈A, βj ∈B, βk=α0, niZ+ ∆i−1)i), k≤m}. In what follows we assume that the Fourier coefficients(γ,β)α,−n} generate the whole V from v0. In this case the above construction gives a filtration E = {Em(V)}m=0 on V, which we refer to as the S-filtration. In our examples, V is a representation of the Virasoro or thesl2 algebra, and φα(z) are vertex operators from the corresponding conformal field theory.

Let us consider the associated graded space grEV := m=0Em(V)/

Em−1(V). Note that the space grEV is bi-graded by the grading grEV =

m=0grEmV and that ofV(β)=

iVi(β). Now there are two natural questions:

(i) Find the bi-graded character chq,vgrEV(α):=

m,n

qnvmdim

Vn(α)∩Em(V))/(Vn(α)∩Em−1(V)).

(ii) Find a monomial basis ofV which is compatible with the filtrationE. This means that one needs to construct a basis ofV of the form

φα01)

1,−n1φα12)

2,−n2. . . φαk−1k)

k,−nk v0k =α0)

with certainαi, ni, βj such that the images of the basis vectors withk≤m form a basis ofEm(V).

In the case whenV is a Virasoro minimal model andSconsists of one field φ2,1(z), these questions have been studied in [FJMMT1], [FJMMT2] under

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certain conditions. In this paper we consider theφ1,3(z) field. The correspond- ing filtration on the Virasoro modules is called the (1,3)-filtration. We also clarify the connection between the φ1,3 case and the fusion filtration on the representations ofsl2. Let us describe our results.

Letp < pbe relatively prime positive integers, and letMr,s(p,p)(1≤r < p, 1≤s < p) be the irreducible representations of the Virasoro algebra with the central chargec= 136(t+ 1/t) and highest weight ∆r,s = ((rt−s)2(t 1)2)/4t, wheret=p/p. We consider the (1,3) primary field (s−s=2,0,2)

φ(s,s)(z) =

n∈Z+∆r,s−∆r,s

φ(s−n,s)zn−∆1,3, φ(s−n,s):Mr,s(p,p)→Mr,s(p,p).

One of our goals is to construct a monomial basis of Mr,s(p,p)compatible with the (1,3)-filtration under the condition 1< t < 2. In this paper we propose a set of vectors as the monomial basis. Let us describe it. Fix a highest weight vector|r, sofMr,s(p,p). Forrands, we determineb(r, s) by the equality

r,b(r,s)= min

1≤a≤p−1 a≡smod 2

r,a.

Next we define some rational numbersw(a, b, c) for 1 ≤a, b, c ≤p such that

|a−b| and |b−c| are 0 or 2. In the case of 1 < t 5/3, they are given as follows:

w(s±2, s, s2) = 2 t,

w(s, s+ 2, s+ 2) =w(s+ 2, s+ 2, s) = 2 s+ 1

t

, w(s, s, s+ 2) =w(s+ 2, s, s) = 1 +

s+ 1 t

, w(s, s+ 2, s) =2

s+ 1 t

+x(s), w(s, s, s) = 2,

w(s, s−2, s) = 2 s+ 1

t

4

t + 5−x(s).

Here{u}:=u−[u], where [u] is the integer part ofu, and x(s) =

2 (1≤s < p/2) 3 (p/2< s≤p1).

In the case of 5/3 < t < 2 we have some choices of the definition of the valuesw(s, s±2, s) and w(s, s, s) under the condition described as (2.23) (see

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(2.16)–(2.21) and the text below for the precise definition). It is a part of our conjecture that, for any choice, the vectors given in the following constitute a basis. We call a vector of the form

(1.2) φ(s−n0,s1)

1 . . . φ(s−nm−1,sm)

m |r, b(r, s) (s0=s, sm=b(r, s)) admissible monomialif it satisfies the condition

(1.3) ni−ni+1≥w(si−1, si, si+1) (i= 1, . . . , m1).

Now we state our conjecture:

Conjecture. The set of admissible monomials form a basis ofMr,s(p,p). To support the conjecture, we prove two statements:

A. The character of the proposed basis coincides with the character ofMr,s(p,p). B. In the unitary casep=p+ 1, the bi-graded character of the proposed basis and that of grEMr,s(p,p)coincide, where grEMr,s(p,p)is the associated graded space with respect to the (1,3)-filtration onMr,s(p,p).

The proof of the statement A is based on combinatorics and the Rocha- Caridi character formula. Namely we show that the character of admissible monomials with fixedmcan be written in the form

(1.4) qr,s

(q)mIm(q), (q)m= m i=1

(1−qi),

where Im(q) is an alternating sum of the characters for the fusion products [FL]. We also prove that the Rocha-Caridi formula for the character ofMr,s(p,p) can be rewritten as

m≥0qr,s

(q)mIm(q). In order to prove that the admissi- ble monomials form a basis of Mr,s(p,p), it is enough to rewrite any monomial φ(s−n0,s1)

1 · · ·φ(s−nm−1,sm)

m |r, b(r, s)in terms of admissible monomials of length less than or equal to m. This was done in [FJMMT1], [FJMMT2] in the case of the (2,1) field using quadratic relations for its Fourier components. Similar quadratic relations can be written also for the (1,3) field using the results of [DF]. Still it is not clear to us how to rewrite an arbitrary monomial in terms of admissible ones using these relations.

The proof of the statement B is based on the coset construction. For i= 0,1 and integersr, kwith 0≤r−1≤k, let

(1.5) Li,1⊗Lr−1,k=

1≤s≤k+2 s:s+r+ieven

Mr,s(k+2,k+3)⊗Ls−1,k+1

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be the decomposition of the tensor product of the irreducible highest weight representations of sl2. Here Lj,k denotes the level k module with the highest weightj with respect tosl21 sl2. We denote by vj,k the highest weight vector. Using a result of [L], we establish the connection between the (1,3) filtration and the fusion filtration on the left hand side of (1.5). Namely, con- sider the action of the algebra sl2 onLi,1 and the corresponding S-filtration Gm(Li,1), whereS={e(z), h(z), f(z)},x(z) =

(x⊗tn)z−n−1, and{e, h, f} is the standard basis ofsl2. We call this filtration the Poincar´e-Birkhoff-Witt (PBW) filtration. We show that

(1.6) U(sl2)·(Gm(Li,1)⊗vr−1,k) =

1≤s≤k+2 s:s+r+ieven

Em(Mr,s(k+2,k+3))⊗Ls−1,k+1,

whereEm(Mr,s(k+2,k+3)) is the (1,3)-filtration. Thus the study of this filtration can be reduced to the study of the left hand side of (1.6).

We recall that, for two cyclic g-modulesV1 and V2 with cyclic vectorsv1 andv2, the fusion filtration on the tensor productV1(z1)⊗V2(z2) of evaluation representations ofgC[u] is defined by

Fm(V1(z1)⊗V2(z2)) = span{(g(1)⊗ui1)· · ·(g(s)⊗uis)·(v1⊗v2)}, whereg(i)gandi1+. . .+is≤m. One can easily show that

Fm(Li,1⊗Ll,k) =U(sl2)·(Gm(Li,1)⊗vl,k).

Using (1.6) we express the bi-graded character of the (1,3)-filtration via that of the PBW-filtration onLi,1. We thus get the bi-graded version (1.4) of the Rocha-Caridi formula as an alternating sum of the bi-graded characters of weight subspaces ofLi,1.

We finish the introduction with a discussion of possible generalizations.

Note that the integer level k in the coset construction (1.5) can be replaced by a fractional one. This generalization leads to the coset realization of the generalMr,s(p,p). We expect that the above construction can be applied to the general case.

Now consider the (1,2) fieldφ1,2(z). In this case the decomposition (1.5) should be replaced by

(L0,1⊕L1,1)⊗Lr−1,k=

1≤s≤k+2

Mr,s(k+2,k+3)⊗Ls−1,k+1,

and the algebrasl2by the vertex (intertwining) operatorC2(z) acting onL0,1

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L1,1. This vertex operator induces the filtrationGm(L0,1⊕L1,1). Then U(sl2)·[Gm(L0,1⊕L1,1)⊗Lr−1,k] =

1≤s≤k+2

Hm (Mr,s(k+2,k+3))⊗Ls−1,k+1,

where Hm (Mr,s(k+2,k+3)) is the (1,2)-filtration. As in the (1,3)-case, the bi-

graded character

m≥0

vmchqHm (Mr,s(k+2,k+3))

can be expressed as an alternating sum of the bi-graded characters ofC2(z)- filtration onL0,1⊕L1,1.

Our paper is organized as follows: In Section 2, the character of the Vi- rasoro moduleMr,s(p,p) is written as an alternating sum by using the character of the weightl component of the fusion productπ∗m2 (Lemma 2.2). The main result in this section is a proof of the statement that for 1< p/p <2 the alter- nating sum with fixedmis the character of the admissible monomials of length m (Proposition 2.3). Section 3 prepares some exact sequences of the fusion products and vanishing of the homology groups of the Lie subalgebran+sl2 generated byf⊗1 ande⊗t−1with coefficients in the tensor product of certain finite dimensional modules and irreducible highest weightsl2-modules. This is used in Section 5 when the characters of the highest weight vectors in integrable sl2-modules are computed. In Section 4, by using Lashkevich’s construction of vertex operators in the GKO construction, an isomorphism is given between the fusion product of level 1 and levelkirreducible highest weightsl2-modules and the associated graded module with respect to the filtration defined by the (1,3) primary field (Proposition 4.4). Section 5 is devoted to the calculation of the characters for them-th graded components of the Virasoro unitary se- ries with the (1,3)-filtration (Theorem 5.15). The result coincides with the combinatorial characters computed in Section 2.

Throughout the text,e, f, hdenotes the standard basis ofsl2, andπj de- notes the (j+ 1)-dimensional irreducible representation.

§2. Conjectural Monomial Basis by (1,3) Field

§2.1. Formulation

In this section, we consider Virasoro modules in the minimal series which are not necessarily unitary.

Let Vir be the Virasoro algebra:

[Lm, Ln] = (m−n)Lm+n+ c

12m(m21)δm+n,0.

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Throughout this section, we fix relatively prime positive integersp, psatisfying 3≤p < p. We denote byMr,s(p,p)(1≤r≤p−1, 1≤s≤p1) the irreducible Vir-module with central chargec= 136(t+1t) and highest weight

r,s= (rt−s)2(t1)2

4t ,

where

t= p p .

We fix 1≤r≤p−1, and consider the (1,3) primary field φ(s,s)(z) =

n∈Z+∆r,s−∆r,s

φ(s−n,s)zn−∆1,3.

The Fourier coefficients φ(s−n,s) are operators acting as

Mr,s(p,p)

d

Mr,s(p,p)

d+n, where

Mr,s(p,p)

d

= {|v ∈ Mr,s(p,p) | L0|v = d|v} stands for the graded component. They are characterized by the intertwining property

[Ln, φ(s,s)(z)] =zn

z d

dz + (n+ 1)∆1,3

φ(s,s)(z).

A non-trivial (1,3) primary field exists if and only ifs=s, s±2 and (s, s)= (1,1),(p1, p1). Moreover, it is unique up to a constant multiple. We fix the highest weight vector|r, s ∈

Mr,s(p,p)

r,s

and use the normalization φ(s,s)

r,s−∆r,s|r, s=|r, s.

Our problem is to construct a basis of the representationMr,s(p,p)by using the operatorsφ(s−n,s). In this paper we restrict to the case

1< t <2, (2.1)

and give a partial answer to this problem.

The form of the basis we propose is similar to the one studied in [FJMMT1], [FJMMT2] using the (2,1) primary field. We define a set of weights

w(a, b, c)∈r,a2∆r,b+ ∆r,c+Z, (2.2)

and consider vectors of the form φ(s−n0,s1)

1 · · ·φ(s−nm−1,sm)

m |r, sm (2.3)

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satisfying

ni−ni+1≥w(si−1, si, si+1). (2.4)

The actual form of the weights (2.2) is a little involved, and we postpone their definition to subsection 2.3 (see (2.16)–(2.21) and paragraphs following them).

The vectors (2.3) are parametrized by spin-1 and level-p restricted paths, i.e., sequences of integers

s= (s0, s1, . . . , sm) satisfying

1≤si≤p1, si=si+1 orsi+1±2,

(si, si+1)= (1,1),(p1, p1).

We call them simply paths. The non-negative integer m is called the length of the path. We denote byP(pa,b,m) the set of paths (s0, s1, . . . , sm) of lengthm satisfyings0 =aandsm=b. Note that the parity of si is common with each path. In particular, we havea≡bmod 2 ifP(pa,b,m) is non-empty.

The set of rational numbers

(n1, . . . , nm)

in the expression (2.3) is called a rigging associated with the path (s0, s1, . . . , sm). A rigging satisfies

niZ+ ∆r,si−1r,si. A path with rigging is called a rigged path.

A rigged path of lengthmis called admissible if and only if (2.4) and the following condition hold:

nmr,sm−1r,sm+δsm−1,sm. (2.5)

We denote byR(p,pr,a,b,m) the set of admissible rigged paths of lengthmsuch that s0=aandsm=b.

Finally, we fix the boundary of a pathsm =b(r, s) by the following rule:

b=b(r, s) is the unique integer satisfying 1≤b≤p1,b≡amod 2 and

r,b= min

1≤s≤p−1 s≡amod 2

r,s. (2.6)

We will comment on this choice in the next subsection 2.2.

Now we put forward the

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Conjecture 2.1. The set of vectors (2.3), where (n1, . . . , nm) runs through the set

m≥0R(p,pr,a,b,m) and b = b(r, a) as given in (2.6), constitute a basis ofMr,a(p,p).

Note that the meaning of the condition (2.5) is clear. If ∆r,sm +nm <

r,sm−1, the vector φ(s−nm−1,sm)

m |r, sm is zero because

Mr,s(p,pm−1)

r,sm+nm

= {0}. If sm−1 =sm and nm = 0, the vector φ(s0m,sm)|r, sm is proportional to

|r, smbecause it belongs to

Mr,s(p,pm)

r,sm =C|r, sm. We have also ∆1,1= 0 and

M1,1(p,p)

1 ={0}. This is not in contradiction to the condition (2.5) for r=sm=sm−1= 1 because the casesm=sm−1= 1 is prohibited.

In order to support the conjecture, we will show that the set of admissible monomials has the same character as that of Mr,a(p,p) (Theorem 2.4). The conjecture will follow if we show further that the above set of vectors span the space Mr,a(p,p). So far we have not been able to check the latter point in full generality.

§2.2. A character identity

In this subsection, we rewrite the character of Mr,a(p,p) in a form suitable for comparison with the set of paths.

We use the q-supernomial coefficients introduced in [SW]. They are a q- analog of the weight multiplicities of tensor products of variousπk, where πk denotes the irreduciblesl2-module of dimensionk+ 1. As shown in [FF1], they can be defined as the coefficients of zl of the graded character of the fusion product (for the definition and properties of fusion product, see Section 3).

chq,z

π∗L1 1∗ · · · ∗πN∗LN

=

l∈Z+12PN

j=1jLj

L1, . . . , LN l

q

zl.

Here we will need only the special caseN = 2,L1= 0. Set Sm,l(q) :=

0, m

l

q−1

. (2.7)

Formula (2.9) in [SW] gives Sm,l(q) =

ν∈Z

q(ν+l−m)(ν+l)+ν(ν−m)

m

ν

q

ν m−l−ν

q

.

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In the right hand side

L a

q

=

(qL−a+1)a

(q)a (aZ≥0, L∈Z),

0 (otherwise),

stands for theq-binomial symbol, and (x)n= (x)/(qnx), (x)=

i=0(1 qix).

Recall that the characterχ(p,pr,s )(q) ofMr,s(p,p)is given by [RC]

q−∆(p,pr,s)χ(p,pr,s )(q) =

λ∈Z

1

(q)qλ2pp+λ(pr−ps) (2.8)

λ∈Z

1

(q)qλ2pp+λ(pr+ps)+rs.

Lemma 2.2. For anyb∈Zsatisfyingb≡amod 2, the character (2.8) can be written in terms of (2.7)as

q−∆(p,p

)

r,a χ(p,pr,a )(q) =

m≥0

1

(q)mIr,a,b,m(p,p) (q), (2.9)

where

Ir,a,b,m(p,p) (q) =

λ∈Z

qλ2pp+λ(pr−pa)+m2−((a−b)/2−pλ)2Sm,(a−b)/2−pλ(q) (2.10)

λ∈Z

qλ2pp+λ(pr+pa)+ra+m2−((a+b)/2+pλ)2Sm,(a+b)/2+pλ(q).

Proof. The q-supernomial coefficients satisfy the recurrence relations ([SW, Lemma 2.3])

L1, L2

a

q

=qL1+L2−1

L12, L2 a

q

+

L12, L2+ 1 a

q

L1,0 a

q

=

L1 a+L1/2

q

.

Iterating thisktimes, we find

L1, L2 a

q

= k m=0

q(k−m)(L1+L2−k)

k m

q

L12k, L2+m a

q

.

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ChoosingL1 = 2N, L2 = 0, k=N, changing q→q−1 and lettingN → ∞we obtain for alll∈Z

1

(q) =

m≥0

qm2−l2

(q)m Sm,l(q).

In each summand of the first (resp. second) sum of (2.8), replace 1/(q) by the right hand side of the above identity, choosingl= (a−b)/2−pλ(resp.

(a+b)/2 +pλ). The desired identity follows.

Though (2.9) is an identity valid for any b∈Z, in most cases the polyno- mial (2.10) comprises negative coefficients. We prove in subsection 2.4 that, if 1< t <2 and (r, b) satisfies (2.6), then the coefficients ofIr,a,b,m(p,p) (q) are non- negative integers. In fact we will show that it can be written as a configuration sum over the set of pathsP(pa,b,m) . Define the weight of a pathsP(pa,b,m) by

E(s) =

m−1

i=1

iw(si−1, si, si+1).

Proposition 2.3. Under the conditions 1 < t <2 and(2.6), we have an equality

Ir,a,b,m(p,p) (q) =

s∈P(pa,b,m)

qE(s)+m(∆r,sm−1−∆r.bsm−1,b)+∆r,b−∆r,a. (2.11)

We give a proof in Section 2.4. Note that the exponent of qin (2.11) is an integer because of (2.2).

From Proposition 2.3 immediately follows

Theorem 2.4. Notation being as above, we have an identity for the character

χ(p,pr,a)(q) =

m≥0

1 (q)m

s∈P(pa,b,m)

qE(s)+m(∆r,sm−1−∆r.bsm−1,b)+∆r,b.

§2.3. Definition of w(a, b, c) In this subsection we introduce our weightw(a, b, c).

In (2.11), we fixed b = b(r, a) by the condition (2.6). Conversely, for a givenb, rfor which (2.6) is valid is eitherr1(b) =b+1

t

or r2(b) =b−1

t

+ 1.

Set

τ(b) = b+ 1

t

b−1

t

.

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We haveτ(b) = 1 or 2. If τ(b) = 1 we haver1(b) =r2(b), and if τ(b) = 2 we haver1(b) =r2(b) + 1. We list a few other properties ofτ(s).

τ(1) = 1, (2.12)

τ(2) =

2 if 1< t < 32; 1 if 32 < t <2, (2.13)

τ(p1) = 2, (2.14)

τ(s) =τ(p−s) if 1< s < p1.

(2.15) Set

{x}=x−[x]

where [x] is the integer part ofx.

We define the weightw(a, b, c) in the following form:

w(s±2, s, s2) = 2 t, (2.16)

w(s, s+ 2, s+ 2) =w(s+ 2, s+ 2, s) = 2 s+ 1

t

, (2.17)

w(s, s, s+ 2) =w(s+ 2, s, s) = 1 + s+ 1

t

, (2.18)

w(s, s+ 2, s) =2 s+ 1

t

+x(s), (2.19)

w(s, s, s) = 3−τ(s), (2.20)

w(s, s−2, s) = 2 s+ 1

t

4

t +y(s).

(2.21)

Herex(s), y(s) are integers defined as follows. If τ(s) = 2, then x(s) = 2 andy(s) = 4. In the case ofτ(s) = 1 we have two possibilities: (x(s), y(s)) = (2,3) or (3,2). In the following we give the precise choice of the values x(s) andy(s) whenτ(s) = 1, and discuss the motivation for it. To simplify the text we use the notationC(s) = 1A,1B to signify the choice atsas indicated in the following table.

C(s) τ(s) x(s) y(s)

1A 1 2 3

1B 1 3 2

2 2 2 4

For example, the wording “C(s) = 1B” means that τ(s) = 1 and we set (x(s), y(s)) = (3,2). In addition to it we writeC(s) = 2, which meansτ(s) = 2 and hence (x(s), y(s)) = (2,4).

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Now we definex(s) andy(s) fors such thatτ(s) = 1. If 1< t≤5/3 and τ(s) = 1, then we take

C(s) =

1A (1≤s < p2), 1B (p2 < s≤p1).

If 5/3< t <2, then we determineC(s)’s so that the sequenceC(1), C(2), . . . , C(p1) becomes in the form

1A,1B,1A,1B, . . . ,1B,1A,2,1B,1A,1B, . . . ,1B,1A,2, 1B,1A,1B, . . . ,1A,2,1B,1A,1B, . . . ,1B,1A,2,

where 1’s between successive 2’s come always with an even number. Below we will motivate the above assignment and show that it can be made consistently.

Let us seek for the weights w(a, b, c) in the above form (2.16)–(2.21). We will take them independently of the choice r = r1(b) or r2(b). We demand further the following.

(i) (2.11) holds for m= 2,

(ii) left-right symmetry w(a, b, c) =w(c, b, a), (iii) symmetry reflectingMr,s(p,p)Mp−r,p(p,p)−s,

w(a, b, c) =w(p−a, p−b, p−c), (2.22)

(iv)

ifC(s) = 1B thenC(s+ 2)= 1A. (2.23)

The last condition turns out to be necessary in the course of the proof of (2.11), see subsection 2.4.

The validity of (2.11) form= 2 gives a linear constraint on the weights.

There are three cases: a =4, b±2, b. In the first two cases, the relevant weights are (2.16)–(2.18). They are independent of τ(s). It is easy to check that the constraint is satisfied in these cases. In the third case, the relevant weights are (2.19)–(2.21). Here the value of τ(s) matters. If τ(s) = 2, the weights are uniquely given by (2.19)–(2.21), and they satisfy the constraint. If τ(s) = 1, we must specifyC(s) = 1A or 1B. For s= 1 and 2, see (2.12) and (2.13). In these cases, the constraint implies

C(1) = 1A, (2.24)

C(2) = 1B for 3

2 < t <2.

(2.25)

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If 2< s < p2, the constraint is satisfied for either choice.

The left-right symmetry (ii) is automatically satisfied by the formulas (2.16)–(2.21).

Symmetry (2.22) is also valid for (2.16)–(2.18) and (2.20). It is obvious for (2.16); and follows from

s t

+ −s

t

= 1 if 1< s < p for (2.17)(2.18), and for (2.20).

We determine the choice of 1A/B so that the symmetry (2.22) for (2.19)

(2.21) is valid. SinceC(1) = 1A andC(p1) = 2, we have w(1,3,1) =w(p1, p2, p1) = 44

t. For 1< t <3/2, we haveC(2) =C(p2) = 2, and

w(2,4,2) =w(p2, p4, p2) = 66 t;

for 3/2< t <2, we haveC(2) = 1B. SettingC(p2) = 1A, we have w(2,4,2) =w(p2, p4, p2) = 56

t.

For 2 < s < p2, the symmetry is valid if τ(s) = τ(p−s) = 2; ifτ(s) = τ(p−s) = 1, we need to choose 1A/B in such a way that

C(s) = 1A,1B C(p−s) = 1B,1A. (2.26)

The last requirement would be inconsistent ifτ(s) = 1 fors=p/2. However, we have

Lemma 2.5. If p is even, we haveτ(p/2) = 2.

Proof. Ifp is even, then pis odd. We have τ(p/2) = [α]−[β] whereα= p

2+ p

p, β=p 2 p

p.

Sinceα+β=pis odd and 1< α−β= 2p/p <2, we haveτ(p/2) = 2.

We need also to satisfy the condition (2.23). Let us show the consistency of (2.24), (2.25), (2.26) and (2.23). Suppose thatτ(s) = 1. We chooseC(s) = 1A or 1B as follows.

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If 1< t <3/2, we haveC(1) = 1AandC(2) = 2. Therefore, the following choice of 1Aor 1B forssuch thatτ(s) = 1 satisfies all the constraints.

C(s) =

1A ifs < p/2;

1B ifs > p/2.

(2.27)

If 3/2< t≤5/3, we haveC(1) = 1A,C(2) = 1B andτ(4) = 2. Therefore, the same choice (2.27) will do.

Before going to the case 5/3< t <2, we prepare a few lemmas.

Lemma 2.6. If 3/2< t <2, we do not have the sequence(τ(s), τ(s+ 1)) = (2,2).

Proof. Since 12 < 1

t <1, the increment [s+1

t ][s

t] is either 0 or 1. There- fore, ifτ(s) =τ(s+ 1) = 2, we have

m≤s−1

t < m+ 1, m+ 1 s

t < m+ 2, m+ 2≤s+ 1

t < m+ 3, m+ 3 s+ 2

t < m+ 4 for some integerm. From s−1

t < m+ 1 andm+ 3 s+2t followst < 32. Lemma 2.7. Suppose thats

t

=m,s+1

t

=m. Then,s+2

t

=m+ 1.

Proof. The statement follows from 1< 2t <2.

Lemma 2.8. Suppose thatτ(s) = 2,τ(s+ 1) =· · ·=τ(s+k) = 1 and τ(s+k+ 1) = 2. Then,k is even.

Proof. By Lemma 2.7, we have the sequence s−1

t

=m,s

t

=m+ 1,s+1

t

=m+ 2,s+2

t

=m+ 2, s+3

t

=m+ 3,s+4

t

=m+ 3, . . .s+k−1

t

=m+l,s+k

t

=m+l, s+k+1

t

=m+l+ 1,s+k+2

t

=m+l+ 2.

Therefore,k= 2l.

If 5/3 < s <2, we have C(1) = 1A andC(2) = 1B. If we determine the choice for s < p/2, the rest is determined by (2.26). The constraint (2.23) together with the symmetry (2.26) implies that if s+ 2 < p/2 and τ(s) = τ(s+ 2) = 1 we have

C(s) = 1A,1B C(s+ 2) = 1A,1B.

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We start from C(1) = 1A and C(2) = 1B and continue as 1A,1B,1A,1B, . . . until 2 appears. By a similar argument as in the proof of Lemma 2.8, we can show that the first appearance of 2 is for evens. Therefore, from Lemmas 2.6 and 2.8 we can define the sequenceC(1), C(2), . . .as

1A,1B,1A,1B, . . . ,1B,1A,2,1B,1A,1B, . . . ,1B,1A,2,1B,1A,1B, . . . This sequence does not contain (C(s), C(s+ 2)) = (1B,1A). The constraint (2.23) is also satisfied.

§2.4. Proof of Proposition 2.3

In this subsection we fix the weightsw(a, b, c) as in the previous section, and prove (2.11). To that end we consider the configuration sum

Xa,b,c,m(q) :=

s∈P(pa,b,c,m)

qE(s),

whereP(pa,b,c,m) is the set of paths (s0, . . . , sm+1) satisfyings0=a, sm=b and sm+1 = c. From the definition, Xa,b,c,m(q) = 0 unless 1 a, b, c p1, b=c, c±2 and (b, c)= (1,1),(p1, p1). Note thatXa,b,c,m(q) is uniquely determined from the initial condition Xa,b,c,0(q) = δa,b and the recurrence relation

Xa,b,c,m+1(q) =

d=b, b±2

q(m+1)w(d,b,c)Xa,d,b,m(q).

(2.28)

Let us give an explicit formula forXa,b,c,m(q). As an ingredient we intro- duce the functionSm,l(q) defined by

Sm,l(q) :=

ν∈Z

q(ν+l−m)(ν+l−1)+ν(ν−m)

m

ν

q

ν m−l−ν

q

. The functionsSm,l(q) andSm,l(q) are related to each other as follows.

Lemma 2.9. The following formulae hold: Sm,−l(q) =Sm,l(q), Sm,−l(q) =qlSm,l(q), (2.29)

Sm+1,l(q) =q−m−l−1Sm,l+1(q) +Sm,l(q) +q−m+l−1Sm,l−1(q), (2.30)

Sm+1,l(q) =q−m−l−1Sm,l+1(q) +q−mSm,l(q) +Sm,l−1(q), (2.31)

Sm+1,l(q) =q−mSm,l+1(q) +Sm,l(q) +q−m+l−1Sm,l−1(q), (2.32)

Sm+1,l(q) =q−lSm,l+1(q) +Sm,l(q) +q−m+l−1Sm,l−1(q), (2.33)

Sm+1,l(q) =q−lSm,l+1(q) +q−mSm,l(q) +Sm,l−1(q), (2.34)

Sm+1,l(q) =q−m−lSm,l+1(q) +q−lSm,l(q) +Sm,l−1(q). (2.35)

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Proof. We use the notation ofq-trinomial

n a b c

:= (q)n

(q)a(q)b(q)c for a+b+c=n.

Then the product ofq-binomials in the definition ofSm,landSm,lis rewritten as

m

ν

ν m−ν−l

=

m

m−ν m−l−ν 2ν+l−m

From this expression it is easy to check (2.29). The other formulae except (2.31) and (2.34) follow directly from theq-trinomial identity:

n a b c

=

n−1 a−1 b c

+qa

n−1 a b−1c

+qa+b

n−1 a b c−1

. (2.36)

In the following we prove (2.31). The proof of (2.34) is similar.

We start from Sm+1,l(q)

=

ν∈Z

q(ν+l−m−1)(ν+l)+ν(ν−m−1)

m+ 1

m+ 1−l−ν 2ν+l−m−1 m+ 1−ν

. Decompose the right hand side above into three parts by applying (2.36) to the q-trinomial. Then by changingν→ν+ 1 we see that the first part is equal to q−m−l−1Sm,l+1(q). In the third part we rewrite theq-trinomial as follows:

m

m+ 1−l−ν 2ν+l−m−1 m−ν

=

qm+1−l−ν+ (1−qm+1−l−ν)

m

m+ 1−l−ν 2ν+l−m−1 m−ν

=qm+1−l−ν

m

m+ 1−l−ν 2ν+l−m−1 m−ν

+ (1−q2ν+l−m)

m

m+l−ν 2ν+l−m m−ν

.

From the first term in the right hand side above, we obtainq−mSm,l(q). Then

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