XXZ-type Bethe ansatz equations and quasi-polynomials

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XXZ -type Bethe ansatz equations and

quasi-polynomials

Jian Rong Li and Vitaly Tarasovlowast Department of Mathematics Lanzhou University Lanzhou 730000 PRChinalowast Department of Mathematical Sciences Indiana University ndashPurdue University Indianapolis 402North Blackford St Indianapolis IN 46202-3216 USAlowast St Petersburg Branch of Steklov Mathematical Institute Fontanka 27 St Petersburg 191023Russia

Abstract We study solutions of the Bethe ansatz equation for the XXZ -type integrable modelassociated with the Lie algebra slN We give a correspondence between solutions of the Betheansatz equations and collections of quasi-polynomials This extends the results of EMukhin andAVarchenko for the XXX-type model and the trigonometric Gaudin model

1 IntroductionIn this paper we study solutions of the Bethe ansatz equation for the XXZ -type integrable modelsassociated with the Lie algebra slN see (1) These equations arise in the Bethe ansatz methodof computing eigenvalues and eigenvectors of commuting Hamiltonians of integrable models Themethod gives the eigenvalues and eigenvectors by evaluating explicit rational functions on solutionsof the Bethe ansatz equations see for instance [KBI] and references therein

Solutions of the Bethe ansatz equations for the Gaudin model both rational and trigonometricas well for the XXX-type model have been studied by E Mukhin and AVarchenko in [MV1] ndash[MV3] They established a correspondence between solutions of the Bethe ansatz equations andspaces of polynomials quasi-polynomials or quasi-exponentials with certain properties In thispaper we extend the results of [MV2] [MV3] to the case of the XXZ-type Bethe ansatz equations

Our method to construct a collection of quasi-polynomials corresponding to a solution of theBethe ansatz equations is not completely analogous to that of [MV2] [MV3] As a result wemanaged to weaken technical assumptions on solutions of the Bethe ansatz equations comparedwith those imposed in [MV3] Another advantage is that our method works withut restrictions forthe root of unity case

The plan of the paper is as follows In Section 2 we describe the XXZ-type Bethe ansatzequations and refine the problem In particular we define regular and admissible solutions of theBethe ansatz equations In Section 3 we introduce regular collections of quasi-polynomials andshow that each such a collection gives a regular solution of the Bethe ansatz equations We alsoformulate there the main result of the paper Theorem 34 which says that every admissible regularsolution of the Bethe ansatz equations comes from a regular collection of quasi-polynomials Weprove Theorem 34 in Section 4 In Section 5 for a collection of quasi-polynomials U we consider

XXth International Conference on Integrable Systems and Quantum Symmetries (ISQS-20) IOP PublishingJournal of Physics Conference Series 411 (2013) 012020 doi1010881742-65964111012020

Published under licence by IOP Publishing Ltd 1

the monic difference operator DU whose kernel is generated by U We show that the operator

DU has rational coefficients if and only if there is a regular collection of quasi-polynomials U suchthat D

U= DU Also for each solution t of the Bethe ansatz equations we define a difference

operator Dt and show that the collection of quasi-polynomials U associated with t gives a basisof the kernel of Dt that is Dt = DU The Appendix contains necessary technical identities

The authors thank E Mukhin for helpful discussions

2 Bethe ansatz equations and quasi-polynomials21 Bethe ansatz equationsThroughout the paper we fix a complex number q such that q 6= 0plusmn1 More precisely we fix thevalue of log q and for any complex number α set qα = exp(α log q)

Fix an integer N gt 2 Given a collection l = ( l1 lNminus1) of nonnegative integers consider thespace C l = C l1oplus oplus C lNminus1 We label coordinates on this space by two indices the superscriptenumerating the summands and the subscript enumerating coordinates within a summand Thatis we write

t = (t(1)1 t

(1)l1 t

(Nminus1)1 t

(Nminus1)lNminus1

) isin C l

The product of the symmetric groups Sl = Sl1times times SlNminus1acts on C l by permuting coordinates

with the same superscript The Sl -orbit of a single point will be called an elementary Sl-orbit

Let T1(x) TNminus1(x) be monic polynomials in one variable and λ1 λN be complex

numbers Consider a system of equations on the variables t(a)b a = 1 N minus 1 b = 1

la

Ti(t(i)j q

2)

liminus1prodr=1

(t(i)j q minus t

(iminus1)r qminus1)

li+1prods=1

(t(i)j minus t

(i+1)s )

liprodk=1k 6=j

(t(i)j qminus1minus t(i)k q) = (1)

= q 2(λi+1minusλi+ li)minus liminus1minus li+1 Ti(t(i)j )

liminus1prodr=1

(t(i)j minus t

(iminus1)r )

li+1prods=1

(t(i)j qminus1minus t(i+1)

s q)

liprodk=1k 6=j

(t(i)j q minus t

(i)k qminus1)

i = 1 N minus 1 j = 1 li Here and later we assume that l0 = lN = 0 unless otherwisestated Equations (1) are called the XXZ -type Bethe ansatz equations associated with the datal = (l1 lNminus1) T = (T1 TNminus1) λ = (λ1 λN )

The group Sl acts on solutions of equations (1) mdash for every solution t of (1) all points of theSl -orbit of t are solutions of (1) as well

Equations (1) arise in the Bethe ansatz method for the quantum integrable model defined

on the irreducible finite-dimensional representation of the quantum affine algebra Uq(slN )

The polynomials T1(x) T2(xqminus2) TNminus1(xq

minus2(Nminus1)) are the Drinfeld polynomials of therepresentation and the parameters q 2λ1 q 2λN describe the algebra of commuting Hamiltoniansof the model Notice that in the literature equations (1) are usually written in the form

Ti(t(i)j q

2)

Ti(t(i)j )

liminus1prodr=1

t(i)j q minus t

(iminus1)r qminus1

t(i)j minus t

(iminus1)r

li+1prods=1

t(i)j minus t

(i+1)s

t(i)j qminus1minus t(i+1)

s q

liprodk=1k 6=j

t(i)j qminus1minus t(i)k q

t(i)j q minus t

(i)k qminus1

= (2)

= q 2(λi+1minusλi+ li)minus liminus1minus li+1 i = 1 N minus 1 j = 1 li


2

22 Regular solutions of Bethe ansatz equationsFor i = 1 N minus 1 define the polynomials

pi(x t) =

liprodj=1

(xminus t(i)j ) (3)

and

Pi(x t) = q 2λi+1pi(xq2 t) piminus1(x t) pi+1(xq

minus2 t)Ti(x) + (4)

+ q 2λipi(xqminus2 t) piminus1(xq

2 t) pi+1(x t)Ti(xq2)

Then equations (1) take the form

Pi(t(i)j t) = 0 i = 1 N minus 1 j = 1 li (5)

A solution t of equations (1) is called regular if for any i = 1 N minus 1 and any subset I sub 1 li the following condition holds suppose that all the coordinates t

(i)j j isin I are equal with

the common value t(i)I then

dk

dxkPi(x t)|

x=t(i)I

= 0 k = 0 |I | minus 1 (6)

where and |I | is the cardinality of I Clearly if for each i = 1 N minus 1 all the coordinates

t(i)j j = 1 li are distinct the solution t is regular

If t is a regular solution of equations (1) then all points of the Sl -orbit of t are regularsolutions of equations (1) as well

The next proposition is valid by definition of a regular solution of equations (1)

Proposition 21 A point t is a regular solution of equations (1) if and only if for any i = 1 N minus 1 the polynomial pi(x t) divides the polynomial Pi(x t)

3 Quasi-polynomials31 Quasi-polynomialsA quasi-polynomial f(x) of type α is an expression of the form f(x) = xαp(x log x) whereα isin C and p(x s) is a polynomial Call the quasi-polynomial f(x) log-free if p(x s) does notdepend on s Say that a polynomial r(x) divides f(x) if r(x) divides p(x s) The product ofquasi-polynomials of types α and β is supposed to be of type α+β Clearly the product of twononzero quasi-polynomials is not zero

We will use only algebraic properties of quasi-polynomials The key relation is if f(x) =xαp(x log x) then

f(xqβ) = qαβxαp(xqβ log x+ β log q)

Given functions g1(x) gk(x) their discrete Wronskian Wk[g1 gk] is defined by the rule

Wk[g1 gk](x) = det(gi(xq

minus2(jminus1)))kij=1

(7)

Let λ = (λ1 λN ) be a collection of complex numbers By definition a collection of quasi-polynomials of type λ is a sequence of quasi-polynomials U = (u1 uN ) such that for every


3

i = 1 N the quasi-polynomial ui has type λi and WN [u1 uN ] 6= 0 We call thecollection U semiregular if WN [u1 uN ] is log-free

For a semiregular collection of quasi-polynomials U = (u1 uN ) of type λ a sequence of

monic polynomials T = (T1 TN ) is called a preframe of U if for any k = 1 N minus1 and any

subset i1 ik sub 1 N the productkprodi=1

kminusiprodj=0

TNminusi+1(xqminus2j) divides the quasi-polynomial

Wk[ui1 uik ](x) of type λi1 + + λik and

WN [u1 uN ](x) = const xλ1++λNNprodi=1

Nminusiprodj=0

TNminusi+1(xqminus2j) (8)

DenoteQTk (x) =

kprodi=1

kminusiprodj=0

TNminusi+1(xqminus2j) k = 1 N (9)

Say that a preframe T is stronger than a preframe T = (T1 TN ) if QTk (x) divides QT

k (x) forany k = 1 N minus 1

Until the end of this section we fix λ = (λ1 λN ) and assume that every collection of quasi-polynomials is of type λ

Lemma 31 Let U = (u1 uN ) be a semiregular collection of quasi-polynomials There is apreframe T = (T1 TN ) of U such that for every k = 1 N the polynomial QT

k (x) see(9) is the greatest common divisor of the quasi-polynomials Wk[ui1 uik ](x) where i1 ik runs over all k-element subsets of 1 N and Wk[ui1 uik ](x) has type λi1 + +λik

Proof The proof is similar to that of Lemma 49 in [MV2]

The preframe T defined in Lemma 31 is clearly the strongest preframe of U It is called theframe of U

For a semiregular collection of quasi-polynomials U = (u1 uN ) and a preframe T = (T1 TN ) of U define quasi-polynomials y0 yNminus1 by the rule uN = yNminus1Q

T1

WNminusi [ui+1 uN ](x) = yi(x)QTNminusi(x) i = 0 N minus 2 (10)

where QT1 Q

TN are given by (9) Notice that y0(x) is proportional to xλ1++λN Call the

semiregular collection U regular if the quasi-polynomials y1 yNminus1 are log-free

Assume that the collection U is regular For i = 1 N minus 1 let li be the degree of the

polynomial xminusλi+1minusminusλN yi(x) and t(i)1 t

(i)li

be its roots That is

yi(x) = ci xλi+1++λN

liprodj=1

(xminus t(i)j ) i = 1 N minus 1 (11)

for some nonzero complex numbers c1 cNminus1 Set l = lUT = ( l1 lNminus1) and denote by

XUT the Sl -orbit of the point (t(i)j ) i=1Nminus1 j=1 li in C l


4

Theorem 32 Let U be a regular collection of quasi-polynomials of type λ T = (T1 TN ) bea preframe of U and l = lUT = ( l1 lNminus1) Then XUT is an elementary Sl-orbit of regularsolutions of equations (1) associated with the data l1 lNminus1 T1 TNminus1 λ1 λN

Proof Define the quasi-polynomials y1 yNminus1 by the rule

WNminusi [ui ui+2 uN ](x) = yi(x)QTNminusi(x) (12)

By Lemmas Appendix A1 Appendix A2 and formula (9) we have

W2 [ yi yi](x) = yiminus1(x) yi+1(xqminus2)Ti(x) (13)

that isyi(x) yi(xq

minus2)minus yi(xqminus2) yi(x) = yiminus1(x) yi+1(xqminus2)Ti(x)

Therefore

yi(x)(yi(xq

2) yi(xqminus2)minus yi(xqminus2) yi(xq 2)

)= (14)

= yi(xq2) yiminus1(x) yi+1(xq

minus2)Ti(x) + yi(xqminus2) yiminus1(xq

2) yi+1(x)Ti(xq2)

If the point t = (t(i)j ) is defined by (10) (11) the polynomials pi(x t) Pi(x t) are given by (3)

(4) and the polynomial pi(x) equals xminusλiminusλi+2minusminusλN y(x) then relation (14) is equivalent to

Pi(x t) = pi(x)(q 2(λiminusλi+1) pi(xq

2) pi(xqminus2)minus q 2(λi+1minusλi) pi(xq

minus2) pi(xq2))

By Proposition 21 this proves Theorem 32

Remark By Lemma Appendix A1 a collection of quasi-polynomials U = (u1 uN ) has

a preframe T = (T1 TN ) if and only if U = (u1TN uNTN ) is a collection of quasi-

polynomials and has a preframe T = (T1 TNminus1 1) Clearly XUT = XUT

and for both U and

U equations (1) are the same since they involve only the polynomials T1 TNminus1 Thus discussingrelations between collections of quasi-polynomials and solutions of Bethe ansatz equations (1) wecan restrict ourselves without loss of generality to preframes of the form (T1 TNminus1 1)

Say that a point t = (t(i)j ) is generic with respect to the polynomials T = (T1 TNminus1) if

t(i)j 6= t

(i+1)k for all i = 1 N minus 2 j = 1 li k = 1 li+1 and Ti(t

(i)j ) 6= 0 for all i = 1

N minus 1 j = 1 li Clearly all points of the Sl -orbit of a generic point t are generic

Lemma 33 Let U be a regular collection of quasi-polynomials of type λ T = (T1 TN ) bea preframe of U and l = lUT = ( l1 lNminus1) Assume that XUT is the Sl-orbit of a genericpoint with respect to the polynomials T1 TNminus1 Then T is the frame of U

Proof Let pi(x) = xminusλi+1minusminusλN yi(x) Suppose T = (T1 TN ) is a preframe of U strictly

stronger than T Let i be the largest number such that Ti 6= Ti Notice that i gt 1 Thereis a number a such that Ti(a) = 0 and Ti(a) 6= 0 This implies that piminus1(a) = 0 andpiminus2(a)Timinus1(a) = 0 see (9) (10) However if XUT is the Sl-orbit of a generic point thenfor every j = 1 N minus 1 the polynomials pjminus1 Tj and pj are coprime The claim follows


5

32 Main resultA point t = (t

(i)j ) i=1Nminus1 j=1 li is called admissible if

t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)

In particular t(i)j 6= 0 for all i = 1 N minus 1 j = 1 li Clearly all points of the Sl -orbit of

an admissible point t are admissible

Fix collections of complex numbers λ = (λ1 λN ) nonnegative integers l = (l1 lNminus1)and monic polynomials T = (T1 TNminus1) The following theorem is the main result of the paper

Theorem 34 Let t be an admissible regular solution of equations (1) associated with the datal T λ Then there is a regular collection of quasi-polynomials U of type λ such that T = (T1 TNminus1 1) is a preframe of U and the Sl-orbit of t equals XUT

Theorem 34 will be proved in Section 4 The proof is going in three steps First we prove thetheorem for N = 2 The obtained statement is employed then at the second step to construct therequired collection of quasi-polynomials U for general N The final step is to show that T = (T1 TNminus1 1) is a preframe of the constructed collection U

Corollary 35 Let t be a generic admissible regular solution of equations (1) associated withthe data l T λ Then there is a regular collection of quasi-polynomials U of type λ such thatT = (T1 TNminus1 1) is the frame of U and the Sl-orbit of t equals XUT

Proof The statement follows from Theorem 34 and Lemma 33

4 Proof of Theorem 3441 Proof of Theorem 34 for N = 2Let y(x) = xαp(x) be a log-free quasi-polynomial of type α We call y(x) admissible if thepolynomials p(x) and p(xqminus2) are coprime In particular this implies that p(0) 6= 0 Thusfor an admissible quasi-polynomial y(x) the number α and the polynomial p(x) are determineduniquely

Let y(x) = xαp(x) be an admissible quasi-polynomial of type α Since p(x) and p(xqminus2) arecoprime it is known that there are unique polynomials r(x) and s(x) of degree at most deg psuch that r(x)p(x) + s(x)p(xqminus2) = 1 Define the quasi-polynomials A[y ] and B[y ] of type minusαby the rule

A[y ](x) = xminusαr(x) B[y ](x) = xminusαs(x) (16)

so thaty(x)A[y ](x) + y(xqminus2)B[y ](x) = 1 (17)

For a polynomial P (s) and a number c let I [P c ] be the unique polynomial such that

I [P c ](s)minus c I [P c ](sminus 2 log q) = P (s)

deg I [P c ] = deg P if c 6= 1 and deg I [P 1] = 1 + deg P I [P 1](0) = 0 For example

I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6

For a quasi-polynomial f(x) = xαsumixiPi(log x) of type α define the quasi-polynomial I[f ] of

type α by the rule I[f ](x) = xαsumixi I [Pi q

minus2α ](log x) so that

I[f ](x)minus I[f ](xqminus2) = f(x) For example

I[1 ] =log x

log q 2 I[xα ] =

xα

1minus qminus2α qminus2α 6= 1

For a quasi-polynomial f(x) = xαp(x log x) of type α and a log-free quasi-polynomialg(x) = xβr(x) of type β define the quasi-polynomial 〈f(x)g(x)〉+ = xαminusβh(x log x) oftype α minus β by requiring that h(x s) is the polynomial part of the ratio p(x s)r(x) that isdegx

(p(x s) minus r(x)v(x s)

)lt deg r(x) If f(x) = g(x) [f(x)g(x)]+ we say that g(x) divides

f(x)

For an admissible quasi-polynomial y(x) = xαp(x) of type α and a quasi-polynomial V (x)of type β define the quasi-polynomial F[yV ] of type β minus α as follows Let a = A[y ] andb = B[y ] see (16) Consider the quasi-polynomial

v(x) =

langa(x)V (x) + b(xqminus2)V (xqminus2)

y(xqminus2)

rang+

of type β minus 2α Set J[yV ] = I[v ] and

F[yV ](x) = V (x)B[y ](x) + y(x) J[yV ](x) (18)

Proposition 41 Let y(x) be an admissible quasi-polynomial of type α V (x) be a quasi-polynomial of type β Let Y = F[yV ] Assume that y(x) divides the quasi-polynomialy(xq 2)V (x) + y(xqminus2)V (xq 2) of type α+ β Then W2[Y y ] = V

Proof Let a = A[y ] b = B[y ] f = J[yV ] By (17)

a(x)y(x) + b(x)y(xqminus2) = 1 (19)

anda(xq 2)b(x)

(y(xq 2)V (x) + y(xqminus2)V (xq 2)

)=

= a(xq 2)V (xq 2) + b(x)V (x)minus(a(x)a(xq 2)V (xq 2) + b(x)b(xq 2)V (x)y(x)

so y(x) divides the quasi-polynomial a(xq 2)V (xq 2) + b(x)V (x) of type β minus α Hence

y(xqminus2)(f(x)minus f(xqminus2)

)= a(x)V (x) + b(xqminus2)V (xqminus2)

and W2[Y y ] = V by an easy simplification using (19)

Proof of Theorem 34 for N = 2 Let t = (t(i)j ) be an admissible regular solution of equations

(1) and p1(x t) be the polynomial given by (3) Define an admissible quasi-polynomial y(x) =

xλ2p1(x t) of type α = λ2 and a quasi-polynomial V (x) = xλ1+λ2 T1(x) of type β = λ1 + λ2

Let u2 = y and u1 = F[yV ] Proposition 21 shows that y(x) divides the quasi-polynomialy(xq 2)V (x) + y(xqminus2)V (xq 2) of type α+ β Hence by Proposition 41

W2[u1 u2] = xλ1+λ2 T1(x)

that is U = (u1 u2) is the required collection of quasi-polynomials of type λ = (λ1 λ2)


7

42 Construction of a collection of quasi-polynomials

Let t = (t(i)j ) be an admissible regular solution of equations (1) associated with the data l T λ

Let

pi(x) =

liprodj=1


cf (3) and p0(x) = pN (x) = 1 Then for every i = 0 N

yi(x) = xλi+1++λN pi(x)

is an admissible quasi-polynomial of type λi+1 + + λN Set ai = A[yi] and bi = B[yi] see(16) so that

ai(x)yi(x) + bi(x)yi(xqminus2) = 1 (21)

The next lemma is equivalent to Proposition 21

Lemma 42 For any i = 1 N minus 1 the quasi-polynomial yi(x) divides the quasi-polynomial

Ai(x) = yi(xq2)yiminus1(x)yi+1(xq

minus2)Ti(x) + yi(xqminus2)yiminus1(xq

2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0


cf (9) We will construct a collection of quasi-polynomials U = (u1 uN ) of type λ such thatuN = yNminus1 and

WNminusi [ui+1 uN ](x) = yi(x)QNminusi(x) i = 0 N minus 2 (24)

cf (10) We employ the recursive procedure described below Given the quasi-polynomials ui+1 uN we obtain the quasi-polynomial ui by formula (31) and verify relation (24) in Proposition45

For the first step of the process set uNminus1 = F[yNminus1 yNminus2TNminus1 ] Then Lemma 42 for i = Nminus1

and Proposition 41 yield

W2[uNminus1 uN ] = W2[uNminus1 yNminus1] = yNminus2TNminus1 = yNminus2Q2 (25)

which is relation (24) for i = N minus 2

Lemma 43 The quasi-polynomial yNminus2(x) divides the quasi-polynomial

B(x) = yNminus2(xq2) yNminus3(x)uNminus1(xq

minus2)TNminus2(x) + yNminus2(xqminus2) yNminus3(xq

2)uNminus1(x)TNminus2(xq2)

of type λNminus2 + 3λNminus1 + 2λN


8

Proof By Lemma 42 yNminus2(x) divides the quasi-polynomial ANminus2(x) see (22) of type λNminus2 +2λNminus1 + 3λN Relations (21) and (25) yield

B(x) = ANminus2(x)(aNminus1(x)uNminus1(x) + bNminus1(x)uNminus1(xq

minus2))

+

+ yNminus2(x)TNminus1(x)(bNminus1(x)minus aNminus1(x)

)

which proves Lemma 43

Assume that the quasi-polynomials ui+1 uN are constructed already and the followingproperties hold

A For any j = i N

WNminusj [uj+1 uN ](x) = yj(x)QNminusj(x) (26)

B For any j = i+ 1 N there is a quasi-polynomial wi+1j(x) of type λi+1+ +λN minusλjsuch that

WNminusiminus1 [ui+1 ujminus1 uj+1 uN ](x) = wi+1j(x)QNminusiminus1(x) (27)

In particular wi+1i+1(x) = yi+1(x) see (26) for j = i+ 1

C For any j = i+ 1 N the quasi-polynomial yi(x) divides the quasi-polynomial

Bij(x) = yi(xq2) yiminus1(x)wi+1j(xq


2)wi+1j(x)Ti(xq2)

of type λi minus λj + 3(λi+1 + + λN ) In particular Bi+1i+1(x) = Ai+1(x) see (22)

For i = N minus 2 property A coincide with formula (25) property B is straightforwardwNminus1Nminus1 = uN wNminus1N = uNminus1 and property C follows from Lemma 42 for i = N minus 2 and Lemma 43

Define the quasi-polynomials wij(x) j = i N by the rule

wij = F[yi yiminus1 wi+1j Ti ] (28)

see (18) where wi+1j(x) = wi+1j(xqminus2) Property C and Proposition 41 yield

W2[wij yi ](x) = yiminus1(x)wi+1j(xqminus2)Ti(x) (29)

Lemma 44 The quasi-polynomial yi(x) divides the quasi-polynomialNsum

j=i+1(minus1)jwij(x)uj(x)

of type λi + + λN

Proof Let bi = B[yi ] see (16) By (28) and (18)

Nsumj=i+1

(minus1)j wij(x)uj(x) = yi(x)

Nsumj=i+1

(minus1)j J[yi yiminus1 wi+1j Ti ](x) +

+ bi(x)Nsum

j=i+1

(minus1)j yiminus1(x)Ti(x)wi+1j(xqminus2)uj(x)


9

Then formula (27) Lemma Appendix A3 and formula (26) for j = i give

Nsumj=i+1

(minus1)jminusiminus1wi+1j(xqminus2)uj(x) =

WNminusi [ui+1 uN ](x)

QNminusiminus1(xqminus2)= yi(x)

Nminus1prodk=i+1

Tk(x) (30)

which proves the lemma

Set cij(x) = wij(x)yi(x) j = i N and

ui(x) =

Nsumj=i+1

(minus1)jminusiminus1 cij(x)uj(x) (31)

By Lemma 44 ui(x) is a quasi-polynomial of type λi

Proposition 45 WNminusi+1 [ui uN ](x) = yiminus1(x)QNminusi+1(x)

Proof By (29) and (27)

cij(x)minus cij(xqminus2) =yiminus1(x)wi+1j(xq

minus2)Ti(x)

yi(x) yi(xqminus2)

=yiminus1(x)Ti(x)WNminusiminus1 [ui+1 ujminus1 uj+1 uN ](xqminus2)

yi(x) yi(xqminus2)QNminusiminus1(xqminus2)

So Lemma Appendix A3 yields

Nsumj=i+1

(minus1)j(cij(x)minus cij(xqminus2)

)uj(xq

minus2l) = 0 l = 1 N minus iminus 1

andNsum

j=i+1

(minus1)Nminusj(cij(x)minus cij(xqminus2)

)uj(xq

2(iminusN)) =

=yiminus1(x)Ti(x)WNminusi [ui+1 uN ](xqminus2)

yi(x) yi(xqminus2)QNminusiminus1(xqminus2)=

yiminus1(x)QNminusi+1(x)


where the last equality also uses formula (26) for j = i and formula (23) These relations togetherwith (31) give

ui(xqminus2l) =

Nsumj=i+1

(minus1)jminusiminus1 cij(x)uj(xqminus2l) l = 1 N minus iminus 1 (32)

and

ui(xq2(iminusN)) =

(minus1)Nminusi yiminus1(x)QNminusi+1(x)

WNminusi [ui+1 uN ](x)+

Nsumj=i+1

(minus1)jminusiminus1 cij(x)uj(xq2(iminusN)) (33)

Using equalities (32) (33) in the definition of WNminusi+1 [ui uN ] see (7) completes the proof ofthe proposition


10

Proposition 46 For any j = i N

WNminusi [ui ujminus1 uj+1 uN ](x) = wij(x)QNminusi(x)

and the quasi-polynomial yiminus1(x) divides the quasi-polynomial

Biminus1j(x) = yiminus1(xq2) yiminus2(x)wij(xq

minus2)Timinus1(x) + yiminus1(xqminus2) yiminus2(xq

2)wij(x)Timinus1(xq2)

of type λiminus1 minus λj + 3(λi + + λN )

Proof Using (32) in the definition of WNminusi [ui ujminus1 uj+1 uN ](x) we get

WNminusi [ui ujminus1 uj+1 uN ](x) = cij(x)WNminusi [ui+1 uN ](x) = wij(x)QNminusi(x)

By Lemma 42 yiminus1(x) divides the quasi-polynomial Aiminus1(x) see (22) of type λiminus1 + 2λi +

3(λi+1 + + λN ) Relations (21) and (29) yield

Biminus1j(x) = Aiminus1(x)(ai(x)wij(x) + bi(x)wij(xq

minus2))

+

+ yiminus1(x)Ti(x)wi+1j(xqminus2)(bi(x)minus ai(x)

)

which proves the second part of the proposition

Propositions 45 and 46 shows that properties Andash C with i replaced by iminus 1 are valid So wecan construct recursively all quasi-polynomials u1 uN satisfying relations (24)

43 Proof of Theorem 34In this section we will show that the sequence of monic polynomials T = (T1 TNminus1 1) is apreframe of the collection of quasi-polynomials U = (u1 uN ) constructed in Section 42

It is shown in Section 42 that the quasi-polynomials u1 uN satisfy relations (24) seeProposition 45 Moreover for any 1 6 i 6 j 6 N there is a quasi-polynomial wij(x) of typeλi + + λN minus λj such that

WNminusi [ui ujminus1 uj+1 uN ](x) = wij(x)QNminusi(x) (34)

see Proposition 46 and


cf (29)

Lemma 47 For any 1 6 i 6 j lt k 6 N there is a quasi-polynomial wijk(x) of type λi+ +λN minus λj minus λk such that

W2[wij wik ](x) = yiminus1(x) wijk(x)Ti(x)


11

Proof For j = i we have wii(x) = yi(x) and wiik(x) = minuswi+1k(xqminus2) by (35) so the statement

holds For j gt i by Lemma Appendix A3 and formula (21) we have

W2[wij wik ](x) =(ai(x)wik(x) + bi(x)wik(xq

minus2))W2[wij yi ](x)minus

minus(ai(x)wij(x) + bi(x)wij(xq

minus2))W2[wik yi ](x)


Proposition 48 For any 1 6 i 6 j1lt lt jk 6 N there is a quasi-polynomial wij1jk(x)of type λi + + λN minus λj1minus minus λjk such that

Wk[wij1 wijk ](x) = wij1jk(x)

kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))

Proof We prove the statement by induction with respect to k The case k = 2 is the base ofinduction see Lemma 47

Set fij = W [yi wij ] Assume that j1 = i By Lemma Appendix A2 and formula (35)

Wk[yi wij2 wijk ](x) =Wkminus1[fij2 fijk ](x)

yi(xqminus2) yi(xqminus2k)=

= (minus1)kminus1Wkminus1[wi+1j2 wi+1jk ](xqminus2)

kminus2prodl=0

yiminus1(xqminus2l)Ti(xq

minus2l)

yi(xqminus2(l+1))

= (minus1)kminus1 wi+1j2jk(x)

kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))

where for the last equality we use the induction assumption For j1 gt i by Lemma Appendix A3and formula (21) we have

Wk[wij1 wijk ](x) =

=ksuml=1

(minus1) lminus1(ai(x)wijl(x) + bi(x)wijl(xq

minus2))Wk[yi wij1 wijlminus1

wijl+1 wijk ](x)

which proves the proposition

To complete the proof of Theorem 34 we should show that for any k = 1 N minus 1 and anyk-element subset i1 ik sub 1 N the polynomial Qk(x) divides the quasi-polynomialWk[ui1 uik ](x) of type λi1 + + λik

Let j1 jNminusk be the complement of i1 ik in 1 N Then by formula (34)Lemmas Appendix A1 Appendix A4 and formula (24)

Wk[ui1 uik ](x) = const middot x(k+1minusN)(λ1++λN ) times

times WNminusk[w1j1 w1jNminusk](xq 2(Nminuskminus1)) QNminus1(x)

Nminuskminus1prodl=1

QNminus1(xq2l)

QN (xq 2l)


12

Then by Proposition 48

Wk[ui1 uik ](x) = const middot w1j1jNminusk(xq 2(Nminuskminus1))Qk(x)

where the quasi-polynomial w1j1jNminuskhas type λi1 + + λik Theorem 34 is proved

5 Difference operators51 Difference operator of a collection of quasi-polynomialsRecall that q 6= 0plusmn1 Denote by τ the multiplicative shift operator that acts on functions of xby the rule

(τf)(x) = f(xqminus2)

A function f(x) is called a quasi-constant if τf = f

An operator D = a0(x) + a1(x) τ + + aN (x) τN where a0 aN are functions and aN isnot identically zero is called a difference operator of order N The functions a0 aN are thecoefficients of D If aN = 1 the operator D is called monic

Recall that for any collection of quasi-polynomials U = (u1 uN ) by definition in Section31 WN [u1 uN ] 6= 0

Lemma 51 For any collection of quasi-polynomials U = (u1 uN ) there exists a unique monicdifference operator DU such that DU ui = 0 for all i = 1 N

Proof Define the operator DU by the rule

DU f =WN+1 [u1 uN f ]

WN [u1 uN ] (36)

Then clearly DU ui = 0 for all i = 1 N On the other hand write DU = a0 + +aNminus1 τ

Nminus1+τN Then equalities DU ui = 0 i = 1 N amount to a system of linear equationson a0 aNminus1

a0 ui + a1 τui + + aNminus1 τNminus1ui = minusτNui i = 1 N (37)

Since WN [u1 uN ] 6= 0 the matrix (τ jminus1ui)Nij=1 is invertible and solution of system (37) has

is unique

The operator DU is called the fundamental difference operator of the collection U Notice thatby (36)

a0(x) = (minus1)NWN [u1 uN ](xqminus2)

WN [u1 uN ](x) (38)

Lemma 52 Let U = (u1 uN ) be a collection of quasi-polynomials Any solution f ofthe difference equation DU f = 0 is a linear combination of u1 uN with quasi-constantcoefficients

Proof Letci(x) = minusWN [u1 uiminus1 f ui+1 uN ](x)

WN [u1 uN ](x) i = 1 N (39)


13

By Lemma Appendix A4 f = c1u1 + + cN uN If ci 6= 0 consider a collection Ui = (u1 uiminus1 f ui+1 uN ) Lemma 51 implies that D

Ui= DU and by formula (38)

WN [u1 uiminus1 f ui+1 uN ](xqminus2)

WN [u1 uiminus1 f ui+1 uN ](x)=

WN [u1 uN ](xqminus2)

WN [u1 uN ](x)

Thus ci(xqminus2) = ci(x) for all i = 1 N

Lemma 53 For a collection of quasi-polynomials U = (u1 uN ) set vN = uN and

vi =WNminusi+1 [ui uN ]

WNminusi [ui+1 uN ] i = 1 N minus 1 (40)

ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)

Proof By applying repeatedly Lemma Appendix A2 we get

(τ minus τvi

vi

) (τ minus τvN

vN

)f =

WNminusi+2 [ui uN f ]

WNminusi+1 [ui uN ] i = 1 N

Comparing this formula with formula (36) completes the proof

Corollary 54 Let U be a regular collection of quasi-polynomials Then the operator DU hasrational coefficients

Proof For a regular collection of quasi-polynomials U the expressions τvivi in formula (41) arerational functions which proves the claim

Proposition 55 Let U = (u1 uN ) be a collection of quasi-polynomials such that the operatorDU has rational coefficients Then the collection U is semiregular that is the quasi-polynomialWN [u1 uN ] is log-free

We will prove Proposition 55 as well as Propositions 56 ndash 59 below in Section 53

Proposition 56 Let U be a collection of quasi-polynomials such that the operator DU has

rational coefficients Then there is a regular collection of quasi-polynomials U such that DU

= DU

Say that λ = (λ1 λN ) is dominance-free if q 2(λiminusλj) 6= q 2s for all 1 6 i lt j 6 N ands isin Zgt1

Proposition 57 Let U be a collection of quasi-polynomials of type λ such that the operator DU

has rational coefficients Assume that λ is dominance-free or q is a root of unity Then there is

a regular collection of quasi-polynomials U of type λ such that DU

= DU

Example Let U = (u1 u2) be the collection of quasi-polynomials of type λ = (1 0)

u1(x) = x u2(x) =1

1minus qminus2+q2x log x

2 log q

Then

W2[u1 u2 ](x) = xminus x2 DU = τ2minus(qminus2+

1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14

and U = (u2 u1) is a regular collection of quasi-polynomials such that DU

= DU However if q is

not a root of unity DU is not the fundamental difference operator of any regular collection of quasi-polynomials of type λ = (1 0) If q 2` = 1 for a positive integer ` set uprime2(x) = x`u2(x) Then

Uprime= (uprime2 u1) is a regular collection of quasi-polynomials of type λ = (1 0) such that DUprime

= DU

Say that λ = (λ1 λN ) is generic if q 2(λiminusλj) 6= q 2s for all 1 6 i lt j 6 N and s isin Z

Proposition 58 Let U = (u1 uN ) be a collection of quasi-polynomials of type λ such thatthe operator DU has rational coefficients Assume that λ is generic Then the quasi-polynomialsu1 uN are log-free In particular the collection U is regular

Proposition 59 Let U = (u1 uN ) and U = (u1 uN ) be collections of quasi-polynomialsof type λ such that DU = D

Uand the operator DU has rational coefficients Assume that λ is

generic Then there are quasi-constants c1 cN such that ui = ciui i = 1 N If q isnot a root of unity then the quasi-constants c1 cN are constants

52 Difference operator of a solution of Bethe ansatz equationsFix collections of complex numbers λ = (λ1 λN ) nonnegative integers l = (l1 lNminus1) andmonic polynomials T = (T1 TNminus1)

Given t isin C l define the polynomials p1(x) pNminus1(x) by the rule

pi(x) =

liprodj=1


cf (20) and set p0(x) = pN (x) = 1 Let

Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i

Tj(xq2(iminusj)) i = 1 N (43)

Define the fundamental difference operator Dt of the point t by the rule

Dt =(τ minus qminus2λ1

R1(xqminus2)

R1(x)

) (τ minus qminus2λN

RN (xqminus2)

RN (x)

) (44)

Theorem 510 Let t isin C l be a solution of equations (1) associated with the data l T λ LetU be a collection of quasi-polynomials of type λ such that T = (T1 TNminus1 1) is a preframe ofU and the Sl-orbit of t equals XUT Then Dt = DU

Proof Let U = (u1 uN ) Since the Sl-orbit of t equals XUT formulae (9) (10) imply

xλiRi(x) =WNminusi+1 [ui uN ](x)


Thus Dt = DU see formulae (40) (41) (44)

Recall that a point t isin C l is admissible if it satisfies conditions (15)


15

Corollary 511 Let t isin C l be an admissible solution of equations (1) associated with the datal T λ Then there exists a collection of quasi-polynomials U = (u1 uN ) of type λ such thatDt = DU and

WN [u1 uN ](x) = xλ1++λNNprodi=1

Nminusiprodj=0


Proof By Theorem 34 the collection of quasi-polynomials U = (u1 uN ) constructed inSection 42 has a preframe T = (T1 TNminus1 1) and the Sl-orbit of t equals XUT Henceformula (45) holds cf (8) and Dt = DU by Theorem 510

Theorem 512 Let t isin C l be an admissible solution of equations (1) associated with the datal T λ Let U be a collection of quasi-polynomials of type λ such that Dt = DU Assume thatq is not a root of unity and λ is generic Then T = (T1 TNminus1 1) is a preframe of U andthe Sl-orbit of t equals XUT

Proof Let U = (u1 uN ) and U = (u1 uN ) be the collection of quasi-polynomials of typeλ constructed in the proof of Theorem 34 see Section 42 By Theorem 510 D

U= Dt = DU

Then by Lemma 59 for any i = 1 N the quasi-polynomials ui and ui are proportionalwhich proves the theorem

Theorem 512 shows that for q being not a root of unity generic λ and an admissible solutiont of the Bethe ansatz equations (1) solving the difference equation Dtf = 0 allows one to producea collection of quasi-polynomials U of type λ such that the Sl-orbit of t equals XUT This isan alternative way to the construction of such a collection of quasi-polynomials given in Section42 Moreover Theorem 512 and Lemma 59 imply that such a collection of quasi-polynomials isunique up to rescaling of individual quasi-polynomials

53 Proofs of Propositions 55 ndash 59Recall that q 6= 0plusmn1 Most of the technicalities in this section are related to the fact that q canbe a root of unity

Lemma 513 Let c(x) be a quasi-constant of the form c(x) = xα r(x log x) where α isin C andr(x y) is a rational function in x and a polynomial in y Then r(x y) does not depend on y

Proof Let r(x y) = r0(x) + r1(x) y + + rk(x) yk and k gt 1 The equality c(x) = c(xqminus2)imply that

rk(x) = qminus2α rk(xqminus2) rkminus1(x)minus qminus2α rkminus1(xqminus2) = minus2k rk(x) log q (46)

Denote by di the order of ri(x) at x = 0 that is limxrarr0

(xminusdi ri(x)

)6= 0infin Then the first

equality in (46) gives q 2(α+dk) = 1 which makes impossible matching the orders of the left andright sides in the second equality

For a quasi-polynomial f(x) = xα(P0(x) + P1(x) log x + + Pk(x)(log x)k

) where P0

Pk are polynomials Pk 6= 0 denote by 〈f 〉(x) = xαPk(x) the top part of f(x)

Lemma 514 Let D be a difference operator with rational coefficients If a quasi-polynomialf(x) satisfies the equation Df = 0 then D〈f 〉 = 0 too

Proof Since τ(xαxi(log x)j

)= qminus2(α+i)xαxi r(log x) for some monic polynomial r(s) of degree

j the claim follows


16

Proof of Proposition 55 Let c(x) = WN [u1 uN ](x)langWN [u1 uN ]

rang(x) By Lemma 514

c(x) is a quasi-constant and c(x) satisfies the assumption of Lemma 513 Since the quasi-polynomial

langWN [u1 uN ]

rangis log-free Proposition 55 follows

Proof of Propositions 56 57 We will prove the Propositions by induction on the number of quasi-polynomials in the collection U = (u1 uN )

If the quasi-polynomial uN is not log-free let f = 〈uN 〉 By Lemma 514 DU f = 0

If WN [u1 uNminus1 f ] 6= 0 set U = (u1 uNminus1 f ) Then the collection U has type λ

DU

= DU and the quasi-polynomial f is log-free

If WN [u1 uNminus1 f ] = 0 then WN [u1 uiminus1 f ui+1 uN ] 6= 0 for some i = 1

Nminus1 see Lemma 39 Then for the collection U = (u1 uiminus1 uN ui+1 uNminus1 f ) DU

= DU

and the quasi-polynomial f is log-free

The quasi-constants ci(x) given by (39) have the form ci(x) = xλNminusλi ri(x) for some rationalfunctions ri(x) Let di be the order of ri(x) at x = 0 Since ci(x) is a quasi-constantq 2(λNminusλi+di) = 1 For a dominance-free λ we have di 6 0 and we set ` = 0 If q is a root ofunity we find an integer ` gt di such that q 2` = 1 Then xλiminusλN+`minusdi is a quasi-constant and

ui(x) = xλiminusλN x`minusdiuN (x) is a quasi-polynomials of type λi Thus Uprime = (u1 uiminus1 ui ui+1 uNminus1 f ) is a collection of quasi-polynomials of type λ D

Uprime= DU and the quasi-polynomial

f is log-free

Assume now that the quasi-polynomial uN is log-free so r(x) = uN (xqminus2)uN (x) is a rationalfunction Let the functions v1 vN be given by (40) Set

D =(τ minus τv1

v1

) (τ minus τvNminus1

vNminus1

)= τNminus1+ bNminus2 τ

Nminus2+ + b0

By Lemma 53 the operator DU factors

D middot(τ minus r(x)

)= DU = τN+ aNminus1 τ

Nminus1+ + a0

so the coefficients b0 bNminus2 are determined by the equations

ai(x) = biminus1(x)minus bi(x) r(xqminus2i) i = 0 N minus 1

where bminus1(x) = 0 and bNminus1(x) = 1 Therefore b0(x) bNminus1(x) are rational functions

Let uprimei = W2[ui uN ] and Uprime = (uprime1 uprimeNminus1) By Lemmas 53 and 51 the operator DUprime

acts as follows

DUprime f(x) =1

uN (xq 2(1minusN))D(f(x)uN (x)

)

and hence has rational coefficients By the induction assumption there is a regular collection ofquasi-polynomials Uprimeprime= (uprimeprime1 u

primeprimeNminus1) such that DUprimeprime = DUprime Set

cij(x) = minusWNminus1 [uprime1 u

primejminus1 u

primeprimei uprimej+1 u

primeNminus1 ](x)

WNminus1 [uprime1 uprimeNminus1 ](x)

i j = 1 N minus 1 (47)

Similarly to the proofs of Proposition 55 and Lemma 52 one can show that

uprimeprimei (x) = ci1(x)uprime1(x) + + ciNminus1(x)uprimeNminus1(x) i = 1 N minus 1 (48)


17

the Wronskians in (47) are log-free quasi-polynomials and cij(x) are quasi-constants of the formcij(x) = xαij rij(x) for some rational functions rij(x) Let dij be the order of rij(x) at x = 0

Since cij(x) is a quasi-constant q 2(αij+dij) = 1 and the function xminusdij rij(x) is also a quasi-constant Denote by P (x) the least common denominator of the functions xminusdij rij(x) i j = 1 N minus 1 The polynomial P (x) is a quasi-constant as well

Define the collection of quasi-polynomials U = (u1 uN ) by the rule uN = uN and

ui(x) = P (x)(ci1(x)u1(x) + + ciNminus1(x)uNminus1(x)

) i = 1 N minus 1

Then W2[ ui uN ](x) = P (x)uprimeprimei (x) i = 1 N minus 1 see (48) By Lemmas Appendix A1Appendix A2

WNminusi+1 [ ui uN ](x)

Nminusiminus1prodj=1

uN (xqminus2j) = WNminusi [uprimeprimei u

primeprimeNminus1 ](x)

Nminusiprodk=0

P (xqminus2k)

i = 1 N minus 2 which implies that the quasi-polynomials WNminusi+1 [ ui uN ] are log-free and

U is a regular collection of quasi-polynomials Since DU

= DU by Lemma 51 Proposition 56 isproved

Proof of Proposition 59 By Proposition 55 the quasi-polynomial WN [u1 uN ] is log-freeSince DU ui = 0 the quasi-polynomials WN [u1 ujminus1 ui uj+1 uN ] are log-free as wellSet

cij(x) = minusWN [u1 ujminus1 ui uj+1 uN ](x)

WN [u1 uN ](x) i j = 1 N

Similarly to the proof of Lemma 52 cij(x) are quasi-constants of the form cij(x) = xλiminusλj rij(x)

for some rational functions rij(x) Since λ is generic that is q 2(λiminusλj) 6= q 2s for i 6= j and anys isin Z we get that cij = 0 for i 6= j and ui(x) = rii(x)ui(x) i = 1 N where rii(x) arerational functions and quasi-constants

If q is not a root of unity then the only rational functions that are quasi-constants are constantfunctions

Proof of Proposition 58 Let ui = 〈ui〉 i = 1 N Then U = (u1 uN ) is a collection ofquasi-polynomials of type λ By Lemmas 514 and 51 D

U= DU By the proof of Proposition 59

there are rational functions rii(x) such that ui(x) = rii(x)ui(x) i = 1 N Since the quasi-polynomials u1 uN are log-free the quasi-polynomials u1 uN are log-free as well

Appendix A The Wronskian identitiesLemma Appendix A1 Given functions f1(x) fk(x) and g(x) we have

Wk[gf1 gfk ](x) = Wk[f1 fk ](x)kminus1prodi=0

g(xqminus2i)

The proof is straightforward


18

Lemma Appendix A2 Given functions f1(x) fj(x) g1(x) gk(x) let

hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then

Wk[h1 hk ](x) = Wj+k[g1 gk f1 fj ](x)kminus1prodl=1

Wj [f1 fj ](xqminus2l)

The proof is similar to that of Lemma 94 in [MV2]

Lemma Appendix A3 Given functions f1(x) fk(x) we have

ksumi=1

(minus1)iWkminus1[f1 fiminus1 fi+1 fk ](x) fi(xqminus2l) = 0 l = 0 k minus 2

ksumi=1

(minus1)kminusiWkminus1[f1 fiminus1 fi+1 fk ](x) fi(xq2minus2k) = Wk[f1 fk ](x)

Proof Consider a ktimes k matrix M with entries Mij = fi(xqminus2(jminus1)) for j lt k and Mil =

fi(xqminus2l) If l = 0 kminus 2 two rows of M are the same hence det M = 0 If l = kminus 1 then

det M = Wk[f1 fk ] Expanding the determinant in the last row yields the claim

Lemma Appendix A4 Given functions f1(x) fk(x) let

gi(x) = Wkminus1[f1 fiminus1 fi+1 fk ](x) i = 1 k

Then

Wj [gj g1 ](x) = Wkminusj [fj+1 fk ](xq 2minus2j)kminus2prodl=0

Ws[f1 fs](xqminus2l)


References[KBI] V EKorepin NMBogoliubov and AG Izergin Quantum inverse scattering

method and correlation functions Cambridge University Press 1993[KS] P PKulish and EK Sklyanin Quantum spectral transform method Recent developments Lect Notes in

Phys 151 (1982) 61ndash119[MV1] EMukhin AVarchenko Critical points of master functions and flag varieties Commun Contemp Math

6 (2004) no 1 111ndash163[MV2] EMukhin AVarchenko Solutions to the XXX type Bethe ansatz equations and flag varieties Cent Eur

J Math 1 (2003) no 2 238ndash271[MV3] EMukhin AVarchenko Quasi-polynomials and the Bethe ansatz Geom Topol Monogr 13 (2008) 385ndash

420


19

XXZ -type Bethe ansatz equations and

quasi-polynomials

Jian Rong Li and Vitaly Tarasovlowast Department of Mathematics Lanzhou University Lanzhou 730000 PRChinalowast Department of Mathematical Sciences Indiana University ndashPurdue University Indianapolis 402North Blackford St Indianapolis IN 46202-3216 USAlowast St Petersburg Branch of Steklov Mathematical Institute Fontanka 27 St Petersburg 191023Russia

Abstract We study solutions of the Bethe ansatz equation for the XXZ -type integrable modelassociated with the Lie algebra slN We give a correspondence between solutions of the Betheansatz equations and collections of quasi-polynomials This extends the results of EMukhin andAVarchenko for the XXX-type model and the trigonometric Gaudin model

1 IntroductionIn this paper we study solutions of the Bethe ansatz equation for the XXZ -type integrable modelsassociated with the Lie algebra slN see (1) These equations arise in the Bethe ansatz methodof computing eigenvalues and eigenvectors of commuting Hamiltonians of integrable models Themethod gives the eigenvalues and eigenvectors by evaluating explicit rational functions on solutionsof the Bethe ansatz equations see for instance [KBI] and references therein

Solutions of the Bethe ansatz equations for the Gaudin model both rational and trigonometricas well for the XXX-type model have been studied by E Mukhin and AVarchenko in [MV1] ndash[MV3] They established a correspondence between solutions of the Bethe ansatz equations andspaces of polynomials quasi-polynomials or quasi-exponentials with certain properties In thispaper we extend the results of [MV2] [MV3] to the case of the XXZ-type Bethe ansatz equations

Our method to construct a collection of quasi-polynomials corresponding to a solution of theBethe ansatz equations is not completely analogous to that of [MV2] [MV3] As a result wemanaged to weaken technical assumptions on solutions of the Bethe ansatz equations comparedwith those imposed in [MV3] Another advantage is that our method works withut restrictions forthe root of unity case

The plan of the paper is as follows In Section 2 we describe the XXZ-type Bethe ansatzequations and refine the problem In particular we define regular and admissible solutions of theBethe ansatz equations In Section 3 we introduce regular collections of quasi-polynomials andshow that each such a collection gives a regular solution of the Bethe ansatz equations We alsoformulate there the main result of the paper Theorem 34 which says that every admissible regularsolution of the Bethe ansatz equations comes from a regular collection of quasi-polynomials Weprove Theorem 34 in Section 4 In Section 5 for a collection of quasi-polynomials U we consider


Published under licence by IOP Publishing Ltd 1








t = (t(1)1 t

(1)l1 t

(Nminus1)1 t

(Nminus1)lNminus1

) isin C l





la

Ti(t(i)j q

2)

liminus1prodr=1

(t(i)j q minus t

(iminus1)r qminus1)

li+1prods=1

(t(i)j minus t

(i+1)s )

liprodk=1k 6=j



liminus1prodr=1

(t(i)j minus t

(iminus1)r )

li+1prods=1


s q)

liprodk=1k 6=j

(t(i)j q minus t

(i)k qminus1)







Ti(t(i)j q

2)

Ti(t(i)j )

liminus1prodr=1

t(i)j q minus t

(iminus1)r qminus1

t(i)j minus t

(iminus1)r

li+1prods=1

t(i)j minus t

(i+1)s


s q

liprodk=1k 6=j


t(i)j q minus t

(i)k qminus1

= (2)



2


pi(x t) =

liprodj=1

(xminus t(i)j ) (3)

and










dk

dxkPi(x t)|

x=t(i)I

= 0 k = 0 |I | minus 1 (6)












(7)



3





kminusiprodj=0




Nminusiprodj=0


DenoteQTk (x) =

kprodi=1

kminusiprodj=0










T1


where QT1 Q





(i)li



liprodj=1





4






that isyi(x) yi(xq


Therefore

yi(x)(yi(xq


)= (14)



2) yi+1(x)Ti(xq2)





minus2) pi(xq2))





and for both U and



t(i)j 6= t


(i)j ) 6= 0 for all i = 1






5



t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)















I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19








t = (t(1)1 t

(1)l1 t

(Nminus1)1 t

(Nminus1)lNminus1

) isin C l





la

Ti(t(i)j q

2)

liminus1prodr=1

(t(i)j q minus t

(iminus1)r qminus1)

li+1prods=1

(t(i)j minus t

(i+1)s )

liprodk=1k 6=j



liminus1prodr=1

(t(i)j minus t

(iminus1)r )

li+1prods=1


s q)

liprodk=1k 6=j

(t(i)j q minus t

(i)k qminus1)







Ti(t(i)j q

2)

Ti(t(i)j )

liminus1prodr=1

t(i)j q minus t

(iminus1)r qminus1

t(i)j minus t

(iminus1)r

li+1prods=1

t(i)j minus t

(i+1)s


s q

liprodk=1k 6=j


t(i)j q minus t

(i)k qminus1

= (2)



2


pi(x t) =

liprodj=1

(xminus t(i)j ) (3)

and










dk

dxkPi(x t)|

x=t(i)I

= 0 k = 0 |I | minus 1 (6)












(7)



3





kminusiprodj=0




Nminusiprodj=0


DenoteQTk (x) =

kprodi=1

kminusiprodj=0










T1


where QT1 Q





(i)li



liprodj=1





4






that isyi(x) yi(xq


Therefore

yi(x)(yi(xq


)= (14)



2) yi+1(x)Ti(xq2)





minus2) pi(xq2))





and for both U and



t(i)j 6= t


(i)j ) 6= 0 for all i = 1






5



t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)















I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19


pi(x t) =

liprodj=1

(xminus t(i)j ) (3)

and










dk

dxkPi(x t)|

x=t(i)I

= 0 k = 0 |I | minus 1 (6)












(7)



3





kminusiprodj=0




Nminusiprodj=0


DenoteQTk (x) =

kprodi=1

kminusiprodj=0










T1


where QT1 Q





(i)li



liprodj=1





4






that isyi(x) yi(xq


Therefore

yi(x)(yi(xq


)= (14)



2) yi+1(x)Ti(xq2)





minus2) pi(xq2))





and for both U and



t(i)j 6= t


(i)j ) 6= 0 for all i = 1






5



t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)















I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19





kminusiprodj=0




Nminusiprodj=0


DenoteQTk (x) =

kprodi=1

kminusiprodj=0










T1


where QT1 Q





(i)li



liprodj=1





4






that isyi(x) yi(xq


Therefore

yi(x)(yi(xq


)= (14)



2) yi+1(x)Ti(xq2)





minus2) pi(xq2))





and for both U and



t(i)j 6= t


(i)j ) 6= 0 for all i = 1






5



t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)















I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19






that isyi(x) yi(xq


Therefore

yi(x)(yi(xq


)= (14)



2) yi+1(x)Ti(xq2)





minus2) pi(xq2))





and for both U and



t(i)j 6= t


(i)j ) 6= 0 for all i = 1






5



t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)















I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19



t(i)j 6= t

(i)k q

2 i = 1 N minus 1 j k = 1 li (15)















I [1 1](s) =s

2 log q I [1 c ](s) =

1

1minus c c 6= 1


6





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19





I[1 ] =log x

log q 2 I[xα ] =

xα





f(x)


v(x) =


y(xqminus2)

rang+






anda(xq 2)b(x)


)=










W2[u1 u2] = xλ1+λ2 T1(x)



7



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19



Let

pi(x) =

liprodj=1










2)yi+1(x)Ti(xq2) (22)

of type λi + 2λi+1 + 3(λi+2 + + λN )

Set Q1(x) = 1 and

Qk(x) =kprodi=2

kminusiprodj=0















8



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19



minus2))

+


)



A For any j = i N








2)wi+1j(x)Ti(xq2)









of type λi + + λN


Nsumj=i+1


Nsumj=i+1


+ bi(x)Nsum

j=i+1



9


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19


Nsumj=i+1




Nminus1prodk=i+1

Tk(x) (30)



ui(x) =

Nsumj=i+1






minus2)Ti(x)

yi(x) yi(xqminus2)




Nsumj=i+1


)uj(xq


andNsum

j=i+1


)uj(xq

2(iminusN)) =






ui(xqminus2l) =

Nsumj=i+1


and

ui(xq2(iminusN)) =



Nsumj=i+1




10













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19













minus2))

+


)








cf (29)




11










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19










kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))






kminus2prodl=0


minus2l)

yi(xqminus2(l+1))


kminus2prodl=0

(yiminus1(xq

minus2l)

i+lprodm=i

Tm(xqminus2l))


Wk[wij1 wijk ](x) =

=ksuml=1



wijl+1 wijk ](x)







QNminus1(xq2l)

QN (xq 2l)


12












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19












WN [u1 uN ] (36)





is unique



WN [u1 uN ](x) (38)



WN [u1 uN ](x) i = 1 N (39)


13






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19






WN [u1 uN ](x)





ThenDU =

(τ minus τv1

v1

) (τ minus τvN

vN

) (41)


(τ minus τvi

vi

) (τ minus τvN

vN

)f =










= DU





= DU


u1(x) = x u2(x) =1


2 log q

Then


1minus xqminus2

1minus x

)τ + qminus2

1minus xqminus2

1minus x


14





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19





= DU








pi(x) =

liprodj=1



Ri(x) =piminus1(x)

pi(x)

Nminus1prodj=i




R1(xqminus2)

R1(x)


RN (xqminus2)

RN (x)

) (44)








15



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19



Nminusiprodj=0





U= Dt = DU








(xminusdi ri(x)




) where P0





j the claim follows


16




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19




langWN [u1 uN ]





DU




= DU





f is log-free


D =(τ minus τv1

v1


vNminus1


Nminus2+ + b0




Nminus1+ + a0





acts as follows

DUprime f(x) =1


)




primejminus1 u


primeNminus1 ](x)


i j = 1 N minus 1 (47)




17





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19





) i = 1 N minus 1






Nminusiprodk=0

P (xqminus2k)






WN [u1 uN ](x) i j = 1 N









g(xqminus2i)



18


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19


hi(x) = Wj+1[gi f1 fj ](x) i = 1 k

Then





ksumi=1


ksumi=1







Then









420


19

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XXZ-type Bethe ansatz equations and quasi-polynomials