Boundary quantum Knizhnik-Zamolodchikov equation Masahiro Kasatani (Univ. of Tokyo) July 30, 2009 Infinite Analysis 09 - New Trends in Quantum Integrable Systems @Kyoto University (Joint work with K. Shigechi)

Boundary qKZ equation

Summary Yang-Baxter and reflection equations (K-matrix) boundary qKZ equation (Def. 1) ⇑ (Prop. 3) qKZ families (families of Laurent polynomials) (Def. 2) m (Thm. 5) Solutions of an eigenvalue problem (Def. 4) ⇑ (Example 1,2) Non-symmetric Koornwinder polynomials (Def. 6) 1

Boundary qKZ equation

Yang-Baxter and reflection equations V = Cv−M ⊕ · · · ⊕ CvM - (2M + 1)-dim. vector space and its basis R(z) : V ⊗ V → V ⊗ V - R-matrix K(z) : V → V - K-matrix µ R1,2

¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ z1 z1 z2 z2 z1 z1 R1,3 R2,3 = R2,3 R1,3 R1,2 z2 z3 z3 z3 z3 z2

on V ⊗3, ³w ´ ³w ´ K2(w)R1,2(zw)K2(z)R1,2 = R1,2 K2(z)R1,2(zw)K2(w) z z on V ⊗2.

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Boundary qKZ equation

The R-matrix R(z) ∈ EndC(V ⊗2) with parameter q is defined by R(z) (v²1 ⊗ v²2 ) =

X ²01 ,²02

R(z)²²01²²02 v²01 ⊗ v²02 , 1 2

where R(z)ii ii

= 1,

R(z)ij ij

(1 − z)q = , 2 1−q z

R(z)ji ij

1 − q 2 θ(i>j) = z 2 1−q z

(i 6= j)

and R(z)ij i0 j 0 = 0 otherwise. Here ½ θ(P ) =

1 if P is true, 0 if P is false. 3

Boundary qKZ equation

The K matrices We consider “partial” reflections: α, β - two non-negative integers such that 0 ≤ α, β ≤ M K(z), K(z) ∈ EndC(V ) with parameters qn, un, q0, u0, (s) i

K(z)v =

X

Kii0 (z)vi0 ,

i

K(z)v =

X

i0

Kii(z)

= 1

Kii(z)

1 − z2 = qn (1 − az)(1 − bz)

i

K i0 (z)vi0 ,

i0

(|i| ≤ α), (|i| > α), 1/2

−1/2

−1 2θ(i<0) (q − q + (un − un n n )z i K−i(z) = −qn (1 − az)(1 − bz)

)z

(|i| > α)

1/2 −1/2 (a = qn1/2u1/2 , b = −q ). n n un 4

Boundary qKZ equation

i

K i(z) = 1 i K i(z) i K −i(z)

|i| ≤ β,

1 − sz −2 = q0 (1 − cz −1)(1 − dz −1)

|i| > β, 1/2

−1/2

(q0 − q0−1)sθ(i>0)z −2θ(i>0) + (u0 − u0 = −ciq0 (1 − cz −1)(1 − dz −1) 1/2 1/2

1/2 −1/2

(c = s1/2q0 u0 , d = −s1/2q0 u0

)s1/2z −1

|i| > β

).

i

and Kji (z) = K j (z) = 0 otherwise. c1, . . . , cM - additional parameters (c−i = c−1 i )

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Boundary qKZ equation

ˇ and K ˇ matrices with signs R ˇ and K-matrices ˇ We define Rwith signs, denoted by ± ± ⊗n (1 ≤ i ≤ n − 1), Q ), as follows: Q± 0 and Qn ∈ End(V i Q+ i (z) = Pi,i+1 Ri,i+1 (z),

Q− i (z) = f (z)Pi,i+1 Ri,i+1 (z),

Q+ n (z) = Pn Kn (z),

n Q− (z) = f (z)PnKn(z), n

Q+ 0 (z) = P0 K 0 (z),

0 Q− 0 (z) = f (z)P0 K 0 (z),

where Pi,i+1(· · · ⊗ u ½ ⊗ v ⊗ ···) = ··· ⊗ v ⊗ u ⊗ ···, ½ · · · ⊗ vi (|i| ≤ α), vi ⊗ · · · (|i| ≤ β), Pn(· · ·⊗vi) = and P0(vi⊗· · · ) = · · · ⊗ v−i (|i| > α), civ−i ⊗ · · · (|i| > β). f, f n, f 0 are rational functions satisfying f (z)f (1/z) = 1, f n(z)f n(1/z) = 1, and f 0(z)f 0(s/z) = 1. 6

Boundary qKZ equation

Explicitly, q 2z − 1 , f (z) = 2 q −z

1/2

n

f (z) =

−qn2

+

z2

−

1/2

f 0(z) =

1 − sq02z −2 − s1/2q0(u0 −q02

+

sz −2

−

−1/2

1 − qn2 z 2 − (un − un

s1/2q

1/2 0 (u0

1/2 (un

−

−1/2

− u0 −

)z

−1/2 un )z

)z −1

−1/2 u0 )z −1

,

.

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Boundary qKZ equation

Boundary qKZ equation Fix three signs σ, σn, σ0. Definition 1. For a V ⊗n-valued function F (z1, . . . , zn), the boundary quantum Knizhnik-Zamolodchikov (qKZ) equaion is a system of s-difference equations given as follows: for 1 ≤ i ≤ n F (z1, . . . , szi, . . . , zn) = Qσi−1(szi/zi−1) · · · Qσ1 (szi/z1)Qσ0 0 (zi) ×Qσ1 (z1zi) · · · Qσi (zizi+1) · · · Qσn−1(znzi)Qσnn (zi) ×Qσn−1(zi/zn) · · · Qσi (zi/zi+1)F (z1, . . . , zn).

8

Boundary qKZ equation

Related works on boundary qKZ eq. Jimbo-Kedem-Konno-Miwa-Weston - dim V = 2 (sl2 case), K is diagonal Kojima-Quano - dim V = n (sln case), K is diagonal Cherednik, Kato - on a vector space (⊂ V ⊗n) of dim = |W fin| = n!2n, connection to Macdonald eigenvalue problem Mimachi - qKZ of type C, sum of components of a special solution=symmetric C Macdonald poly. P(λ,0,...,0) of type C Di Francesco-Zinn-Justin - K is diagonal, a special solution of product form de Gier-Ponsaing-Shigechi - K : full-reflection, special solutions at root of unity Takeyama - rational case 9

Boundary qKZ equation

The affine Hecke algebra Hn and Noumi representation We introduce the following system of equations: Qσi (zi/zi+1)F (. . . , zi, zi+1, . . .) = F (. . . , zi+1, zi, . . .) (1 ≤ i ≤ n − 1)(1) Qσnn (zn)F (. . . , zn) = F (. . . , , 1/zn)

(2)

Qσ0 0 (z1)F (z1, . . .) = F (s/z1, . . .).

(3)

Action of the affine Weyl group W = hs0, . . . , sni of type Cn on n-variable functions: sif (. . . , zi, zi+1, . . .) = f (. . . , zi+1, zi, . . .) snf (. . . , zn) = f (. . . , 1/zn) s0f (z1, . . .) = f (s/z1, . . .). 10

Boundary qKZ equation

Noumi representation 1/2 1/2

1/2 −1/2

1/2 1/2 −1/2 Put a := t1/2 , c := s1/2t0 u0 , d := −s1/2t0 u0 n un , b := −tn un

.

Define linear operators on C[z1±1, . . . , zn±1] as follows: Tb0±1 =

±1/2 t0

+

−1/2 (1 t0

− cz1−1)(1 − dz1−1) (s0 − 1) −2 1 − sz1

−1 1 − t z z i i −1/2 i+1

±1/2 Tbi±1 = ti + ti

1−

−1 zizi+1

(si − 1)

(1 ≤ i ≤ n − 1)

±1 ±1/2 −1/2 (1 − azn )(1 − bzn ) b Tn = tn + tn (sn − 1). 1 − zn2

Then the map Ti 7→ Tˆi (0 ≤ i ≤ n) gives a representation of the affine Hecke algebra Hn of type Cn. 11

Boundary qKZ equation

The affine Hecke algebra Hn 1/2

1/2

The affine Hecke algebra Hn = Hn(t1/2, tn , t0 ) of type Cn is a unital associative algebra generated by T0, . . . , Tn with defining relations as follows: 1/2

−1/2

(T0 −t0 )(T0 +t0

) = 0,

(Ti −t1/2)(Ti +t−1/2) = 0

1 ≤ i ≤ n − 1,

−1/2 (Tn −t1/2 ) = 0, n )(Tn +tn

T0 T1 T0 T1 = T1 T0 T1 T0 , TiTi+1Ti = Ti+1TiTi+1

1 ≤ i ≤ n − 2,

Tn−1TnTn−1Tn = TnTn−1TnTn−1, Ti Tj

= Tj Ti

|i − j| ≥ 2.

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Boundary qKZ equation

qKZ family By definition of R, K, and Q± matrices, we can decompose V ⊗n to orbits of the actions (transposition and reflection). LabelPeach orbit by non-negative integers d−M , d−M +1, . . . , dγ:=min(α,β) such γ that i=−M di = n. Let δ

:= ((−M )d−M , . . . , . . . , (−γ − 1)d−γ−1 , (−γ)d−γ , . . . , γ dγ )

Id := {(m1, . . . , mn) ∈ Zn; ]{j; mj = i} = di

(4) (5)

(−γ ≤ i ≤ γ)

]{j; mj = i} + ]{j; mj = −i} = di

(−M ≤ i ≤ −γ − 1)}.

Definition 2 (qKZ family). A family of Laurent polynomials {f²; ² ∈ Id} is called a qKZ family with signs (σ, σn, σ0) and exponents c1, . . . , cM if 13

Boundary qKZ equation

for 1 ≤ i ≤ n − 1 Tˆif² = qf² if ²i = ²i+1 Tˆif² = f...,²i+1,²i,... if ²i > ²i+1 for i = n Tˆnf² = qnf² if |²n| ≤ α Tˆnf² = f...,²n−1,−²n if ²n > α for i = 0 Tˆ0f² = q0f² if |²1| ≤ β Tˆ0f² = c−²1 f−²1,²2,... if ²1 < −β σ /2

(6) (7) (8) (9) (10) (11)

σ /2

where (q, qn, q0) = (σtσ/2, σntnn , σ0t0 0 ).

14

Boundary qKZ equation

Proposition 3. Let {f²; ² ∈ Id} be a qKZ family of signs (σ, σn, σ0) with exponents (c1, . . . , cM ). F (z1, . . . , zn) :=

X

f²1,...,²n (z1, . . . , zn) v²1 ⊗ · · · ⊗ v²n .

(²1 ,...,²n )∈Id

Then F (z1, . . . , zn) is a solution of the qKZ equation. Proof. (1) is equivalent to (6) and (7). (2) is equivalent to (8) and (9). (3) is equivalent to (10) and (11).

15

Boundary qKZ equation

Equivalence to the eigenvalue problem Fix the signs (σ, σn, σ0), and Pγ take non-negative integers d−M , d−M +1, . . . , dγ such that i=−M di = n. Put δ and Id as above (4),(5). −1 Put Yi := Tj . . . Tn−1(Tn . . . T0)T1−1 . . . Tj−1 .

σ /2

σ /2

Definition 4 (Eigenvalue problem). Set (q, qn, q0) = (σtσ/2, σntnn , σ0t0 0 ). Consider the following eigenvalue problem for E(z1, . . . , zn): YiE = χiE TiE = qE

if δi < − max(α, β) if δi = δi+1 (more...)

16

Boundary qKZ equation

−1 Ti−1 · · · T1T0T1−1 · · · Ti−1 E = q0 E −1 Ti · · · Tn−1TnTn−1 · · · Ti−1E = qnE

if |δi| ≤ β if |δi| ≤ α

−1 −1 −1 (Tn−1 · · · T1T0T1−1 · · · Ti−1 ) Tn(Tn−1 · · · T1T0T1−1 · · · Ti−1 )E

= qn E

if −α ≤ δi < −β

−1 −1 −1 −1 (T1−1 · · · Tn−1 Tn−1Tn−1 · · · Ti−1)−1T0(T1−1 · · · Tn−1 Tn−1Tn−1 · · · Ti−1)E

= q0 E

if −β ≤ δi < −α.

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Boundary qKZ equation

−1 YiE := Ti . . . Tn−1(Tn . . . T0)T1−1 . . . Ti−1 E = χiE if δi < − max(α, β)

M

M α

β

δ

−β −α −M

−M 18

Boundary qKZ equation

−1 Ti · · · Tn−1TnTn−1 · · · Ti−1E = qnE

if |δi| ≤ α

α β

δ

−β −α

19

Boundary qKZ equation

−1 −1 −1 (Tn−1 · · · T1T0T1−1 · · · Ti−1 ) Tn(Tn−1 · · · T1T0T1−1 · · · Ti−1 )E = qn E if −α ≤ δi < −β

α β

δ

−β −α

20

Boundary qKZ equation

−1 Ti−1 · · · T1T0T1−1 · · · Ti−1 E = q0 E

if |δi| ≤ β

α β

δ

−β −α

21

Boundary qKZ equation

Construction of the qKZ family Theorem 5. Let E be a solution of the eigenvalue problem in Definition 4. Define a family {f²; ² ∈ Id} as follows. For the anti-dominant δ ∈ Id, put fδ = E. For any ² ∈ Id, take a “δ-good” element w² ∈ W such that w² · δ = ². Then (δ) Tw² E does not depend on a choice of w². (δ)

Putting f² := Tw² E, the family {f²; ² ∈ Id} forms a qKZ family with exponents c−δi := χiq −n(δ,>i)+n(δ,
where

n(δ, < i) := ]{j; j < i, δj = δi}

and n(δ, > i) := ]{j; j > i, δj = δi}.

22

Boundary qKZ equation

Solution to the eigenvalue problem Definition 6 (non-symmetric Koornwinder polynomials). For λ ∈ Zn, the non-symmetric Koornwinder polynomial Eλ is defined by the following properties: YiEλ

=

Eλ

=

y i Eλ

(1 ≤ ∀i ≤ n) X λ z + cλµz µ µ≺λ

yi := sλi tρ(λ)i (tnt0)σ(λ)i/2. Eλ depends on 6 parameters s, t, tn, t0, un, u0. cλµ is a rational function of these parameters. • Well-definedness of Eλ — for certain type of specialization, there is a combinatorial condition to guarantee well-definedness. 23

Boundary qKZ equation

• The individual action of T1, . . . , Tn on Eλ — a formula is known Especially, TiEλ = t1/2Eλ if λi = λi+1. 1/2 TnEλ = tn Eλ if λn = 0. (but actions of products of them are complicated in general.) It is rather easier to solve the eigenvalue problem for the case σ = + or σn = +. Unless the parameters are specialized, Eλ cannnot be an eigenfunction of Ti −1/2 (1 ≤ i ≤ n − 1) or Tn of the eigenvalue −t−1/2 or −tn .

24

Boundary qKZ equation

A factorized solution for specialized parameters : Example 1 Suppose n = M m. Specialize the parameters at s = t−M −1. For µ = (m {z 2, . . . , 1, 0}, m | − 1, m − {z 2, . . . , 1, 0}, . . . , |m − 1, m − {z 2, . . . , 1, 0}), | − 1, m − Eµ is well-defined. Moreover, Eµ(z1, . . . , zn; s = t

−M −1

)=

k Y

Y

(zi − t

`=1 m(`−1)
and TiEµ = −t−1/2Eµ

−1

t`−M −1 ) zj )(1 − zi zj

if i ∈ {1, . . . , n} \ {m, 2m, . . . , M m}.

Hence Eµ is a solution of the eigenvalue problem for the case α = β = 0, σ = −, (d−M , . . . , d−1, d0) = (m, . . . , m, 0). 25

Boundary qKZ equation

A solution for specialized parameters : Example 2 Specialize the parameters at tn = −s−`. E = E(`,...,`)(z1, . . . , zn; tn = −s−`) is well-defined, and it satisfies: Ti E

= t1/2E

Tn E

= −t−1/2 E, n

T0 E

= t0 E.

(1 ≤ i ≤ n − 1),

1/2

Hence E is a solution of the eigenvalue problem for the case (σ, σn, σ0) = (+, −, +).

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