Remark 2. In [6], the solution of the Knizhnik-Zamolodchikov equation for the. SU(2) Wess-Zumino-Witten model is given in terms of the Selberg type integral with.
Homological representations of the Iwahori-Hecke algebra associated with a Selberg type integral Katsuhisa Mimachi Abstract We realize a series of irreducible representations of the Iwahori-Hecke algebra on the space of homology group with coefficients in the local system associated with a Selberg type integral. This gives an affirmative answer to the conjecture by R. Lawrence [11].
A Selberg type integral is (ti − tj )ν γ(z1 ,...,z n ) 1≤i for ti > zn . On the other hand, Lemma tσ(2) tσ(1)
tσ(2) on the path and arg(ti − zn ) = 0 4 below gives the equality ⊗ u(tσ(1) , tσ(2) ) ∞
zi+1 zi
ν = e(− ) 2
tσ(2)
tσ(1)
∞
zi+1 zi
⊗ u(tσ(2) , tσ(1) ).
Consequently, the right hand side of (4) turns out to be (1 − e(λi+1 +
ν ν )) γi,i (. . . , zi+1 , zi , . . .) + e(λi+1 + ) γi,i+1 (. . . , zi+1 , zi , . . .). 2 2
This is a second relation of Proposition 3. For the third relation, we start with the following:
σ∈S2
τ
i
−→
tσ(1)
zi
σ∈S2
tσ(2)
∞
zi+1
⊗ u(tσ(1) , tσ(2) )
tσ(1) tσ(2)
∞
zi+1 zi
17
⊗ u(tσ(1) , tσ(2) ).
By applying the second relation of Proposition 1 to the path in the t σ(2) -space, the right hand side turns out to be
(1 − e(λi+1 ))
σ∈S2
+ e(λi+1 )
σ∈S2
tσ(1)
tσ(2)
∞
zi+1 zi
tσ(1)
tσ(2)
...
... ∞
zi+1 zi
⊗ u(tσ(1) , tσ(2) )
⊗ u(tσ(1) , tσ(2) ).
Apply the second relation of the present Proposition to the path in t σ(1) -space of the first term and the second relation of Proposition 1 to the path in t σ(1) -space of the second term. Then we have the third desired relation. Other relations which remains to be proved are induced by Proposition 1. This completes the proof of Proposition 3. Lemma 4. The loaded cycle t2 t1
zj1
∞
zj2
⊗ u(t1 , t2 ),
where arg(t1 − t2 ) = 0 on the path for t1 > t2 and arg(ti − zn ) = 0 for ti > zn , is homologous to t 2 t 1 ν ⊗ u(t2 , t1 ), e(− ) 2 zj1 zj2 ∞ where arg(t2 − t1 ) = 0 on the path for t2 > t1 and arg(ti − zn ) = 0 for ti > zn , that is, γj1 j2 (t2 , t1 ; z). 18
Proof. The left hand side is considered as t2 -space t -space 1 arg(t − t ) = −π 1 2 arg(t1 − t2 ) = 0 ⊗ u(t1 , t2 ). zj1 zj2 ∞ zj1 zj2 ∞ t1 Hence, by considering u(t) only as a function of t 2 , this turns out to be ν e(− ) 2
t1 -space
zj1
zj2
t2 -space arg(t2 − t1 ) = 0 arg(t2 − t1 ) = π
∞
zj1
zj2
t1
⊗u(t2 , t1 ), ∞
which is the right hand side of the desired equality. This complets the proof of Lemma 4. Next, linear relations among γ j1 j2 (z) are given as follows, which is Theorem 2 in the case of m = 2. Proposition 4. Suppose that λ1 + · · · + λn + ν ∈ / Z. Then, for 1 ≤ j ≤ n, we have j−1
(e(λ1 + · · · + λs−1 ) − e(λ1 + · · · + λs )) γ s, j (z)
s=1
+ (e(λ1 + · · · + λj−1 ) − e(λ1 + · · · + λj +
+
n
(e(λ1 + · · · + λs−1 +
s=j+1
ν )) γ j, j (z) 2
ν ν ) − e(λ1 + · · · + λs + )) γ j, s (z) = 0. 2 2
Proof. We have the equality
19
j−1
(e(λ1 + · · · + λs−1 ) − e(λ1 + · · · + λs ))
s=1
+ e(λ1 + · · · + λj−1 )
+
tσ(1)
zs
tσ(2) tσ(1)
− e(λ1 + · · · + λj + ν)
n
zj
∞
tσ(2)
zj
tσ(2)
∞
zj
tσ(1) ∞
(e(λ1 + · · · + λs−1 + ν) − e(λ1 + · · · + λs + ν))
s=j+1
tσ(2)
=
tσ(1)
zj
zs
∞
tσ(2) -space
j−1
∞
zj
tσ(1) -space
⊗ u(tσ(1) , tσ(2) )
tσ(2) zs
zj
∞
(e(λ1 + · · · + λs−1 ) − e(λ1 + · · · + λs ))
s=1
+ e(λ1 + · · · + λj−1 )
20
tσ(1) -space tσ(2) zj
∞
tσ(1) -space tσ(2) − e(λ1 + · · · + λj + ν) zj n
+
∞
(e(λ1 + · · · + λs−1 + ν) − e(λ1 + · · · + λs + ν))
s=j+1
t -space σ(1)
tσ(2)
zj
∞
zs
⊗ u(tσ(1) , tσ(2) ) = 0,
where arg(tσ(1) − tσ(2) ) = 0 for tσ(1) > tσ(2) on the path. Under the assumption / Z, the last equality follows from the same argument as that λ1 + · · · + λn + ν ∈ in the proof of Proposition 2 (apply it to the path in t σ(1) -space). On the other hand, Lemma 4 implies that the loaded cycles
tσ(2)
zj
tσ(2)
tσ(1)
∞
⊗ u(tσ(1) , tσ(2) )
and
tσ(1)
zj
∞
zs
⊗ u(tσ(1) , tσ(2) ),
where arg(tσ(1) − tσ(2) ) = 0 for tσ(1) > tσ(2) on the path, are homologous to
ν e(− ) 2
tσ(2)
tσ(1) ∞
zj
⊗ u(tσ(2) , tσ(1) )
and ν e(− ) 2
tσ(2) zj
tσ(1)
zs
∞ 21
⊗ u(tσ(2) , tσ(1) ),
where arg(tσ(2) − tσ(1) ) = 0 for tσ(2) > tσ(1) , respectively. Therefore, by taking a symmetric sum for σ ∈ S2 , we obtain the desired result.
4
General m case.
Let Tz be a complex manifold Cm \Dz , where Dz = ∪1≤i zn . In what follows, we also use the symbol 1a1 2a2 ... nan (z) γ1a1 2a2 ... nan (z) or γ in stead of γ1, · · · , 1,2, · · · , 2,··· ,n, · · · , n (z) or γ 1, · · · , 1,2, · · · , 2,··· ,n, · · · , n (z)
a1
a2
an
a1
a2
an
with a1 + · · · + an = m. Then we have the following, which implies Theorem 1 in the general m case.
22
Proposition 5. For 1 ≤ i ≤ n − 1, we have τi γ1a1 ···nan (z) =
ai s=0
ν ν ν ai e(λi+1 + ai+1 ); e( ) e(s(λi+1 + ai+1 )) s e( ν ) 2 2 ai −s 2 2
×γ ···iai+1 +ai −s ,(i+1)s ··· (. . . , zi+1 , zi , . . . ), where (a; q)n =
n−1
i
(1 − aq )
and
i=0
m n
= q
(q; q)m . (q; q)n (q; q)m−n
Proof. Without loss of generality, we prove ik ,(i+1)m−k (z) τi γ =
k ν ν ν k e(λi+1 + (m − k) ); e( ) e(s(λi+1 + (m − k) )) s e( ν ) 2 2 k−s 2 s=0 2
× γim−s ,(i+1)s (. . . , zi+1 , zi , . . . ), for 0 ≤ k ≤ m. We prove this by induction on k. The case k = 0: τ i γ (i+1)m (z) = γim (. . . , zi+1 , zi , . . . ) is trivially derived. For the case of k + 1, we start with the following: τi γ ik+1 ,(i+1)m−k−1 (z)
=
σ∈Sm
tσ(k+2)
tσ(m)
tσ(k+1) tσ(1)
zi+1
∞
zi
⊗ u(tσ(1) , . . . , tσ(m) ). (6)
Apply the relation in the case of k to the paths in tσ(i) -space for 2 ≤ i ≤ m.
23
Then the right hand side of (5) turns out to be k s=0
ν ν ν k e(s(λi+1 + (m − 1 − k) )) e(λi+1 + (m − 1 − k) ); e( ) s e( ν ) 2 2 k−s 2 2
×
σ∈Sm
tσ(2) tσ(1)
tσ(m−s) tσ(m−s+1) tσ(m)
∞
zi+1 zi
⊗ u(tσ(1) , . . . , tσ(m) ).
By the same way as the proof of the second relation of Proposition 3 (along with Lemma 5 below), the sum over the symmetric group S m of the right hand side is shown to be ν γ m−s ,(i+1)s (. . . , zi+1 , zi , . . . ) (1 − e(λi+1 + (m − s − 1) )) 2 i ν + e(λi+1 + (m − s − 1) ) γ m−s−1 ,(i+1)s+1 (. . . , zi+1 , zi , . . . ). 2 i Hence τi γ ik+1 ,(i+1)m−k−1 (z) =
k ν ν ν e(λi+1 + (m − 1 − k) ); e( ) e(s(λi+1 + (m − 1 − k) )) 2 2 2 k−s+1 s=0
×
+
k s
k s=0
e( ν2 )
γ im−s ,(i+1)s (. . . , zi+1 , zi , . . . )
ν ν ν e(λi+1 + (m − 1 − k) ); e( ) e((s + 1)(λi+1 + (m − 1 − k) )) 2 2 k−s+1 2
ν × e((k − s) ) 2
k s
e( ν2 )
γ im−s−1 ,(i+1)s+1 (. . . , zi+1 , zi , . . . ).
Change of the index s into s − 1 of the sum of the second term of the right hand
24
side leads to τi γ ik+1 ,(i+1)m−k−1 (z) =
k+1 s=0
×
=
×
k s
k+1 s=0
ν ν ν e(λi+1 + (m − 1 − k) ); e( ) e(s(λi+1 + (m − 1 − k) )) 2 2 k−s+1 2
ν + e( )k−s+1 2 ν e( ) 2
k s−1
γ im−s ,(i+1)s (. . . , zi+1 , zi , . . . )
e( ν2 )
ν ν ν e(λi+1 + (m − 1 − k) ); e( ) e(s(λi+1 + (m − 1 − k) )) 2 2 k−s+1 2
k+1 s
e( ν2 )
γim−s ,(i+1)s (. . . , zi+1 , zi , . . . ).
Here we have used the relation k k+1 k + q k−s+1 = . s q s s−1 q q This completes the proof of Proposition 5. Lemma 5. For σ ∈ Sm , a loaded cycle t t σ(1)
σ(2)
tσ(m) zj1
zj2
zjm
∞
⊗ u(t1 , . . . , tm ),
where arg(ti − tj ) = 0 for ti > tj with 1 ≤ i < j ≤ m, is homologous to tσ(1) tσ(2) ν ⊗ u(tσ(1) , . . . , tσ(m) ), e(−l(σ) ) tσ(m) 2 zj1 zj2 zjm ∞ where arg(tσ(i) − tσ(j) ) = 0 for tσ(i) > tσ(j) with 1 ≤ i < j ≤ m, and l(σ) stands for the inversion number of σ ∈ Sm : l(σ) = Card{ (i, j) ; σ(i) > σ(j) and i < j }. Proof. Apply Lemma 4 repeatedly. Then we obtain the result.
25
Next, the follwing linear relations among γ j1 j2 ...jm (z) (1 ≤ j1 ≤ j2 ≤ · · · ≤ jm ≤ n) are obtained in the same way as in the proof of Proposition 4. / Z. Then, Proposition 6 (Theorem 2).Suppose that λ1 + · · · + λn + (m − 1)ν ∈ we have, for 1 ≤ j1 ≤ j2 ≤ · · · ≤ jm−1 ≤ n, j 1 −1
(e(λ1 + · · · + λs−1 ) − e(λ1 + · · · + λs )) γ s j1 ... jm−1 (z)
s=1
+(e(λ1 + · · · + λj1 −1 ) − e(λ1 + · · · + λj1 + +
j 2 −1
(e(λ1 + · · · + λs−1 +
s=j1 +1
ν )) γj1 j1 ... jm−1 (z) 2
ν ) 2 ν )) γ j1 s j2 ... jm−1 (z) 2
− e(λ1 + · · · + λs + +···
ν +(e(λ1 + · · · + λjm−1 −1 + (m − 2) ) 2 ν j1 ... jm−1 jm−1 (z) − e(λ1 + · · · + λjm−1 + (m − 1) )) γ 2 +
n s=jm−1
ν (e(λ1 + · · · + λs−1 + (m − 1) ) 2 +1 ν − e(λ1 + · · · + λs + (m − 1) )) γ j1 ... jm−1 s (z) = 0. 2
5
Proof of Theorem 3
In what follows, we suppose the condition, called n-regular, q(1 + q) · · · (1 + q + · · · + q n−1 ) = 0.
(7)
Let Ψ be a H(Sn )-module spanned by the elements ψj1 ···jm for 1 ≤ j1 < · · · < jm ≤ n endowed with the action determined by ψ··· js −1 ··· , if i = js−1 , i + 1 = js , τi ψj1 ···jm −→ (1 − q)ψj1 ···jm + qψj1 ··· js +1 ··· jm , if i = js , i + 1 = js+1 , ψj1 ···jm , if (i, i + 1) = (js , js+1 ) or i, i + 1 = j 1 , . . . , jm . It is seen that Ψ=
1≤j1
