Contributions à l'Etude du Vertex Topologique en Théorie ... - Toubkal

L.B. Drissi et al. / Nuclear Physics B 801 [FS] (2008) 316–345 343we obtain the relationΓ (3)−(z)|0〉=Γ(2)−(3)(z)Γ − (qz)|0〉as claimed in Section 6 Eq. (6.18).(A.26)Higher levelsFrom the above analysis, it is not difficult to check that the explicit expression of the vertexoperators Γ (p)− (z) acting on the vacuum is given by, (A.27)( ∞Γ (p)− (z)|0〉=exp ∑n=1Moreover using the identity,z n p−1(1 − q n ) p−1 = ∑ (qz) n(1 − q n ) k−1 ,k=1)iz n J −nn(1 − q n ) p−1 |0〉, p 1.we can decompose the above relation as followsΓ (p)−(z)|0〉=Γ(1)−p−1(z) ∏k=2Γ −(k) (qz)|0〉.Thenextstepistousetherelationz n(1 − q n ) p−1 = z n(1 − q n ) p−2 + (qz) n(1 − q n ) p−1 ,that imply the equalityΓ (p)−(p−1)(z)|0〉=Γ (z)Γ (p) (qz)|0〉.−−Doing the same for the first term for the right-hand sidez n(1 − q n ) p−2 = z n(1 − q n ) p−3 + (qz) n(1 − q n ) p−2 ,Eq. (A.30) can be brought to the formz n(1 − q n ) p−1 = z n(1 − q n ) p−3 + (qz) n(1 − q n ) p−2 + (qz) n(1 − q n ) p−1 ,leading then toΓ (p)−(z)|0〉=Γ(p−2)−(z)Γ (p−1)−(qz)Γ (p)− (qz)|0〉.We can repeat this operation successively to end with the two following:(i) the expression of level p vertex operator as we have used it in Section 6 Eq. (6.32),(A.28)(A.29)(A.30)(A.31)(A.32)(A.33)(A.34)withZ pd =〈0|T |0〉,∏p jT = Γ + (1) (z) Γ (j+1) (− q j z ) ,i=1(A.35)(A.36)