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

340 L.B. Drissi et al. / Nuclear Physics B 801 [FS] (2008) 316–345or equivalently like( ∑ i y n )Ψ − (y) = expn (1 − q n ) J −n ,n1(Ψ + (x) = exp − ∑ i x −n )n (1 − q n ) J n .n1(A.3)Let us compute the algebra of these vertex operators.First, we have,q L 0Ψ ± (z)q −L 0= Ψ ± (qz),showing that q L 0 acts as a translation operator. We also have(A.4)Ψ ± (x)Ψ ± (y) = Ψ ± (y)Ψ ± (x).(A.5)To get the commutator between Ψ + (x) and Ψ − (y), we can do it in two ways which, by theircomparison, allow us to get a new identity:(i) Computation by using products of Γ ± .Wehave,Ψ + (x)Ψ − (y) = ∏ (Γ + q l x ) ∏ (Γ − q k y )l0 k0∞∏(= 1 − q s y ) −(s+1)Ψ − (y)Ψ + (x),(A.6)xs=0in particular(∏ ∞(Ψ + (1)Ψ − (q) = 1 − qt ) )−tΨ − (q)Ψ + (1).Noticet=1Γ + (1)Ψ − (y) = Γ + (1) ∏ k0Γ −(q k y )(A.7)= ∏ k0(1 − q k y ) −1 Ψ− (y)Γ + (1)(A.8)and alsoΓ + (x)Ψ − (y) = Γ + (x) ∏ k0Γ −(q k y )= ∏ (1 − q k y ) −1Ψ − (y)Γ + (x).xk0(A.9)We also haveΨ + (x)Γ − (1) = ∏ k0Γ +(q −k x ) Γ − (1)= ∏ k0(1 − q k x −1) −1 Γ− (1)Ψ + (x).(A.10)