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

336 L.B. Drissi et al. / Nuclear Physics B 801 [FS] (2008) 316–345we obtain at a first stageΓ (4)−(q) = Γ(3)−(q)Γ(3)−(q2 ) Γ (3)−(q3 ) Γ (4) (q4 ) .At a second stage, we substitute Γ − (3)(3)(q) and Γ − (q2 ) by Eq. (6.18), we getΓ − (3) (2) (3)( (q) = Γ − (q)Γ − q2 ) ,Γ −(3) (q2 ) = Γ −(2) (q2 ) Γ −(3) (q3 ) .Putting back into Eq. (6.24), we obtainΓ − (4) (2) (2)( (q) = Γ − (q)Γ − q2 ) Γ −(2) (q2 ) Γ −(3)Next replacing Γ (2)−(q3 ) Γ (3)−(q3 ) Γ (3)−(q3 ) Γ (4) (q4 ) .−(6.24)(6.25)− (q) by Eq. (6.13), we end with the following, (q2 ))( 3∏ (q3 )) Γ (4) (q4 ) .(6.26)( 3∏Γ − (4) (1)(q) = Γ − (q) l=1Comparing withwe obtainΓ (2)−l=1Γ (3)−G 5 =〈0|O 0 (x 0 )O 1 (x 1 )O 2 (x 2 )O 3 (x 3 )O 4 (x 4 )|0〉,O 0 (x 0 ) = Γ + (1),O 1 (x 1 ) = Γ −(1)( (q),3∏O 2 (x 2 ) = Γ (2) (− q2 )) ,l=1−(6.27)( 3∏O 3 (x 3 ) = Γ (3) (− q3 )) ,l=1O 4 (x 4 ) = Γ −(4) (q4 ) .With these results on lower values of p, it is straightforward to derive the generic picture.6.2. Generic result(6.28)We start from the expression ofG p (q) =〈0|Γ + (1)Γ (p)− (q)|0〉, p 1.Then we use the identity,withΓ (p)−p k =p−1(q) = ∏k=0( pk∏l k =1Γ −(k+1) (qk+1 )) ,(p − 1)!k!(p − k − 1)! , 0 k p − 1,whichisprovedinAppendix A,Eq.(A.37), we can bring G p (q) into the form,(6.29)(6.30)(6.31)