SLIDE 8 Example xi Γj Λa pl 1 1 1 1 1 1 1 2 2 2 −2 −2 −4 −5 1 2 1 1 6 −3 −8 (10) Here ||G1|| = (2, 4), ||G2|| = (2, 5), ||F 1
1 || = (2, 8), ||F 1 2 || = (3, 7), ||F 1 3 || = (3, 7), ||F 1 4 || = (1, 2).
Sum of third and fourth F equals sum of two G’s. Redefine: ˜ Γ1 = p1Λ3, ˜ Γ2 = p1Λ4, ˜ Λ3 =
Γ1 p1, ˜
Λ4 =
Γ2 p1,
˜ G = F 1
3 , ˜
G2 = F 1
4 , ˜
F 1
3 = G1, ˜
F 1
4 = G2
then the new geometry is given by: || ˜ G1|| = (3, 7), || ˜ G2|| = (1, 2), || ˜ F 1
3 || = (2, 4), ||F 1 4 || = (2, 5)
xi Γj Λa pl 1 1 1 1 1 1 1 2 2 2 −3 −1 −7 −2 1 1 1 1 4 3 −3 −8 (11)
He Feng (Virginia Tech) New Evidence for (0, 2) Target Space Duality 8 / 35