String Theory and M-Theory

398 String geometry
For these choices the Kähler potential is
K = − log i dz ∧ d¯z = − log(2τ2).
This gives the metric
Gτ ¯τ = ∂τ ∂¯τ K = 1
4τ 2 .
2
Under a change in complex structure τ → τ + dτ the metric components
change by
δgzz = dτ
2τ 2 2
and δg¯z¯z = d¯τ
2τ 2 .
2
Using the definition of the metric on moduli space (9.97) we find the modulispace
metric
ds 2 = 2Gτ ¯τ dτd¯τ = 1
(g
2V
z¯z ) 2 √ 2 dτd¯τ
δgzzδg¯z¯z gd x =
2τ 2 2
in agreement with the computation based on the Kähler potential. ✷
EXERCISE 9.10
Prove that ∂αΩ = KαΩ + χα, where the χα are the (2, 1)-forms defined in
Eq. (9.99).
SOLUTION
By definition
so the derivative gives
Ω = 1
6 Ωabcdz a ∧ dz b ∧ dz c ,
∂aΩ = 1 ∂Ωabc
6 ∂tα dza ∧ dz b ∧ dz c + 1
2 Ωabcdz a ∧ dz b ∧ ∂(dzc )
.
∂tα The first term is a (3, 0)-form, while the derivative of dz c is partly a (1, 0)form
and partly a (0, 1)-form. Since the exterior derivative d is independent
of t α , ∂Ω/∂t α is closed, and hence
∂Ω/∂t α ∈ H (3,0) ⊕ H (2,1) .
Now we are going to show that the (2, 1)-form here is exactly the χα in
Eq. (9.99). By Taylor expansion we have
z c (t α + δt α ) = z c (t α ) + M c αδt α ,