String Theory and M-Theory

322 M-theory and string duality
SOLUTION
By definition τ = C0 + ie−Φ and
M = e Φ
|τ| 2 −C0
−C0 1
As a result,
So
Also,
M −1 = e Φ
1 C0
C0 |τ| 2
1
4 tr(∂µ M∂µM −1 ) = 1
2 ∂µ
Φ 2
e |τ| ∂ µ e Φ − 1
2 ∂µ
C0e Φ ∂ µ C0e Φ
.
.
= − 1 µ
∂ Φ∂µΦ + e
2
2Φ ∂ µ
C0∂µC0 .
− ∂µ τ∂µ¯τ
= −1
2(Imτ) 2 2 e2Φ∂ µ C0 + ie −Φ
∂µ C0 − ie −Φ
= − 1 µ
∂ Φ∂µΦ + e
2
2Φ ∂ µ
C0∂µC0 .
This establishes the required identities. The SL(2, ¡ ) symmetry is manifest
for tr(∂ µ M∂µM −1 ), because when one substitutes M → (Λ −1 ) T MΛ −1 the
constant Λ factors cancel using the cyclicity of the trace. ✷
EXERCISE 8.4
Verify that the action in Eq. (8.70) agrees with Eq. (8.53).
SOLUTION
First we need the action (8.53) in the Einstein frame. Using Eqs (8.68) and
(8.69), it is given by S = SNS + SR + SCS, where
SNS = 1
2κ2
d 10 x √
−g R − 1
2 ∂µΦ∂ µ Φ − 1
2 e−Φ |H3| 3
SR = − 1
4κ2
d 10 x √
−g e 2Φ |F1| 2 + e Φ | F3| 2 + 1
2 | F5| 2
SCS = − 1
4κ2
C4 ∧ H3 ∧ F3.
We only need to rewrite the first two terms in Eq. (8.70) and compare them