V = eK Gρ¯Dρ W Dρ W ’ 3|W |2

ρ

(10.190)

¯

10.3 Moduli stabilization 505

V(σ)

σ0

σ

V0

Fig. 10.7. Form of the potential as a function of the radial modulus. The values of

the potential and the size depend on the values used for A, a and W0 . The ¬gure

displays a minimum at which the potential is negative leading to an AdS vacuum.

has a minimum that is given by

a2 A2 ’2aσ0

V0 = ’ e . (10.191)

6σ0

Here σ0 is the value of σ in the radial modulus ρ = iσ at the minimum of the

potential. Since this potential is negative, the only maximally symmetric

space-time allowed by such a supersymmetric compacti¬cation is AdS space-

time.

One can break supersymmetry explicitly by adding anti-D3-branes. This

gives an additional term in the scalar potential of the form

D

δV = , (10.192)

σ2

where D is proportional to the number of anti-D3-branes.

It can be chosen to make the vacuum energy density positive, so that a

compacti¬cation to dS space becomes possible. Including the anti-D3-brane

contribution results in the scalar potential

aAe’aσ 1 D

σaAe’aσ + W0 + Ae’aσ

V (σ) = + . (10.193)

2σ 2 σ2

3

The form of V (σ) is displayed in Fig. 10.8. It shows that a vacuum with a

positive cosmological constant can be obtained. Strictly speaking, the vacua

obtained in this way are only metastable. However, the lifetime could be

extremely long.

506 Flux compacti¬cations

V(σ )

σ

Fig. 10.8. Form of the potential as a function of the radial modulus after taking

anti-D3-branes into account. The ¬gure displays a minimum at which the potential

is positive leading to a de Sitter vacuum.

EXERCISES

EXERCISE 10.7

Derive the extremum of the potential in Eq. (10.191).

SOLUTION

The only solution is supersymmetric, so let us assume it from the outset.

Using

W = W0 + Aeiaρ ,

the solution for ρ = iσ in the ground state, which we denote by σ0 , is the

solution of

Dρ W = ‚ρ W + ‚ρ K W = 0 K = ’3 log [’i(ρ ’ ρ)] .

with ¯

This gives

2

aσ0 + 1 e’aσ0 ,

W0 = ’A

3

or

2

W = ’ Aaσ0 e’aσ0 .

3

So the minimum of the potential

V = eK Gρ¯Dρ W Dρ W ’ 3|W |2

ρ

¯

10.3 Moduli stabilization 507

is

a2 A2 ’2aσ0

V0 = ’ e ,

6σ0

in agreement with Eq. (10.191). 2

EXERCISE 10.8

Show that the potential Eq. (10.181) can be expressed entirely in terms of

the K¨hler-transformation invariant combination

a

K = K + log |W |2 .

SOLUTION

Using this de¬nition, Eq. (10.181) is equal to

¯ Da WD¯W

e

V = e K Gab b

’3 .

W W

However,

Ga¯ = ‚a ‚¯K = ‚a ‚¯K,

b b b

¯

and thus the inverse metric Gab only depends on K. Also,

Da W

= ‚a log W + ‚a K = ‚a K.

W

Therefore,

e ¯

V = eK Gab ‚a K‚¯K ’ 3

b

only depends on K. 2

EXERCISE 10.9

Use dimensional analysis to restore the factors of κ4 in the scalar potential.

Discuss the limit κ4 ’ 0.

SOLUTION

W has dimension three, K has dimension two and the scalar potential V has

dimension four. Therefore, restoring the powers of κ4 , Eq. (10.181) takes

the form

¯

2

κ4 V = eκ4 K κ4 Gab Da W D¯W ’ 3κ6 |W |2 ,

4 4 4

b

508 Flux compacti¬cations