¯

‚„¦ = 0. (10.223)

Note that the norm of „¦, de¬ned by

¯ ¯ ¯

¯

||„¦||2 = „¦a1 a2 a3 „¦¯1¯2¯3 g a1 b1 g a2 b2 g a3 b3 , (10.224)

bbb

is related to the dilaton by

||„¦||2 = e’4(¦+¦0 ) , (10.225)

where ¦0 is an arbitrary constant.

The existence of the holomorphic (3, 0)-form implies the vanishing of the

¬rst Chern class, that is, c1 = 0. Together with Eq. (10.211) this implies that

the background has SU (3) holonomy. However, since the internal manifolds

are not K¨hler they cannot be Calabi“Yau. Note that even though the

a

background is not K¨hler, it still satis¬es the weaker condition

a

d e’2¦ J § J = 0, (10.226)

which means that the background is conformally balanced.

The vanishing of the supersymmetry variation of the gluino, Fµ = 0,

implies that

¯ ¯

(Fab γ ab + Fa¯γ ab + 2Fa¯γ ab )· = 0

¯

(10.227)

¯b b

and hence that the gauge ¬eld satis¬es

¯

g ab Fa¯ = Fab = Fa¯ = 0, (10.228)

b ¯b

which is called the hermitian Yang“Mills equation.

10.4 Fluxes, torsion and heterotic strings 515

Once a solution for the hermitian Yang“Mills ¬eld has been found, the

fundamental form is constrained to satisfy the di¬erential equation

±

¯ [tr(R § R) ’ tr(F § F )] ,

i‚ ‚J = (10.229)

8

which is a consequence of the anomaly cancellation condition.

To summarize, supersymmetry is unbroken if the external space-time is

Minkowski and the internal space satis¬es the following conditions:

• It is complex and hermitian.

• There exists a holomorphic (3, 0)-form „¦ that is related to the fundamental

form by the condition that the background is conformally balanced, that

is,

d(||„¦||J § J) = 0. (10.230)

• The gauge ¬eld satis¬es the hermitian Yang“Mills condition.

• The fundamental form satis¬es the di¬erential equation in Eq. (10.229).

These are the only conditions that have to be imposed. Once a solution

of the above constraints has been found, H and ¦ are determined by the

data of the geometry according to

1

¯

H = i(‚ ’ ‚)J ¦ = ¦0 ’ log ||„¦||.

and (10.231)

2

There exist six-dimensional compact internal spaces that solve the above

constraints and lead to interesting phenomenological models in four dimen-

sions. However, they lie beyond the scope of this book. In the following we

describe a simpler example in which the internal space is four-dimensional.

Conformal K3

Four-dimensional internal spaces for heterotic-string backgrounds with tor-

sion can be constructed by considering an ansatz of the form of a direct

product in the string-frame, as before, with

gmn (y) = e2D(y) gmn (y),

K3

(10.232)

K3

where gmn (y) represents the (unknown) metric of K3, and gmn (y) is the

internal part of the string-frame metric. In this four-dimensional example,

the internal manifold is given by a conformal factor times a Calabi“Yau

manifold.

516 Flux compacti¬cations

In this background the dilatino and gravitino supersymmetry conditions

can be written in the form

1

(‚m ¦ + ‚m h)γ m · = 0 (10.233)

2

and

1 n

m · + ‚n h γm · = 0. (10.234)

4

Here dh = H is the one-form dual to H in four dimensions and the Hodge-

star operator is de¬ned with respect to the metric gmn . The ¬rst equation

implies that

1

¦(y) = ’ h(y) + const. (10.235)

2

In other words, the ¬‚ux is given in terms of the dilaton by H = ’2 d¦. In

terms of the metric rescaled by the factor e2D , Eq. (10.234) takes the form

1 1

+ ‚n D γm n · + ‚n h γm n · = 0.

m· (10.236)

2 4

Therefore, for the choice

D(y) = ¦(y) (10.237)

one ¬nds m · = 0. This is just the Killing spinor equation required to

de¬ne a Calabi“Yau manifold. Since K3 is the only Calabi“Yau manifold in

four dimensions, one is justi¬ed in identifying the rescaled metric with the

K3 metric.

The conformal factor and the dilaton are constrained by the Bianchi iden-

tity for the H ¬‚ux

±

d d¦ = ’ [tr(R § R) ’ tr(F § F )] . (10.238)

8

Solutions can be found if the right-hand side is exact. The conditions

¯

Fa¯ = Fab = g ab Fa¯ = 0 (10.239)

¯b b

are conformally invariant. Therefore, they only need to be solved for K3.

EXERCISES

EXERCISE 10.10

Show that the backgrounds described in Section 10.4 are complex.

10.4 Fluxes, torsion and heterotic strings 517

SOLUTION

In order to prove that the manifold is complex one computes the Nijenhuis

tensor, which was de¬ned in the appendix of Chapter 9 to be

Nmn p = Jm q J[n p ,q] ’ Jn q J[m p ,q] .

Eq. (10.217) implies that the Nijenhuis tensor takes the form

1

Hmnp ’ 3J[m q Jn s Hp]qs

Nmnp =

2

Identities for Dirac matrices, which are listed in the appendix of this chapter,

imply

J[m p Jn] q = 1 g pr g qs (J § J)mrns + 1 Jmn J pq

4 2