and „¦ is a representative of the unique (3, 0) cohomology class. Other

representatives di¬er by a nonzero multiplicative constant.

The existence of a holomorphic (3, 0)-form implies that the manifold has

a vanishing ¬rst Chern class. Indeed, since the holomorphic indices take

three values, „¦abc must be proportional to the Levi“Civita symbol µabc :

„¦abc = f (z)µabc , (9.78)

with f (z) a nowhere vanishing holomorphic function of z1 , z2 and z3 . This

implies that the norm of „¦, de¬ned according to

1 ¯

||„¦||2 = „¦abc „¦abc , (9.79)

3!

382 String geometry

satis¬es

|f |2

√

g= , (9.80)

||„¦||2

where g denotes the determinant of the metric. As a result, the Ricci form

is given by

√

¯ ¯

R = i‚ ‚ log g = ’i‚ ‚ log ||„¦||2 . (9.81)

Since log ||„¦||2 is a coordinate scalar, and therefore globally de¬ned, this

implies that R is exact and c1 = 0. Since the internal spaces are K¨hler

a

manifolds with a vanishing ¬rst Chern class, they are by de¬nition Calabi“

Yau manifolds.

To summarize, assuming H = 0 and a constant dilaton, the requirement

of unbroken N = 1 supersymmetry of the heterotic string compacti¬ed

to four dimensions implies that the internal manifold is K¨hler and has a

a

vanishing ¬rst Chern class. In other words, it is a Calabi“Yau three-fold.

Such a manifold admits a unique Ricci-¬‚at metric. The Ricci-¬‚at metric is

generally selected in the supergravity approximation (analyzed here), while

stringy corrections can deform it to a metric that is not Ricci-¬‚at.15 The

advantage of this formulation is that K¨hler manifolds with vanishing ¬rst

a

Chern class can be constructed using various methods (some of which are

presented in Section 9.3). However, backgrounds in which only the holonomy

is speci¬ed, which in the present case is SU (n), are extremely di¬cult to

deal with.

EXERCISES

EXERCISE 9.4

Given a theory with N = 1 global supersymmetry, show that a supersym-

metric state is a zero-energy solution to the equations of motion.

SOLUTION

A supersymmetric state |Ψ is annihilated by one or more supercharges

Q|Ψ = 0.

15 However, the known corrections to the metric can be absorbed in ¬eld rede¬nitions, so that

the metric becomes Ricci-¬‚at again.

9.4 Calabi“Yau compacti¬cations of the heterotic string 383

For an N = 1 supersymmetric theory there is no central charge, and we can

write the Hamiltonian as

H = Q† Q,

which is positive de¬nite. A supersymmetric state satis¬es

H|Ψ = 0,

and therefore it gives a zero-energy solution of the equations of motion.

The converse is not true, since there are positive-energy solutions of the

equations of motion that are not supersymmetric. In classical terms, this

result means that a ¬eld con¬guration satisfying δµ ψ = 0, for all the fermi

¬elds, as discussed in the text, is a solution of the classical ¬eld equations.2

EXERCISE 9.5

Prove that ·± in Eqs (9.59) are Weyl spinors of opposite chirality, that is,

γ7 has eigenvalues ±1.

SOLUTION

2

Using γ7 ≡ iγ1 γ2 γ3 γ4 γ5 γ6 , one ¬nds γ7 = 1. This is manifest for the repre-

1

sentation presented in Eq. (9.63). We can then de¬ne P± ≡ 2 (1 ± γ7 ), and

·± ≡ P± ·. Therefore,

γ7 ·’ = ’·’ .

γ7 ·+ = ·+ and

In terms of holomorphic and antiholomorphic indices

γ7 = (1 ’ γ¯ γ1 )(1 ’ γ¯ γ2 )(1 ’ γ¯ γ3 ) = ’(1 ’ γ1 γ¯ )(1 ’ γ2 γ¯ )(1 ’ γ3 γ¯ ),

1 2 3 1 2 3

so the conditions γa ·+ = 0 and γa ·’ = 0 also give the same results. 2

¯

EXERCISE 9.6

1

= 4 Rmnpq “pq · used in Eq. (9.53).

Derive the identity [ m, n ]·

SOLUTION

Using Eq. (9.60) and the de¬nition of the covariant derivative in the ap-

pendix,

1 pq

n · = ‚n · + ωnpq γ ·,

4

where ωnpq are the components of the spin connection. Thus

1 1

= [‚m + ωmpq γ pq , ‚n + ωnrs γ rs ]·.

[ m, n ]·

4 4

384 String geometry

In writing this one has used the fact that Christo¬el-connection terms of the

form (“p ’ “p )‚p · cancel by symmetry. The commutator above gives

mn nm

1 1

(‚m ωnrs ’ ‚n ωmrs )γ rs · + ωmpq ωnrs [γ pq , γ rs ]·,

4 16

which simpli¬es to

1 1

(‚m ωnrs ’ ‚n ωmrs + ωmrp ωn p s ’ ωnrp ωm p s )γ rs · = Rmnrs γ rs ·,

4 4

where we have used

[p q]

[γrs , γ pq ] = ’8δ[r γs] .

2

EXERCISE 9.7

Prove that Rmnpq γ pq · = 0 implies that Rmn = 0.

SOLUTION

Multiplying the above equation with γ n gives

γ n γ pq Rmnpq · = 0.

Using the gamma matrix identity

γ n γ pq = γ npq + g np γ q ’ g nq γ p

and the equation

γ npq Rmnpq = γ npq Rm[npq] = 0,

which is the Bianchi identity, we get the expression

2g nq γ p Rmnpq · = 0.

This implies γ p Rmp · = 0. If · = ·+ is positive chirality, for example, this

gives

i·’ γq γ p ·+ Rmp = Jq p Rmp = 0.