¯

As a consequence, the Hodge and the Betti numbers are related by

k

hp,k’p .

bk = (9.275)

p=0

If ω is a (p, q)-form on a K¨hler manifold with n complex dimensions,

a

then the complex conjugate form ω is a (q, p)-form. It follows that

hp,q = hq,p . (9.276)

Similarly, if ω is a (p, q)-form, then ω is a (n ’ p, n ’ q)-form. This implies

that

hn’p,n’q = hp,q . (9.277)

One way of understanding this result is to focus on the harmonic represen-

tatives of the cohomology classes, which are both closed and co-closed. As

in the case of real manifolds, the Hodge dual of a closed form is co-closed

and vice versa, so the Hodge dual of a harmonic form is harmonic.

In terms of complex local coordinates, only the mixed components of the

Appendix: Some basic geometry and topology 451

Ricci tensor are nonvanishing for a hermitian manifold. Therefore, one can

de¬ne a (1, 1)-form, called the Ricci form, by

¯

R = iRa¯dz a § d¯b .

z (9.278)

b

For a hermitian manifold, the exterior derivative of the Ricci form is propor-

tional to the torsion. However, for a K¨hler manifold the torsion vanishes,

a

and therefore the Ricci form is closed dR = 0. It is therefore a represen-

tative of a cohomology class belonging to H 1,1 (M ). This class is called the

¬rst Chern class

1

c1 = [R]. (9.279)

2π

EXERCISES

EXERCISE A.1

Use Stokes™ theorem to verify Poincar´ duality.

e

SOLUTION

Consider a form A ∈ H p (M ). It can be expanded in a basis {w i }, so that

A = ±i wi . Consider also a form B ∈ H d’p (M ), which is expanded in a basis

{v j } as B = βj v j . Therefore,

wi § v j ≡ ±i βj mij .

A § B = ± i βj

M M

Now we de¬ne the dual basis to {v j } as {Zj }, which are (d ’ p)-cycles that

satisfy

v i = δj .

i

Zj

According to Stokes™ theorem, we can integrate B over the (d ’ p)-cycle

N = ±i mij Zj , to get

γ

βγ v γ = ±i βγ mij δj = ±i βj mij = A § B.

B=

mij Z

N ±i M

j

It follows that, for any A ∈ H p (M ), there is a corresponding N ∈ Hd’p (M ).

This implies Poincar´ duality

e

H p (M ) ≈ Hd’p (M )

452 String geometry

. 2

EXERCISE A.2

Consider the complex plane with coordinate z = x + iy and the standard

¬‚at Euclidean metric (ds2 = dx2 + dy 2 ). Compute dz and d¯.

z

SOLUTION

Because we have a Euclidean metric, it is easy to check

dy = ’dx,

dx = dy and

where we have used µxy = ’µyx = 1. Thus

dz = ’idz and d¯ = id¯.

z z

EXERCISE A.3

is a torsion-free connection, which means that “p = “p , show that

If mn nm

Eq. (9.260) is equivalent to

N p mn = J q m p

’ J qn p

’ J pq q

+ J pq q

qJ n qJ m mJ n nJ m.

SOLUTION

By de¬nition

J qm p

+ J pq q

’ J qn p

’ J pq q

qJ n nJ m qJ m mJ n

= J qm (‚q J pn + “q» p J »n ’ “qn » J p» )

+J pq (‚n J qm + “n» q J »m ’ “nm» J q» ) ’ (n ” m).

Because J qm “q» p J »n and J pq “nm» J q» are symmetric in (n, m), if the con-

nection is torsion-free, these terms cancel. To see the cancellation of

J pq “n» q J »m ’ J qm “qn » J p» ,

we only need to exchange the index » and q of the ¬rst term. So all the

a¬ne connection terms cancel out, and the expression simpli¬es to

N pmn = J qm ‚q J pn + J pq ‚n J qm ’ (n ” m),

which is what we wanted to show. 2

Homework Problems 453

HOMEWORK PROBLEMS

PROBLEM 9.1

By considering the orbifold limit in Section 9.3 explain why the 22 harmonic

two-forms of K3 consist of three self-dual forms and 19 anti-self-dual forms.

PROBLEM 9.2

Show that the Eguchi“Hanson space de¬ned by Eq. (9.24) is Ricci ¬‚at and

K¨hler and that the K¨hler form is anti-self-dual.

a a

PROBLEM 9.3