massive level of the E8 — E8 heterotic string.

SOLUTION

The mass formula for the heterotic string is

16

12 1

(pI )2 .

M = NR = NL ’ 1 +

8 2

I=1

For the ¬rst massive level, M 2 = 8, there are three possibilities:

(i)

16

(pI )2 = 0

NR = 1, NL = 2,

I=1

290 The heterotic string

There are 324 possible left-moving states:

˜i ˜j

˜I ˜J ˜I ˜i ˜i ˜I

±’1 ±’1 |0 L , ±’2 |0 L , ±’1 ±1 |0 L , ±’2 |0 L , ±’1 ±’1 |0 L

(ii)

16

(pI )2 = 2

NR = 1, NL = 1,

I=1

In this case there are 24 — 480 possible left-moving states:

16 16

±’1 |pJ ,

˜I J2

±’1 |pI ,

˜i (pJ )2 = 2 L .

(p ) = 2 L ,

J=1 I=1

(iii)

16

(pI )2 = 4

NR = 1, NL = 0,

I=1

In this case there are 129 — 480 possible left-moving states:

16

I

(pI )2 = 4 L .

|p ,

I=1

The total number of left-moving states is 73 764. In each case the right-

movers have NR = 1, so these are the 256 states

i i a a

±’1 |j ±’1 |a ; S’1 |i S’1 |b

R, R, R, R.

The spectrum of the heterotic string at this mass level is given by the tensor

product of the left-movers and the right-movers, a total of almost 20 000 000

states. 2

Appendix: The Poisson resummation formula

Let A be a positive de¬nite m — m symmetric matrix and de¬ne

exp ’πM T AM .

f (A) = (7.122)

{M }

Here M represents a vector made of m integers M1 , M2 , . . . , Mm each of

which is summed from ’∞ to +∞. The Poisson resummation formula is

1

f (A’1 ).

f (A) = √ (7.123)

det A

Homework Problems 291

To derive this formula it is convenient to add dependence on m variables

xi and de¬ne

exp ’π(M + x)T A(M + x) .

f (A, x) = (7.124)

{M }

This function is periodic, with period 1, in each of the xi . Therefore, it must

have a Fourier series expansion of the form

CN (A) exp(2πiN T x).

f (A, x) = (7.125)

{N }

The next step is to evaluate the Fourier coe¬cients:

1

Tx

f (A, x)e’2πiN dm x.

CN (A) = (7.126)

0

Inserting the series expansion of f (A, x) in Eq. (7.124) gives

∞

exp(’πN T A’1 N )

T T m

√

exp(’πx Ax ’ 2πiN x)d x =

CN (A) = . (7.127)

det A

’∞

Note that the summations in Eq. (7.125) have been taken into account by

extending the range of the integrations. It therefore follows that

1

f (A’1 )

CN (A) = √

f (A) = (7.128)

det A

{N }

as desired.

HOMEWORK PROBLEMS

PROBLEM 7.1

Section 7.1 discussed several possibilities for generating nonabelian gauge

symmetries in string theory. Show that in the context of toroidally com-

pacti¬ed type II superstring theories, the only massless gauge ¬elds are

abelian.

PROBLEM 7.2

It is possible to compactify the 26-dimensional bosonic string to ten di-

mensions by replacing 16 dimensions with 32 Majorana fermions. The 32

left-moving fermions and the 32 right-moving fermions each give a level-one

292 The heterotic string

SO(32) current algebra. Making the same GSO projection as in the left-

moving sector of the heterotic string, ¬nd the ground state and the massless

states of this theory.

PROBLEM 7.3

Exercise 7.1 introduced free-fermion representations of current algebras and

showed that fermions in the fundamental representation of SO(n) give a

level-one current algebra.

(i) Find the level of the current algebra for fermions in the adjoint rep-

resentation of SO(n).

(ii) Find the level of the current algebra for fermions in a spinor repre-