String Theory and M-Theory

Homework Problems 291
To derive this formula it is convenient to add dependence on m variables
xi and define
f(A, x) =
exp −π(M + x) T A(M + 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
f(A, x) =
CN (A) exp(2πiN T x). (7.125)
{N}
The next step is to evaluate the Fourier coefficients:
CN (A) =
1
0
f(A, x)e −2πiN T x d m x. (7.126)
Inserting the series expansion of f(A, x) in Eq. (7.124) gives
CN(A) =
∞
−∞
exp(−πx T Ax − 2πiN T x)d m x = exp(−πN T A−1N) √ . (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
as desired.
f(A) =
CN(A) =
{N}
HOMEWORK PROBLEMS
1
√ det A f(A −1 ) (7.128)
PROBLEM 7.1
Section 7.1 discussed several possibilities for generating nonabelian gauge
symmetries in string theory. Show that in the context of toroidally compactified
type II superstring theories, the only massless gauge fields are
abelian.
PROBLEM 7.2
It is possible to compactify the 26-dimensional bosonic string to ten dimensions
by replacing 16 dimensions with 32 Majorana fermions. The 32
left-moving fermions and the 32 right-moving fermions each give a level-one