String Theory and M-Theory

7.4 Bosonic construction of the heterotic string 289
is invariant under τ → τ + 1. In order to check how the partition function
behaves under the second transformation, we rewrite it in terms of a vector
N with components ni and the matrix
where
Aij = −iτGij,
Gij = e I ieI j
and e I i are the basis vectors that appear in Eq. (7.117). This gives
ΘΓ16 (τ) =
p∈Γ16
exp −πN T AN .
Applying the Poisson resummation formula yields
1
√ exp
det A
−πN T A −1 N .
p∈Γ16
Since the lattice is self-dual det G = 1. Also, replacing the matrix G by its
inverse corresponds to replacing the basis vectors by the dual basis vectors,
which span the same lattice. Therefore, the result simplifies to
−8
τ exp −πN T A −1 N
−8
= τ exp − iπ
τ p2
,
p∈Γ16
p∈Γ16
which is exactly the transformation obtained from (7.119) for τ → −1/τ. ✷
EXERCISE 7.10
Use the bosonic formulation of the heterotic string to construct the first
massive level of the E8 × E8 heterotic string.
SOLUTION
The mass formula for the heterotic string is
1
8 M 2 = NR = NL − 1 + 1
2
16
(p I ) 2 .
For the first massive level, M 2 = 8, there are three possibilities:
(i)
NR = 1, NL = 2,
I=1
16
(p I ) 2 = 0
I=1