String Theory and M-Theory

7.2 Fermionic construction of the heterotic string 261
to a spectrum that is inconsistent due to gauge anomalies. Therefore, only
the n = 16 case remains to be considered.
The n = 16 case
This is the case of most interest. It would naively appear to have an
SO(16) × SO(16) gauge symmetry, but it turns out that each SO(16) factor
is enhanced to an E8. The AP and PA sectors have ã = 0 for n = 16.
This value makes it possible to contribute states to the massless spectrum,
a fact that proves to be very important in understanding the symmetry
enhancement.
Let us now examine the massless spectrum in the n = 16 case. The rightmovers
have NR = 0 (in the light-cone GS description) and contribute a
vector supermultiplet, which should be tensored with the massless states of
the left-moving sectors. The left-movers can have the boundary conditions:
• The PP sector does not contribute to the massless spectrum, as before.
• The AA sector, on the other hand, does contribute states with NL = 1.
These include states of the form
and
˜α i −1|0〉L
(7.41)
λ A −1/2 λB −1/2 |0〉L. (7.42)
The eight states (7.41), when tensored with the right-moving vector multiplet
give the N = 1 gravity supermultiplet, just as in the case of the
SO(32) theory. The 496 states in (7.42) are exactly those that gave the
SO(32) gauge supermultiplets previously. They will do so again unless
some of them are projected out. To see what is required, let us examine
how they transform under SO(16) × SO(16):
(120, 1) if A, B = 1, . . . , 16,
(1, 120) if A, B = 17, . . . , 32,
(16, 16) if A = 1, . . . , 16, B = 17, . . . , 32.
(7.43)
Here 120 = 16 × 15/2 denotes the antisymmetric rank-two tensor in the
adjoint representation of SO(16) and 16 denotes the vector representation.
Clearly, if we want to keep only the SO(16) × SO(16) gauge fields,
then we need a rule that says that the (120,1) and the (1,120) multiplets
are physical, while the (16, 16) multiplet is unphysical. The way to do
this is to require that the number of λ A excitations involving the first
set of 16 components and the second set of 16 components should each
be even. This is more restrictive than just requiring that their sum is