String Theory and M-Theory

196 T-duality and D-branes
Chan–Paton charges, Wilson lines and multiple branes
In the preceding construction a single Dp-brane appeared naturally after
applying T-duality to an open string with Neumann boundary conditions.
This section shows that, when several Dp-branes are present instead of a
single one, something rather interesting happens, namely nonabelian gauge
symmetries emerge in the theory.
An open string can carry additional degrees of freedom at its end points,
called Chan–Paton charges. These are degrees of freedom that were originally
introduced, when string theory was being developed as a model for
strong interactions, to describe flavor quantum numbers of quarks and antiquarks
attached to the ends of an open string. The original idea was to describe
the global SU(2) isotopic spin symmetry acting on a quark–antiquark
pair located at the ends of the string, but it was eventually realized that the
construction actually gives a gauge symmetry.
n
Fig. 6.3. Chan–Paton charges at the ends of an open string.
The Chan–Paton factors associate N degrees of freedom with each of the
end points of the string. For the case of oriented open strings, which is the
case we have discussed so far, the two ends of the string are distinguished,
and so it makes sense to associate the fundamental representation N with
the σ = 0 end and the antifundamental representation N with the σ = π
end, as indicated in Fig. 6.3. In this way one describes the gauge group
U(N).
For strings that are unoriented, such as type I superstrings, the representations
associated with the two ends have to be the same, and this forces the
symmetry group to be one with a real fundamental representation, specifically
an orthogonal or symplectic group. Each state is either symmetric
or antisymmetric under orientation reversal, an operation that interchanges
the two ends. If the massless vectors correspond to antisymmetric states,
then there are N(N − 1)/2 of them and the group is SO(N). On the other
hand, if they are symmetric, there are N(N + 1)/2 of them and the group is
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