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When putting fermions on the string, we have to choose boundary conditions for our spinor fields - Ramond or Neveu-Schwarz. NS conditions on the closed string have antiperiodic conditions such as $\psi_L(\sigma) = -\psi_L(\sigma+\pi)$ (where $\pi$ is the periodicity of the string). I'm confused by this - shouldn't $\psi$ be single-valued?

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As a follow-up, I wonder why we can't impose (anti-)periodicity as the condition on the open string - instead, we insist that the variation of the fermionic field vanishes at both endpoints. Why is that? – user3888 Jun 5 '11 at 18:50
Dear @James, if fields are periodic, $X_{or \,\,\psi}(\sigma+\pi)=X_{or\,\,\psi}(\sigma)$, then the topology is a circle because $\sigma$ is effectively a periodic variable, and it's called the closed string. For both open strings and closed strings, the condition constraining boundary condition is the vanishing of the boundary terms in $\delta S$. – Luboš Motl Jun 5 '11 at 19:50
The boundary terms in $\delta S$, coming from integration by parts, are $\psi_\mu \delta \psi^\mu |^{\pi}_0-\tilde\psi_\mu \delta \tilde\psi^\mu |^{\pi}_0$ . This can vanish either by making $\psi$ and $\tilde\psi$ periodic or by setting $\psi=\pm\tilde\psi$ at both boundary points $\sigma=0,\pi$. The relative sign in these $\pm$ among both endpoints indeed gives an open string version of (anti)periodicity that leads to integral or half-integral modes. But it is (anti)periodicity over a doubled range only. – Luboš Motl Jun 5 '11 at 19:53

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Dear James, there is no reason why $\psi$ should be periodic. First, if you have a problem, imagine that $\psi$ are just auxiliary variables but the true ones are the bilinears $\psi_i \psi_j$ which are still periodic. Only things such as the world sheet stress-energy tensor $T_{++}$ and $T_{--}$ have to be periodic and they are because they're bilinear in fermions and their derivatives.

Take an analogy with the winding strings. On those strings, $X$ at $\sigma+\pi$ is $X$ at $\sigma$ shifted by the winding number times the circumference. It is not single-valued, either.

In a similar way, the fields on the string may be twisted by the action of an orbifold. The only condition is that the operation equivalent to the monodromy around the closed string is a symmetry of the theory. It means that it commutes with the Hamiltonian - so that $T_{ab}$ constructed out of the fields remains periodic.

In fact, the winding numbers as well as general periodic/antiperiodic conditions for fermions may be viewed as special cases of an orbifold. For each possible group action allowed when you go around the closed string, there is a corresponding GSO-like projection on the string.

In fact, the antiperiodic NS boundary conditions on the string are the "primary ones", the more natural ones than the periodic R boundary conditions. If you map the closed string or a cylinder to a complex plane with an operator at the center by conformal transformations, the NS sector gets mapped to normal operators such as identities that make the fermions single-valued around the origin, while the R boundary conditions periodic on the cylinder become antiperiodic on the plane, and are associated with spin fields that create branch cuts in the plane.

So the NS sector is the normal one while the R sector is the twisted one. But the R sector is allowed, in fact, it is required for modular invariance. In type 0, only NS NS and R R sectors exist. In type II theories, there is a separate orbifold and separate GSO projection on left-moving and right-moving side.

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Thanks - the analogy to string winding really helped. – James Jun 3 '11 at 14:26

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