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My question is about the Neveu-Schwarz and the Ramond sector in the free fermion CFT. The setup is as follows.

We consider two dimensional Minkowski space with a point removed $M = \mathbb{R}^{1, 1}-\{0\}$. We have $Spin(1, 1) = GL(1, \mathbb{R})\rightarrow SO(1, 1) = GL(1, \mathbb{R})_e$ and the pseudo-Riemannian manifold $M$ admits two spin structures $F_i\rightarrow SO(M)$ ($i\in \{0, 1\}$) which are the trivial and the Moebius bundle over $M$ respectively.

The even part of the Clifford algebra $Cl_{1, 1}$ is isomorphic to the diagonal matrices in $M_2(\mathbb{R})$ and we denote by $V = \mathbb{R}^2$ the tautological representation. A Majorana spinor $\Psi$ is then a section of $S_i = F_i\times_{Spin(1, 1)} V\rightarrow M$. The two cases $i = 0, 1$ are usually referred to as Neveu-Schwarz and Ramond boundary conditions respectively.

Writing $\Psi = (\psi_1, \psi_2)$, the action for the free fermion reads $$ \int dx^0dx^1 \psi_1(\partial_0-\partial_1)\psi_1 + \psi_2(\partial_0+\partial_1)\psi_2. $$

In order to define a CFT, one performs a Wick rotation (cf. Blumenhagen Plauschinn 2.9.2) $x_1\mapsto ix_1$ which has the effect $\partial_1\mapsto -i\partial_1$ and thus the Wick rotated action reads (up to a factor) $$ \int dzd\bar{z} \psi_1 \bar{\partial} \psi_1 + \psi_2\partial\psi_2 $$ with equations of motion $\bar{\partial} \psi_1 = 0 = \partial \psi_2$.

I don't know how to understand this mathematically. It seems like you have to complexify the bundle $S_i\rightarrow M$ and then view $\Psi$ as a Dirac spinor on $\mathbb{C}-0$. On the one hand, all complex vector bundles over $\mathbb{C}-0$ are trivial, so that there is no distinction between Neveu-Schwarz and Ramond boundary conditions after complexification. On the other hand, both sectors appear and are distinct in the Ising CFT. How is this possible?

Another approach would be to define the free fermion on the double cover of $\mathbb{C}-0$, which would make it possible to define NS and R boundary conditions. But then what happens if you do the same with the free boson theory?

Many thanks!

Edit: Yet another interpretation would be that the above manipulation is to be understood as formal. Mathematically one uses analytic continuation to obtain the euclidean theory. As far as I see this yields the correct correlation functions.

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    $\begingroup$ Neveu-Schwarz and Ramond boundary conditions arise when you try to put the free fermion CFT on a torus, e.g. closed strings, and I think basically a choice of spin structures for Dirac spinors. For bosons there is no need to specify any spin structures, so just periodic boundary conditions. But a similar issue arises if you consider orbifolding of boson CFT. $\endgroup$ – Meng Cheng Feb 6 '15 at 21:26
  • $\begingroup$ You said "On the one hand, all complex vector bundles over $\mathbb{C}-0$ are trivial, so...."For fermions, you need to consider spin bundle instead of complex vector bundle, which does depend on the spin structure you picked. $\endgroup$ – Yingfei Gu Dec 11 '15 at 1:53

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