Polchinski equation 11.2.7 In Polchinski's string theory volume 2, when discussing the GSO projection for the heterotic string he says:

In the IIA and IIB superstrings the GSO projection acted separately on the left- and right-moving sides. This will be also true in any supersymmetric heterotic theory. The world-sheet current associated with spacetime symmetry is $\mathcal{V}_\mathbf{s}$ as in eq. (10.4.25), with $\textbf{s}$ in the $\textbf{16}$. In order for the corresponding charge to be well deﬁned, the OPE of this current with any vertex operator must be single-valued. For the right-moving spinor part of the vertex operator, the spin eigenvalue $\mathbf{s'}$ must then satisfy
$$ \mathbf{s} \cdot \mathbf{s'} + \frac{l}{2} \in \mathbb{Z} \tag{11.2.7}$$
for all $\mathbf{s} \in \mathbf{16}$, where $l$ is $-1$ in the NS sector and $-\frac{1}{2}$ in the R sector.

The equation he refers to (10.4.25) is
$$ \mathcal{V}_\mathbf{s} = e^{-\phi/2}\Theta_\mathbf{s}$$
$$ \Theta_\mathbf{s} \cong \exp \left[i\sum_a s_aH^a\right]$$
I do not understand the origin of eq. (11.2.7). I gather that if it takes on a non-integer value a branch cut appears and the OPE is no longer single valued. But how does that particular combination appear?
 A: The world-sheet current associated to space-time supersymmetry is $\mathcal{V}_s$ and all the different pictures ($\phi$-charge) representations of him. The supersymmetry "current algebral" does not closes with fixed picture, it is required to work all the pictures at the same time. The supersymmetry generator in this picture $-1/2$ is:
$$
Q_{s}=\oint_C\frac{dz}{2\pi i}\mathcal{V}_s
$$
Now, if you want to have states in your theory $|\psi\rangle$ that transform under supersymmetry you need to do a projection. The projection should make the OPE of the vertex operators $\mathcal{V}_{\psi}$ associated with the state $|\psi\rangle$ and $\mathcal{V}_s$ single-valued in order to make $Q_s|\psi\rangle$ well defined, i.e. to perform the contour integral. 
$$
Q_{s}|\psi\rangle\cong \oint_C\frac{dz}{2\pi i}:\mathcal{V}_s(z)::\mathcal{V}_{\psi}(0):
$$
if the OPE is not single-valued this countour will not close, and we will not pick the single pole of the OPE.
For NS vertex operators $\mathcal{V}_{NS,\,s'}=e^{-\phi}e^{is'_aH^a}$ with $s'$ an integer, the single-valuedness requires that $s.s'-1/2$ be a integer since the OPE will be of the form
$$
:e^{-\phi/2}e^{is_aH^a}(z)::e^{-\phi}e^{is'_aH^a}:\,=z^{-1/2+s.s'}:e^{-\phi/2}e^{is_aH^a}(z)e^{-\phi}\mathcal{V}_{s'}(0):
$$
For R vertex operators $\mathcal{V}_{R,\,s'}=e^{-\phi/2}e^{is'_aH^a}$ with $s'$ a half integer, the single-valuedness requires that $s.s'-1/4$ be a integer since the OPE will be of the form
$$
:e^{-\phi/2}e^{is_aH^a}(z)::e^{-\phi/2}e^{is'_aH^a}:\,=z^{-1/4+s.s'}:e^{-\phi/2}e^{is_aH^a}(z)e^{-\phi}\mathcal{V}_{s'}(0):
$$
pay attention to the ghost sector here. What Polchinski is doing here is projecting the theory in order to have space-time supersymmetry ant noting that this is precisely the GSO projection on the left-movers.
