Let's consider string theory (open or closed) on $AdS_3\times S^3\times T^4$ and let's also define a chiral primary vertex operator in the NS sector in the -1 picture and let's call it $\mathcal{W}_{(-1)}$. We can also construct spacetime supercharges, $Q^i_p$ with $i$ a generic index that labels the total number of supercharges and $p$ labels the picture. So we should have, for half of the supercharges $$ [Q^i_p,\mathcal{W}_{(-1)}]=0 $$

Now, the questions are

1) Is the above formula valid for each picture $p$ of the supercharges?

2) In the bosonized formulation we have $Q=\oint S(z)e^{-\phi/2}$ with $S(z)=\prod_{i=1}^5e^{\frac{i}{2}\epsilon_iH_i}$. GSO projection turns out to be $\prod_i^5\epsilon_i=+1$ (that kills half of supercharges), but I cannot see the connection with the standard projection using gamma matrices. How does it work ?


The space-time supersymmetry on left-handed modes is generated by

$$ Q_{\alpha}^{q}=\oint\frac{dz}{2\pi i} \mathcal{V}_{\,\alpha}^{\,q}(z) $$

where the $\mathcal{V}_{\,\alpha}^{\,q}(z)$ is the $q$-picture vertex operator of the Ramond ground state. This means that this charge will generate a symmetry on the states of the theory. The problem here is that there are an infinite tower of redundancies, namely the different pictures that represent the same state.

A given property of a state, e.g. the fact that is annihilated by half the super-symmetries, does not depend on the picture you are using to represent it. You can see it as some sort of a residual gauge symmetry that comes from the gauge fixation of the superconformal world-sheet symmetry. The only thing that is fixed by the theory is the total picture of a given Riemann surface.

This is analogous to the case in bosonic string theory. The gauge fixing of the conformal world-sheet symmetry gives a residual redundancy: what operator should be integrated and what should be fixed with $c\tilde{c}$ insertion. Since the $c$ ghost is nilpotent, we have a more restricted redundancy than the bosonic ghosts on superconformal gauge.

The answer for your first question is yes. The only novelty of this "current algebra" is that the picture number is not conserved by the algebra, so you should work with all pictures. At the end you can always doing a picture change back to what your operator.

Related to your second question the GSO projection $(-1)^{F}$ is at the center-of-mass the $\Gamma$ matrix. Acting in the Ramond Ground State $(-1)^{F}$ will be the same as $\Gamma$.


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