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Is a CFT$_2$ always holomorphically factorizable? I had this idea because that's what we usually see is taken in string theory e.g (taking $z$ and $\bar{z}$ as independent variables). E.g. Ginsparg explains in "Applied Conformal field theory" paper page 7, that due to the emergence of Virasoro symmetry group, it is "useful" to consider them as independent, though he said that we should take $z^*=\bar{z}$ later according to our convenience. But it would be nice to understand what's going on. E.g, does holomorphic factorizability becomes an essential in some CFT$_2$'s? E.g. if we are considering some field theory in AdS$_3$ (no Chern Simons term say), should we expect the boundary CFT to be factorizable (its then not a pure gravity any more)?

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  • $\begingroup$ What is your definition of the term "holomorphically factorizable"? I think the common definition has something to do with the partition function of the theory factoring into a sum over terms that can be written as products of holomorphic and anti-holomorphic pieces. $\endgroup$ – joshphysics Nov 27 '13 at 4:25
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No, two-dimensional CFTs are not all holomorphically factorizable. They can even be fully solvable without being holomorphically factorizable. A good example is the $GL(1|1)$ WZW model. In this model, the symmetry algebra factorizes (as in all WZW models), but the space of states does not.

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