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In 2d CFT, we usually say that the contribution of $z$ and $\bar{z}$ factorize, and then we just consider the $z$ part. In what sense is this true? For example, can the 4-point function be factorized into the form $f(z)g(\bar{z})$? What about the global conformal blocks? The conformal blocks is given in this paper, https://arxiv.org/pdf/0807.0004v2.pdf, equation 3.15, $$ g_{O}\left(u,v\right)\equiv g_{\Delta,l}\left(u,v\right)=\frac{\left(-1\right)^{l}}{2^{l}}\frac{z\overline{z}}{z-\overline{z}}\left[k_{\Delta+l}\left(z\right)k_{\Delta-l-2}\left(\overline{z}\right)-\left(z\leftrightarrow\overline{z}\right)\right] $$ $$ k_{\beta}\left(x\right)\equiv x^{\frac{\beta}{2}}\ _{2}F_{1}\left(\frac{\beta}{2},\frac{\beta}{2},\beta,x\right) $$

$$ u=z\overline{z},v=\left(1-z\right)\left(1-\overline{z}\right) $$ It seems to me that $g_{\Delta,l}(u,v)$ is not factorized into the form $f(z)g(\bar{z})$. I think I have some serious misunderstanding of these stuff, can anybody point out my misunderstanding?

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  • $\begingroup$ What context do you have in mind whit factorisation? It is not true in general that everything always factorises in products of $z,z^*$ $\endgroup$
    – gented
    Aug 26 '16 at 7:54
  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0807.0004 $\endgroup$
    – Qmechanic
    Dec 7 '16 at 15:00
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The conformal blocks you cited correspond to the 4d blocks. In that case the contributions of $z$ and $\bar{z}$ do not factorize. For the 2d conformal blocks see 2.9 of https://arxiv.org/abs/hep-th/0309180. They are given by, in the notation you've used:

$\big(\frac{-1}{2}\big)^{\ell}[k_{\Delta+\ell}(z)k_{\Delta-\ell}(\bar{z})+(z\leftrightarrow \bar{z})]$.

These are the conformal blocks for four point functions of identical external scalars, for the general 2d case see http://arxiv.org/abs/1205.1941. The conformal blocks for external scalars have a similar form in any even dimension (see also http://arxiv.org/abs/1108.6194).

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