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?

  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.