I am wondering whether correlation functions e.g. in 2D CFTs such as

$$\langle \phi\phi\phi\rangle=C\frac{1}{z_{12}^{h_1+h_2-h_3}z_{23}^{h_2+h_3-h_1}z_{31}^{h_3+h_1-h_2}}\frac{1}{\bar z_{12}^{\bar h_1+\bar h_2-\bar h_3}\bar z_{23}^{\bar h_2+\bar h_3-\bar h_1}\bar z_{31}^{\bar h_3+\bar h_1-\bar h_2}}$$

where the weights $h,\bar h$ are conformal dimensions $\pm$ spins of the fields

$$h=\frac{\Delta+s}{2}~~~,~~~\bar h=\frac{\Delta-s}{2}\,,$$

are always supposed to be single valued functions?

I am asking because of the following consideration. Writing e.g.

$$z_{12}=r e^{i\theta}~~~,~~~\bar z_{12}=r e^{-i\theta}$$

we have

$$z_{12}^{h_1+h_2-h_3}\bar z_{12}^{\bar h_1+\bar h_2-\bar h_3}=r^{\Delta_1+\Delta_2-\Delta_3}e^{i\theta(s_1+s_2-s_3)}$$

As long as $s_1+s_2-s_3$ is an integer the phase factor stays single valued under a full rotation


However, the above assumes that all conformal dimensions $\Delta_i$ are real! In case if they are complex numbers (like in non-unitary CFTs) the part $r^{\Delta_1+\Delta_2-\Delta_3}$ could also become non-single valued! Is that OK, and the correlation function is just not single valued in such a case? Or perhaps we have to generalize the weights as

$$h=\frac{\Delta+s}{2}~~~,~~~\bar h=\frac{\Delta^*-s}{2}\,,$$

in such a case, where $\Delta^*$ is the complex conjugate of $\Delta$, to preserve single-valuedness? How to decide which one it is? Thanks for any suggestion!


1 Answer 1


$r^\Delta$ is single-valued even if $\Delta$ is complex, so long $r$ is real positive. Moreover, there are plenty of non-unitary CFTs where all conformal dimensions are real, starting with minimal models.

Some CFTs have multivalued correlation functions, but this is typically because spins are fractional, as in parafermions.

  • $\begingroup$ Now that you mention it, I can't think of a way to make $r^\Delta$ wander over its branch cut either. Thanks for pointing it out! $\endgroup$
    – Kagaratsch
    Commented Jun 21, 2018 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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