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The primary field in a 2D CFT is defined by the transformation property \begin{align} \phi^{'}(w) = \left(\frac{dw}{dz}\right)^{-h} \left(\frac{d\bar{w}}{d\bar{z}}\right)^{-\bar{h}} \phi(z) \end{align} and the conformal dimension of the field is $\Delta=h+\bar{h}$ and the spin is $s=h-\bar{h}$. In the BPZ paper it's claimed the spin can only take the value of an integer or half-integer, I cannot find this statement in some other books/reviews on CFT.

My question is as follows. We know in 4D the spin of a particle can only take the value of an integer or half-integer because rotation by $4\pi$ is null-homotopic, that is, it can be continuously deformed to the identity transformation (not rotating at all), while rotation by $2\pi$ may not be null-homotopic. In 2D, the rotation group is $U(1)$ with universal cover being the real line $\mathbb{R}$, wouldn't that imply the spin can take an arbitrary value?

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    $\begingroup$ For the spin of excitations in low-dimensional systems, a search term is “anyon.” $\endgroup$
    – rob
    Dec 13, 2022 at 14:08
  • $\begingroup$ Abstract page link to BPZ paper? Which page? $\endgroup$
    – Qmechanic
    Dec 13, 2022 at 14:49
  • $\begingroup$ The footnote under the equation (1.18) "The spin $s_n$ of a local field can take an integer or half-integer value only" $\endgroup$ Dec 13, 2022 at 14:52
  • $\begingroup$ Related: physics.stackexchange.com/q/170627/2451 $\endgroup$
    – Qmechanic
    Dec 13, 2022 at 15:41
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    $\begingroup$ BPZ is referring to the fact that the 3pt function of primaries is fixed by global conformal invariance up to a constant. It's a good exercise to show that integer or half integer spins are required to make this function single valued. $\endgroup$ Dec 13, 2022 at 16:24

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The two-point function of a conformal primary operator is $$ \langle O(z,{\bar z}) O ( 0 ,0 ) \rangle = \frac{1}{z^{2h} {\bar z}^{2{\bar h} } } $$ We find that $$ \langle O( e^{2\pi i} z,e^{-2\pi i} {\bar z}) O ( 0 ,0 ) \rangle = e^{- 4\pi i (h - {\bar h } ) } \frac{1}{z^{2h} {\bar z}^{2{\bar h} } } $$ If the CFT is defined on the plane, then $e^{2\pi i} z$ and $z$ represent the same point so in this case, we must have $ h - {\bar h} \in \frac{1}{2} {\mathbb Z}$.

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As demonstrated in the answer by Prahar, the spin must be half-integer for correlation functions to be single-valued. If however we relax single-valuedness, then the spin needs not be half-integer. There exist parafermionic CFTs that include fields with fractional spins. Unfortunately, good references on parafermions are scarce, see https://ncatlab.org/nlab/show/parafermion .

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  • $\begingroup$ Thank you very much, Professor! Since the form of correlation functions is obtained from symmetry transformations including rotation, the single-valuedness is a tautology of the restriction of spin to half integers. A priori, there is no physical principle or mathematical consistency condition that rules out multi-valuedness of correlation functions, for example, the parafermion you mentioned. Am I correct? $\endgroup$ Dec 15, 2022 at 4:22
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    $\begingroup$ A multivalued function on a space $X$ is a function that lives on some covering of $X$, nothing inconsistent here. But if your physical system lives on $X$, you need single-valuedness. $\endgroup$ Dec 15, 2022 at 7:57

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