# Is the spin of a primary field in 2D necessarily integer or half-integer?

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?

• For the spin of excitations in low-dimensional systems, a search term is “anyon.”
– rob
Commented Dec 13, 2022 at 14:08
• Abstract page link to BPZ paper? Which page? Commented Dec 13, 2022 at 14:49
• The footnote under the equation (1.18) "The spin $s_n$ of a local field can take an integer or half-integer value only" Commented Dec 13, 2022 at 14:52
• Commented Dec 13, 2022 at 15:41
• 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. Commented Dec 13, 2022 at 16:24

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 $$h - {\bar h} \in \frac{1}{2} {\mathbb Z}$$.
• 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. Commented Dec 15, 2022 at 7:57