# Planar spin in two-dimensional CFT

I have several questions regarding the definition of planar spin.

I was reading the big yellow book (by Di Francesco et. al.)

Section 5.1.5 looks a little mysterious. Look at 5.25, which is the two-point correlation function:

$$<\phi_1(z_1,\bar{z}_1)\phi_2(z_2,\bar{z}_2)> = \frac{C_{12}}{(z_1-z_2)^{2h}(\bar{z}_1-\bar{z}_2)^{2\bar{h}}}\tag{5.25}$$

if $$h_1=h_2= h$$ and $$\bar{h}_1=\bar{h}_2= \bar{h}$$. I have several question about this formula.

First of all, the authors mention that this formula is just the standard two-point function (Look at 4.55), just written in complex coordinates. I think they're not quite clear about the assumptions here, since by $$\Delta_1 = \Delta_2$$, we can just conclude that $$h_1+\bar{h}_1 = h_2 + \bar{h}_2$$, and the relations $$h_1=h_2= h$$ , $$\bar{h}_1=\bar{h}_2= \bar{h}$$ are not the most general ones. Can anyone help me to derive this relation and clarify the hidden assumptions?

Second, they argue that sum of spins within a correlation functions must vanish. I don't know how this condition is implied mathematically, and I think there are counterexamples! We all know that $$<\partial_z \phi(z,\bar{z}) \partial_w \phi(w,\bar{w})> = \frac{1}{(z-w)^2},$$ while $$h = 1, \bar{h} = 0$$ and $$s = h-\bar{h} = 1$$. This means that sum of spins in the correlator is 2, not zero. $$$$ is other counterexample to this rule.

Third, due to equation 5.25, two-point functions have always a chance of being complex numbers (note that $$\bar{h}$$ is not a complex number, or complex conjugate of $$h$$, look at 5.21). Isn't it a problem? or there are any additional assumptions that render it real?

Finally, in section 5.1.4, the authors introduce the idea of planar spin $$s$$. I think it's a little vague to me. Does this "spin" has anything to do with representation of Lorentz group in two dimensions $$SO(1,1)$$? I know that in two-dimensional Euclidean space, there's a rotation on the plane and in 2d Minkowskian spacetime, there's just one boost generator. I want to make sure does this "spin" has anything to do with irreps of Lorentz group?

You should have one question per post. Here is the answer:

1. $$h_1+{\bar h}_1 = h_2 + {\bar h}_2$$ is just one of the conditions. Lorentz invariance implies that the spin of the two fields must also be equal, so $$h_1 - {\bar h}_1 = h_2 - {\bar h}_2$$.

2. "Second, they argue that sum of spins within a correlation functions must vanish." where do they say this? What is the exact page number where this sentence appears?

3. Yes, correlators can be complex. If we work on the complex plane, they must be single-valued, so we must have $$h - {\bar h} \in {\mathbb Z}$$.

4. It has to do with the rotation group in 2 dimensions, $$SO(2)$$. He introduces the idea of planar spin earlier below equation (2.118).

• Thanks for the answer. 1) There's nothing such as Lorentz invariance in 2d Euclidean space. So do you mean invariance under rotation would imply that condition? 2) In page 117, below equation (5.25) they say "the sum of spins within a correlator should be zero". So how do you interprete this? 3) Seems very complicated! Why correlator could be complex? 4) Thanks for the answer. Commented Feb 10 at 12:45
• Yes, I mean rotation group of course. Sum of spins just means rotational invariance. You can work out the Ward identity for rotational invariance to see what that means. Why do you believe correlators have to be real in the first place? Commented Feb 10 at 12:49
• For the Ward identity, according to $\varepsilon_{\mu\nu} <T^{\mu\nu}X> = -i \sum_{i=1}^n s_i \delta(x-x_i) <X>$, there are two fields at $x_1,x_2$, so the RHS has two terms $(-is_1 \delta(x-x_1) -is_2 \delta(x-x_2))<\phi_1\phi_2>$. I'm confused about the interpretation. would you please do me a favour and help me out? Commented Feb 10 at 12:59