# Implications of null-fields for operator algebra in CFT

In the context of Conformal Field Theory (CFT), I have a primary field $$\phi_{(r,s)}$$ with a level 1 null-descendant, i.e. $$(r,s)=(1,1)$$ and $$h_{(1,1)}=0$$. My goal is to understand how this condition constraints the operator algebra, especially regarding the OPE (Operator Product Expansion).

Looking at the 3-point function, I see the following:

$$\langle L_{-1} \phi_{(1,1)} \phi_1 \phi_2 \rangle = \partial_z \langle \phi_{(1,1)} \phi_1 \phi_2 \rangle, \tag{1}$$

but we know that:

$$L_{-1} \phi_{(1,1)} = 0, \tag{2}$$

which in turn means

$$\langle L_{-1} \phi_{(1,1)} \phi_1 \phi_2 \rangle = 0. \tag{3}$$

We also know that in CFT, the general form of a 3-pt function is (with $$h_{(1,1)}=0$$):

$$\langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = c_{12}^h (z-z_1)^{-h_1+h_2} (z_1-z_2)^{-h_1-h_2} (z_2-z)^{-h_2+h_1}. \tag{4}$$

Thus equating $$(1)$$ with $$(3)$$, and using $$(4)$$, we find:

$$\left( \frac{-h_1+h_2}{z-z_1} - \frac{-h_2+h_1}{z_2-z} \right) \langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = 0, \tag{5}$$

and therefore that $$h_1 = h_2$$ or $$\langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = 0$$.

So far so good. Now the claim is that, as a consequence, we can write the following OPE:

$$\phi_{(1,1)} \phi_1 = \sum c_{12}^h \phi_2 + \text{descendants} \tag{6}$$

when the condition $$h_1 = h_2$$ is satisfied. I don't see that. Can someone explain how to obtain eq. $$(6)$$ from eq. $$(5)$$?

Edit:

Here is a picture of the script of my professor, maybe his intentions are clearer then? • What are you summing over in eq $(6)$? – MannyC Jul 8 '19 at 21:46
• @MannyC I guess that I understand my stupid confusion now thanks to your comment. Actually, the notation of the script of my professor is $\phi_{(1,1)} \phi_i = \sum c_{ij}^h \phi_j \text{descendants}$, but because he used a $j$ instead of a $2$ all along during the derivation, and did not write over what he was summing over, I assumed that this field $\phi_j$ was the same as the field that I called $\phi_2$ in the OP, which is obviously wrong. So I guess that the condition below eq. ($6$) would also change to $h_1 = h_j$ for all $j$? – Jxx Jul 8 '19 at 22:52
• @MannyC Well, he wrote explicitly $h_i = h_j$ next to the equation (although he did not write "for all $j$"). I've added a screenshot of the script in the OP. Eq. (223) is the one that I don't understand. – Jxx Jul 9 '19 at 8:39

The note is simply saying this: the three-point function $$\langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = c^{h = 0}_{12}\, (z-z_1)^{-h_1+h_2} (z_1-z_2)^{-h_1-h_2} (z_2-z)^{-h_2+h_1} \,,$$ satisfies a differential equation that implies $$c_{12}^0 \,(h_1 - h_2) = 0\,.$$ So this means either $$c_{12}^h = 0$$ or $$h_1 = h_2$$. The first trivial consequence is that in the OPE $$\phi_1 \times \phi_2$$ $$\phi_1 \times \phi_2 = \sum_j c^{h_j}_{12} \,\phi_j$$ you can find the operator $$\phi_{(1,1)}$$ (i.e. $$c^{0}_{12} \neq 0$$) only if $$h_1 = h_2$$.
But you can say other things: using the associativity of the OPE you can flip this around and look at the OPE $$\phi_{(1,1)} \times \phi_1$$ $$\phi_{(1,1)} \times \phi_1 = \sum_j c_{0 1}^j \phi_j \underset{\mathrm{Associativity}}{=}\sum_j c_{1j}^0 \,\phi_j \,.$$ If $$h_j = h_1$$ (for example if $$\phi_j$$ is $$\phi_2$$ or $$\phi_1$$ itself), then the operator appears in the OPE, otherwise it does not.
Indeed $$\phi_{(1,1)}$$ can be thought of as an "identity element" in the OPE algebra: $$\phi_j \times \phi_{(1,1)} \sim \phi_j$$, which makes this thing very intuitive.