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. $(5)$?
Thanks in advance!
Edit:
Here is a picture of the script of my professor, maybe his intentions are clearer then?

 A: 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.
