Lightcone singularity of a 3 point function in CFT I had a quick question regarding the title of the question. In e.g. 2D CFT (for simplicity), the three point function of three operators with conformal dimension $a$, $b$ and $c$ are given as
$$
\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\rangle~=~\frac{c_{123}}{(x_1-x_2)^{a+b-c}(x_2-x_3)^{-a+b+c}(x_1-x_3)^{a-b+c}}
$$
Now, I will expect that this correlator shows UV divergences when any two of these operators are coincident. But from the RHS, it seems that e.g. for $c>a+b$, there's no light cone singularity for $x_1=x_2$. What am I missing here?
 A: You need to refine your intuition a little bit. If you bring two operators together, you indeed get singular behaviour, which is taken into account by the OPE. The unit operator is the most singular, and the higher the scaling dimension of an operator $\phi$ in the $O_1 \times O_2$ OPE, the less singular its contribution will be. (In your case, the contribution of $O_3$ is regular if $c > a+b$). 
In the limit $x_1 \rightarrow x_2$, any 3-pt function
$$ < O_1(x_1) O_2(x_2) \phi(x_3) >$$
will measure the overlap of the $O_1 \times O_2$ OPE with $\phi$ inserted 'far away' at $x_3$. According to the above paragraph, to see singular behaviour, you need $\phi$ to be a light operator. If $\phi$ is too high in the spectrum (as in your case), you will only measure a term that lives in the tail of the OPE so you'll get a small, regular result.
I hope that this is more or less what you expected - otherwise I can add more details. 
EDIT to answer the remaining questions. You have hopefully learned that two-point functions are diagonal, so 
$$ < \phi(x) \phi'(y) > = 0$$
unless $\phi = \phi'.$ So the 3-point function with $O_3$ can only measure the contribution of $O_3$ in the OPE - that's the overlap. It's blind to the presence of any other operators.
To see how singular the contribution of some operator $\phi$ is, just write the leading OPE term explicitly:
$$O_1(x_1) O_2(x_2) \sim \frac{1}{|x_1 - x_2|^z} \left\{\phi(x_2) + \text{descendants} \right\}$$
and we want to determine $z$. But you can just apply a dilatation $x \mapsto \lambda x$ and compare the left and right hand sides, and you'll see that $z = a+b-\Delta$ where $\Delta$ is the dimension of $\phi$. The unit operator has dimension zero, all other operators (in a unitary theory) have positive scaling dimensions. 
