# OPE of three operators

Consider a product of 3 local operators in a 2d CFT:

$$X(x) Y(y) Z(z) = \sum_{n=-N}^{\infty} A_n(x) Z(z) (x-y)^n,$$

where we have substituted $$X(x) Y(y)$$ for the $$XY$$ OPE. This expression contains the singular terms for $$x = y$$.

Now because by definition of OPE $$A_n(x)$$ is a local operator at $$x$$, we can use the $$A_n Z$$ OPE again:

$$X(x) Y(y) Z(z) = \sum_{n=-N}^{\infty} \sum_{m=-M}^{\infty} B_{nm} (x) (x-y)^n (x-z)^m.$$ This expression contains the singular terms for $$x = y$$ and $$x = z$$.

Question: where did the $$y = z$$ singular terms go?

This is likely related to the convergence of the series, but I wasn't able to formulate a convincing argument.

Let me redo the calculation, while explicitly writing OPE coefficients. Let $$(A_n(z))_n$$ be a basis of operators at $$z$$. We use the two OPEs $$Y(y)Z(z) = \sum_n c_n(y,z) A_n(z)$$ and $$X(x)A_n(z) = \sum_m d_{m,n}(x,z) A_m(z)$$ We end up with the result $$X(x)Y(y)Z(z) = \sum_{m,n} c_n(y,z)d_{m,n}(x,z) A_m(z)$$ where the coefficient of $$A_m(z)$$ is $$\sum_n c_n(y,z)d_{m,n}(x,z)$$. This coefficient is an infinite sum, and it can very well be singular as $$x\to y$$, although this is not manifest. For instance, $$\frac{1}{x-y} = \sum_{n=0}^\infty (y-z)^n (x-z)^{-n-1}$$ You can recover a similar result in your calculation by distinguishing more clearly the operator basis from the OPE coefficients. Your operators $$B_{m,n}(x)$$ should not all be linearly independent, and you should rewrite them in terms of a basis of operators.
• That sum that you wrote down contains infinite negative powers of $x-z$. Doesn’t this contradict the axiom that OPE starts at the finite order $-N$? – Prof. Legolasov May 21 at 17:59
• No. Each OPE $XA_n$ starts at a finite order. But we are taking a linear combination of infinitely many such OPEs. – Sylvain Ribault May 22 at 6:50