# Does Operator Product Expansion form an algebra?

The operator product algebra in CFT is defined as $$\mathcal{O}_i(z,\bar{z})\mathcal{O}_j(\omega,\bar{\omega}) = \sum_{k} C^k_{ij}(z-\omega,\bar{z}-\bar{\omega})\mathcal{O}_k(\omega,\bar{\omega}).$$

This definition reminds me of the definition of the adjoint representation of a Lie algebra, where $C^k_{ij}$ is like the structure constant, and operators are like basis vectors. Does such a correspondence exist?

• yes a commutator is needed but in the left hand side. Mar 11, 2016 at 17:22
• Yes. This is indeed the form of an algebra. The products of operators corresponds directly to the commutator of the charges constructed out of contour integrals out of such operators. See Chapter 2 of Polchinski for details of this relationship. Mar 11, 2016 at 19:43

Generically, however, one cannot take the coincident limit on the left, and is forced to work with operators at different points. You could try to treat operators at different points as if they were independent, but then you would lose the conformal invariance which constraints the structures of the OPE, since it also relates the operator values of $\mathcal{O}$ at different points.