# 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. – halfmetal Mar 11 '16 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. – Prahar Mar 11 '16 at 19:43

## 1 Answer

Although the identity you write does look like an algebra relation, and as Prahar mentions in the comment it is related to Lie algebra of charges, it is not an algebra in a conventional sense. For example, the relation Prahar mentions is not an algebra isomorphism, but rather the statement that commutation relations of certain modes of the operators are encoded in the singular terms of the operator product expansions.

The reason that it is not very useful to try to put the OPE in the conventional algebraic context is that on the left hand side you necessarily have operators at different spacetime points -- this is because the OPE's are generally singular. This is notably not so for certain classes of protected operators in supersymmetric field theories, where one can show that there are no singular terms in the OPE on the right, and one can take the limit of coincident points. It then turns out that in this limit only the operators from the same class remain on the right hand side, and then the OPE literally translates into multiplication in what is called a chiral ring.

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.

The framework used by mathematicians to deal with this kind of structures is basically a formalization of CFT, and goes by the name vertex operator algebra. I am not very familiar with this construct, but the wiki article should get you started.