# How to show OPE coefficients are symmetric in three indices ?

May it is very trivial, but I am stuck here, given (I have suppressed the conjugate coordinates) $$\phi_i(x) \phi_j(y) \sim \sum_{k} c_{ijk} (x-y)^{h_k - h_i - h_j} \phi_k(y)$$

$$\langle \phi_i(x) \phi_j(y)\rangle = \delta_{ij} \dfrac{1}{(x-y)^{2h_i}}$$

Show that $c_{ijk}$ is symmetric in three indices, (i,j) is straightforward how to go about (j,k) ?

Hint: The fusion rule Clebsch-Gordan-like coefficients $c_{ij}{}^k=c_{ijk}$ are related to the 3-point function $\langle \phi_i(x) \phi_j(y)\phi_k(z)\rangle$ of 3 primary fields, which in turn is totally symmetric.
• Correct me if am wrong, consider $$\langle \phi_1(z_1) \phi_2(z_2) \phi_3 (z_3) \rangle = \dfrac{c_{123}}{z_{12}^{h_2 + h_1 - h_3} z_{23}^{h_2 + h_3 - h_1} z_{13}^{h_1 + h_3 -h_2}}$$ which is same has $$\langle \phi_1(z_1) \phi_3(z_3) \phi_2 (z_2) \rangle = \dfrac{c_{132}}{z_{13}^{h_1 + h_3 - h_2} z_{32}^{h_3 + h_2 - h_1} z_{12}^{h_1 + h_2 -h_3}}$$ owning to radial ordering, hence $c_{123} = c_{132}$. Feb 15, 2015 at 18:43
• $\uparrow$ Yes. Feb 15, 2015 at 22:32