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May be 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) ?

This post imported from StackExchange Physics at 2015-02-16 11:40 (UTC), posted by SE-user Jaswin

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.

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