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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) ?

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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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  • $\begingroup$ 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}$. $\endgroup$
    – Jaswin
    Feb 15, 2015 at 18:43
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Feb 15, 2015 at 22:32
  • $\begingroup$ Btw the word symmetric should be interpreted in a superized sense, i.e. with appropriate Grassmann sign factors. $\endgroup$
    – Qmechanic
    Feb 16, 2015 at 13:45

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