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I am trying to reproduce result results from an article about genus 2 partition function in conformal field theory.

At some point, the author state that, for a general conformal field theory, the currents satisfy the usual commutation relations, with a normalization of the current :

$[J_m^a,J_n^b] = if_{abc}J_{m+n}^c + m\delta_{ab}\delta_{m,-n}$

Just after, he states that one find, in terms of those structure constant,

$\sum_{a}Tr_{\mathcal{H}_{1}}(J_{-1}^aJ_1^a)=N = dim(\mathcal{H_1})$

And

$\sum_{a}Tr_{\mathcal{H}_{1}}(J_{0}^aJ_0^a)=-\sum_{abc}f_{abc}f_{acb}$

I tried many different things in order to recover those results but I don't find a convincing way. The most promising way was to use the OPE of two CFT currents inside a double integral as :

$J_n^aJ_m^a = \oint\frac{dz}{2\pi i}z^nj^a(z)\oint\frac{dw}{2\pi i}w^mj^a(w)$

with the OPE given by

$j^a(z)j^b(w)\sim\frac{1}{(z-w)^2}+\sum_{c}\frac{if^{abc}}{z-w}j^c(w)$

I arrived to $J_n^aJ_m^a = m\delta_{m+n,0}$ but I am really not sure about the result. Nevertheless, I really don't understand how a sum over two structure constant can appear .

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  • $\begingroup$ Are you unsure how to derive the algebra $[ J_m^a , J_n^b] = \cdots$ or only the equations that follow it? $\endgroup$ – Prahar Nov 10 '16 at 13:58
  • $\begingroup$ The two results that follows $\endgroup$ – Ezareth Nov 10 '16 at 14:45

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