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I'm following the notes "Ginsparg - Applied Conformal Field Theory" (https://arxiv.org/abs/hep-th/9108028) and I'm stuck on a proof at page 140 about Kac-Moody algebras. I would like to prove that $\beta = 2\tilde{k} +C_A$ using the given definition of the stress-energy tensor $$T(z) = \frac{1}{\beta}\lim_{z\to z'} \left\{ \sum\limits_{a=1}^{|G|} J^a(z) J^a(z') - \frac{\tilde{k}|G|}{(z-z')^2} \right\}$$ and the Operator Product Expansion (OPE) of two conserved currents $$J^a(z) J^b(w) = \frac{\tilde{k} \delta^{ab}}{(z-w)^2}+\frac{i f^{abc} J^c(w)}{(z-w)}.$$

I'm following this way: starting from $$T(z) J^b(w) = \frac{1}{\beta}\left\{\lim_{z\to z'} \sum\limits_{a=1}^{|G|} J^a(z) J^a(z') J^b(w) - \frac{\tilde{k}|G|}{(z-w)^2}J^b(w) \right\}.$$ I wish to demonstrate that this is equal to the OPE of the stress-energy tensor with a primary field $$T(z) J^b(w) = \frac{J^a(w)}{(z-w)^2}+\frac{\partial J^b(w)}{(z-w)}$$ if and only if $\beta = 2\tilde{k} +C_A$, making use of the quadratic Casimir eigenvalue in the adjoint representation $f^{abc} f^{acd} = \delta^{bd} C_A$.

Could someone explain to me all the passages?

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Rather than explicitly subtracting the singular term when writing $T(z)$, you could write it as a normal-ordered product $T(z)\propto \sum_a (J^aJ^a)(z)$, and use Wick's theorem to compute the OPE of $(J^aJ^a)$ with $J^b$. This is done in some detail in Exercise 4.4 of my review article https://arxiv.org/abs/1406.4290 .

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