# Kac-Moody algebra, proof of parameters calculation

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?

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 .