# Kac-Moody primary OPE

I am reading a paper and on page 13-14 (PDF page 15-16), they say that,

The fermionic generators [$$G^\pm$$ and $$\tilde{G}^\pm$$] are Virasoro and affine Kac-Moody primaries with weights $$h= 3/2$$ and $$j=1/2$$.

I understand the first statement to mean, with respect to the stress-energy tensor $$T$$, being a primary with weight $$h = 3/2$$ implies,

$$T(z)G^\pm(0) = \frac{3/2}{z^2}G + \frac{\partial G}{z} + \mathrm{non-singular \, terms}$$

for the OPE. Now, in Ketov's CFT book, he writes that a field $$\phi$$ being a primary with respect to an affine Kac-Moody current implies the OPE,

$$J^a(z)\phi(0) = \frac{t^a_{(r)}}{z}\phi(0)$$

where $$t^a$$ are generators of a matrix representation labelled $$(r)$$. For the small $$N=4$$ superconformal algebra, the $$J$$ generators are for an affine $$\mathfrak{sl}(2)$$, but I am not sure how to decipher the exact OPE for $$J$$'s with $$G$$'s?

Note that while $$J^a = J^0, J^\pm$$ are generators of an affine Lie Algebra, their zero modes are generators of the finite and simple $$\mathfrak{sl}(2)$$.
The fact that $$G^\pm$$ and $$\tilde G^\pm$$ have Kac-Moody weight $$j=1/2$$ means that they are in the $$j=1/2$$ highest representation of $$\mathfrak{sl}(2)$$. In other words, $$j=1/2$$ is their weight under the Cartan element $$J^0_0$$.
So in order to compute the $$J^a(z)G^\pm(0)$$ and $$J^a(z)\tilde G^\pm(0)$$ OPEs, you just need $$t^a$$ matrices in the $$j=1/2$$ representation of the simple $$\mathfrak{sl}(2)$$ algebra. These are essentially given by the Pauli matrices and should be easy to compute.
• The issue I have is the Pauli matrices are matrices, so you've got $t^a_{ij}$ i.e. two extra indices you need to contract with something, since $J^a(z)G^+(0)$ for example only has one index. Commented Feb 27, 2020 at 15:28
• If $G^\pm$ are in the $j=1/2$ highest-weight representation, then they must have extra indices $G^\pm_j$ ($(t^aG^\pm)_i = t^a_{ij}G^\pm_j$). This index in usually suppressed in books when talking about Kac-Moody primaries (including in the equation you wrote above). The other option is that they are talking about a $U(1)$ subgroup. Commented Feb 27, 2020 at 15:33
• Aren't the sl(2) indices just the $\pm$ superscripts of $G$? Commented Feb 27, 2020 at 22:26
• I hadn't looked at the paper, but looking at it now it's clear that Sylvain is right. The $\pm$ are the $j=1/2$ $\mathfrak{sl}(2)$ indices while $(\pm, 0)$ are the $j=1$ indices. In equation (2.7), they even introduce auxiliary variables to avoid dealing with these $\mathfrak{sl}(2)$ indices. Commented Feb 28, 2020 at 10:38