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?


1 Answer 1


I haven't looked at this stuff for many years, so take my answer with caution.

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.

  • $\begingroup$ 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. $\endgroup$ Commented Feb 27, 2020 at 15:28
  • $\begingroup$ 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. $\endgroup$
    – Heidar
    Commented Feb 27, 2020 at 15:33
  • 1
    $\begingroup$ Aren't the sl(2) indices just the $\pm$ superscripts of $G$? $\endgroup$ Commented Feb 27, 2020 at 22:26
  • $\begingroup$ 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. $\endgroup$
    – Heidar
    Commented Feb 28, 2020 at 10:38

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