3

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?

user2062542
  • 739

1 Answers1

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.

Heidar
  • 5,541