0

In a few articles that I read namely this one: The Hitchhiker’s Guide to 4d $\cal N = 2$ Superconformal Field Theories

people talk about the so-called null states. The discussion continues till it introduces the short multiplet at the threshold where it is mentioned in equation 2.8 that $l=1,2$. Given equation 2.5, where does the restriction on $l$ come from?

Oppositely, in the following paper: Multiplets of Superconformal Symmetry in Diverse Dimensions in equation 1.11 the authors say that $l$ can be any non-negative integer. Why is that? Is this the particular case of $\mathfrak{su}(2,2|2)$ that limits $l$ to 1,2? if so how?

Bastam Tajik
  • 1,398
  • 8
  • 26
  • The (super)conformal algebra dictates that certain descendants have to vanish if the scaling dimension of the primary becomes too small. See https://arxiv.org/abs/1612.00809 and https://physics.stackexchange.com/questions/563571/recombination-phenomena-in-cft/644041 – Connor Behan Aug 21 '22 at 22:58
  • @ConnorBehan Thanks for the paper. Honestly, I read also the same paper but in equation 1.11 the range of $l$ is all non-negative integers in contrast to the paper I attached. What's the reason? Does it have to do with the particular case of $\mathfrak{su}(2,2|2)$? Then how? – Bastam Tajik Aug 21 '22 at 23:05
  • CDI were not trying to claim that all non-negative integer values for $l$ would be realized. As page 33 shows, there are two A-type shortening conditions in 4d. The highest this number can go across all dimensions is 4. – Connor Behan Aug 22 '22 at 19:47
  • I see, but I can't still see the reason for $l$ to be at most 2 in this particular case of $\mathfrak{su}(2,2|2)$ @ConnorBehan – Bastam Tajik Aug 22 '22 at 23:25
  • 1
    What does it mean for $l = 3$ to not be allowed? It means that for all states of the schematic form $Q^3 \left | \Delta \right >$, zeros of its norm occur at values of $\Delta$ which make $Q^2 \left | \Delta \right >$ or $Q^1 \left | \Delta \right >$ type norms strictly negative. I'm not sure there is a more succinct reason. – Connor Behan Aug 22 '22 at 23:58
  • No. Long multiplets allow you to act with all the $Q$s. So there are certainly large enough values of $\Delta$ such that $Q^1 \left | \Delta \right >$, $Q^2 \left | \Delta \right >$ and $Q^3 \left | \Delta \right >$ all have positive norm. When you are looking for shortening conditions though it's a different story. – Connor Behan Aug 23 '22 at 10:43

0 Answers0