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  Why the bosonic part of the superconformal group $SU(2,2|1)$ is $SO(4,2) \times U(1)_R$?

+ 5 like - 0 dislike

Why in $d=4$ $\mathcal{N}=1$ SCFT the bosonic part of the superconformal group $SU(2,2|1)$ is $SO(4,2) \times U(1)_R$?

More generally how can I determine the such a thing in other theories? Is there some specific way to think about how to find such a subgroup? Say in $d=4$ $\mathcal{N}=2$ theory. I know this is known, I want to know how one finds it though.

This post imported from StackExchange Physics at 2014-12-25 23:28 (UTC), posted by SE-user Marion

asked Dec 25, 2014 in Theoretical Physics by Marion Edualdo (250 points) [ revision history ]
edited Jan 2, 2015 by conformal_gk

1 Answer

+ 4 like - 0 dislike

Firstly the bosonic part of superconformal algebra of this theory is $SO(4,2)\times U(1)_R$ not $SO(4,1)\times U(1)_R$. I have corrected this mistake that I guess it was a typo. Now, let us consider the conformal algebra generators.

  1. The momenta $P^{\mu}$ that generate spacetime translations. 
  2. The angular momenta $J^{\mu \nu}$ that generate spacetime rotations.
  3. The dilatation operator $D$ that generates uniform rescalings to coordinates and fields.
  4. The special conformal transformations $K^{\mu}$.

There are many places where you can check the algebra they create, namely the superconformal algebra $\mathfrak{so}(4,2)$. How can this be true? You can verify this by introducing indices $a,b=-1,0,1,2,3,4$ and defining the operators $J^{ab}= -J^{ba}$ as following $J^{ab}= J^{\mu \nu}$ for $a,b=0,1,2,3$, $J^{4\mu}=-J^{\mu4} = \frac{P^{\mu} + K^{\mu} }{2}$, $J^{-1\mu}=-J^{\mu-1} = \frac{P^{\mu} - K^{\mu} }{2}$ and $J^{4-1}=-J^{-14} = D$. Then these operators follow the Lorentz algebra (up to some signs that might be wrong in my notes and I will not try to verify now)

$$ [J^{ab}, J^{cd}] = i( g^{bc}J^{ad} - g^{ac}J^{bd} - g^{bd}J^{ac} + g^{ad}J^{bc} ) $$

if and only iff $g^{-1-1}=g^{00}=1$, $g^{11}=g^{22}+g^{33}=-1$, $g^{ab}=0$ for $a\neq b$. In other words the operators $J^{ab}$ create the Lorentz algebra in a spacetime of two time dimensions and 4 space dimensions. Thus $SO(4,2)$ . Now, note that this is the symmetry group of $AdS$ embedded into 2+4 dimensions as a hypersurface $g_{ab}=X^aX^b=R^2$ where $R$ is thr $AdS_5$ radius. If you try to find the isometries of $AdS_5$ via the very lengthy computation of Killing vectors etc, you will find $SO(4,2)$ and this should give you  the information you want via $AdS$/CFT  . More generally, $AdS_D$ has the symmetry group $SO(p+1,q+1)$ for $D=p+q$ which isomorphically matches with the one of $\mathbb{R}^{p,q}$. The $U(1)_R$ part is easier to understand. Holographically it is what survives after the Killing spinor equations come in action. It is the $R$ symmetry group for $\mathcal{N}=1$.

answered Jan 2, 2015 by conformal_gk (3,625 points) [ revision history ]
edited Jan 3, 2015 by conformal_gk
Hi conformal_gk, potentially VERY dumb question:

Why are there no SUSY generators (charges) in your list of generators of the superconfofmal algebra? Looking at the list, it seems to me these are just the generators of an "ordinary" conformal group?

Anyway, many thanks for giving so many great answers here recently here, and happy new year to you and everybody ;-)

@Dilaton I have written down only the conformal algebra. The susy generators generate the $U(1)_R$ symmetry. I had written superconformal in my main text but I corrected it. 

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