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

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

2. 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.

OP's questions are quite broad. Here we will focus on OP's first question, but hopefully the reader gets some idea how this can be generalized.

1. Consider the super inner product space $$V~:=~\mathbb{C}^{2,2|1}~=~V_0\oplus V_1, \qquad V_0~:=~\mathbb{C}^{2,2|0}, \qquad V_1~:=~\mathbb{C}^{0|1},$$ which has 2+2=4 bosonic and 1 fermionic dimensions, and which is endowed with the standard metric $$\eta ~=~ {\rm diag}(1,1,-1,-1|1) ~\in~ {\rm End}(\mathbb{C}^{2,2|1}).$$

2. Supermatrices
$$m~=~\begin{pmatrix} m_{00} & m_{01} \cr m_{10} & m_{11} \end{pmatrix},$$ corresponding to the super matrix algebra $$A~:=~{\rm End}(V)~=~A_0 \oplus A_1, \qquad A_0~:=~{\rm End}(V_0)\oplus {\rm End}(V_1), \qquad A_1~:=~{\cal L}(V_0;V_1)\oplus {\cal L}(V_1;V_0),$$ of endomorphisms can be decomposed in two diagonal bosonic blocks $$m_{00}\in{\rm End}(V_0)\qquad\text{and}\qquad m_{11}\in{\rm End}(V_1),$$ and two off-diagonal fermionic blocks $m_{01}$ and $m_{10}$. The fermionic sector $A_1$ contains linear maps between $V_0$ and $V_1$.

3. The super Lie group $$U(2,2|1) ~~:=~~ \{U\in {\rm End}(\mathbb{C}^{2,2|1}) \mid U^{\dagger}\eta U = \eta \}$$ has corresponding super Lie algebra $$u(2,2|1) ~~:=~~ \{m\in {\rm End}(\mathbb{C}^{2,2|1}) \mid m^{\dagger} =-\eta m \eta^{-1} \}.$$ (Warning: The super-Hermitian conjugation "$\dagger$" involves appropriate sign-factors.) The bosonic part of the super Lie algebra is $$A_0\cap u(2,2|1) ~\cong~ u(2,2) \oplus u(1)_R$$ $$~\cong~ su(2,2) \oplus u(1) \oplus u(1)_R ~\cong~ so(4,2) \oplus u(1) \oplus u(1)_R.$$ Here subscript $R$ stands for the $R$-charge in the fermionic sector. The bosonic part of the super Lie group is $$A_0 \cap U(2,2|1) ~\cong~ U(2,2) \times U(1)_R$$ $$~\cong~\frac{ SU(2,2)\times U(1)}{\mathbb{Z}_4} \times U(1)_R ~\cong~\frac{ SPIN(4,2)\times U(1)}{\mathbb{Z}_4} \times U(1)_R,$$ cf. e.g. this Phys.SE post and this Math.SE post.

4. Now let us return to OP's first question. The super Lie group $$SU(2,2|1) ~~:=~~ \{U\in U(2,2|1) \mid {\rm sdet} (U) =1\}$$ has corresponding super Lie algebra $$su(2,2|1) ~~:=~~ \{m\in u(2,2|1) \mid {\rm str} (m) =0 \}.$$ The bosonic part of the super Lie algebra becomes $$A_0\cap su(2,2|1) ~\cong~ su(2,2) \oplus u(1)_R ~\cong~ so(4,2) \oplus u(1)_R.$$ This is the answer to OP's first question at the Lie algebra level.

5. Concerning conformal groups without SUSY, see e.g. this Phys.SE post. In 3+1D the connected component that contains the identity element is $${\rm Conf}_0(3,1)~\cong~ SO^+(4,2)/\mathbb{Z}_2~\cong~SU(2,2)/[\mathbb{Z}_2\times \mathbb{Z}_2].$$ For superconformal groups, see e.g. Wikipedia and nLab.

• Stupid question but what is $\mathcal{L}$? I assume it has to do with even and odd parts. – Marion Sep 27 '17 at 22:44
• ${\cal L}(V;W)$ is the set of linear maps: $V\to W$. – Qmechanic Sep 27 '17 at 22:57