This may be a silly question, and so I apologize in advance. But it stems from a reading of section 2 (page 5) of the physics paper, "Counting chiral primaries in N=1 d=4 superconformal field theories".

My question is: What does the representation of $U(1)$ labeled as $\frac{1}{2}$ indicate?

My question is a group-theory question, and is as such a math question, but for those unfamiliar with the context, here is some background.

  1. The superconformal group for $\mathcal{N} = 1$ supersymmetry in $d = 4$ spacetime dimensions is $SU(2,2|1)$.
  2. We focus our attention on a particular subgroup of $SU(2,2|1)$, called the maximal bosonic subgroup: $SU(2,2) \times U(1)_R$. The $U(1)_R$ is known as an R-symmetry group in physics.
  3. The generators of supersymmetry (the ''supercharges'') $Q$ and $Q^\dagger$ belong to representations $4_1$ and $\bar{4}_{-1}$ of $SU(2,2) \times U(1)_R$. The subscript denotes the $U(1)_R$ representation ($1$ is the fundamental, $-1$ is the anti-fundamental) and the $4$ and $\bar{4}$ are $SU(2,2)$ representations. So, so far we are labeling everything in terms of irreps of the maximal bosonic subgroup.
  4. The conformal group in $d = 4$ spacetime dimensions is $SO(4,2)$, which has a covering group $SU(2,2)$.
  5. We want to study a quantum field theory not in 4-dimensional Minkowski ("flat") space but on the space $\mathbb{R} \times S^3$. So one is interested in the Killing spinors of this space, and the isometries.
  6. Based on (4) and (5), we restrict our attention to the subgroup $U(1) \times SO(4)$ of the conformal group, which is the isometry group of $\mathbb{R} \times S^3$.
  7. The idea then is to decompose the generators in terms of representations of the isometry group $SU(2)_l \times SU(2)_r \times U(1)$ and the R-symmetry $U(1)_R$. Here we've used the fact that $\mathbb{so(4)} = \mathbb{su}(2)_l \times \mathbb{su}(2)_r$.

The claim is

$$4_1 \longrightarrow (2,1)_{\frac{1}{2},1} \oplus (1,2)_{-\frac{1}{2},1}$$ $$\bar{4}_{-1} \longrightarrow (2,1)_{-\frac{1}{2},-1} \oplus (1,2)_{\frac{1}{2},-1}$$

There are now two $U(1)$ subscripts: the first is for the $U(1)$ which is part of the isometry group, and the second is for the $U(1)_R$ which is the R-symmetry group. Note that the $U(1)_R$ subscript is the same on each term on the right hand side and carries over from the left.

The question posed above pertains to the first subscript in the above decomposition, i.e. the representation of the $U(1)$ which is part of the isometry group of the manifold.

I know that an element in $U(1)$ is represented by $e^{i\theta}$ and moreover, $U(1)$ is isomorphic to $SO(2)$. The latter made me think of the irreducible spinor representation of $SO(2)$, but that too is a real 1-dimensional representation (2d Dirac spinor with 2 complex components, but the Majorana-Weyl condition brings it down to 1 real component).

Also, if $U(1)$ is parametrized by $\left(\begin{array}{cc}\cos n\theta & \sin n\theta\\-\sin n\theta & \cos n\theta\end{array}\right)$, does $\frac{1}{2}$ simply mean that $n = 1/2$ in this representation, so that effectively, its a half rotation?

  1. TL;DR: The label $q\in\mathbb{Q}$ (e.g. $q=\frac{1}{2}$) refers to the following (possibly multi-valued) 1-dimensional representation $$U(1)~\in~\alpha~~\stackrel{R_q}{\mapsto}~~\alpha^q ~\in~GL(1,\mathbb{C})/\sim \tag{1}$$ of the group $U(1)$.

  2. The connected component of the (global) conformal group that contains the identity element is $$ {\rm Conf}_0(3,1)~\cong~SO^+(4,2)/ \mathbb{Z}_2~\cong~SU(2,2)/ \mathbb{Z}_4~\cong~PSU(2,2) ,\tag{2}$$ where $$ \mathbb{Z}_2~\cong~\{\pm {\bf 1}_{6\times 6} \}, \qquad\mathbb{Z}_4~\cong~\{\pm {\bf 1}_{4\times 4},\pm i{\bf 1}_{4\times 4}\},\tag{3} $$ see e.g. Ref. 2.

  3. One can make a Lie group homomorphism $$ G~:=~SU(2)_{\ell} \times SU(2)_r \times U(1)~\ni~ (g_{\ell},g_r,\alpha)~~\stackrel{\Phi_q}{\mapsto}~~ \begin{pmatrix} \alpha^q g_{\ell} & 0 \cr 0 & \alpha^{-q} g_r \end{pmatrix}~\in~ PSU(2,2), \tag{4}$$ where $$q~\in~\frac{1}{2}\mathbb{Z}\tag{5} $$ is a half integer (or integer).

  4. If one chooses $q=\frac{1}{2}$, one may show that the Lie group homomorphism (4) is injective. The product group $G$ is hence a subgroup of $PSU(2,2)$. This means that representations of $PSU(2,2)$ can be decomposed as representations of $G$.

  5. Examples: In a hopefully obvious notation, the fundamental (multi-valued) representation decomposes as $$ {\bf 4}~=~ ({\bf 2}_{\ell}, {\bf 1}_r)_{\frac{1}{2}} \oplus ({\bf 1}_{\ell}, {\bf 2}_r)_{-\frac{1}{2}}, \tag{6}$$
    while the complex conjugate representation decomposes as $$ \overline{\bf 4}~=~ ({\bf 2}_{\ell}, {\bf 1}_r)_{-\frac{1}{2}} \oplus ({\bf 1}_{\ell}, {\bf 2}_r)_{\frac{1}{2}}, \tag{7}$$
    cf. eq. (2.1) in Ref. 1.


  1. C Romelsberger, arXiv:hep-th/0510060.

  2. R. Penrose & W. Rindler, Spinors and Space-Time Vol. 2, 1986; p. 303-304.

  • $\begingroup$ Thank you @Qmechanic! It'll take me some time to go through all this, but I understand now that my question wasn't quite as trivial as I originally thought! $\endgroup$ – leastaction Sep 14 '16 at 14:46

I believe this just has to do with differing conventions used by physicists and mathematicians. In this context, I think the $1/2$ and $-1/2$ are $U(1)$ R-charges, and not the dimensions of the representations of $U(1)$, which obviously are all 1-dimensional.


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