This is taken from arXiv:1910.14051, pg 32:

Decomposing this $SO(d+ 1, 1)$ representation into $SO(1, 1)× SO(d)$ representations as in (A.4), we find

$$\square \underset{\operatorname{SO}(1,1) \times \operatorname{SO}(d)}{\longrightarrow}(\bullet)_{-1} \oplus(\square)_{0} \oplus(\bullet)_{1}\tag{A.7}$$

where • denotes the trivial spin representation, and the subscripts are the dilation weight.

Can anyone please explain decomposing a representation into representations of its subgroup through Young's tableaux?


1 Answer 1


This follows because the $d$-dimensional (global) conformal group ${\rm Conf}(d)$ is locally isomorphic to the proper Lorentz group $G:=SO(d\!+\!1,1)$ in Minkowski space $$\mathbb{R}^{d+1,1}~\cong~\mathbb{R}^d\times \mathbb{R}^{1,1}, \qquad \mathbb{R}^{1,1} ~\cong \underbrace{\mathbb{R}}_{\ni x^+}\times \underbrace{\mathbb{R}}_{\ni x^-},\tag{1} $$ cf. e.g. this Phys.SE post. We can clearly embed the product subgroup $$\begin{align}H~:=~SO(d)\times SO(1,1)~\cong~&\begin{pmatrix} SO(d) \cr & SO(1,1) \end{pmatrix}_{(d+2)\times (d+2)}\cr \subseteq ~&SO(d\!+\!1,1)~=:~G.\end{align}\tag{2}$$ Here the proper Lorentz group in 1+1D becomes diagonal $$SO(1,1)~=~\left\{\begin{pmatrix} b \cr & b^{-1} \end{pmatrix} \in{\rm Mat}_{2\times 2}(\mathbb{R}) ~\mid~ b\in\mathbb{R}\backslash\{0\}\right\}~\cong~\mathbb{R}\backslash\{0\}\tag{3}$$ if we use light-cone (LC) coordinates $x^{\pm}$. The LC coordinates $x^{\pm}$ have weights $\pm 1$ because we identify boosts with conformal dilations. The sought-for eq. (A.7) is just the following branching rule $$ \underline{\bf d+2}~\cong~ \underline{\bf d}_0\oplus \underline{\bf 1}_{1}\oplus \underline{\bf 1}_{-1} \tag{4}$$ of Minkowski space (1). The single box $\Box$ on the LHS and the RHS of eq. (A.7) denotes the defining representation of $SO(d\!+\!1,1)$ and $SO(d)$, respectively.

  • $\begingroup$ I am most grateful. $\endgroup$ Jun 14, 2020 at 17:34

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