# Young tableaux for conformal group representation

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?

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.