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A tensor product of group representations has character equal to the product of its factor characters. Multiplying the rows of the character table in solution, using $s^2=2+t$, $t^2=2+s$ and $st=-1$, gives all the Clebsch-Gordan decomposition of group representations:For example, the character of $R_2\otimes R_2$ is $(4,s^2,t^2,0)=(4,2+t,2+s,0)$, the sum of the characters of $R_0$, $R_1$ and $R_3$. Likewise $R_2\otimes R_3$ has character $(4,-1,-1,0)={\chi }_2+{\chi }_3$. The trivial representation and $R_1\otimes R_1=R_0$ account for the remaining entries.

Use the basis of solution for the two doublets, treating the barred left-handed fields as the components to which the printed representation assignment applies. Write $x=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $y=({\psi }_R^1,{\psi }_R^2{)}^T$. The doublet contribution to the Yukawa interaction is $H{x}^{T}Ly$, so invariant tensors obeyThe first equation forces $L_{11}=L_{22}=0$, since their factors are ${\alpha }^{\pm 2}\ne 1$. The second sets $L_{12}=L_{21}$. Thus the doublet invariant tensor is $x_1{y}_2+x_2{y}_1$. The product of either doublet with $R_1$ has no trivial representation, whereas $R_1\otimes R_1=R_0$. ThereforeIf instead the unbarred left-handed doublet is chosen to transform by $D_1$, its barred components transform by $D_1^{\ast }$; in that equivalent basis the upper block is $y{I}_2$. Exchanging the two left components relates these conventions. The singular value decomposition gives the same physical masses in either basis.