According to Cahn's book [1] (pg. 148), given an irreducible representation (irrep) of a simple algebra of rank $r$, to construct the irreps of its regular subalgebras we can consider an extended Dynkin diagram. The weight $\Lambda$ can be written in Dynkin basis as $$\Lambda=(a_1,a_2,...,a_r)$$ before extension, where the Dynkin coefficients/labels/components $a_i$ for $i=1,2,...,r$ are defined by $$a_i=2\frac{\langle \Lambda, \alpha_i\rangle}{\langle \alpha_j, \alpha_j \rangle}$$ where $\alpha_i$s are simple roots. This is extended by adding a new Dynkin coefficient calculated w.r.t. the extra root in the extended Dynkin diagram which is the negative of the highest root $\gamma$. Then the extended weight will be $(a_1,a_2,...,a_r,a_{r+1}$ where $$a_{r+1}=-2 \frac{\langle \Lambda,\gamma \rangle}{\langle \gamma, \gamma \rangle}$$. This can be calculated in Dynkin basis using the metric tensor $G_{ij}=\frac{1}{2}(C_{ij})^{-1} \langle \alpha_j, \alpha_j \rangle$. Now the usual rules of constructing weights from the highest one by "laddering down" are employed: subtract the row of the Cartan matrix whose row number matches with the positive entries in Dynkin basis of the weight $\Lambda$. We get a list of weights from which we strike out the Dynkin coefficients corresponding to the node removing which we got the subalgebra. Finally, we choose those weights which have all non-negative Dynkin coefficients as they can be the highest weight irreps of the subalgebra whose dimensions can be easily found out by Weyl's dimensionality formula.
Using this algorithm, $B_3 \to A_3$ and $B_3 \to A_1 + A_1 + A_1$ is achieved in [1]. In both calculations, we have some weights of the form $(0,0,...,0)$ called singlets. But somewhat mysteriously in the first case, the singlets are kept and in the 2nd case they are not. Indeed the decompositions are $$7 \to 1+6$$ $$21 \to 15+6$$ where the dimensionalities are indicated in the above equation. The $1$ in the first equation is the dimension of $(0 0 0)$ in $A_3$ which arises by the weight generation algorithm in both cases but kept in the decomposition only in one case. I have tried some other examples where similar things happen. E.g. In $E_8 \to E_6 \times SU(3)$ there is a well known decomposition
$$248 \to (3,27)+(\overline{3},\overline{27})+(8,1)+(1,78)$$ See [2] pg.113 for validity of this. But following the above algorithm we have a $(1,1)$ arising from multiple $(00000000)$ weights which appear when laddering down from $(00000010)$ which is the $248$ irrep (cf. [2] pg 113). Ofcourse by dimensionality matching of both sides we can decide whether to keep the $(1,1)$. But what is its origin and how to decide when to keep the singlet weights and when not?
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