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I would like to apply branching rule to decompose a representation of $SO(10)$ into that of $SU(4)$. So I did the following, and I'm not sure if it's correct. I use the following chain of decomposition:

$$SO(10)\hspace{1mm}\rightarrow\hspace{1mm}SU(2)\times SU(2) \times SU(4) \hspace{1mm} \rightarrow \hspace{1mm}SU(4).$$

In particular, I'm interested in the adjoint $\mathbf{45}$ of $SO(10)$: \begin{eqnarray} \mathbf{45} &\rightarrow &\mathbf{(3,1,1) + (1,3,1) + (1,1,15) + (2,2,6)} \nonumber\\ % &\rightarrow & \mathbf{(3\times 1) + (3\times 1) + 15 + (4\times 6)}. \nonumber\\ % \end{eqnarray} My concern is in the second line of the above equation. Since I'm only interested in the $SU(4)$ part and not the $SU(2)\times SU(2)$ part, can I just simply convert the $SU(2)\times SU(2)$ representation into the coefficients counting the number of representation of $SU(4)$?

Qmechanic
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user195583
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1 Answers1

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Note that the branching rule $$so(10)\quad \supseteq \quad so(4) \oplus so(6)\quad \cong \quad su(2)\oplus su(2) \oplus su(4)$$ only holds at the level of Lie algebras, not Lie groups. But apart from that then Yes, the number of $su(4)$-representation in OP's example is given by the product of dimensions of the two $su(2)$-representations.

Qmechanic
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