We can split up $SU(5)$ into real and imaginary parts as $U=U_R+iU_I$ and in doing so embed this in $SO(10)$ as $\begin{pmatrix} U_R & -U_I \\ U_I &U_R\end{pmatrix}$.

Hence we know that $SU(5)$ is a subgroup of $SO(10)$. We can also show that the embedding above is diagonalisable and hence hence branches to $5\oplus\bar{5}$.

The question then asks how the adjoint representation of $SO(10)$ branches under this embedding. So the dimension of the adjoint representation is the number of elements in the lie algebra so for $SO(10)$ the dimension is 45.

How do I work this out?


An object that is in the fundamental of $SO(10)$ will carry the index $V^m$, with $m=1$ to $10$. This will split into $(V^a,V_a)$, with $a=1$ to $5$, the fundamental and anti-fundamental of $U(5)$ subgroup.

$$ 10=5\oplus \bar{5} $$

The $SO(10)$ generator will be $M_{mn}=-M_{nm}$ and so splits into $(M_{ab}, M^{ab}, M_{a}^{b})$, where $M_{a}^{b}$ is the generator of the $U(5)$, so they satisfy $(M^a_b)^*=M^b_a$, and have a total of 25 real components. The $M^{ab}$ and $M_{ab}$ are both antisymmetric and are complex conjugate of each other $(M_{ab})^*=M^{ab}$ so they have a total of 20 real components. In total we have 45 real components as desired.

$$ 45=10\oplus 25\oplus \bar{10} $$

where $25$ is the adjoint of $U(5)$ and $10$ is the antisymmetric part of $5\otimes 5$:

$$ 5\otimes 5 = 10 \oplus 15 $$

$10$ is the antisymmetric part and $15$ is the symmetrical part. The same goes for $\bar{10}$.


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