# How does the adjoint of $SO(10)$ branch under $SU(5)$

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 $$\bar{10}$$.