How do I decompose an $SO(10)$ antisymmetric tensor in $SO(4) \times SU(3) \times U(1)$? My guess is that If I denote the $SO(4)$ indices $\mu, \nu = 1,...4$  and the $SU(3)$ indices by $I,J=1,2,3$, I think $N^{mn}$ should decompose as $N^{\mu \nu}, N^{IJ}, N^{I}_J, N_{IJ}$ plus other terms with mixed indices $I,\mu$, which I don't know how to determine.
I would appreciate if someone could give an explanation of an honest way to decompose $SO(10)$ in $SO(4) \times SU(3) \times U(1)$.
 A: Hints:

*

*Firstly,
$$SO(10) ~\supseteq~ SO(4)\times SO(6),\tag{1}$$
so we get the branching rules
$$ {\bf 10}~\stackrel{(1)}{\cong}~({\bf 4},{\bf 1}) \oplus ({\bf 1},{\bf 6}),\tag{2}$$
and
$${\bf 10}\wedge{\bf 10}~\stackrel{(2)}{\cong}~ ({\bf 4}\wedge{\bf 4},{\bf 1})\oplus ({\bf 4},{\bf 6})\oplus ({\bf 1},{\bf 6}\wedge{\bf 6}).\tag{3}$$
Here the space of 2-forms
$$ {\bf 4}\wedge{\bf 4}~\cong~{\bf 3}_+\oplus{\bf 3}_-\tag{4}$$
splits into selfdual/antiselfdual parts under 4D Hodge duality.


*Secondly,
$$SO(6)~\supseteq~ U(3)~\cong~[SU(3)\times U(1)]/\mathbb{Z}_3,\tag{5}$$
cf. e.g. this Phys.SE post. Up until now the vector spaces under point 1 could in principle be real. In particular the representations are real. Now we complexify the vector spaces. We get the branching rules
$$ {\bf 6}~\stackrel{(5)}{\cong}~{\bf 3}_1 \oplus \bar{\bf 3}_{-1},\tag{6}$$
and
$$\begin{align} {\bf 6}\wedge{\bf 6}
~\stackrel{(6)}{\cong}~&({\bf 3}\wedge{\bf 3})_2 
\oplus (\bar{\bf 3}\wedge\bar{\bf 3})_{-2}
\oplus ({\bf 3}\otimes\bar{\bf 3})_0 \cr
~\cong~&\bar{\bf 3}_2 \oplus {\bf 3}_{-2}
\oplus ({\bf 8}_0 \oplus {\bf 1}_0) 
,\end{align}\tag{7}$$
cf. e.g. this Phys.SE post.
