How do I decompose an $SO(10)$ antisymmetric tensor in $SO(4) \times SU(3) \times U(1)$?

Question

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 $SO(4) \times SU(3) \times U(1)$.

Due diligence: have you absorbed the totality and methods of Slansky's review? — Cosmas Zachos, Aug 26 '21 at 13:03
$S^2\boldsymbol{10}$ decomposes as $(\boldsymbol1,\boldsymbol1)0+(\boldsymbol4,\boldsymbol3){-1}+(\boldsymbol4,\overline{\boldsymbol3}){+1}+(\boldsymbol9,\boldsymbol1)_0+(\boldsymbol1,\boldsymbol6){-2}+(\boldsymbol1,\overline{\boldsymbol6})_{+2}+(\boldsymbol1,\boldsymbol8)_0$ and $A^2\boldsymbol{10}$ decomposes as — AccidentalFourierTransform, Aug 26 '21 at 13:13
$(\boldsymbol1,\boldsymbol1)0+(\boldsymbol1,\boldsymbol3){+2}+(\boldsymbol1,\overline{\boldsymbol3}){-2}+(\boldsymbol3,\boldsymbol1){0}+(\overline{\boldsymbol3},\boldsymbol1){0}+(\boldsymbol4,\boldsymbol3){-1}+(\boldsymbol4,\overline{\boldsymbol3}){+1}+(\boldsymbol1,\boldsymbol8)_0$ under $\mathfrak{so}{10}\supset\mathfrak{so}_4+\mathfrak{su}_3+\mathfrak u_1$, unless I messed it up. So double check. In any case this is a math question. — AccidentalFourierTransform, Aug 26 '21 at 13:13
@AccidentalFourierTransform What do you mean by $S^2 \textbf{10}$ and $A^2 \textbf{10}$ ? — Slayer147, Aug 26 '21 at 13:20
Ignoring Slansky if you are studying group theory in GUTs is not an option. Start from table 43. — Cosmas Zachos, Aug 26 '21 at 13:26
@CosmasZachos It seems very useful, but it doesn't have $SO(10)$ in terms of $SO(4) \times SU(3) \times U(1)$. — Slayer147, Aug 26 '21 at 13:44
SO(4) ~ SU(2)xSU(2) at the algebra level. Do due diligence. Surely, you don't want a one liner for your homework? — Cosmas Zachos, Aug 26 '21 at 13:46

Qmechanic · Answer 1 · 2021-08-28T15:12:29.783

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.

How do I decompose an $SO(10)$ antisymmetric tensor in $SO(4) \times SU(3) \times U(1)$?

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