# Branching rules for $SU(3)$

How does one compute the branching rules for $SU(3)\to SU(2)\times U(1)$.?

In particular, I do not know how to put the abelian charges.

Take for example the adjoint $\mathbf{8}$ of $SU(3)$.

I can see it decomposes as $\mathbf{8}\to\mathbf{3}+2\cdot\mathbf{2}+\mathbf{1}$.

But how do I figure out the representations of the $U(1)$ factor?

Furtheremore, does a general procedure, or a specific computer program exist in order to compute these branching rules? For example, how would one go with computing the branching rules of $SU(5)$?

• Related: math.stackexchange.com/q/587761/11127, which mentions a software package LieArt. See also the software package LiE. – Qmechanic Jan 15 '15 at 17:23
• This question (v2) seems like an archetype of a math problem encountered in many areas of physics, e.g. QCD, and which the community consistently wants to not migrate to Math.SE, cf. this meta post. – Qmechanic Jan 15 '15 at 17:30
• @Qmechanic Agreed, I'd like to see this answered here. – JamalS Jan 15 '15 at 18:00
• @Qmechanic. Sorry, I don't understand what you said. Should I have posted this in math.SE? – Federico Carta Jan 16 '15 at 10:38
• Well, people on Phys.SE are divided over the issue. Do you want me to migrate it to Math.SE? – Qmechanic Jan 16 '15 at 11:43

I'll use a notation that probably you know. I'll denote group representations by Dynkin labels. Consider the following: \begin{equation*} [1,0]_3 = _2 q^1 + _2q^0 \ , \end{equation*} and \begin{equation*} [0,1]_3 = _2 q^{-1} + _2q^0 \ . \end{equation*} Where $q$ is the $U(1)$-charge of the representation. By Dynkin diagrams the $SU(3)$ algebra can be represented as 0---0 and the reflection symmetry correspond to the complex conjugation. Thus, we know the adjoint rep $[1,1]_3$ to be real, because it takes $\boldsymbol{3}$ and its conjugate $\boldsymbol{\bar{3}}$ on the same foot; this reasoning applies also to the Abelian charge. Thus, this manifests into decomposition as follows \begin{equation*} [1,1]_3 = _2 q^0 + _2 \left(q^1 + q^{-1}\right) + q^0 \ . \end{equation*}
• I may be mistaken, but the $U(1)$ charges here do not look quite right: they should sum to $0$ in any representation $R$ of $SU(3)$, since $\mbox{Tr}_R(T)=0$ for any generator $T$ in $SU(3)$, in particular for the generator of $U(1)$. The $U(1)$ charges given here seem to be charges under a linear combination of the $U(1)$ in $SU(3)$ and the $U(1)\simeq U(3)/SU(3)$. That piece cancels out in real representations such as the adjoint, but not in the fundamental/antifundamental representations. (Oh, and I fixed a minor typo.) – Bruno Le Floch Jun 8 '16 at 0:33