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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 $SU(5)$?

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    $\begingroup$ Related: math.stackexchange.com/q/587761/11127, which mentions a software package LieArt. See also the software package LiE. $\endgroup$
    – Qmechanic ♦
    Jan 15, 2015 at 17:23
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic ♦
    Jan 15, 2015 at 17:30
  • $\begingroup$ @Qmechanic Agreed, I'd like to see this answered here. $\endgroup$
    – JamalS
    Jan 15, 2015 at 18:00
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    $\begingroup$ @Qmechanic. Sorry, I don't understand what you said. Should I have posted this in math.SE? $\endgroup$
    – Federico Carta
    Jan 16, 2015 at 10:38
  • $\begingroup$ Well, people on Phys.SE are divided over the issue. Do you want me to migrate it to Math.SE? $\endgroup$
    – Qmechanic ♦
    Jan 16, 2015 at 11:43

1 Answer 1

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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 = [1]_2 q^1 + [0]_2q^0 \ , \end{equation*} and \begin{equation*} [0,1]_3 = [1]_2 q^{-1} + [0]_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]_2 q^0 + [1]_2 \left(q^1 + q^{-1}\right) + [0]q^0 \ . \end{equation*}

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  • $\begingroup$ 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.) $\endgroup$
    – Bruno Le Floch
    Jun 8, 2016 at 0:33

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