6
$\begingroup$

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 $SU(5)$?

||cite|improve this question
$\endgroup$
  • $\begingroup$ Related: math.stackexchange.com/q/587761/11127, which mentions a software package LieArt. See also the software package LiE. $\endgroup$ – Qmechanic♦ Jan 15 '15 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 '15 at 17:30
  • $\begingroup$ @Qmechanic Agreed, I'd like to see this answered here. $\endgroup$ – JamalS Jan 15 '15 at 18:00
  • 1
    $\begingroup$ @Qmechanic. Sorry, I don't understand what you said. Should I have posted this in math.SE? $\endgroup$ – Federico Carta Jan 16 '15 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 '15 at 11:43
3
$\begingroup$

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*}

||cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '16 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.