I have long been unable to follow section 12.3 of Georgi - Lie algebras in particle physics. This section deals with how irreps of $SU(3)$ decompose as irreps of subgroups $H \subset SU(3)$ and is later generalised to $SU(N)$ in section 13.5. Although I understand the concept and have seen other treatments that I follow, I would like some help understanding Georgi's treatment.

First Georgi says (as far as I can see without justification) that the fundamental, $\mathbf{3}$, of $SU(3)$ decomposes as an $SU(2)\times U(1)$ doublet with hypercharge $1/3$ and a singlet of hypercharge $-2/3$. Is it clear why? How does this generalise for the $\mathbf{N}$ decomposing in arbitrary subgroups $H \subset SU(N)$?

Following this Georgi considers an arbitrary Young Tableau (i.e. irrep) of $SU(3)$ with $n$ boxes; I believe with arbitrary symmetry of the indices. He assumes that $j$ indices of the tensor transform as $SU(2)$ doublets and (n-j) as singlets -- but why is this the only possibility? Is it because any irrep of $SU(2)$ can be formed from tensor products of the doublet?

We go on to represent the $n-j$ singlets by a Young Tableau of $n-j$ boxes in a row: does Georgi mean $SU(3)$ Tableaux or $SU(2)$ Tableaux? From figure 12.6 it seems the are $SU(3)$ Tableaux but then why must they be rows?

For the actual algorithm Georgi says, without proof,

To determine whether a given $SU(2)$ rep, $\alpha$, appears in the decomposition we take the tensor product of $\alpha$ with the $n-j$ boxes.

I need some help with this. Firstly does this mean writing the $SU(2)$ rep as a Young Tableau and taking the $SU(3)$ tensor product with $n-j$ boxes in a row? And what value of $j$ do we choose?

Now in the examples (12.6 onwards) I don't understand the notation. In (12.6) we are looking for how the $6$ (two boxes in a row) decomposes. What is the notation below? Is it $\left( SU(2) \textrm{ irrep } \, \, \, SU(3) \textrm{ irrep } \right)$ where the $SU(3)$ irrep is row of some number $n-j$ of boxes for different $j$? In that case how was $n$ chosen?


1 Answer 1


OP asks many questions, so we will be somewhat sketchy.

  1. Actually the Lie group $$G~:=~SU(2) \times U(1)$$ is not a subgroup of $SU(3)$ but $G/\mathbb{Z}_2$ is. This is because the group homomorphism $$G~\ni~ (g,\alpha)~~\stackrel{\Phi}{\mapsto}~~ \begin{pmatrix} \alpha g & \mathbb{0}_{2\times 1} \cr \mathbb{0}_{1\times 2} & \alpha^{-2}\end{pmatrix}_{3\times 3}~\in~SU(3)$$ has kernel $${\rm Ker}(\Phi)~=~\{\pm (\mathbb{1}_{2\times 2},1)\}~\cong~\mathbb{Z}_2.$$

  2. Here we will argue at the level of Lie algebras $$su(2) \oplus u(1)\subseteq su(3).$$ In detail, we identify $$su(3)~\cong~ {\rm span}_{\mathbb{R}}(\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6,\lambda_7,\lambda_8)$$ with Hermitian traceless $3\times 3$ matrices; the isospin $$su(2)~\cong~ {\rm span}_{\mathbb{R}}(\lambda_1,\lambda_2,\lambda_3)$$ with Hermitian traceless $2\times 2$ block matrices in rows/columns 1,2; while the hypercharge $$u(1)~\cong~ {\rm span}_{\mathbb{R}}(\lambda_8)$$ is generated by the diagonal traceless matrix ${\rm diag}(1,1,-2)$ times an real number. (Here $\lambda_a$ denotes Gell-mann matrices). In other words, up to normalization of the hyperchange, the fundamental representation decomposes as $${\bf 3}~\to~{\bf 2}_{1/3}\oplus {\bf 1}_{-2/3}.$$

  3. Concerning OP's questions on arbitrary $su(3)$ representations, for intuition purposes, the tensor picture is perhaps helpful: Each tensor index (corresponding to a box) takes values $1,2,3$. The index values $1,2$ correspond to an $su(2)$ dublet, while the index value $3$ corresponds to an $su(2)$ singlet. This leads to a distributive property, where a box (i.e. a $su(3)$ triplet) can turn into an $su(2)$ dublet or an $su(2)$ singlet. This is what Georgi tries to indicate in eqs. (12.16-19). As OP already anticipates, the $j$th $su(2)$ irrep can be realized as a symmetric tensor product $({\bf 2j+1})\cong {\bf 2}^{\odot 2j}$ of the $su(2)$ dublet ${\bf 2}$ alone.

  4. Concerning the generalization to $SU(n) \times SU(m) \times U(1)$ in Georgi's section 13.5, see e.g. my Phys.SE answer here.

  • 1
    $\begingroup$ given the OP asks about the more general case, it would be nice to expand a bit on the cosetting by $\mathbb{Z}_2$ given the literature (see for instance Hagen, C. R., and A. J. Macfarlane. "Reduction of representations of SUm+ n with respect to the subgroup SUm⊗ SUn." Journal of Mathematical Physics 6.9 (1965): 1366-1371. ) tends to omit this subtle point. $\endgroup$ Jan 7, 2020 at 16:49
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jan 8, 2020 at 16:00
  • $\begingroup$ Very helpful - what about the quoted part from georgi... The algorithm is given without proof so even though now I understand it I'd like to know how the combinatorics works out correctly $\endgroup$
    – nox
    Jan 11, 2020 at 23:16

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