# Georgi - decomposition of representations into subgroups

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?

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.

• 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. – ZeroTheHero Jan 7 at 16:49
• I updated the answer. – Qmechanic Jan 8 at 16:00
• 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 – lux Jan 11 at 23:16