This question comes from Srednicki's textbook "Quantum Field Theory". On pages 514-515, it states:

Under the unbroken $\rm SU(3)\times SU(2) \times U(1)$ subgroup, the $5$ representation of $\rm SU(5)$ transforms as \begin{equation} 5 ~\rightarrow~ \left(3, 1, -\frac{1}{3}\right) \oplus \left(1, 2, +\frac{1}{2}\right) .\tag{84.12/97.2} \end{equation}

I wonder ─ how is this decomposition derived?


1 Answer 1

  1. Actually the Lie group $$G~:=~SU(3)\times SU(2) \times U(1)$$ is not a subgroup of $SU(5)$. However the standard model gauge group $G/\mathbb{Z}_6$ is a subgroup of the GUT gauge group $SU(5)$, cf. e.g. this & this Phys.SE posts.

  2. Here we will argue at the level of Lie algebras $$su(3)\oplus su(2) \oplus u(1)\subseteq su(5).$$ In detail, we identify $su(5)$ with anti-Hermitian traceless $5\times 5$ matrices; $su(3)$ with anti-Hermitian traceless $3\times 3$ block matrices in rows/columns 1,2,3; and $su(2)$ with the anti-Hermitian traceless $2\times 2$ block matrices in rows/columns 4,5; while $u(1)$ is generated by the diagonal traceless matrix ${\rm diag}(-2,-2,-2,3,3)$ times an imaginary number.

  3. The vectorspace $V_5=V_3\oplus V_2$ of the defining/fundamental representation $\underline{\bf 5}$ of $su(5)$ is decomposed into the defining/fundamental representation $\underline{\bf 3}$ of $su(3)$ in rows 1,2,3; and the defining/fundamental representation $\underline{\bf 2}$ of $su(2)$ in rows 4,5.

  4. On the other hand, the first three rows $V_3$ are a singlet under $su(2)$; while the last two rows $V_2$ are a singlet under $su(3)$.

  5. Also note that the generator of $u(1)$ has the same weak hypercharge/eigenvalue $-2i$ and $3i$ on $V_3$ and $V_2$, respectively. The overall normalization of the weak hypercharge depends on conventions/choice of the $u(1)$ generator.

  6. Altogether, the decomposition of the $su(5)$ representation becomes $$\underline{\bf 5}~~\cong~~(\underline{\bf 3}\otimes\underline{\bf 1})_{-\frac{1}{3}}~~\oplus~~ (\underline{\bf 1}\otimes\underline{\bf 2})_{\frac{1}{2}}.$$


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