7
$\begingroup$

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} \end{equation}

I wonder ─ how is this decomposition derived?

$\endgroup$
7
$\begingroup$
  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 & 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 $$\underline{\bf 5}~~\cong~~(\underline{\bf 3}\otimes\underline{\bf 1})_{-\frac{1}{3}}~~\oplus~~ (\underline{\bf 1}\otimes\underline{\bf 2})_{\frac{1}{2}}.$$

$\endgroup$

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.