Same $U(1)$ charge for the $SU(2)$ doublet

Question

Consider the electroweak gaugesymmetry $SU(2)_L\times U(1)_Y$. The entries of $SU(2)_L$ doublet will have same $U(1)$-charge. How can this be shown mathematically?

Answer 1

1. If a theory is declared to have a symmetry group $G$, it means more abstractly that the group $G$ acts on the constituents (fields etc.) according to some rules and the theory (Lagrangian etc) stays invariant under such transformations.

2. Often the constituents (fields etc.) form a (linear) representations $V$ of the group $G$. If the representation is (completely) reducible we can decompose it in irreps. The fundamental objects (fields etc) [that we consider] are for this reason often chosen to transform as irreps of the theory.

3. Now an irrep $V$ of a product group $G=G_1\times G_2$ is of the form of a tensor products $V\cong V_1 \otimes V_2$ of irrep $V_1$ and $V_2$ for the groups $G_1$ and $G_2$, respectively.

4. The irreps of the Abelian group $U(1)$ are all $1$-dimensional and labelled by an integer $n\in \mathbb{Z}$ called the charge.

5. So to return to OP's question, in the electroweak theory with group $G=SU(2)\times U(1)$, the field transform by definition as an irrep $V\cong V_1 \otimes V_2$ of $SU(2)\times U(1)$. In particular, the irrep $V$ carries a $U(1)$ charge, which (modulo various normalization conventions) is the weak hypercharge. To summarize: The main point is that the weak hypercharge is fixed by definition/construction.

6. Perhaps the following comment is helpful: If we are given a tensor product $V=V_1\otimes V_2$, where we assume that

   • (i) $V$ is a (completely) reducible representation of $SU(2)\times U(1)$,

   • (ii) $V_1$ is an irrep of $SU(2)$, and

   • (iii) $V_2$ is a $1$-dimensional representation of $U(1)$,

   then it follows that $V_2$ (and $V$) must be irreps as well. And hence $V_1$ carries a fixed weak hypercharge, cf. OP's title question.