# Representation in $SU(2)_{L}\times U(1)_{Y}$

I'm reading the book "Massive neutrinos in physics and astrophysics" by R. N. Mohapatra and P. B. Pal, and have a question about its notation. The Equation 2.10, it says: $$q_{L}=\binom{u_{L}}{d_{L}} : (2,\frac{1}{3})\\ u_{R}:(1,\frac{4}{3})\\ d_{R}:(1,-\frac{2}{3})\\ \psi_{L}=\binom{\nu_{eL}}{e_{L}} : (2,-1)\\ e_{R}:(1,-2)\\ \tag{2.10}$$ I don't understand where the terms $(2,\frac{1}{3}),(1,\frac{4}{3}),(1,-\frac{2}{3}),(2,-1),(1,-2)$ come from and what they represent. Also, it would be nice if anyone can tell me why left handed particles form doublets and right handed particles are singlets.

## 1 Answer

To answer first to your last question, it is an experimental fact that only left-handed fermions are affected by weak interactions. The couplings of the fermions to the $W$ bosons is proportional to their weak isospin $T_3$. Therefore, right-handed fermions must have $T_3=0$. Left-handed fermions are observed to have the same isospin (charged currents universality), and charged currents only connect two types of fermions. The conclusion is that left-handed fermions must come in doublets of weak isospin, with $T= \frac{1}{2}$, $T_3 = \pm \frac{1}{2}$.

Now, the representations of the $SU(2)_L$ group are labelled by the dimension of the representation, which is given by $2 T+1$. This is the reason that the representations of left-handed fermions are of the form $(2, Y)$ and those of right-handed fermions are $(1, Y)$.

The group $U(1)_Y$ is analogous to the group of electromagnetism $U(1)_{em}$. Its representations are labelled by the weak hypercharge $Y$. So, what determines the value of $Y$ for each fermion?

• After the electroweak spontaneous symmetry breaking, only the subgroup $U(1)_{em} \subset SU(2)_L\times U(1)_Y$ remains unbroken. The generator for this group is the electric charge $$Q = T_3 + \frac{Y}{2}\,,$$ so working backwards, you can determine the value of $Y$ if you know the values of $Q$ and $T_3$.
• Hypercharge assignments aren't as arbitrary as it seems. To prevent chiral anomalies caused by triangle diagrams (at one loop) that would render the electroweak sector non-renormalizable, the choice of hypercharge is determined up to a twofold ambiguity.