# 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.

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 $T_3$.