TITLE: What are the Generators of the electroweak interaction after symmetry breaking. (SM)
QUESTION [1 upvotes]: In the standard model (omitting the QCD part), we start off with the set of generators 
$T_1$, $T_2$, $T_3$, $Y$ 
for the four-parametric gauge group $SU(2)_L \times U(1)_Y$.
We then define a new generator $Q= T_3+Y$ and make the transition to the four-parametric gauge group $SU(2)_? \times U(1)_Q$.
What are, aside from $Q$, the new generators for this "new" gauge group?
$?$ , $?$ , $?$ , $Q$ 
Do we still use the $T$'s we used in $SU(2)_L$? That means the left factor in the group product is still the same as before the symmetry breaking?
My motivation for asking is the observation that in $SU(2)_L \times U(1)_Y$, the four generators are orthogonal and a basis for the space of all complex self-adjoint matrices.
The set of $T_1$, $T_2$, $T_3$, $Q$, while still a basis, is however not orthogonal, since
$( T_3| Q )$=$(T_3|T_3+Y)$=$(T_3|T_3) \neq 0$ 
It would seem that we would probably want to preserve that orthogonality property and thus not use $T_3$ as a generator after symmetry breaking.

REPLY [3 votes]: Well, after symmetry breaking, all that remains is electromagnetic $U(1)$, so the only generator that is truly a symmetry generator is $Q$.
The fermions couple to the "Higgs" via the Yukawa coupling:
$\mathcal{L}_y = -y_e^{ij} \bar L_{L,i} \Phi e_{R,j} - y_u^{ij} \bar Q_{L,i} \tilde{\Phi} u_{R,j} - y_d^{ij} \bar Q_{L,i} \Phi d_{R,j} + h.c.\,$
which mixes left and right handed fermions. Here $L$ is the left-handed doublet $(e_L, \nu_L)$, and $e_R$ is the right-handed singlet. Because both $L$ and $\Phi$ transform under $SU(2)_L$, there is a symmetry. After symmetry breaking, 
$\mathcal{L}_m = -\frac{y_e^{ij} v}{\sqrt{2}} \bar e_{L,i} e_{R,j} -\frac{y_u^{ij} v}{\sqrt{2}} \bar u_{L,i} u_{R,j} -\frac{y_d^{ij} v}{\sqrt{2}} \bar d_{L,i} d_{R,j} + h.c.$
where $v$ is the Higg's vev. This is not invariant under $SU(2)_L$.
The same thing happens with the gauge bosons that become massive, although there the interaction term comes from the covariant derivative acting on $\Phi$. 
Finally, the potential for $\Phi$, (the Mexican hat) is symmetric under SU(2), but the vacuum is not, because for the vacuum state, $\langle 0 | \Phi | 0\rangle = (0,v/\sqrt{2})$, which is not invariant.
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