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Jan 25, 2015 at 14:03 comment added ACuriousMind @silrfück: Yes. The $W^\pm$ and the $Z$ still act as if they were $\mathrm{SU}(2)$ bosons, though they are exactly the combinations of which you speak. I am quite sure they form a $\mathrm{SU}(2)$ subgroup of the electroweak group.
Jan 25, 2015 at 7:53 comment added Yossarian why do you say "we are breaking an $SU(2)\in{}SU(2)_L\times{}U(1)_Y$ completely". Isn't it broken all of $SU(2)_L\times{}U(1)_Y$ except a $U(1)_{em}$ which is a combination of generators of $SU(2)_L$ and $U(1)_Y$? Do the broken generators also form a $SU(2)$?
Dec 9, 2014 at 14:10 vote accept phy_math
Dec 8, 2014 at 14:11 history answered ACuriousMind CC BY-SA 3.0