Let me first say that I'm a layman who's trying to understand group theory and gauge theory, so excuse me if my question doesn't make sense.
Before symmetry breaking, the Electroweak force has 4 degrees of freedom ($B^0$, $W^1$, $W^2$, and $W^3$ right?) and after symmetry breaking, we are left with the weak $SU(2)$ bosons ($W^+$, $W^-$, $Z$) and the photon. Why, then, is the symmetry group before Electroweak symmetry breaking $SU(2) \times U(1)$ and not just $U(2)$ (Since a Unitary group of $n$ dimension contains $n^2$ degrees of freedom?) What am I missing?
Best Answer
Actually we have the following Lie algebra isomorphism
$$\tag{1}u(2)~\cong~ u(1)\oplus su(2),$$
and there exists the following Lie group isomorphism
$$\tag{2} U(2)~\cong~[U(1)\times SU(2)]/\mathbb{Z}_2 ,$$
cf. e.g. this Phys.SE post.
In other words, there is a two-to-one map from $U(1)\times SU(2)$ to $U(2)$.
So in that sense the Glashow-Salam-Weinberg (GSW) $U(1)\times SU(2)$ model already contains a $U(2)$ gauge group!
Of course, the various matter and gauge fields transform naturally in the weak hypercharge and weak isospin language, i.e., under $U(1)\times SU(2)$.
On the other hand, if we construct the GSW electroweak model with the smaller gauge group $U(2)$ [rather than the full $U(1)\times SU(2)$ gauge group], then a matter field transforming in a (hypercharge,isospin) irrep $(Y,I)$ may become multi-valued, depending on the fractional value of the hypercharge $Y$. This does not happen for the standard model field content, cf. this Phys.SE post. So in that sense the electroweak gauge group is $U(2)$ [via the Lie group isomorphism (2)].