$SU(2)$ to $U(1)$ symmetry breaking Why does electroweak theory have $SU(2) \times U(1)$ to $U(1)$ symmetry breaking. Is it not possible to have simply $SU(2)$ to $U(1)$ symmetry breaking?
 A: It is technically possible to have a theory where and $SU(2)$ symmetry spontaneously breaks to a $U(1)$ symmetry. The simplest example is when you have a field that transforms in the adjoint (spin-1) representation of $SU(2)$ and it gets a non-zero vacuum expectation value. If the $SU(2)$ symmetry were gauged this would leave you with one massless gauge boson and one charged massive gauge boson.
However, the Standard Model needs spontaneous gauge symmetry breaking for two reasons. The first is to give mass the to the $W$ boson, but the second is to give mass to the fermions. The weak gauge bosons treat left- and right-handed fermions differently, as scattering experiments teach us, but mass terms mix left- and right-handed fermions, so gauge symmetry forbids fermion mass terms. The solution is to have fermions couple to the Higgs field in a gauge-invariant way that then produces an effective mass term after spontaneous symmetry breaking.
The $SU(2)\rightarrow U(1)$ breaking pattern cannot do this without introducing additional fermions. Weak interaction phenomenology suggests that left-handed leptons and neutrinos transform in an $SU(2)$ doublet while right-handed fermions do not transform under $SU(2)$ at all. To have a gauge-invariant Yukawa term we then need the Higgs field to also transform as a doublet.
However, a doublet with a vacuum expectation value breaks $SU(2)$ to the trivial group (no remaining symmetry), leaving us without a massless boson to act as the photon.
The next simplest solution is to increase the symmetry to $SU(2)\times U(1)$. Now the Higgs doublet vacuum expectation value can break $SU(2)\times U(1)$ to $U(1)$ as long as the doublet's $U(1)$ value has the right value.
A: For the SU(2) gauge Group you have only 3 Generators of a gauge field. This is because from the 2*2 =4 Matrix entries you remove one due to the condition that the determinant of the Group must be equal to 1. In SU(2) group are encoded the force carriers of the weak interaction; These are $W_+,W_-$ (W-bosons in positive and negative charge) and $Z$ (Z-bosons).
In the Group $SU(2) \times U(1)$ you have also a $U(1)$ symmetry with a single Group generator and therefore you have 3+1=4 gauge fields for this group.
When symmetry is broken, you will get 3 massive weak bosons and the photon.
A: There is the Pati-Salam model that has an electroweak field $SU(2)_L\times SU(2)_R$. With the strong nuclear interactions it is $SU(4)\times SU(2)_L\times SU(2)_R$. There are some problems with the idea of an $SU(4)$ strong nuclear interaction theory. The recent results by the LLNL lattice group might breath some life into this. The $SU(4)$ has $4$ colors with $7$ additional gluons that change color with an additional color neutral gluon. Maybe the additional stuff is associated with dark matter. The symmetry breaking of this theory involves the decomposition of the $SU(2)_R~\rightarrow~U(1)$ that recovers the standard electroweak field theory. The $\sigma_{\pm}$ elements from this broken $SU(2)_R$, which in the $SU(2)_L$ go on to define the $W^{\pm}$, might have some additional physics. These might go into the generation of additional photons or combine to form a sort of $Z_\gamma$. Whether this is the case or bit will ultimately come from experimental data.
