Symmetry breaking in cosmology

Question

In A. Vilenkin and E. P. S. Shellard's "Cosmic strings and other topological defects", section 1.1.2, they say the following.

"The basic premise of grand unification is that the known symmetries of the elementary particles resulted from a larger symmetry group G after a series of spontaneous symmetry breakings,

$G \to H \to SU(3) \times SU(2) \times U(1) \to SU(3) \times U(1)_{em}$"

Could someone explain please what occurs during the final transition, that is $SU(3) \times SU(2) \times U(1) \to SU(3) \times U(1)_{em}$. Specifically, what happens to $SU(2)$ and the weak interactions? Is this electroweak symmetry breaking?

Answer 1

Yes. Electroweak symmetry breaking proceeds as: $SU(2)_{I_3} \times U(1)_Y \rightarrow U(1)_{em}$, where $I_3$ is the z-component of the weak isospin and $Y$ is weak hypercharge. The electric charge, $q$, is related to these generators via $q = I_3 + Y/2$.

Before symmetry breaking, the generators of $SU(2)_{I_3} \times U(1)_Y$ are the three massless weak isospin fields $W_i$, $i = 1,2,3$ and weak hypercharge field, $B$. Electroweak symmetry breaking mixes these fields according to $\gamma = B\cos \theta_{W} + W_3 \sin \theta_{W} $ and $Z^0 = -B\sin \theta_W + W_3\cos \theta_W $ where the angle $\theta_W$ determines the mixing. Of course, the photon $\gamma$ remains massless while $Z^0$ picks up a mass; likewise for $W^{\pm}$ which are linear combinations of the $W_{1,2}$.