# If $SU(2)_{L} \times U(1)_{Y}$ breaks to $U(1)_{em}$ when a non-zero mass for the Higgs boson is chosen, why do we still have weak interactions?

As I understand it, when we say that the $SU(2)_{L} \times U(1)_{Y}$ is broken via the Higgs mechanism, this is because the symmetry acts on the Higgs mass in a way that would change it's value. If we want to pick a particular model we need to pick a fixed value of the Higgs mass, and this is only possible if we say $SU(2)_{L} \times U(1)_{Y}$ is broken to $U(1)_{em}$. The Lagrangian is always invariant to $SU(2)_{L} \times U(1)_{Y}$ (even if the Higgs mass changes under $SU(2)_{L} \times U(1)_{Y}$ the Lagrangian is such that it remains invariant). The symmetry breaking is just neccesary in choosing a theory with one particular value of the Higgs vev.

Real life does appear to have a fixed Higgs mass so we require that $SU(2)_{L} \times U(1)_{Y}$ breaks to $U(1)_{em}$. But then aren't we saying that observable physics is described by $SU(3)_{C} \times U(1)_{em}$, why does the weak interaction still work in our universe that contains a fixed Higgs mass?

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