# Custodial symmetry of the standard model symmetry group $SU(2)_L \times SU(2)_R$

I am studying the standard model including the Higgs sector and electroweak interactions. Here, all of my terms have their usual meanings. Therefore my symmetry group is $$SU(2)_L \times SU(2)_R \times U(1)_Y$$. Here I have included $$U(1)_Y$$ within the larger group of $$SU(2)_L \times SU(2)_R \approx O(4)$$ as I want to describe electroweak interactions. Spontaneous symmetry breaking (SSB) and the Higgs mechanism reduce this group to $$SU(2)_C \times U(1)_{EM}$$. What I am finding confusing is understanding what happens to all of these generators when SSB occurs? In my understanding:

• $$SU(2)_L$$ has 3 generators, all of which are broken
• $$SU(2)_R$$ has 3 generators, but only $$T^3_R$$ is broken. The R subscript is just to note that this is a generator for the right-handed fermions
• $$U(1)_Y$$ has 1 generator which remains unbroken and is the gauge photon

In my understanding, the broken generators from $$SU(2)_L \times SU(2)_R$$ combine in some way to give us the three gauge bosons for weak interactions: $$W^+, W^-, Z^0$$, and the remaining two unbroken generators of $$SU(2)_R$$ are 'eaten' by the gauge bosons. So what are the generators for the remaining $$SU(2)_C \times U(1)_{EM}$$ symmetry group?

I understand the model when we just include $$SU(2)_L \times U(1)_Y$$ which results in just $$U(1)_{EM}$$ after SSB. What I am struggling to understand is the inclusion of the group $$SU(2)_R$$ and how all of this leads to the custodial symmetry? I also know that the different masses of the up and down quarks lead to an explicit custodial symmetry breaking, but it confuses me on how to incorporate this also.

• Review 1, 2, 3, 4, Commented Apr 9 at 17:34
• and 5. Commented Apr 9 at 17:34
• "the remaining two unbroken generators of $SU(2)_R$ are 'eaten' by the gauge bosons": who told you UNBROKEN generators can be eaten by the gauge bosons? Commented Apr 9 at 18:01
• "what are the generators for the remaining $SU(2)_C \times U(1)_{EM}$ symmetry group?": If by $SU(2)_C$ you mean the custodial symmetry, then $U(1)_{EM}$ is part of $SU(2)_C$, rather than in addition to $SU(2)_C$. Commented Apr 9 at 18:20

I have included 5 links, including a "primary" one, which should help clear up the trail map for you; I think you've gone off it, but I can't be sure, as weird misconceptions have crept in.

The big picture, which it behooves you to work out the explicit transformation laws for, for all fields, "before SSB" is:

1. The gauge & Yukawa sectors of the SM have an $$SU(2)_L \times U(1)_Y$$ gauge symmetry.

2. The non-Yukawa fermion & Higgs potential sectors, however, have a larger global symmetry, with two more "R", generators scrambling the four Higgs d.o.f., beyond the hypercharge $$U(1)_Y$$, namely $$SU(2)_L \times SU(2)_R\sim SO(4)$$; you may think of these two new $$SU(2)_R$$ generators, explicitly broken by the Yukawa sector, as "custodial", as they are an approximate symmetry of the model, i.e., explicitly broken by the capricious array of Yukawa coupling constants y. These will enter as spoilers in radiative corrections, then, (unless the Yukawa couplings and hence masses of the fermions were equal among themselves).

• "After" SSB, these symmetries persist as symmetries, but three are realized nonlinearly ("SSBroken") and their goldstons Higgs-eaten by three of the four gauge bosons, with the Weinberg angle twist that reconfigures $$T^3_L$$ and $$U(1)_Y$$ into the cockeyed neutral current Z and the surviving vector $$U(1)_{EM}$$. The two (global) custodial generators and the charge one are still around, not SSBroken, and scrambling the (pseudo)-goldstons in the belly of the Ws and the Z among themselves. But for Weak mixing (set $$\theta_W=0$$ to keep track of them), these goldstons transform as a triplet of the surviving (imperfect) custodial SO(3).

I have accounted for all 6 generators―I'm not sure where you got the 7th from.

Geeky

I also know that the different masses of the up and down quarks lead to an explicit custodial symmetry breaking, but it confuses me on how to incorporate this also.

The ideal $$SU(2)_L \times SU(2)_R\sim SO(4)$$ Yukawa sector would have only a common coupling y for both ups and downs, cavalierly in normalizations, $$y(\overline{u_L}, \overline{d_L}) (\tilde\Phi, \Phi) \begin{pmatrix} u_R \\ d_R \end{pmatrix}=y(\overline{u_L}, \overline{d_L}) \begin{pmatrix} \phi^{0~~*} &\phi^+ \\ -\phi^- & \phi^0 \end{pmatrix}\begin{pmatrix} u_R \\ d_R \end{pmatrix}.$$

The v.e.v. of the middle Higgs matrix is $$v{\mathbb I}$$, yielding equal masses, $$yv (\overline{u_L} u_R + \overline{d_L}d_R),$$ so you see that for different Yukawas you loose the full custodial symmetry, here (plain, strong) isospin symmetry.

• Thank you so much for your help! And for the links! Commented Apr 9 at 21:26