If the gauge symmetry is not broken by spontaneous symmetry breaking, what symmetry is broken?

In this post, the answer by buzhidao showed that the $$U(1)$$ gauge symmetry is not broken by spontanous symmetry broken and Higgs mechanism.

What role does "spontaneously symmetry breaking" played in the "Higgs Mechanism"?

If we generalize this to electro-weak theory in SM, it seems like that before and after SSB and HM, the symmetry of the gauge lagrangian is always the $$SU(2)_{L}\times U(1)_Y$$.

But peole also say that $$SU(2)_{L}\times U(1)_Y$$ is broken to $$U(1)_{em}$$. What does $$U(1)_{em}$$ refer to if $$SU(2)_{L}\times U(1)_Y$$ is still a symmetry of the Lagrangian?

So what is the particular breaking symmetry people are talking about when they are talking SSB? I have the follwing answers I came up with, not sure which one is the one.

1. SSB is the process where the global symmetry being $$SU(2)_{L}\times U(1)_Y$$ of the total lagrangian is broken by non-zero VEV of Higgs field. So when people talk about SSB, there are essentially talking about this process.So, $$U(1)_{em}$$ is the global symmetry of the lagrangian.

2. SSB has only to do with the higgs. To be precise, when people talk about SSB, they are only talking about the global symmetry of the lagrangian of Higgs is broken by non-zero VEV. If so, what is $$U(1)_{em}$$ here?

3. SSB is the process where the symmetry of the groud state of the higgs is broken by non-zero VEV. That is, the ground state of higss does not has the same symmetry as that of the Lagrangian. So, $$U(1)_{em}$$ is the global symmetry of the ground state of the higgs.

I am also very confused when people talk about gauge symmetry breaking, which is not in any case.

So I guess in this the case of "breaking the gauge symmetry ", people are talking about choosing a perticular gauge such that the redundant phase is "rotated away"?

PS: I know how global $$U(1)$$ symmetry breaking works as it's the case in the answer in the link. I am not sure how global $$SU(2)_L\times U(1)_Y$$ is broken in the lagrangian(or is it?). But I know how the ground state is broken into $$U(1)_{em}$$by VEV (This is what confuses me, is $$SU(2)_L\times U(1)_Y$$ the symmetry of the ground state also?) The reason I am confused is because sometimes people talking about the symmetry of the lagrangian, however sometimes people talk about the symmetry of the ground states. Which one is exactly broken, and which one is not?

• This sounds a bit confused. Before doing anything else, do you understand how global symmetry breaking works? Suppose $SU(2)_L \times U(1)_Y$ was not gauged. Do you know the answer to your question in that case? – knzhou May 5 at 15:35
• @knzhou Thank you for your reply! I know how global $U(1)$ symmetry breaking works as it's the case in the answer in the link. I am not sure how global $SU(2)_L\times U(1)_Y$ is broken in the lagrangian(or is it?). But I know how it's broken into $U(1)_{em}$by VEV. The reason I am confused is because sometimes people talking about the symmetry of the lagrangian, however sometimes people talk about the symmetry of the ground states. – Universe Maintainer May 5 at 18:44