Why do we conclude $SU(2)\times U(1)$ is broken into $U(1)_{em}$ when local gauge symmetry is not broken?

Recently, I spent some time reading some articles on the concept of SSB and I found that after SSB the local gauge symmetry will not be broken.What are broken are the global symmetries by non-zero VEV of Higgs field. But this is a little bit confusing, since I remembered that $$SU(2)\times U(1)$$ was broken into $$U(1)_{em}$$ by SSB and Higgs mechanism. But in principle, the $$SU(2)\times U(1)$$ local gauge symmetry should also be presented after SSB and Higgs mechanism. How do we conclude that local $$SU(2)\times U(1)$$ is broken when it is not? Or are we also talking about the global symmetry instead of local symmety here?

Also, I am wondering why there are four componets for Higgs field in SM. Are they all necessary?

• I don't have an understanding of the spontaneous symmetry breaking but I wonder how can one not have global symmetry despite having local symmetry. If one has local symmetry, one can choose the phase to be independent of position and that would automatically mean there is also a global symmetry, right? – Feynmans Out for Grumpy Cat May 3 at 23:06
• @DvijMankad physics.stackexchange.com/questions/449496/… – Universe Maintainer May 3 at 23:17
• @DvijMankad under the global transformation, the gauge fields remain the same. But for local transformations, the gauge fields have to transform accordingly to compensate the effect. – Universe Maintainer May 3 at 23:19
• A propos of your separate, logically disjoint last paragraph question: one only uses a 4d.o.f. Higgs field for renormalizability. One mostly needs 3 Goldstone bosons for the Higgs mechanism, and if a "light/medium" Higgs particles had not been discovered, Higgsless models were around to explain the fact, essentially. – Cosmas Zachos May 4 at 0:32
• Near duplicate. – Cosmas Zachos May 4 at 0:34