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In the typical treatment of electroweak theory, the gauge group is $G = \mathrm{SU}(2)_I \times \mathrm{U}(1)_Y$. This group is broken by the Higgs mechanism, while the combination of generators $Q = I_3 + Y$ generates an unbroken $\mathrm{U}(1)_Q$ subgroup.

I've wondered whether it is possible to represent the gauge group in a way that the embedding of $\mathrm{U}(1)_Q$ is obvious. The Lie bracket of $I_1$, $I_2$, $I_3$ and $Q$ seems to have the structure of a semidirect product.

Is it correct to say that the gauge group is a semidirect product $G \cong \mathrm{SU}(2)_I \rtimes \mathrm{U}(1)_Q$?

Qmechanic
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w123
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1 Answers1

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  1. Note that the group monomorphism $$U(1)_Q\quad \stackrel{i}{\hookrightarrow}\quad G~:=~ N\times U(1)_Y, \qquad N~:=~SU(2)_I,$$ depends on the Weinberg angle, cf. e.g. my Phys.SE answer here. Here the subscript $I=T$ denotes weak isospin, the subscript $Y$ denotes weak hypercharge, and the subscript $Q$ denotes electric charge. Let $i_1=\pi_1\circ i$ and $i_2=\pi_2\circ i$ denote the projection to 1st and 2nd group, respectively.

  2. We can use the embedding $i$ to define a semidirect product $N\rtimes U(1)_Q$ with elements $$\begin{align} N\rtimes U(1)_Q~\ni~(g,h)~:=~&(gi_1(h),i_2(h))~\in~N\times U(1)_Y,\cr (g_1,h_1)\bullet(g_2,h_2)~=~&\left(g_1i_1(h_1)g_2i_1(h_1^{-1}),h_1h_2\right).\end{align}$$ Returning to OP's title question, it is less clear whether this constitutes a bijection between Lie groups $$N\rtimes U(1)_Q\quad\to \quad N\times U(1)_Y~=:~G,$$ although they are certainly isomorphic at the Lie algebra level.

Qmechanic
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