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I know that $SU(2)\times U(1)$ is the product of each element of $SU(2)$ by an element of $U(1)$. But looking at the Lagrangian, such a product exists only in covariant derivatives of Higgs or fermion fields. It turns out that the gauge fields $SU(2)$ and $U(1)$ exist on their own without interacting directly? Or am I misunderstanding something?

ACuriousMind
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Peter
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1 Answers1

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The statement "The symmetry group is $G\times H$" really just means that both $G$ and $H$ are symmetry groups of the theory and these two symmetries do not have anything to do with each other, i.e. they commute. This is because the natural inclusions $G\mapsto G\times \{1\}$ and $H\mapsto \{1\}\times H$ into $G\times H$ commute. Applying some element $(g,h)\in G\times H$ just means "apply $g$, then $h$, or $h$, then $g$, it doesn't matter".

Each object transforms separately under $G$ and $H$ in representations $V_G$ and $V_H$. This is equivalent to saying it transforms in the tensor representation $V_G\otimes V_H$ of $G\times H$.

So instead of enumerating all the independent symmetry groups, one just says that their direct product is "the" symmetry group. It's just language.

However, pay attention to a subtlety in the case of the electroweak interaction: While there are independent $\mathrm{SU}(2)$ and $\mathrm{U}(1)$ in the unbroken electroweak phase, the $\mathrm{U}(1)$ there is not the $\mathrm{U}(1)$ of electric charge, but the $\mathrm{U}(1)$ of weak hypercharge. The electromagnetic $\mathrm{U}(1)$ sits in the electroweak $\mathrm{SU}(2)\times\mathrm{U}(1)$ via $\phi\mapsto (\phi T_3, \frac{\phi}{2})$ (the 2 might be absent in some scaling conventions), where $T_3$ is the third generator of the isospin $\mathrm{SU}(2)$.

ACuriousMind
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