If $ϕ^† ϕ$ is invariant under $SU(2) \times U(1)$, a Phys.SE question I recently posted, then does that mean $ϕ^† ϕ$ is invariant under rotations and translations?

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    $\begingroup$ I guess by $\phi$ you are talking about a collection of scalar fields (in some representation of the gauge group $SU(2)\times U(1)$). If so, then the word scalar means that $\phi$ transforms in the trivial representation of the space-time symmetries (such as rotations and translations). In other words, $\phi$ is invariant by definition. $\endgroup$ – Heidar Jul 19 '13 at 14:18
  • $\begingroup$ @Qmechanic: Wow, you're fast $\endgroup$ – curiousGeorge119 Jul 19 '13 at 14:21
  • $\begingroup$ @Heidar: Thanks! So this is true only for scalars, is that correct? $\endgroup$ – curiousGeorge119 Jul 19 '13 at 14:24
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    $\begingroup$ Yes that's correct. In order to write down Lagrangians we usually pick fields which transform in particular nice ways under space-time symmetries. The simplest are spin-$0$ fields, which are the scalars $\phi$ (which do not transform). Then there are spin-$\frac 12$ fields, the spinors $\psi_\alpha$. There are different kinds of these, where the Dirac spinor is the one people usually see in textbooks. Then there are spin-$1$ fields, called vectors $A_\mu$. The fact that spinors and vectors have space-time indices mean they transform under space-time transformations. $\endgroup$ – Heidar Jul 19 '13 at 14:33
  • $\begingroup$ @Heidar: Great explanation!!! Thank you very much! $\endgroup$ – curiousGeorge119 Jul 19 '13 at 19:46

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