Constructing gauge symmetry for $SU(2)\times U(1)$

I'm currently reading the book "Classical Theory of Gauge Fields" by Rubakov, therefore I will use his convention in this question. In the following we assume that:

• $$\phi$$ comprises columns consisting of complex scalar fields.
• $$G$$ is a simple compact Lie Group.
• The representations $$\rho:G\rightarrow \mathrm{Gl}(V)$$ are unitary.
• $$\rho_*: \mathfrak{g} \rightarrow \mathfrak{gl}(V):A\mapsto \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon}\right|_{\epsilon=0}\rho\left(\exp[\epsilon A] \right)$$
are the corresponding Lie Algebra representations and therefore anti-hermitian.

Now, we have global gauge symmetry of the Lagrangian

$$\begin{equation} \mathcal{L} = \partial_{\mu}\phi^{\dagger}\partial^{\mu}\phi - m^2\phi^{\dagger}\phi-\lambda\left(\phi^{\dagger}\phi\right)^2 \end{equation}$$

under the action

$$\phi(x) \mapsto \phi'(x) = \rho(\omega)\phi(x)$$

We can generalize this global symmetry to a gauge symmetry by introducing the gauge field $$A_{\mu}:\mathbb{R}^4 \to \mathfrak{g}$$ transforming as

$$A_{\mu} \mapsto A'_{\mu}(x) = \omega(x)A_{\mu}(x)\omega^{-1}(x) + \omega(x)\partial_{\mu}\omega^{-1}(x),$$

where $$\omega:\mathbb{R}^4 \to G$$ and additionally introducing the covariant (In the sense of the gauge transformation) derivative

$$D_{\mu}\phi(x) := \left[\partial_{\mu} + \rho_*(A_{\mu}(x))\right] \phi(x) \Rightarrow \left( D_{\mu}\phi \right)' = \rho(\omega(x))\left( D_{\mu}\phi \right)$$.

Then, the invariant Lagrangian of the Field is given by

$$\mathcal{L}_{\phi} = \left( D_{\mu}\phi \right)^{\dagger} \left( D^{\mu}\phi \right) -m^2 \phi^{\dagger} \phi -\lambda\left(\phi^{\dagger} \phi\right)^2$$

So far so good. But what if we have the case of a non-simple compact Lie Group, say $$SU(2) \times U(1)$$. Let's assume we have two doublets $$\phi, \chi$$ under $$SU(2)$$ (fundamental representation) and $$\epsilon$$ a singlet under $$SU(2)$$ (trivial representation):

\begin{align} \phi &\mapsto \omega \phi,\\ \chi &\mapsto \omega \chi,\\ \epsilon &\mapsto \epsilon, \end{align}

where $$\omega \in SU(2)$$. The fields $$\phi, \chi$$ are two components complex columns, $$\epsilon$$ is a one component complex scalar field. Furthermore, we assume the fields to transform under $$U(1)$$ as

\begin{align} \phi & \mapsto \exp\left[ iq_{\phi} \alpha\right] \phi, \\ \chi & \mapsto \exp\left[ iq_{\chi} \alpha\right] \chi, \\ \epsilon & \mapsto \exp\left[ iq_{\epsilon} \alpha\right] \epsilon, \\ \end{align}

where $$q_{\phi}, q_{\chi}, q_{\epsilon} \in \mathbb{R}$$. The kinetic term in the Lagrangian has the standard form and the interaction can be chosen as

\begin{align} \lambda [ (\phi^{\dagger} \epsilon) \chi + \chi^{\dagger}(\epsilon^* \phi) ]. \end{align}

It is invariant under these global gauge transformations if $$q_{\chi}+ q_{\epsilon} - q_{\phi}=0$$.

Question: So, how would one approach the generalization to a gauge symmetry of this system as described above? I am really lost here, so every hint would be highly appreciated.

• I don't understand your question. Since your interaction term doesn't contain any derivatives, what problem do you see in just introducing a gauge field via the covariant derivative like in the first case? – ACuriousMind Mar 9 at 16:48
• Electroweak theory is gauged $SU(2)\times U(1)$ so that's an example of how it works. – MannyC Mar 9 at 17:28
• @ACuriousMind : My question is a purely technical question. I tried applying the above construction but I failed as I don't really understand how to apply it in this case. My problem is that I don't understand how to apply the construction to the fields $\phi$ and $\chi$. Here, we do not only have the fundamental representation of $SU(2)$ acting on them, but additionally the fundamental representation of $U(1)$. How do we account for this? – Muphys Mar 9 at 19:27
• @CosmasZachos : Of course you are totally right, thank you for pointing out this typo! – Muphys Mar 9 at 19:28
• Basically you need to introduce a gauge field for each of the two groups. In forming the covariant derivative these just add. – Bruce Greetham Mar 9 at 20:52