field theory - How can I tell that a Lagrangian has an $SU(2)times SU(2)$ symmetry?

this is a very basic question and it probably has a very simple answer.

I was reading through some handouts when I came over something that I did not understand. One considered the simple Lagrangian

$$\mathscr{L}=\partial_\mu\phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi - \lambda\left(\phi^\dagger\phi\right)^2$$ where $\phi$ is a complex doublet. The author then said that the symmetry of this Lagrangian is $SU(2)\times SU(2)$. However, I thought it was just $SU(2)$ or $U(1)\times U(1)$? I then googled this and found that the Lorentz group is isomorphic to $SU(2) \times SU(2)$, which I guess is one explanation. However, I was wondering if one could show that the Lagrangian is invariant under $SU(2) \times SU(2)$ by acting with some transformation, without taking the shortcut with Lorentz invariance?

Any help would be greatly appreciated (:

Answer

Answer of this question is quite subtle. First let us consider the most general Higgs potential which is renormalizable and invariant under $SU(2)_{L}\otimes U(1)_{Y}$ gauge transformations, which has the form \begin{equation} V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2} \end{equation} Where \begin{equation} \phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+i\phi_{2} \\ \phi_{3}+i\phi_{4} \end{pmatrix} \end{equation} In terms of $\phi_{i}(i=1,2,3,4)$, Higgs potential can be expressed as \begin{equation} V = \frac{\lambda}{4}(\phi^{2}_{1}+\phi^{2}_{1}+\phi^{2}_{3}+\phi^{2}_{4}-2\mu^{2})^{2} \end{equation} If we define \begin{equation}\Phi=\begin{pmatrix} \phi_{1}\\ \phi_{2}\\ \phi_{3}\\ \phi_{4}\\ \end{pmatrix}\end{equation} Then, Higgs potential is invariant under rotations of the four fields which lead to $SO(4)$ as the global symmetry group. This group is isomorphic to $SU(2)\otimes SU(2)$, because both have the same Lie algebra. This symmetry is global and it does not necessary to introduce gauge fields.

When the symmetry is broken, the scalar field $\phi_{4}$ get a v.e.v different from zero, and it can be redefined as follows $\phi_{4}=H+v$. where H gets its mass and is called the Higgs particle. Moreover, it has a v.e.v equal to zero. This field is a physical degree of freedom and its mass is proportional to the $λ$ parameter which is unknown in the model. The other scalar fields remain massless. They are the would-be Goldstone bosons and correspond to the degrees that the gauge fields ‘eat‘ in order to get mass or longitudinal component. The Higgs potential can be written as a function of the new fields as follows \begin{equation} V = \frac{\lambda}{4}(\phi^{2}_{1}+\phi^{2}_{1}+\phi^{2}_{3}+H^{2}+2Hv+v^{2}-2\mu^{2})^{2} \end{equation} In this new potential the global symmetry is broken to $SO(3)$, which only rotates three scalar fields. It is isomorphic to $SU(2)_V$, the diagonal part of $SU(2)\otimes SU(2)$. Which is also known as Custodial symmetry.