Given
$ \Phi=\left(\begin{array}{c} \phi_1\\ \phi_2\\ \phi_3\\ \phi_4 \end{array}\right) $
where $\Phi$ is real, I have the following Lagrangian density:
$ \mathcal{L}=\frac{1}{2}(\partial_\mu \Phi)^T\partial^\mu \Phi -\frac{1}{2} \Phi^T M^2 \Phi-\frac{\lambda}{2m^2}(\Phi^TM^2 \Phi)^2 $
where
$ M^2 =m^2\left(\begin{array}{cccc} 1 & 0 & 0& 0\\ 0 & 1 & 0& 0\\ 0 & 0 & -1& 0\\ 0 & 0 & 0& -1 \end{array}\right) $
I think that the fields have a symmetry $SO(2)\times SO(2)$, since I can rotate indipendently the first two fields or the second two fields. However then I expect a symmetry breaking only for the second two fields because they have the minus sign on the mass term. Is it this correct? Is there an easy way to find the minima of the potential?