 Solution

Let $r_1={\mu }_1^2$, $r_2={\mu }_2^2={\mu }_3^2$, and ${\rho }^2={\varphi }_2^2+{\varphi }_3^2$. With every $g_{ij}=g>0$, the uniform potential isFor $r_1>0,r_2>0$, the minimum is the origin and ${\mathbb{Z}}_2\times O(2)$ is unbroken. If $r_1<\min (0,r_2)$, then ${\varphi }_1^2=-r_1/(4g)$ and $\rho =0$; the ${\mathbb{Z}}_2$ factor is broken, $O(2)$ remains, and there is no Goldstone mode. If $r_2<\min (0,r_1)$, then ${\varphi }_1=0$ and ${\rho }^2=-r_2/(4g)$; the first ${\mathbb{Z}}_2$ remains while $O(2)$ is broken to the reflection fixing the chosen direction, producing one Goldstone mode.

The positive coordinate half-axes are continuous transition lines. For $r_1=r_2<0$, the potential has an enhanced $O(3)$ symmetry and a sphere of minima; $O(3)\to O(2)$ gives two Goldstone modes on this line. Crossing the negative diagonal exchanges the two ordered phases and gives a first-order line at mean-field level.