= Solution
Let $r_1=\mu_1^2$, $r_2=\mu_2^2=\mu_3^2$, and $\rho^2=\phi_2^2+\phi_3^2$. With every $g_{ij}=g>0$, the uniform potential is
$$
V=\frac12r_1\phi_1^2+\frac12r_2\rho^2+g(\phi_1^2+\rho^2)^2.
$$
For $r_1>0,r_2>0$, the minimum is the origin and $\mathbb Z_2\times O(2)$ is unbroken. If $r_1<\min(0,r_2)$, then $\phi_1^2=-r_1/(4g)$ 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 $\phi_1=0$ 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.
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