 Solution

For ${\mu }^2<0$ and $g>0$, minimizing the $O(N)$-invariant potential givesThe free energy has the full $O(N)$ symmetry, but choosing one point on this sphere leaves only the rotations $O(N-1)$ fixing that point. This is spontaneous symmetry breaking, $O(N)\to O(N-1)$. The $N-1$ tangent directions along the sphere cost no potential energy and are the massless Goldstone modes predicted by the Goldstone theorem.