 Global triplet scalar symmetry breaking

For a real scalar triplet with $V=\frac{1}{2}{m}^2{\Phi }^2+\frac{\lambda }{4}({\Phi }^2{)}^2$, $\lambda >0$ and $m^2<0$, the vacuum sphere has radius $v^2=-m^2/\lambda$. At ${\Phi }_0=v{e}_3$, the Hessian is $\mathrm{diag}(0,0,2\lambda v^2)$. Global $O(3)\to O(2)$ breaking therefore gives two physical massless Goldstone bosons and one radial scalar with squared mass $2\lambda v^2$. The angular modes cannot be gauged away in a theory with only global rotations.