 Solution

The constant term in the spin sum cancels. Linearizing the saddle point equation givesThe paramagnetic auxiliary-field curvature isConsequently the Stoner criterion for the loss of local paramagnetic stability isThe printed $\nu (0)$ uses energy measured from the Fermi level. For a smooth density of states, $-n_F^{\prime }$ becomes a narrow unit-area peak at $\mu$; a singular density of states or appreciable temperature requires the integral form. The curvature establishes local instability, not the order or location of every possible thermodynamic transition.

With exact spin-rotation symmetry, itinerant ferromagnetism selects a magnetization direction. The two broken transverse generators form one quadratic type-B Goldstone boson, a long-wave ferromagnetic magnon with $E_q\sim D{q}^2$. Longitudinal amplitude motion is not symmetry-protected to be gapless. In a metallic ordered state, particle-hole excitations remain near each surviving exchange-split Fermi surface; these are distinct from the collective magnon. The itinerant spin-flip particle-hole continuum starts at exchange scale $2|M|$ at $q=0$ in this simple mean-field spectrum, while its finite-$q$ boundaries differ. Spin anisotropy, spin-orbit coupling or an external field can gap the transverse mode; none is present in the stated model.