= Solution
The negative sign of the exchange term and $J>0$ make this a <Heisenberg ferromagnet>. Each classical bond minimizes its energy when its two fixed-length spins are parallel. On the connected <cubic lattice> all ground states are therefore uniform:
$$
\boxed{\mathbf S_j=S\mathbf n,\qquad |\mathbf n|=1,}
$$
using length $S$ for the classical large-spin convention. Any two orientations are related by a common spatial rotation of every spin. The global spin-rotation symmetry is $SO(3)$, and a chosen direction retains its $SO(2)$ subgroup of rotations about that axis.
<Spontaneous symmetry breaking> means selecting one member of this continuous set in the <thermodynamic limit>, for example by taking volume to infinity before sending an aligning field to zero. The zero-field <Hamiltonian operator> remains rotationally invariant even though the selected state is not. There are gapless transverse <spin waves>. Although two rotation generators are broken, an isotropic ferromagnet has one quadratic <magnon> branch, whose two transverse spin components form a conjugate dynamical pair. It is not necessary to invent two independent linear-dispersion branches from the number of broken generators. The long-wavelength dispersion is derived in <2/d>.
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