= Solution
Use <bosonic Bogoliubov diagonalization> with $A_k=2JS$, $B_k=2JS\gamma_k$. Away from the zero modes, let
$$
a_k=u_k\alpha_k+v_k\alpha_{-k}^\dagger,\qquad
u_k=\sqrt{\frac{A_k/\omega_k+1}{2}},\qquad
v_k=-\operatorname{sgn}(B_k)\sqrt{\frac{A_k/\omega_k-1}{2}},\qquad
\omega_k=\sqrt{A_k^2-B_k^2}.
$$
Here $u_k^2-v_k^2=1$ preserves the <canonical commutation relations>, while the sign of $v_k$ cancels anomalous pairing. In hyperbolic notation $u_k=\cosh\theta_k$, $v_k=\sinh\theta_k$, this condition is $\tanh(2\theta_k)=-B_k/A_k$. Combining the <normal ordering> constants gives
$$
\boxed{H=-NJS(S+1)+\sum_k\omega_k(\alpha_k^\dagger\alpha_k+1/2)+O(S^0),\qquad\omega_k=2JS|\sin k|.}
$$
The dispersion vanishes linearly near $k=0$ and $\pi$, with <spin-wave velocity> $2JS$ in unit lattice spacing. The exact zero modes make the Bogoliubov coefficients singular; use a small infrared regulator or a symmetry-selected reference and treat the global rotations separately. A finite transformation at $\omega_k=0$ is not asserted.
On a <hypercubic lattice> of coordination $z=2d$, the same <hypercubic antiferromagnetic spin-wave dispersion> uses
$$
\gamma_{\mathbf k}=\frac1d\sum_{j=1}^d\cos k_j,\qquad
\boxed{\omega_{\mathbf k}=2dJS\sqrt{1-\gamma_{\mathbf k}^2}.}
$$
The corresponding constant is $-dNJS(S+1)$, and $\omega_{\mathbf k}\sim2JS\sqrt d\,|\mathbf k|$ near a Goldstone point. Restoring lattice spacing $a$ multiplies this velocity by $a$.
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