= Solution
For the <XXZ spin chain>, write $\Delta=1+\alpha$ and replace each <spin> by a classical <vector> of length $S$. A bond is $JS^2\mathbf n_m^{\mathsf T}\operatorname{diag}(1,1,\Delta)\mathbf n_{m+1}$, whose minimum is $-JS^2\max(1,|\Delta|)$. For an even periodic chain all bonds can attain that bound simultaneously. The <classical ground states of an XXZ spin chain> therefore give \b[$\alpha>0$ gives the two easy-axis Néel states] $\mathbf S_m=(-1)^m S\widehat{\mathbf z}$ and its reversal, with energy $-JNS^2(1+\alpha)$.
For \b[$-2<\alpha<0$ the <ground states> are planar Néel configurations], $\mathbf S_m=(-1)^m S(\cos\varphi,\sin\varphi,0)$, with arbitrary $\varphi$: their degeneracy is a circle and their energy is $-JNS^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha<-2$, $\Delta<-1$ and the lowest states are instead \b[the two uniform $z$-polarized ferromagnetic configurations], with energy $JN\Delta S^2$. At $\alpha=-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic <ferromagnet>: $\mathbf S_m=S((-1)^m n_x,(-1)^m n_y,n_z)$ for any <unit vector> $\mathbf n$ is a <ground state>. This special degeneracy is a sphere. At the other boundary, $\alpha=0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum <ground state>.
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