= Solution
Let $k$ count parallel bonds. Then $N-k$ bonds are antiparallel and the <energy> is
$$
H=Jk-J(N-k)=-NJ+2Jk.
$$
The <periodic boundary conditions> impose
$$
\prod_{j=1}^N(S_jS_{j+1})=\prod_{j=1}^NS_j^2=1.
$$
Hence $N-k$ must be even. For odd $N$, $k$ is odd, so its minimum is one: one bond must be parallel. All other bonds alternate. Choosing that bond in $N$ ways and then choosing either value of $S_1$ determines every <Ising spin>, giving
$$
\boxed{E_0=-(N-2)J,\qquad g_0=2N.}
$$
This is an <odd antiferromagnetic Ising ring>: <geometrical frustration> forces one defect, whose location is unrestricted. The <ground-state degeneracy> includes both its possible locations and global <spin inversion symmetry>. The formula also includes the self-bond case $N=1$ under the stated <Hamiltonian> convention.
Back to article page
