Two-domain-wall degeneracy of a Potts ring ($d_1=Q(Q-1)\left(\genfrac{}{}{0ex}{}{L}{2}\right)$)

For $Q,L\ge 2$, a periodic Potts chain has $Q$ uniform ground states with energy $-JL$. A single domain wall cannot close around the ring. Its first excited states have two domain walls and energy $-JL+2J$. Choose the wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways and assign distinct states to the two resulting arcs in $Q(Q-1)$ ways. This proves the displayed degeneracy factor without identifying globally relabelled configurations.