Peierls contour probability bound

For $n\ge 1$, let $C_{\gamma }$ be the parity-inside primal sites from [F1]; that lemma gives $C_{\gamma }\subseteq B_n$ and identifies its boundary bonds with the edges crossed by $\gamma$. For $n=0$, $B_0=\{0\}$: the plus spin has no disagreement edges, and the minus spin has exactly the four dual edges forming the unit square centered at $0$. This is the only possible simple disagreement cycle; its ray-parity inside is $\{0\}$ and its crossed bonds are exactly the four bonds incident to $0$. Thus the same containment and boundary identification hold also for $n=0$. The map that flips every spin in $C_{\gamma }$ and leaves all other spins fixed is an involution on the finite configuration set, preserves the fixed plus exterior, and is injective.

Every bond crossing $\gamma$ has opposite spins on $A_{\gamma }$ and becomes aligned after the flip; its contribution to $H_{B_n}^{+}$ changes from $+J$ to $-J$, a decrease of $2J$. Every other bond has both endpoints inside or both outside $C_{\gamma }$, so its spin product is unchanged. Therefore $H_{B_n}^{+}(\sigma )=H_{B_n}^{+}({\mathrm{f}\mathrm{l}\mathrm{i}\mathrm{p}}_{C_{\gamma }}\sigma )+2J\mathrm{\mid }\gamma \mathrm{\mid }$ on $A_{\gamma }$.

By [F2], summing $e^{-\beta H(\sigma )}$ over $A_{\gamma }$ equals $e^{-2\beta J\mathrm{\mid }\gamma \mathrm{\mid }}$ times the weight sum over the image of the injective flip map. That image is a subset of all box configurations, so its total weight is at most the full partition function. Dividing by that positive partition function proves the bound.