Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 37 / 5 / b

The law is a ferromagnetic Ising model with unit couplings and no field. Its masses are strictly positive. Put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, so $\pi (\sigma )=Z^{-1}{e}^{H(\sigma )}$. The order is coordinatewise with $-1<1$, which identifies the configuration space with a Boolean lattice.

Use the permitted reduction to pairs that disagree at no more than two graph vertices. If such a pair is comparable, its join and meet are the two original configurations, so the lattice inequality is equality. In the incomparable case, let the two differing graph vertices be $x,y$, with one configuration taking $(1,-1)$ there and the other $(-1,1)$. Their join takes $(1,1)$ and their meet $(-1,-1)$.

Every edge not joining $x$ to $y$ cancels ina fixed edge has the same contribution throughout, while an edge from a changed graph vertex to a fixed graph vertex has the same sum of its two endpoint-spin products before and after. An edge between $x$ and $y$ contributes $1+1-(-1)-(-1)=4$. Thus the difference is nonnegative, whether or not the two changed graph vertices are adjacent. Exponentiation, with the normalizing constants cancelling, proves the FKG lattice condition for these pairs, and hence for all pairs by the given reduction.

Equivalently one can see the full ferromagnetic Ising lattice inequality directly. Let $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. Exactly the edges between $U$ and $W$ contribute, each by four. ThereforeThe preceding part now givesso the measure is positively associated. The sign of the coupling is essential to this argument: positive couplings favor agreement and give the nonnegative lattice difference.