Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 30 3 c Solution Created 2026-10-03 Updated 2026-10-14
With the inner boundary specified in the PDF, the Ising boundary condition pins . Explicitly is conditioned on those Ising spins. Its partition function sums over unpinned interior Ising spins only; edges between two pinned sites contribute a constant that cancels on normalization.
Pinning coordinates preserves the FKG lattice condition, since meet in a lattice and join in a lattice still have the same pinned values. It also gives monotonicity in the pins. For any increasing observable , positive association of random variables with an unpinned Ising spin givesBoth conditional probabilities are positive. The same inequality holds after other Ising spins have been pinned. Changing pins one at a time proves stochastic domination of the measures as increases.
Now let and take an increasing observable depending on a fixed set inside the smaller box. Given the Ising spins on , the larger-box measure induces precisely the smaller-box interior law with those boundary values: nearest-neighbor interactions cannot see farther once this boundary is fixed. Its conditional law is therefore between the all-zero and all-one smaller-box laws. Averaging yieldsThese are bounded monotone sequences along any increasing box exhaustion.
For every finite set , the event that all Ising spins in are one is increasing, so its probability converges in both boundary sequences. Probabilities of arbitrary finite zero/one patterns are finite inclusion-exclusion combinations of these all-one probabilities, and hence converge too. The limits are consistent finite-dimensional distributions; they define measures on and give weak convergence of probability measures in the product topology. Thus both extremal infinite-volume Ising measures exist, with below in stochastic domination. This argument proves existence even when the two limits are different.