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Positive Ising transfer matrix and stationary Markov law

Statement

Assume Countable Choice. For finite $β > 0$ and $J, h \in R$ , set $K = β J$ , $b = β h$ , and define the positive symmetric transfer matrix on ${- 1, + 1}$ by $T_{s, t} = e^{K s t + b (s + t) / 2} .$ Let $λ_{+}$ be its larger eigenvalue and let $r$ be its strictly positive unit eigenvector, with coordinates $r_{s}$ . Then $Q_{s, t} = \frac{T_{s, t} r_{t}}{λ_{+} r_{s}}, π_{s} = r_{s}^{2}$ define a stochastic matrix $Q$ and invariant probability vector $π$ . Their consistent consecutive-block laws $π_{s_{ℓ}} \prod_{i = ℓ}^{r - 1} Q_{s_{i}, s_{i + 1}}$ extend uniquely to a stationary probability measure on ${- 1, + 1}^{Z}$ .

Facts & Assumptions

Given: A one-dimensional finite-alphabet lattice and the zero-field-or-field nearest-neighbor Ising transfer matrix specified above, with $β > 0$ and finite real $J, h$ .

[A1]

Adopt Countable Choice for the consistent-cylinder extension step.

[F1]

The finite-alphabet product-measure lemma extends a consistent family of finite-dimensional probabilities uniquely to a probability on the cylinder sigma-algebra, under Countable Choice (Finite-alphabet product and probability compactness).

[F2]

The one-dimensional Ising transfer matrix has eigenvalues $λ_{\pm} = e^{K} \cosh b \pm \sqrt{e^{2 K} \sinh^{2} b + e^{- 2 K}}$ with $λ_{+} > ∣ λ_{-} ∣$ (One-dimensional periodic Ising-chain limit).

Proof

1.1A1F2algebra

The off-diagonal entries of $T$ are positive. For a positive symmetric $2 \times 2$ matrix, its larger eigenvalue exceeds both diagonal entries; explicitly, if $T = (\begin{array}{cc} a & c \\ c & d \end{array})$ with $c > 0$ , then $λ_{+} - a > 0$ and the vector $(c, λ_{+} - a)$ is strictly positive and satisfies $T r = λ_{+} r$ by the characteristic equation. Normalize it to length one.

2.1step 1.1algebra

Since $T r = λ_{+} r$ , $\sum_{t} Q_{s, t} = 1$ for each $s$ . Symmetry of $T$ gives $\sum_{s} π_{s} Q_{s, t} = r_{t} \sum_{s} T_{t, s} r_{s} / λ_{+} = r_{t}^{2} = π_{t}$ . Thus $π$ is invariant.

3.1step 2.1algebra

The finite block probabilities in the Statement are normalized. Summing over the right endpoint uses the row sums of $Q$ ; summing over the left endpoint uses invariance of $π$ . Summing over omitted sites therefore gives a consistent probability family on all finite coordinate sets.

4.1A1F1step 3.1∎

Apply [F1] to that consistent family. The resulting measure is stationary because its cylinder probabilities are shift invariant, and uniqueness on cylinders makes it the unique measure with the stated block laws.

Positive Ising transfer matrix and stationary Markov law

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources