Math-2120 · mathematics · Dynamical systems

Math-2120

Markov random fields and Bernoulli shifts

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Statement

Definitions and mathematical statement

Let $A$ be a finite alphabet and let $X\subseteq A^{\mathbb Z^d}$ be a closed shift-invariant space. A probability measure $\mu$ on $X$ is a Markov random field if, for each finite set $V\subset\mathbb Z^d$, the conditional distribution of the coordinates in $V$, given all coordinates outside $V$, depends only on the coordinates in the nearest-neighbor boundary of $V$. Fix a consistent collection $\gamma$ of these conditional probability kernels on admissible boundary configurations.

Assume $\mu$ is translation invariant and is the unique probability measure satisfying the kernels $\gamma$. Uniqueness is among all such measures on $X$, including those that are not translation invariant. The question is whether $(X,\mu,\sigma)$ must be measurably isomorphic to a Bernoulli $\mathbb Z^d$ shift.

Explicitly, a Bernoulli shift is an independent product process $(B^{\mathbb Z^d},\nu^{\mathbb Z^d},\sigma)$ for a probability space $(B,\nu)$. An isomorphism is a measurable bijection between invariant full-measure subsets, preserving probabilities and commuting with every coordinate shift, with measurable inverse.

The statement asks for an isomorphism of the entire lattice action, not just of a single generating transformation. The uniqueness hypothesis is stronger than uniqueness within the translation-invariant class and cannot be weakened silently. Conversely no uniform mixing rate or exponential decay of conditional influences is assumed; imposing such an additional condition would address only a special case. The specified kernels make precise which competing measures are compared.

Short English statement

Does uniqueness of a Markov random field for its local conditional laws force the resulting lattice action to be Bernoulli?

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