Ergodicity of $\mu^0_\beta$ on a particular $\sigma$-algebra (Ising Model)

Question

Consider the Ising Model on $\mathbb{Z}^d$ with nearest neighbors interaction, free boundary condition,$h=0$,and $\beta>0$. I would like to prove that for all local functions $f$ and $g$ such that $f(-\sigma)=f(\sigma)$ and $g(-\sigma)=g(\sigma)$, we have the following identity $$ \lim_{\|z\|_1 \to \infty}\langle f \cdot (g \circ\theta_z) \rangle^{0}_{\beta} = \langle f \rangle^{0}_{\beta} \langle g \rangle^{0}_{\beta}, $$ where $\theta_z$ is the translation of $\mathbb{Z}^d$ by $z$ and $\langle \cdot \rangle^{0}_{\beta}$ denotes the expected value with respect to the measure $\mu^0_\beta$.

Answer 1

Let me provide one possible proof of this statement. Note that it may be overkill, but I cannot think of a simpler one at the moment.

First, any local function $f$ can be decomposed as $$ f = \sum_{A\subset\mathrm{supp}(f)} \hat f_A \sigma_A, $$ where $\sigma_A = \prod_{i\in A}\sigma_i$ and $\mathrm{supp}(f)$ is the support of the function $f$ (that is, the finite set of spins on which the value of $f$ depends). The coefficients are given by $$ \hat f_A = 2^{-|\mathrm{supp}(f)|} \sum_{\omega} f(\omega) \sigma_A(\omega), $$ and thus the condition $f(\omega)=f(-\omega)$ implies that $\hat f_A = 0$ for all A such that $|A|$ is odd. We conclude that it is sufficient to prove the claim when $f=\sigma_A$ and $g=\sigma_B$ with both $|A|$ and $|B|$ even.

Second, the free state $\mu^0_\beta$ satisfies, at all temperatures, $$ \mu^0_\beta = \tfrac12 (\mu^+_\beta + \mu^-_\beta), $$ where $\mu^+_\beta$ and $\mu^-_\beta$ are the $+$ and $-$ states. When $\beta\neq\beta_c$, this was proved in this paper for $d=2$ and this paper for $d\geq 3$. The result when $\beta=\beta_c$ follows from continuity of the magnetization at this point, which was proved in this paper.

Third, for any $A\Subset\mathbb{Z}^d$ with $|A|$ even, $\langle \sigma_A \rangle^+_\beta=\langle \sigma_A \rangle^-_\beta$ and, therefore, $\langle \sigma_A \rangle^0_\beta=\langle \sigma_A \rangle^+_\beta$.

So, your question reduces to the proof of the claim under $\mu^+_\beta$, which is easy (see Exercise 3.15 in our book).