On generalisation of Aizenman-Higuchi Theorem

Question

Let $\mathbb Z^2$ denote the two-dimensional integer lattice with norm of $i=(i_1,i_2)$ given by $\|i\|=|i_1|+|i_2|$.

For each $x\in\mathbb Z^2$, we assign a uniform random variable, $\sigma_x$ taking values on the set $\{-1,1\}$.

Fix $\omega\in\{-1,0,1\}^{\mathbb{Z}^2}$ and $\beta>0$. For each finite $\Lambda\subset\mathbb{Z}^d$, define a probability measure on the sigma algebra generated by the cylinder sets of $\{-1,1\}^{\mathbb{Z}^2}$, such that for each $\sigma\in\{-1,1\}^{\mathbb{Z}^2}$ the probability of this configuration is given by $$ \mu_{\Lambda}^{\beta,\omega}(\sigma)= \left\{ \begin{array}{rl} \frac{\exp(-\beta H_{\Lambda}^{\omega}(\sigma))}{Z_{\Lambda}^{\omega}},&\text{if}\ \ \sigma_i=\omega_i\ \forall i\in\Lambda^c;\\ \\ \\ 0,& \text{otherwise}, \end{array} \right. $$ where $$ H_{\Lambda}^{\omega}(\sigma)=-\sum_{i,j\in\Lambda}J_{ij}\sigma_i\sigma_j-\sum_{i\in\Lambda, j\in\Lambda^c}J_{ij}\sigma_i\omega_j $$ with $J_{ij}\equiv J(\|i-j\|)\geq 0$ and $J_{ij}=0$ if $\|i-j\|\geq R$, for some positive $R$ and $Z_{\Lambda}^{\omega}$ is a normalizing constant so that $\mu_{\Lambda}^{\beta,\omega}$ is a probability measure.

Question 1: If $\Lambda_n\uparrow\mathbb{Z}^2$ and $\omega_i=0$ for all $i\in\mathbb{Z}^2$, sounds reasonable that any accumulation point of the sequence $\mu_{\Lambda_n}^{\beta,\omega}$, in the weak* topology, is translation invariant. Is this true for any finite $R$ ?

Question 2: Suppose $R$ finite and bigger than one, keeping the setting of Question 1 but $\omega_i=1$ (or $\omega_i=-1$) for all $i\in\mathbb{Z}^2$ is the weak* limit $$w-\lim_{n\to\infty} \mu_{\Lambda_n}^{\beta,\omega}$$
translational invariant ?

Question 3: For finite $R$ bigger than one is it true the Aizenman-Higuchi Theorem
$$w-\lim_{n\to\infty} \mu_{\Lambda_n}^{\beta,\omega}\in [\mu^{\beta,+},\mu^{\beta,-}] ?$$

Answer 1

Concerning questions 1 and 2, if I understand it correctly (i.e., you're simply looking at the finite-range ferromagnetic Ising model with free, resp + or -, b.c.), then the sequences actually converge and the limits are translation invariant. This follows from monotonicity in the volume (GKS for free, GKS of FKG for + or -).

(In all these cases, just fix a local function and take two different increasing sequences of boxes. Show using correlation inequalities that the expectation in both cases necessarily converge to the same limit by sandwiching those in one sequence by those in the other. You can find a version of this classical argument in my lecture notes (in french), see my homepage.)

Concerning 3, the answer is certainly yes, but no proof is known (both the original proofs and the improved one we devised recently with Loren Coquille rely in an essential way on the n.n. nature of the interaction). (There might be some results at very large beta, of course.)

Answer 2

At very large\beta I think the big review of Dobrushin-Shlosman The problem of translation invariance of Gibbs states at low temperatures RL Dobrushin… - Sov. Sci. Rev., Sect. C, Math. Phys. Rev., 1985

gives results for general 2d models.

The fact that you won't get anything with Dobrushin boundary conditions follows from a paper by Bricmont-Lebowitz-Pfister Journal of Statistical Physics Volume 21, Number 5, 573-582, DOI: 10.1007/BF01011169 On the equivalence of boundary conditions That is suggestive, but not quite a proof of what you ask, Aernout van Enter