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Consider the first neighbors Ising model in $\mathbb Z^2$, with the Hamiltonian in the finite volume $\Lambda\subset\mathbb{Z}^2$ given by $$ H_{\Lambda}(\sigma|\omega)=-J\sum_{i,j\in\Lambda\atop{\|i-j\|=1}}\sigma_i\sigma_j-J\sum_{i\in\Lambda, j\in\Lambda^c\atop{\|i-j\|=1}}\sigma_i\omega_j $$ where $\omega\in\{-1,1\}^{\mathbb{Z}^2}$ is a boundary condition.

By the Aizenman-Higuchi Theorem for any $\beta>0$, we have that closed convex hull of the weak limits of Gibbs measures in finite volume is the convex set $ [\mu^{\beta,+},\mu^{\beta,-}]. $

Question: Is there any $\beta>\beta_c$ and $\lambda\in(0,1)$ such that $$ \mu=\lambda\mu^{\beta,+}+(1-\lambda)\mu^{\beta,-} $$ and
$$ \mu\notin \left\{w-\lim_{\Lambda\uparrow\mathbb{Z}^2}\ \ \mu_{\Lambda}^{\beta,\omega}:\omega\in\{-1,1\}^{\mathbb{Z}^2} \right\} \ \ ? $$

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up vote 3 down vote accepted

I don't think so. Just consider Dobrushin boundary conditions (positive spins at vertices with nonnegative second coordinate, negative elsewhere), and a box of the form $$ \Lambda_n=\{-n,\ldots,n\}\times\{-n-[a\sqrt{n}],\ldots,n-[a\sqrt{n}]\}. $$ Then the mixture you'll get in the limit will have $\lambda$ equal to the probability that the open contour passes below $0$, which should go continuously from $1$ to $0$ as $a$ goes from $-\infty$ to $+\infty$ (it is known that the interface converges weakly to a Brownian bridge under diffusive scaling).

Note that this is very much a two-dimensional phenomenon. In 3d, at low enough temperature, I strongly doubt that you can find boundary conditions such that the limiting state is a nontrivial mixture of, say, Dobrushin states. (Of course, it is always true that you can reach extremal states in this way.)

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Thank you very much for help me again. I liked the comment about high dimensional case it was very interesting. – Leandro Dec 2 '10 at 5:30

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