Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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
$$ \mu\notin \left\{w-\lim_{\Lambda\uparrow\mathbb{Z}^2}\ \ \mu_{\Lambda}^{\beta,\omega}:\omega\in\{-1,1\}^{\mathbb{Z}^2} \right\} \ \ ? $$

share|improve this question

1 Answer 1

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.)

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.