Is there an example of Gibbs measure that is not a weak limit of finite volume Gibbs measure ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:22:00Z http://mathoverflow.net/feeds/question/47302 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47302/is-there-an-example-of-gibbs-measure-that-is-not-a-weak-limit-of-finite-volume-gi Is there an example of Gibbs measure that is not a weak limit of finite volume Gibbs measure ? Leandro 2010-11-25T06:11:42Z 2010-12-01T12:08:13Z <p>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 <code>$\omega\in\{-1,1\}^{\mathbb{Z}^2}$</code> is a boundary condition. </p> <p>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,-}]. $</p> <p><strong>Question: </strong> Is there any $\beta>\beta_c$ and $\lambda\in(0,1)$ such that $$ \mu=\lambda\mu^{\beta,+}+(1-\lambda)\mu^{\beta,-} $$ and<br> <code>$$ \mu\notin \left\{w-\lim_{\Lambda\uparrow\mathbb{Z}^2}\ \ \mu_{\Lambda}^{\beta,\omega}:\omega\in\{-1,1\}^{\mathbb{Z}^2} \right\} \ \ ? $$</code></p> http://mathoverflow.net/questions/47302/is-there-an-example-of-gibbs-measure-that-is-not-a-weak-limit-of-finite-volume-gi/47894#47894 Answer by Yvan Velenik for Is there an example of Gibbs measure that is not a weak limit of finite volume Gibbs measure ? Yvan Velenik 2010-12-01T12:08:13Z 2010-12-01T12:08:13Z <p>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).</p> <p>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 <em>extremal</em> states in this way.)</p>