If I understood this correctly, the Gibbs Specification for the Ising model on $ℤ^d$ dos not have a unique Gibbs Measure for β above the critical level. But what about the Ising model on a finite lattice $V⊂ℤ^d$?

And if it is not unique for a finite lattice, how is that possible as I can construct a Gibbs Sampler – a markov chain, that has the measure that fulfills the specification as its stationary distribution – that is clearly irreducible (we can go from anywhere to the $\{-1\}^V$ state and get everywhere from there) and thus must have an unique stationary distribution?

  • $\begingroup$ Too much to parse to me. $\endgroup$ – Włodzimierz Holsztyński Sep 23 '13 at 4:20
  • $\begingroup$ @WlodzimierzHolsztynski ? $\endgroup$ – Chronial Sep 23 '13 at 5:56
  • $\begingroup$ Your second (last) question is difficult to follow to me. (BTW, it'd be nice to see the relevant definitions. I'd like such definition-oriented spirit in general on MO). $\endgroup$ – Włodzimierz Holsztyński Sep 23 '13 at 6:29
  • 1
    $\begingroup$ Isn't the Gibbs measure in a finite lattice (at the given temperature) unique by the very definition? $\endgroup$ – Włodzimierz Holsztyński Sep 23 '13 at 16:53
  • 1
    $\begingroup$ You @Chronial remind me of a Turing secretary. $\endgroup$ – Włodzimierz Holsztyński Sep 23 '13 at 17:09

From the physics point of view, the answer to your question is an immediate "yes": a nonunique Gibbs measure arises if there is a phase transition into a phase with multiple ground states (say, a phase transition into a ferromagnetic state with all spins aligned either up or down). A finite system has no phase transitions, there is a unique equilibrium state, hence a unique Gibbs measure.

Mathematically, one can also see this from the definition of a nonunique Gibbs measure given, for example, by Dror Weitz, Combinatorial Criteria for Uniqueness of Gibbs Measures:

The question of uniqueness can be viewed combinatorially as comparing two finite distributions (conditioned on two different boundary configurations), and asking whether or not their difference goes to zero as the boundary ball recedes to infinity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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