If I understood this correctly, the Gibbs Specification for the [Ising model](http://en.wikipedia.org/wiki/Ising_model#No_phase_transitions_in_finite_volume) on $ℤ^d$ dos not have a unique [Gibbs Measure](http://en.wikipedia.org/wiki/Gibbs_measure) for β above the critical level. But what about the Ising model [on a finite lattice](http://en.wikipedia.org/wiki/Square-lattice_Ising_model) $V⊂ℤ^d$?


And if it is not unique for a finite lattice, how is that possible as I can construct a [Gibbs Sampler](http://en.wikipedia.org/wiki/Gibbs_sampling) – 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?



Uniqueness of Gibbs Measure on Ising model
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:](http://www.stanford.edu/~montanar/TEACHING/Stat375/papers/weitz1.pdf)



> 
> 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.
> 
> 
> 


