Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is possible to prove the existence of Gibbs measures using the Kolmogorov extension theorem? If yes how? If the proof is too long to write here is there any reference?

Thank you.

Edit.

Let $S$ be a countable set and $\mathscr{S}=\mathscr{S}(S)=\{\Lambda: \Lambda \subset S, \quad 0 <| \Lambda | <\infty \} $. To keep in mind we can take $S$ as $ \mathbb{N}, \mathbb{Z}$ or $\mathbb{Z}^2$.

Let $ \mathscr{E}$ a $\sigma$-algebra on the set $\mathbb{E}$. If $(\mathbb{E}, \mathscr{E}) = (\mathbb{E}^i, \mathscr{E}^i), \quad \forall i \in S $ define:$(\mathbb{E}^\Lambda, \mathscr{E}^\Lambda)=\bigotimes_{i\in \Lambda}(\mathbb{E}^i, \mathscr{E}^i).$

If $\Omega \triangleq \mathbb{E}^S = \{\omega=(\omega_i)_{i\in S}: \omega_i \in \mathbb{E}^i, \forall i\in S \} $ then for $ \Lambda, \Gamma \in \mathscr{S} $ with $ \Lambda\subset \Gamma $ define:

$ \Pi_i: \Omega \to \mathbb{E}^i $ as the natural projection of $\Omega \triangleq \mathbb{E}^S $ in $\mathbb{E}^i $,

$ \Pi_\Lambda: \Omega \to \mathbb{E}^\Lambda $ as the natural projection of $ \Omega \triangleq \mathbb{E}^S $ in $ \mathbb{E}^\Lambda $,

$ \Pi_{\Gamma, \Lambda}: \mathbb{E}^\Gamma \to\mathbb{E}^\Lambda$ as the natural projection of $ \Omega \triangleq \mathbb{E}^S$ in $\mathbb{E}^\Lambda $.

Now consider the following $ \sigma $-algebras defined on $\Omega $. $ \mathcal{F}_\Lambda = \sigma \big(\Pi_\Lambda\big) $, $ \mathcal{J}_{\Lambda}=\sigma \big(\Pi_{S/\Lambda}\big)$, $\mathcal{F}=\sigma \big (\{\Pi_\Lambda \}_{\Lambda \in \mathscr{S}}\big)$.

And also consider the $ \sigma$-algebras $ \mathcal{F}_{\Gamma, \Lambda} $ and $ \mathcal{J}_{\Gamma, \Lambda} $ defined on $ \Omega_\Gamma = \mathbb{E}^\Gamma $ respectively by $ \sigma \big(\Pi_{\Gamma, \Lambda} \big) $ and $ \sigma \big(\Pi_{\Gamma/\Lambda, \Lambda} \big)$. In this notation the Kolmogorov extension theorem can be stated as follows.

Definition Given a family of probability measures $\{\mu^{\Gamma}\}_{\Lambda \in \mathscr{S}}$ on $(\mathbb{E}^\Lambda, \mathscr{E}^\Lambda)$ the equations

\begin{equation}\mu^\Lambda(\quad)=\mu^\Gamma\big(\Pi^{-1}_{\Gamma \Lambda} (\;\cdot\;) \big) \quad \forall ,\Gamma, \Lambda \in \mathscr{S} \text{ with } \Lambda \subset \Gamma \end{equation}

are called Kolmogorov consistency condition.

and

Theorem [Kolmogorov extension] If $(\mu_\Gamma)_{\Gamma \in\mathscr{S}} $ is a family of probability measures on $(\mathbb{E}^\Gamma,\mathscr{E}^\Gamma)$, meeting the consistency condition Kolmogorov then there exists a unique probability measure on $(\mathbb{E}^S, \mathscr{E}^S = \mathcal{F})$ such that

\begin{equation} \mu^\Lambda = \mu \big(\Pi_{\Lambda}^{-1}(\; \cdot \;)\big) \quad \forall \;\Lambda \end{equation}

In a brief term as a mean Gibbs measure $ \mu $ on the $(\Omega,\mathcal{F})$ satisfies that the condition Dobrushin-Lanford-Ruelle equivalently that is the same as $$ \mu \Big(\Pi_\Lambda(A) \times \{\Pi_{S/\Lambda}(\omega)\}\Big)=\mu\Big(A|\mathcal{J}_\Lambda \Big)(\omega) $$ for $A\in \mathcal{F}$ and $\mu|_{\mathcal{J}_\Lambda}\mbox{-a.e. }\omega\in\Omega$. Here $\mu|_{\mathcal{J}_\Lambda}$ is the restriction of the measure $\mu$ to $\mathcal{J}_\Lambda$.

In other words it's like $\mu$ to be specified by the probability of kernels $(\Omega,\mathcal{J}_\Lambda)$ to $(\Omega,\mathcal{F})$ given by $$ \mathscr{\mathcal{F}}\times\Omega\ni(A,\omega) \longmapsto \mu\Big(\Pi_\Lambda(A)\!\times\!\{\Pi_{S/\Lambda}(\omega)\}\Big) $$

share|improve this question
    
Can you say precisely what you mean by "Gibbs measures"? –  Nate Eldredge Jun 16 '12 at 20:16
    
@ Nate Eldredge. I add explanations on Gibbs measure and Kolmogorov's extension theorem. Thank's. –  Elias Jun 17 '12 at 14:52
1  
Perhaps you should try mathoverflow.net ? –  leonbloy Jun 19 '12 at 14:43
    
Need not be frightened by the notation. I believe that one need not resort to MathOverflow. But I will consider the suggestion! –  Elias Jun 20 '12 at 14:33

1 Answer 1

up vote 3 down vote accepted
+50

You can, at least for finite state spaces $\mathbb{E}^i$. The link to Kolmogorov's extension theorem is used explicitly in Theorem 5 of these notes:

http://www.stat.yale.edu/~pollard/Courses/606.spring06/handouts/Gibbs1.pdf

The above argument is really using "local convergence". In the case of finite states, local convergence is the same as weak convergence of measures, and the weak topology is known to be compact in this case. To see how local convergence can be used for infinite state spaces, see Section 4 of "Gibbs Measures and Phase Transitions" by H.O. Georgii.

share|improve this answer

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.