# It is possible to prove the existence of Gibbs measures using the Kolmogorov extension theorem?

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

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

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

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

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 $$\mathscr{\mathcal{F}}\times\Omega\ni(A,\omega) \longmapsto \mu\Big(\Pi_\Lambda(A)\!\times\!\{\Pi_{S/\Lambda}(\omega)\}\Big)$$

-
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. – MathOverview Jun 17 '12 at 14:52
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! – MathOverview Jun 20 '12 at 14:33

## 2 Answers

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.

-

In Kolmogorov theorem you get an unique measure corresponding to only one equilibrium state so you are not able to describe phase transitions! Remember in Gibbs measure theory there might be more than one measure for a family of specifications and each measure describe one equilibrium state. For example if you have 2 Gibbs measure your system has 2 equilibrium states which means your system shows phase transitions ( from one to another ). All efforts for modeling a physical system is to discuss about possible phase transitions. if your model is not able to discribe phase transitions like Kolmogorov theory it's useless!

-