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In Velenik's Statistical Mechanics of Lattice Systems, Exercise 6.22 claims that if $\pi =\{ \pi_\Lambda:\Lambda \Subset\mathbb{Z}^d\}$ is a translationally invariant specification with nonempty Gibbs state $\mathscr{G}(\pi)$ (probability measures compatible with $\pi_\Lambda$), then the set of translationally invariant Gibbs state is nonempty.

Velenik provides a hint in which we take $\mu\in \mathscr{G}(\pi)$ and use $$ \mu_n = \frac{1}{|B(n)|} \sum_{j\in B(n)} \theta_j \mu $$

I would assume that we are to take the limit (passing under subsequence by Banach-Alaoglu) and obtain a vague convergence probability measure, but it seems that this proof requires that $\mathscr{G}(\pi)$ be closed under the vague topology, which may not necessarily be true. Is there another way to prove the claim? Or is the statement missing a requirement?

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You are completely correct. We should have added the assumption that the specification $\pi$ is quasilocal. (For instance, using the setting described right before the exercise; I guess that's what we had in mind when stating the exercise, but I don't really remember. Of course, in that case, one can remove the assumption that $\mathcal{G}(\pi)\neq\emptyset$ by Theorem 6.26.)

With the quasilocality assumption, $\mathcal{G}(\pi)$ is closed by Lemma 6.27 and the exercise is proved as described in the book (solutions are given in Appendix C).

I have updated the errata and corrected the "preprint" version of the book. (Thanks for pointing this out!)

(By the way, if you are interested in the corresponding result in a much more general framework than we cover in our book, you should have a look at Corollary 5.16 in Georgii's book.)

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  • $\begingroup$ Thank you for the recommendation! I have seen Georgii's book before, but it is a little too general for me at the moment. $\endgroup$ Sep 30, 2020 at 21:38
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    $\begingroup$ @AndrewYuan The fact that Georgii's book is too abstract to learn the theory from it is one of the reasons we decided to write our book: there was a need for an introductory book on this subject. ;) $\endgroup$ Oct 1, 2020 at 6:03

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