# Elementary equivalence between $n\mapsto n+1$ and its inverse on the Stone-Čech remainder?

Consider structures $$(A,f)$$ encoding a Boolean algebra $$A$$ endowed with an automorphism $$f$$. There is an obvious notion of isomorphism between such structures.

Consider the endomorphism $$\hat{\Phi}$$ of the Boolean algebra $$2^\omega$$ of subsets of $$\omega$$ given by $$A\mapsto \{a\in\omega:a+1\in A\}$$. It induces an automorphism $$\Phi$$ of the quotient Boolean algebra $$2^\omega/\mathrm{fin}$$, where $$\mathrm{fin}$$ is the ideal of finite subsets. (Under Stone duality, this corresponds to the self-homeomorphism of the Stone-Čech remainder of $$\omega$$ induced by $$n\mapsto n+1$$.)

Whether $$(2^\omega/\mathrm{fin},\Phi)$$ and $$(2^\omega/\mathrm{fin},\Phi^{-1})$$ are isomorphic is essentially unknown (see my previous related question for more details).

My question is

Are $$(2^\omega/\mathrm{fin},\Phi)$$ and $$(2^\omega/\mathrm{fin},\Phi^{-1})$$ elementary equivalent?

That is, do they satisfy the same first-order sentences (in the language of Boolean algebras endowed with an automorphism)?

Note that $$\Phi^n$$ has exactly $$2^{|n|}$$ fixed points for $$n\neq 0$$. In particular, $$(2^\omega/\mathrm{fin},\Phi)$$ and $$(2^\omega/\mathrm{fin},\Phi^n)$$ are not not equivalent for $$|n|\ge 2$$.

If the question has a positive answer, it is tempting to ask whether one can characterize simply those $$(A,f)$$ having the same first-order theory as $$(2^\omega/\mathrm{fin},\Phi)$$.

• If they are elementary equivalent, then a back and forth argument shows that they are isomorphic under CH. (And whether they are isomorphic under CH is unknown.) So I think this is open. Feb 19, 2020 at 15:17
• @PaulMcKenney this was indeed my hope to pass from EE to isomorphism, but at the same time if they're not EE, it should be for some "concrete"reason. So I hope this approach leads to something.
– YCor
Feb 19, 2020 at 15:30
• @PaulMcKenney how does the back-and-forth argument work? is $(2^\omega/\mathrm{fin},\Phi)$ $\omega_1$-saturated? since it seems to reduce to my previous question, I think this would be worth an answer.
– YCor
Feb 19, 2020 at 16:44
• That's what I was thinking, but it's no longer clear to me that $(2^\omega / \mathrm{fin}, \Phi)$ is saturated. Feb 19, 2020 at 22:56
• @PaulMcKenney: I was thinking a bit more about this question today and, for what it's worth, I can show that $(2^\omega/\mathrm{fin},\Phi)$ is not $\omega_1$-saturated. I have a somewhat tedious proof of this assertion, but this comment is too short to contain it . . . but feel free to email me if you'd like some details. Aug 11, 2020 at 14:39

Theorem: CH implies that $$\Phi$$ and $$\Phi^{-1}$$ are conjugate to each other in the automorphism group of $$\mathcal P(\omega) / \mathrm{fin}$$.
You can find a proof of this on the arXiv (here), and a short overview of the proof in my answer to this MO question. Let me add a few words explaining why the theorem quoted above implies that the structures $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$$ and $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$$ are elementarily equivalent (even without assuming CH).
The short version is: we can force CH without changing the theory of $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$$ and $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$$, so if they're isomorphic in the forcing extension, and their theories have not changed, they must have been elementarily equivalent to begin with.
In more detail, let $$\mathbb P$$ denote the poset of countable partial functions $$\omega_1 \times \omega \to \omega$$, ordered by extension, i.e., the usual poset for forcing CH with countable conditions. Let $$G$$ be a $$V$$-generic filter on $$\mathbb P$$. Because $$\mathbb P$$ is a countably closed notion of forcing, no new subsets of $$\omega$$ are added by $$\mathbb P$$, which means that $$V \cap \mathcal P(\omega) / \mathrm{fin} = V[G] \cap \mathcal P(\omega) / \mathrm{fin}$$. Furthermore, because $$\Phi$$ is definable by the simple formula $$\Phi([A]) = [A+1]$$, the action of $$\Phi$$ is the same in $$V$$ and in $$V[G]$$. Thus the structures $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle^V$$ and $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle^{V[G]}$$ are one in the same. It follows (by the absoluteness of the satisfaction relation'') the theory of $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$$ is the same in $$V$$ and in $$V[G]$$. The same argument applies to $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$$, so the theory of $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$$ is also the same in $$V$$ and in $$V[G]$$. Because $$V[G] \models$$ CH, the theorem quoted above implies $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$$ and $$\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$$ are conjugate in $$V[G]$$. In particular, they are elementarily equivalent in $$V[G]$$. Because the theories of these two structures is the same in $$V$$ and in $$V[G]$$, this means they have the same theory in $$V$$ as well.