11
$\begingroup$

Let $\beta\omega$ be the Stone-Čech compactification of the discrete infinite countable space $\omega$, and $\beta^*\omega=\beta\omega\smallsetminus \omega$ is the Stone-Čech remainder.

The map $j:n\mapsto n+1$ extends to an self-injection of $\beta\omega$, which itself restricts to a self-homeomorphism $\phi$ of $\beta^*\omega$.

In ZFC+CH, is it true that $\phi$ and $\phi^{-1}$ are not conjugate in $\mathrm{Homeo}(\beta^*\omega)$?

Indeed in Shelah's model ("forcing axiom"), in which CH fails, there exists a homomorphism $\mathrm{Homeo}(\beta^*\omega)\to\mathbf{Z}$ mapping $\phi$ to $1$. So the non-conjugacy of $\phi$ with $\phi^{-1}$ is consistent. But under CH, the group $\mathrm{Homeo}(\beta^*\omega)$ is simple (Rubin) so the non-conjugacy couldn't be attested by a homomorphism to $\mathbf{Z}$ as above.


Note: Boolean algebraic translation through Stone duality: consider the endomorphism 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. Is (under ZFC+CH) $\Phi$ non-conjugate to its inverse in $\mathrm{Aut}_{\mathrm{Ring}}(2^\omega)$?

Indeed Stone duality yields (in ZFC) an isomorphism $\mathrm{Homeo}(\beta^*\omega)\to\mathrm{Aut}_{\mathrm{Ring}}(2^\omega)$ mapping $\phi$ to $\Phi$.


Further comments:

A side question is whether it is consistent with ZFC that $\phi$ and $\phi^{-1}$ are conjugate, I don't know either (but I'm primarily interested in the CH case).

Also in ZFC it is easy to check that $\phi$ is not conjugate to $\phi^n$ for any $n\ge 2$.

$\endgroup$
  • 1
    $\begingroup$ YCor, have you looked at the analogous C*-algebra question: are the unilateral shift and its adjoint related by an automorphism of the Calkin algebra? Will Brian lists several facts about $\beta\omega\setminus\omega$ whose Calkin algebra analogues aren't familiar to me (but maybe experts would know better). $\endgroup$ – Nik Weaver Feb 25 at 14:29
  • $\begingroup$ @NikWeaver Nice question. Farah has results in this direction. For instance (Ann. Math. 2011 link) he proved the consistency of ZFC + all automorphisms are inner. In such a (quite exotic) model, they're non-conjugate since the Fredholm index distinguishes them. I don't know whether CH implies they're conjugate (if true this might be easier than the set-theoretic counterpart). $\endgroup$ – YCor Feb 25 at 15:40
  • $\begingroup$ Yes ... I'm aware of Ilias's paper. It followed a paper by Chris Phillips and me where we showed that CH implies outer automorphisms exist. Our techniques wouldn't be helpful for this problem though. $\endgroup$ – Nik Weaver Feb 25 at 15:43
  • $\begingroup$ @NikWeaver But isn't existence of nontrivial outer automorphisms immediate from Rudin? Rudin proved under CH that $|\mathrm{Aut}(S_\omega/\mathrm{fin})|=2^c$. It immediately implies (since $\mathrm{Aut}(S_\omega/\mathrm{fin})$ embeds into $\mathrm{Aut}$(Calkin)) that $\mathrm{Aut}$(Calkin) has the same cardinality (and hence has Out of the same cardinality, since $\mathrm{Inn}$(Calkin) has cardinal $c$). $\endgroup$ – YCor Feb 25 at 15:46
  • 1
    $\begingroup$ @LSpice "Stone-Cech corona" yields 20 times less Google occurences than "Stone-Cech remainder". One advantage of "Stone-Cech remainder" is that you can guess the meaning assuming that you know what "Stone-Cech compactification" is. I've actually encountered "corona" never in the meaning of this Wikipedia page (to which I'd recommend renaming), but in generalizations such such as the Higson-Roe corona, or binary corona of a metric space. $\endgroup$ – YCor May 8 at 14:45
10
$\begingroup$

This is a great question -- and it's wide open. Here's what I know about it:

$\bullet$ As you mentioned, it is consistent that $\phi$ and $\phi^{-1}$ are not conjugate. This observation was first made by van Douwen, soon after the publication of Shelah's result that you mention in your question. You mentioned forcing axioms, so let me point out that the non-conjugacy of $\phi$ and $\phi^{-1}$ follows from $\mathsf{MA}+\mathsf{OCA}$, which is a weak form of $\mathsf{PFA}$. This is due to Boban Velickovic.

$\bullet$ If it is consistent with $\mathsf{ZFC}$ that $\phi$ and $\phi^{-1}$ are conjugate, then it is consistent with $\mathsf{ZFC}+\mathsf{CH}$. (Proof sketch: If $\phi$ and $\phi^{-1}$ are conjugate in some model, then force with countable conditions to collapse the continuum to $\aleph_1$ and make $\mathsf{CH}$ true. Because this forcing is countably closed, it won't change much about the Boolean algebra $\mathcal P(\omega)/\mathrm{fin}$, and will preserve the fact that $\phi$ and $\phi^{-1}$ are conjugate.)

$\bullet$ Even better, the existence of certain large cardinals implies that if it is possible to force "$\phi$ and $\phi^{-1}$ are conjugate" then this statement is already true in every forcing extension satisfying $\mathsf{CH}$. This follows from a theorem of Woodin concerning what are called $\Sigma^2_1$ statements about the real line (explained further here). The assertion "$\phi$ and $\phi^{-1}$ are conjugate" is an example of such a statement. (Very roughly, this theorem seems to suggest that if this statement is consistent, then it should follow from $\mathsf{CH}$. At any rate, trying to prove it from $\mathsf{CH}$ seems like a reasonable strategy.)

$\bullet$ In fact, Paul Larson has pointed out to me that the statement "$\phi$ and $\phi^{-1}$ are conjugate" is a now very rare example of a $\Sigma^2_1$ statement about the real line whose status we do not know under $\mathsf{ZFC}+\mathsf{CH}$ (plus large cardinal axioms).

$\bullet$ I proved a partial result a few years ago, showing that $\mathsf{CH}$ implies $\phi$ and $\phi^{-1}$ are semi-conjugate:

$\qquad$Theorem: Assuming $\mathsf{CH}$, there is a continuous surjection $Q: \omega^* \rightarrow \omega^*$ such that $$Q \circ \phi = \phi^{-1} \circ Q.$$

The paper is "Abstract $\omega$-limit sets," Journal of Symbolic Logic 83 (2018), pp. 477-495, available here. In the same paper, I show that the forcing axiom $\mathsf{MA}+\mathsf{OCA}$ implies $\phi$ and $\phi^{-1}$ are not semi-conjugate. (Or rather, I show that this is a corollary to a deep structure theorem of Ilijas Farah.)

$\bullet$ Finally, in a more recent paper (to appear in Topology and its Applications, currently available here), I show that there is no Borel set separating the conjugacy class of $\phi$ and the conjugacy class of $\phi^{-1}$ (in the space of self-homeomorphisms of $\omega^*$ endowed with the compact-open topology). Roughly, this shows that if $\phi$ and $\phi^{-1}$ fail to be conjugate, it's not "for any real reason" -- or at least not for any nicely definable topological reason.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks for this detailed answer and for the links! (Side note: I had to open the second link to figure out that "isomorphism class" denotes what I know as "conjugacy class" — I agree that from a certain categorical viewpoint it's also an isomorphism class, namely in the category of spaces endowed with a homeo; actually my colleagues in dynamical systems talk of conjugacy of dynamical systems even when acting on different spaces) $\endgroup$ – YCor Feb 18 at 16:35
  • $\begingroup$ You're welcome! This is a question that I've thought a lot about, so I'm happy to be able to share. $\endgroup$ – Will Brian Feb 18 at 16:50
  • $\begingroup$ Do you know if $(\mathrm{Clopen}(\beta^*\omega),\Phi)$ and $(\mathrm{Clopen}(\beta^*\omega),\Phi^{-1})$ are elementary equivalent (as structures encoding a Boolean algebra endowed with an automorphism)? I might post a separate question if it's not immediate or an immediate consequence of your result (it's not clear to me that the set of homeomorphisms $\alpha$ satisfying some sentence $u(\alpha)$ is Borel). $\endgroup$ – YCor Feb 19 at 13:28
  • $\begingroup$ That's a good question. I don't think it's known. My guess is that the answer is yes, and we might be able to prove they are elementarily equivalent by showing that they're potentially isomorphic (en.wikipedia.org/wiki/Potential_isomorphism). I don't think that the set of all homeomorphisms satisfying some fixed sentence needs to be Borel -- this isn't true for sets of reals, anyway, because unbounded quantifiers push us out of the Borel sets into the projective hierarchy. $\endgroup$ – Will Brian Feb 19 at 13:39
  • $\begingroup$ Thanks, I eventually posted the question mathoverflow.net/questions/353074 $\endgroup$ – YCor Feb 19 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.