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Velickovic proved (Theorem 4.1 of OCA and automorphisms of $\mathcal{P}(\omega)/\mathrm{fin}$) that, assuming

  • OCA (Open Coloring Axiom) and
  • $\rm MA_{\aleph_1}$,

every (Boolean algebra) automorphism of $\mathcal{P}(\omega_1)/\mathrm{ fin}$ is trivial, i.e. for every such automorphism $\varphi$ there is a function $e : \omega_1\to \omega_1$ such that for all $a\subseteq \omega_1$, $\varphi[a] = [e''a]$.

On the other hand, if there is a nontrivial automorphism $\psi$ of $\mathcal{P}(\omega)/\mathrm{fin}$ (as there are in any model of CH), one can easily construct a nontrivial automorphism $\varphi$ of $\mathcal{P}(\omega_1)/\mathrm{fin}$ by just copying $\psi$ on $\omega$ and the identity on $\omega_1\setminus \omega$:

$$ \varphi[x] = \psi[x\cap \omega]\vee [x\setminus \omega] $$

Of course one can replace $\omega$ with any countable set $a$, and the identity with any trivial automorphism of $\mathcal{P}(\omega_1 \setminus a)/\mathrm{fin}$. But this is somewhat unsatisfying; all of these automorphisms seem to be nontrivial "for trivial reasons." Hence the following question:

Is there, consistently, an automorphism of $\mathcal{P}(\omega_1)/\mathrm{fin}$ which is nontrivial on every cocountable set?

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EDIT: The following does not work, as pointed out in the comments. I do not know how to make $\varphi$ a homomorphism.

The answer is yes, but for trivial reasons. Identify $\omega_1$ with $\omega_1\times \omega$. Every automorphism $\psi$ of $P(\omega)/fin$ induces an automorphism of $P(\omega_1\times \omega)/fin$ as follows:

  • If $ A=\bigcup_{i\in \omega_1} \{i\}\times A_i$, then let $\varphi(A) = \bigcup_{i\in \omega_1} \{i\}\times \psi(A_i)$.

Note that $\varphi$ is well-defined on $P(\omega_1\times \omega)/fin$, and is nontrivial on any co-countable set.

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    $\begingroup$ Thanks! It seems clearer now that a "truly nontrivial" automorphism would be one which is nontrivial as an automorphism of $P(\omega_1)/fin$, but trivial when restricted to $P(A)/fin$, for every countable $A\subseteq \omega_1$; but I'll leave that question for someone else to ask. $\endgroup$
    – Paul McKenney
    Oct 12, 2013 at 22:00
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    $\begingroup$ I looked back at this today and I think there might be a problem with your definition of $\varphi$. Each $\psi(A_i)$ is defined only up to a finite set; when taking a union over $i\in\omega_1$, these finite differences may add up to something infinite. To be concrete, let $F : P(\omega)\to P(\omega)$ be a function such that $\psi[A] = [F(A)]$ for all $A$. Formally, you've defined $\varphi[\bigcup_i \{i\}\times A_i] = [\bigcup_i \{i\}\times F(A_i)]$; but if e.g. F(\{0\}) = \emptyset, then \varphi[\omega_1\times \{0\}] = [\emptyset]$. $\endgroup$
    – Paul McKenney
    Oct 14, 2013 at 18:13
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    $\begingroup$ You are right. I thought that I can define $\psi':P(\omega)\to P(\omega)$ by $\psi'(A):=$ any representative of the class $\psi(A)$. I can certainly do that, but then $\varphi$ will not be a homomorphism, unless $\psi'$ is a homomorphism. But I think I can fix that under CH. (to be continued...) $\endgroup$
    – Goldstern
    Oct 14, 2013 at 21:54
  • $\begingroup$ any luck with CH? $\endgroup$
    – Paul McKenney
    Oct 18, 2013 at 11:10
  • $\begingroup$ No. There is no 1-1 homomorphism from $P(\omega)/fin$ into $P(\omega)$ because there are uncountable (incompatibility-)antichains in $P(\omega)/fin$. I have to retract my answer. $\endgroup$
    – Goldstern
    Oct 19, 2013 at 11:27

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