4
$\begingroup$

Consider the situation of a pair $(X,E)$, where $X$ is a standard Borel space and $E$ is an invariant equivalence relation on $X$*. The Friedman-Stanley jump of this pair is an equivalence relation $E^+$ on $X^\omega$ where two sequences $\overline{x}$ and $\overline{y}$ are equivalent if the sets $\{x_n/E:n\in\omega\}$ and $\{y_n/E:n\in\omega\}$ are equal**.

There's a natural inclusion $E\leq_B E^+$ but fixed points are quite rare. If $|X/E|=1$ then $|X^\omega/E^+|=1$, so $E^+\leq_B E$, but for any other Borel $E$, $E^+\not\leq_B E$. However, they do happen: if $E$ is Borel complete (in the sense of Friedman and Stanley) then $E^+\leq_B E$.

With all this in mind, I'm looking for more fixed points. Specifically:

  • Suppose $E^+ \leq_B F^+ \leq_B E$. Can we conclude $F^+\leq_B F$?

Here $(X,E)$ and $(Y,F)$ are not necessarily Borel relations (it's trivial in that case by the theorem stated above).

This seems "obviously true" to me, but the reasoning probably involves set theory that I don't understand.

I would be quite happy to read any sources on this topic. This operation is mentioned here and there across the literature but I haven't seen fixed points discussed. I can't tell if this is because it's a not-well-understood problem, or because it's just a specific case of a well-understood problem so doesn't have many specific references.

* I am interested in the case where $X$ is the set of models on $\omega$, of an $L_{\omega_1,\omega}$-sentence and $E$ is the isomorphism relation for $L$, so the "invariant" notion is "invariant with respect to a closed subgroup of $S_{\infty}$, but other contexts exist.

** In most, but not all, natural examples it is equivalent to have these be sets with multiplicity. The jump has been defined both ways, so I picked one more or less arbitrarily. An answer to the question in either case would be lovely.

$\endgroup$

0

You must log in to answer this question.