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(Throughout, all ultrafilters are nonprincipal.)

Given a property $P$ - really, a sentence in some appropriate logic - say that a ultrafilter $\mathcal{U}$ on a cardinal $\kappa$ averages $P$ iff for every $\kappa$-tuple of structures, the ultraproduct by $\mathcal{U}$ of that tuple has $P$ iff $\mathcal{U}$-many of the structures in the tuple have $P$. For example, Łoś's theorem says that all ultrafilters average all first-order-expressible properties. Meanwhile, non-first-order properties are "usually" not averaged by ultrafilters.

However, even looking at a single strong logic we may see genuinely different "levels of averageing" amongst ultrafilters: e.g. looking at $\mathsf{SOL}$, note that well-foundedness is averaged by exactly the countably complete ultrafilters. This motivates the separate notion of "averageability:" a property $P$ is averageable iff there is some ultrafilter averaging it, and a set $\Gamma$ of properties is averageable iff there is some single ultrafilter which averages each property in $\Gamma$.

Interestingly, I don't know offhand of any examples of divergent behavior in any natural logic. To keep things precise, I'll focus on $\mathsf{SOL}$:

Question 1: Is it consistent with $\mathsf{ZFC}$ + large cardinals that the union of two finite averageable sets of second-order sentences is again averageable?


Now question $1$ is of course extremely broad. We can try to sharpen it by restricting attention to a particular type of ultrafilter. There are, to my mind, two natural ways to do this:

Question 2: What happens if we restrict attention to nonprincipal ultrafilters on $\omega$?

Question 3: What happens if we restrict attention to nonprincipal countably closed ultrafilters (and assume that lots of measurable cardinals exist, to avoid the trivial answer)?

Personally I think that a positive answer to $(1)$ is highly implausible, but a positive answer to $(3)$ might follow from (very) large cardinals. $(2)$ is pretty mysterious to me; large cardinals don't really tame ultrafilters on $\omega$ as far as I know (nor does anything else to be fair), so I'm not really sure where to get started intuition-wise.

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I'm not sure I have a definitive answer, but three nice observations that are too long for comments:

  1. If you relax from ultrafilters to extenders (seen as directed systems of ultrafilters), then there's a much stronger connection with large cardinals. In particular, the moral of Magidor's proof that extendibles are compactness cardinals for $\mathbb{L}^2$ is that the embeddings $V_\alpha \to V_\beta$ are $\mathbb{L}^2$-correct for structures in $V_\alpha$. So then all classes are averageable-by-extenders (although maybe this is slightly off from your definition because you need different extenders based on the rank of the models...)

  2. Say that $\phi \in \mathbb{L}^2$ is existential if all of the second-order quantifiers are existential (and there's no negation outside of them). Then it is the case that existential $\mathbb{L}^2$ properties are averageable upwards, that is, if $U$-many $M_i \vDash \phi$, then $\prod M_i/U \vDash \phi$. So if you say $\phi$ is $\Delta_1$ iff $\phi$ and $\neg \phi$ are existential, then the $\Delta_1$ properties should exactly be the `universally averageable' properties.

  3. Beyond the $\Delta_1$ statements, I don't think there's going to be too much that is averageable within the realm of second-order logic specifically. The example you mention of well-foundedness is really a phenomena where well-roundedness is $\mathbb{L}^2$-definable AND $\mathbb{L}_{\omega_1, \omega_1}$-definable, and this second one is what allows its preservation by countably complete ultrafilters.

To back up this intuition, one way to think about this is to imagine all of the $M_i$ as Henkin structures $(M_i, \mathcal{P}(M_i))$. Then a second-order statement is just first-order in this structure, and so is preserved in moving to the ultra product $\prod (M_i, \mathcal{P}(M_i))/U$. But there is a big gap between

  • $\prod \mathcal{P}(M_i)/U$, the subset sort of the Henkin structure and
  • $\mathcal{P}(\prod M_i/U)$, the actual power set of the ultraproduct that the second order statements will check

This is why universal second-order statements are not `averageable up', because they miss the extra subsets in $\mathcal{P}(\prod M_i/U) - \prod \mathcal{P}(M_i)/U$

I guess a question to help guide any of this intuition is the following: do you have any candidate properties in mind that are averageable but not a) uniformly averageable (by dint of being $\Delta_1$) or b) equivalent to some $\mathbb{L}_{\kappa,\kappa}$-property and are averageable only for $\kappa$-complete ultrafilters?

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  • $\begingroup$ Nice observations, Bill. $\endgroup$
    – Asaf Karagila
    Commented Nov 17, 2021 at 21:53
  • $\begingroup$ Re: (2)/(3) and the end, I don't really have any such interesting candidates but I'll see if I can come up with one. (Maybe something from a fully-compact logic beyond FOL?) Re: (1), I'm happy to generalize to extenders but I really do care about not depending on the input structures, so I'm not fully satisfied with that. $\endgroup$ Commented Nov 18, 2021 at 4:33
  • $\begingroup$ An addendum to the $\Delta_1$ observation above: a better way of looking at it is that existential second-order sentences are preserved by ultraproducts is that the are PC-classes and reduct commutes with ultraproducts. $\endgroup$
    – Will Boney
    Commented Nov 18, 2021 at 22:33
  • $\begingroup$ One more addendum: I think averageability of some honest/unavoidable use of a universal quantifier for all structures is going to be very hard. As M gets big, the different between $P(\prod M)$ and $\prod P(M)$ is going to get bigger and bigger, so you'll keep missing out on more subsets that you need to universally quantify over. This is why I think an example would be very instructive, since you'll start to see where averageability fails. $\endgroup$
    – Will Boney
    Commented Nov 19, 2021 at 17:59
  • $\begingroup$ This is quickly getting out of my realm of competence, but this answer of Farmer S seems relevant. (Today's regret: why didn't I pay more attention in John's seminars?) $\endgroup$ Commented Dec 24, 2021 at 19:42

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