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The following question was posted to MathStackExchange (original here). As there were no comments/answers on the original, I have ported it unedited.

I am interested in determining the cardinality of $j(\kappa)$ when $j\colon V\to M$ is an elementary embedding arising from an ultrapower embedding obtained through $\kappa$, with $\kappa$ an $\aleph_1$-strongly compact cardinal.

We say that $\kappa$ is $\aleph_1$-strongly compact if for all $\lambda\geq\kappa$ there is a fine $\sigma$-complete ultrafilter $\mathcal{U}$ on $\mathscr{P}_\kappa(\lambda)=\{x\subseteq\lambda\mid|x|<\kappa\}$. Here, $\mathcal{U}$ is fine if for all $\alpha<\lambda$, $\{x\in\mathscr{P}_\kappa(\lambda)\mid\alpha\in x\}\in\mathcal{U}$. Here I am taking $\lambda$ to have cofinality at least $\kappa$ and such that $|\mathscr{P}_\kappa(\lambda)|=\lambda$ (though I am very interested in being able to violate either of these rules).

When taking the ultrapower embedding, that is $j\colon V\to M$ obtained from $\mathcal{U}$ (where $M$ is the transitive collapse of the ultrapower), we have that $j``\lambda\subseteq[\operatorname{id}]\in M$, and $M\vDash|[\operatorname{id}]|<j(\kappa)$, so certainly $\lambda\leq j(\kappa)$. Furthermore, since $j(\kappa)=\{[f]\mid f\colon\mathscr{P}_\kappa(\lambda)\to\kappa\}$, we have $|j(\kappa)|<|\kappa^{\mathscr{P}_\kappa(\lambda)}|^+=(2^\kappa)^+$.

This gives us that $\lambda\leq j(\kappa)<(2^\lambda)^+$, so my question is: Can we control this further? Perhaps by imposing more restrictions on $\mathcal{U}$ before implementing the ultrapower.

My hope is that either for any such $\lambda$ we can guarantee that $2^\lambda\leq j(\kappa)$; or that for any such $\lambda$ we can guarantee that $|j(\kappa)|=\lambda$.

$\aleph_1$-strong compactness is not a very strong hypothesis, so if we are unable to control $j(\kappa)$ in any meaningful way, would we be able to do so with a stronger hypothesis? I know that, e.g. strong compactness would be sufficient, so can we do better than that?

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  • $\begingroup$ Your variable names for $\lambda$ and $\kappa$ are backwards from the usual convention, for which one considers fine measures on $P_\kappa\lambda$ rather than $P_\lambda\kappa$. $\endgroup$ Oct 31, 2023 at 13:10
  • $\begingroup$ @JoelDavidHamkins This is an unfortunate consequence of me following some old work of Kunen in which he uses $\lambda$ for the strongly compact cardinal. I probably should have swapped them in the question before posting it. $\endgroup$ Oct 31, 2023 at 13:56
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    $\begingroup$ I think it might be worthwhile still to swap them, since this convention is quite established now. I think many MO readers will appreciate it. $\endgroup$ Oct 31, 2023 at 14:02
  • $\begingroup$ Are you interested in examples where a tighter bound is possible? I can do that. $\endgroup$ Oct 31, 2023 at 14:42
  • $\begingroup$ @JoelDavidHamkins That would be great! $\endgroup$ Oct 31, 2023 at 14:53

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If $2^\gamma \leq \kappa$ for all $\gamma < \kappa$ (e.g., if GCH holds and $\kappa$ is the least $\aleph_1$-strongly compact), then $|j(\kappa)| \geq 2^\lambda$. To see this, let $\sigma = [\text{id}]$. By elementarity, $M\vDash j(\kappa) \geq |P^M(\sigma)|$. But there is an injection $i : P(\lambda)\to P^M(\sigma)$ given by $i(A) = j(A)\cap \sigma$, so $|P^M(\sigma)| \geq 2^\lambda$.

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  • $\begingroup$ GCH seems quite a strong additional assumption on top of the existence of an $\aleph_1$-strongly compact cardinal; do you know if it is forcable from merely 'there exists an $\aleph_1$-strongly compact cardinal'? $\endgroup$ Nov 1, 2023 at 10:11
  • $\begingroup$ In fact, it is an open question whether given a model containing a strongly compact cardinal, it is possible to force GCH and preserve the strongly compact (see Apter-Dimopoulos-Usuba's "Strongly compact cardinals and the continuum function," Question 5.1). I think it is an interesting question whether one can start with an $\omega_1$-strongly compact and force to make the first $\omega_1$-strongly compact a strong limit cardinal; maybe this is possible just by "pushing up" the first $\omega_1$-strongly compact. More information in Gitik's "On $\sigma$-complete uniform ultrafilters." $\endgroup$ Nov 1, 2023 at 14:58
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I want merely to point out an unusual feature of your large cardinal conception. Namely, since you have not required the measure to be $\kappa$-complete, it seems that according to your definition the $\aleph_1$-strongly compact cardinals are closed upward. That is, if $\kappa_0$ is $\aleph_1$-strongly compact, then any larger $\kappa\geq\kappa_0$ is also $\aleph_1$-strongly compact, since $P_{\kappa_0}(\lambda)\subseteq P_{\kappa}(\lambda)$, and we can just use the measure arising from $\kappa_0$, but as a witness for $\kappa$. It will still be fine and $\sigma$-complete.

The embedding $j$ could have critical point $\kappa_0$, or less, rather than $\kappa$.

This kind of situation can be used to make interesting examples that relate to the weakenings of your assumptions that you mention. For example, perhaps although $\lambda^{<\kappa_0}=\lambda$, we might have $\lambda^{<\kappa}$ much bigger, but all the calculations with the measure will arise from $\kappa_0$.

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    $\begingroup$ I guess we have Bagaria-Magidor to thank for the unusual definition, not OP. $\endgroup$ Oct 31, 2023 at 18:44

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