# How can we control the cardinality of $j(\kappa)$ for $\kappa$ an $\aleph_1$-strongly compact cardinal?

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 $$M\vDash|[\operatorname{id}]|, so certainly $$\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?

• 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 $P_\lambda\kappa$. Oct 31, 2023 at 13:10
• @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. Oct 31, 2023 at 13:56
• 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. Oct 31, 2023 at 14:02
• Are you interested in examples where a tighter bound is possible? I can do that. Oct 31, 2023 at 14:42
• @JoelDavidHamkins That would be great! Oct 31, 2023 at 14:53

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 $$|P^M(\sigma)| \geq 2^\lambda$$.
• 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'? Nov 1, 2023 at 10:11
• 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." Nov 1, 2023 at 14:58
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 $$\kappa_0$$.