# Critical points of rank-into-rank embeddings

$$\DeclareMathOperator{\crit}{\operatorname{crit}}$$A rank-into-rank embedding is a non-trivial elementary embedding from a rank initial segment of $$V$$ into itself: $$j:V_\delta\prec V_\delta$$. Define the critical sequence of such an embedding by setting $$\kappa_0=\crit(j)$$ (the first ordinal moved by $$j$$) and $$\kappa_{n+1}=j(\kappa_n)$$. Let $$\lambda=\crit^\omega(j)=\sup_{n<\omega} \langle \kappa_n\rangle$$. It is straightforward to see that $$\lambda$$ is a strong limit cardinal of countable cofinality.

By a theorem of Kunen, if such an embedding can exist, then $$\delta$$ must be the ordinal $$\lambda$$ or $$\lambda+1$$.

It is not hard to see that $$\crit(j)$$ must be measurable. In fact, for any $$n$$, $$\crit(j)$$ is also $$n$$-huge as witnessed by the ultrafilter $$U=\{X\subseteq\mathcal{P}(\kappa_n): j"\kappa_n\in j(X)\}.$$ Further, if we let $$j^n$$ denote $$j$$ composed with itself $$n$$ times, then $$V_\lambda\models \lambda\text{ is supercompact"}.$$ To see this, suppose $$\crit(j)\leq \theta <\kappa_n$$, then $$U=\{X\subseteq\mathcal{P}_{\crit(j)}(\theta): j^n"\theta\in j^n(X)\}$$ winesses the $$\theta$$-compactness of $$\crit(j)$$ (in $$V_\lambda$$).

For the last claim, it is enough that $$\crit(j)$$ is $$<\lambda$$-supercompact, i.e. not fully supercompact in $$V$$. In this case, however, $$\crit(j)$$ could be fully supercompact.

But extendible cardinals are not characterized by the presence of ultrafilters and this motivates my question here.

Question: Can the critical point of a rank-into-rank embedding be extendible?

It may not make sense (I think) to ask for full extendibility of $$\crit(j)$$: Suppose otherwise that $$\crit(j)$$ is fully extendible. Let $$k$$ witness the $$\theta$$-extendibility of $$\crit(j)$$ for some $$\theta>\crit^\omega(j)$$. Then we have $$V_{\crit(j)}\prec V_{\crit^\omega(j)}\prec V_\theta.$$
This looks suspiciously like Woodin's Enormous Cardinal (though his notion is defined in the context of just ZF). See http://logic.harvard.edu/EFI_Woodin_talk.pdf, slide 20. Thus I'm not sure that $$\crit(j)$$ can be fully extendible.

Question: Assume $$j$$ is a rank-into-rank embedding and let $$\lambda=\crit^\omega(j)$$. Can $$\crit(j)$$ be $$<\lambda$$-extendible?

Edit: I should point out (reminded by Carlo Von Shnitzel's comments below) that there is a sort of local intertwining of supercompact cardinals and extendible cardinals that may be relevant. See Kanamori's book, p.316-318.

Also, there may be some subtlety here concerning $$\Sigma_k$$ correctness. Suppose $$j:V_\lambda\prec V_\lambda.$$ I think assuming $$V_\lambda\prec_3 V$$ (or even $$V_\lambda\prec_2 V$$) is a strictly stronger assumption. If $$\crit(j)$$ were extendible, then $$V_{\crit(j)}\prec_3 V$$. But the embedding assumption also gives us that $$V_{\crit^\omega(j)}\prec_3 V$$, even though $$\crit^\omega(j)=\lambda$$ is not itself an extendible cardinal. Similarly if we assume $$\crit(j)$$ is actually supercompact.

• If $\kappa=crit(j)$ is $\theta$-supercompact for some $\theta$ and if we let $j:V \to M$ witness this $\theta$- supercompactness then since $j|V_{\alpha}: V_{\alpha} \to j(V_{\alpha})= M_{j(\alpha)}$ is bounded by $\theta$ so it is in $M$, by the supercompactness, we get that $\kappa$ is $\alpha$-extendible for any $\alpha$ such that $\beth_{\alpha} \leq \theta$. We can get the appropriate supercompactness from embeddings $j:V_{\theta} \to V_{\theta}$, say by $X \in \mu \leftrightarrow j"\delta \in j(X)$ with $X \subset P_{\kappa}(\delta)$ if... Nov 4, 2013 at 7:04
• ...$j(\kappa)>\delta$ and if $P_{\kappa}(\delta) \subset V_{\theta}$. I'm not sure about it and in any case you are asking about the sup of the critical sequence. Nov 4, 2013 at 7:04
• This wrong. We can't get $M_{j(\alpha)}\subseteq M$ from $M^\alpha\subseteq M$; we can only get it from $M^{j(\alpha)}\subseteq M$. For this reason, the least supercompact cardinal is not $1$-extendible, but is a stationary limit of cardinals that $1$-extendible. Oct 7, 2019 at 2:44

Theorem: If $$\kappa$$ is the critical point of $$j\colon V_\lambda\prec V_\lambda$$, then $$\kappa$$ is $$\lambda$$-weakly extendible. Furthermore, if $$\kappa$$ is the critical point of $$j\colon V_\lambda\prec V_\lambda$$, then $$\kappa$$ is $$\lt\lambda$$-strongly extendible.

Proof. First off $$\kappa+\lambda=\lambda$$, as $$\kappa_n$$ is a cardinal for each $$n$$ ($$\kappa_0=\kappa$$), and therefore $$\lambda$$ is a cardinal $$>\kappa$$. By definition, there is an elementary embedding $$j\colon V_\lambda\prec V_\lambda$$ with critical point $$\kappa$$. Similarly, for each $$\alpha\lt\lambda$$ such that $$\alpha\lt j^n(\kappa)$$, $$\kappa$$ is strongly $$\alpha$$-extendible as witnseesed by $$j^{(n)}\restriction V_\alpha: V_\alpha\prec V_{j^n(\alpha)}$$.■

Theorem: If $$\kappa$$ is the critical point of $$j\colon V_\lambda\prec V_\lambda$$ and $$\kappa\in C^{(2)}$$, then $$\kappa\gt$$ the least rank-into-rank cardinal. Furthermore, if $$\kappa$$ is $$\lambda$$-strongly extendible, then $$\lambda>$$ the least rank-into-rank cardinal.

Proof. The statements “$$\lambda$$ is rank-into-rank” and “there exists a rank-into-rank embedding” are both $$\Sigma_2$$. And so if $$V_\kappa\prec_{\Sigma_2} V$$, then $$V_\kappa\vDash\text{There is a rank-into-rank embedding}$$ and if $$V_\kappa\vDash\lambda_0\text{ is rank-into-rank}$$, then $$\lambda_0$$ is rank-into-rank. For the second part, let $$k: V_\lambda\prec V_{k(\lambda)}$$ witness $$\lambda$$-strong extendibility. Then $$k(\kappa)+1\lt k(\lambda)$$ and so $$V_{k(\kappa)+1}\subseteq V_{k(\lambda)}$$ and therefore $$k(\kappa)$$ is inaccessible. Therefore, $$V_{k(\kappa)}\vDash\text{There is a rank-into-rank embedding}$$, and so $$V_\kappa\vDash\text{There is a rank-into-rank embedding}$$ (As $$V_\kappa\prec V_{k(\kappa)}$$), and if $$V_\kappa\vDash\lambda_0\text{ is rank-into-rank}$$, then $$\lambda_0$$ is rank-into-rank, because $$\Sigma_2$$-formulas are upward absolute in $$V_\kappa$$ for inaccessible $$\kappa$$.■

Note then that the consistency strength of “$$\kappa$$ is the critical point of $$j\colon V_\lambda\prec V_\lambda$$, and $$V_\kappa\prec_{\Sigma_2} V$$” is therefore greater than I3. Furthermore, the consistency strength of “$$\kappa$$ is the critical point of $$j\colon V_\lambda\prec V_\lambda$$, and $$\kappa$$ is $$\lambda$$-strongly extendible” is therefore greater than I3.

Theorem: If $$\kappa$$ is I2, then the cardinals which are I3 and extendible in $$V_\kappa$$, are stationary in $$\kappa$$.

Proof. Let $$X$$ be the set of cardinals which are I3 in $$V_\lambda$$ as witnessed by some $$j\subseteq V_\lambda$$. Then, if $$\alpha\in X$$, $$\alpha$$ is I3, because $$V_\alpha^{V_\beta}=V_\alpha$$ whenever $$\alpha\lt\beta$$. Therefore $$\kappa\in j(X)$$, because $$\Sigma_2^1$$-properties are prserved, and so the cardinals $$Y\in D$$, where $$Y$$ is the set of cardinals which are I3 and $$D$$ the measure generated by $$j$$. Similarly, $$\kappa$$ is extendible in $$V_\lambda$$ and so $$V_{j(\kappa)}$$ and so $$Z\in D$$, where $$Z$$ is the set of cardinals extendible in $$V_\kappa$$. Therefore $$Y\cap Z\in D$$, so that $$Y\cap Z$$ is stationary, because every club set $$C$$ has $$j(C)\cap\kappa=C$$ and so $$\kappa\in j(C)$$.■