Critical points of rank-into-rank embeddings

Question

$\DeclareMathOperator{\crit}{\operatorname{crit}}$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.

Answer 1

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 $\kappa\in j(C)$.■

References: Upward reflection of rank-into-rank cardinals