Superextendibles defined analogously to superstrong cardinals: Where are they consistency strength-wise?

Question

A common trend in large cardinal axioms dealing with critical points of elementary embeddings from $V$ into a transitive class $M$ is to make some large cardinal axiom with an ordinal parameter, similar to "$M$ is closed under $\theta$-sequences" or "$V_\theta$ is a subset of $M$". The resulting cardinals from these definitions are the $\theta$-supercompacts and the $\theta$-strongs.

After this, expansions could be made to make the cardinal even larger; specifically, by taking that parameter of $M$ and making it $j^n(\kappa)$. When doing this to $\theta$-strong cardinals, one achieves $n$-superstrong cardinals. When doing this to supercompact cardinals, one achieves $n$-huge cardinals.

Of course, a natural question at this point is to see what happens when one does it to $\eta$-extendible cardinals. Here are the results:

Let $\kappa$ be $n$-superextendible if there is some ordinal $\eta$ such that $\kappa$ is the critical point of some nontrivial elementary embedding $j:V_\theta\prec V_\eta$ such that $\theta>j^n(\kappa)$.

So far, here are the results:

• The $0$-superextendible cardinals are precisely the $0$-extendible cardinals
• I3 cardinals are $n$-superextendible for every $n$ (this property could be called $\omega$-superextendible)
• The $1$-superextendible cardinals, if one lets an arbitrary such be $\kappa$, are $\kappa$-extendible. Of course, every $1$-superextendible cardinal is a supercompact cardinal and is, in fact, a stationary limit of such due to this property.
• If a cardinal is $n+1$-superextendible it is $n$-superextendible

These properties are, yes, very obvious. Are there any other possibly less obvious properties that I am missing? Where do these cardinals fall under consistency strength?

Answer 1

Let $\kappa$ be $n$-Superextendible if there is some ordinal $\eta$ such that $\kappa$ is the critical point of some nontrivial elementary embedding $j:V_\theta\prec V_\eta$ such that $\theta>j^n(\kappa)$.

Without loss of generality $\theta=j^n(\kappa)+1$ and $\eta=j^{n+1}(\kappa)+1$ (otherwise take $j^n\upharpoonright V_{\kappa+1}$). Thus this is the same property that Sato 2007 (Joseph Van Name's link doesn't work) would call $n+1$-fold 1-extendible. As Master's answer points out, $n+1$-fold 1-extendible cardinals are also called n-huge*. By this Mathoverflow answer by Joel David Hamkins, $n$-fold 1-extendible cardinals are $n$-superstrong and by proposition 8.5 of Sato's paper, if $\kappa$ is $n$-fold $2^\kappa$-supercompact, it is a limit of $n$-fold 1-extendible cardinals.

When doing this to $\theta$-Strong cardinals, one achieves n-Superstrong cardinals. When doing this to Supercompact cardinals, one achieves n-Huge cardinals.

A weaker large cardinal property analogous to those has $\theta \ge j^n(\kappa)$ or, without loss of generality, $\theta=j^n(\kappa)$. This property can be called $n+1$-fold 0-extendible. Solovay, Reinhardt and Kanamori 1978 calls the assertion that there exists a 2-fold 0-extedibility embedding $A_2$. By theorems 8.1, 8.2 and 8.3 of that paper, 2-fold 0-extedible cardinals are between almost huge and huge cardinals in the large cardinal hierarchy.

Answer 2

These large cardinals actually already have a name. They are called $n$-huge* cardinals.

Theorem: If $\kappa$ is $n$-huge*, then $\kappa$ is $n$-huge and a limit of $n$-huge cardinals.

Proof. Let $j: V_\alpha\rightarrow V_\beta$ be an $n$-hugeness* embedding, with $\lambda=j^n(\kappa)$. Furthermore, let $Y=\{j(\alpha)|\alpha\lt\lambda\}$ and $D=\{X\subseteq\lambda|Y\in j(X)\}$. It is immediately a normal measure. Now note that $ot(Y\cap j^{i+1}(\kappa))=j^i(\kappa)$. For the rest, note that $D\in V_\beta$.■

Theorem: If $\kappa$ is $n+1$-huge, then $\kappa$ is $n$-huge* and a limit of $n$-huge* cardinals.

Proof. Let $j: V\rightarrow M$ be an $n+1$-hugeness embedding, with $\lambda=j^n(\kappa)$, and let $k=j\restriction V_{\lambda+1}$. Then $k\in M$ and so $M\vDash\kappa\text{ is }n\text{-huge*}$ and so $M\vDash V_{j(\lambda)}\vDash(\kappa\text{ is }n\text{-huge*})$, but $V_{j(\lambda)}\subseteq M$ and so $\kappa$ is actually $n$-huge*. A similar argument goes for the rest .■

Refrences: Virtual large cardinals, Victoria Gitman and Ralf Schindler.