Extendibility vs supercompactness

Question

The following quotes comes from "Cantor's Attic"'s page on supercompacts:

If κ is $|V_{κ+η}|$-supercompact with $η<κ$ then it is preceeded by a stationary set of $η$-extendible cardinals. If $κ$ is $(η+2)$-extendible then it is $|V_{κ+η}|$-supercompact. The least supercompact is not 1-extendible, in fact any cardinal that is both supercompact and 1-extendible is preceeded by a stationary set of cardinals that are both supercompact and limits of supercompact cardinals.

This implies that 1-extendibility is strictly weaker than $2^\kappa$-supercompactness, and weaker than (or as strong as) 3-extendibility. How does 2-extendibility compares to $2^\kappa$-supercompactness? Or more generally $(\eta+1)$-extendibility vs $|V_{\kappa+\eta}|$-supercompactness, $\eta<\kappa$?

Also, is there some known $\eta$ such that $\eta$-extendibility is (consistency-wise) stronger than full supercompactness? By the quote above $\eta\geq\kappa$ is necessary. I'd be surprised to learn that $\eta$-extendibility never gets stronger than supercompactness even though being $\eta$-extendible for all $\eta$ is strictly stronger than supercompactness.

Edit: apparently the quoted results come from Kanamori's "The Higher Infinite", which additionally gives the following interesting result:

If $\kappa$ is $\eta$-extendible and $\delta+5\leq\eta$, then $\kappa$ is $|V_{\kappa+\delta}|$-supercompact.

Answer 1

$2$-extendibility reflects $2^\kappa$-supercompactness. It suffices to show that any $2$-extendible $\kappa$ is $2^\kappa$-supercompact. Then if $\mathcal U$ is a normal fine $\kappa$-complete ultrafilter on $P_\kappa(P(\kappa))$ and $j : V_{\kappa+2}\to V_{\kappa_* + 2}$ witnesses 2-extendibility, $\mathcal U\in V_{\kappa_* + 2}$, and so by the usual reflection argument, $\kappa$ is a limit of $\bar \kappa$ that are $2^{\bar \kappa}$-supercompact.

To demystify Kanamori's result, we show something stronger:

Proposition: If $\kappa$ is $(\eta+1)$-extendible then $\kappa$ is $|V_{\kappa+\eta}|$-supercompact unless $\eta$ is an infinite limit ordinal with $\text{cf}(\eta)< \kappa$.

The case $\eta = 1$ is what we need above. The cofinality restriction is necessary: if $\kappa$ is $|V_{\kappa+\gamma}|$-supercompact where $\gamma$ is a limit ordinal with $\text{cf}(\gamma)< \kappa$, then $\kappa$ is $(\gamma+1)$-extendible but not $|V_{\kappa+\gamma}|$-supercompact in the ultrapower by any $0$-order normal fine $\kappa$-complete ultrafilter on $P_\kappa(|V_{\kappa+\gamma}|)$.

Proof of Proposition. Suppose $j : V_{\kappa+\eta+1} \to V_{\kappa_*+\eta_*+1}$ witnesses $(\eta+1)$-extendibility. We claim that $j$ extends uniquely to an elementary $\hat j : H_{|V_{\kappa+\eta}|^+}\to H_{|V_{\kappa_*+\eta_*}|^+}$. Grant this for now. The cofinality restriction on $\eta$ implies that $|P_\kappa(|V_{\kappa+\eta}|)| = |V_{\kappa+\eta}|$. Therefore $P_\kappa(|V_{\kappa+\eta}|)\in H_{|V_{\kappa+\eta}|^+}$. Hence every subset of $P_\kappa(|V_{\kappa+\eta}|)$ and every function $P_\kappa(|V_{\kappa+\eta}|)\to |V_{\kappa+\eta}|$ are in $H_{|V_{\kappa+\eta}|^+}$. Moreover $\hat j[|V_{\kappa+\eta}|]\in H_{|V_{\kappa_*+\eta_*}|^+}$. So we can set $$\mathcal U = \{X\subseteq P_\kappa(|V_{\kappa+\eta}|) : \hat j[|V_{\kappa+\eta}|]\in \hat j(X)\}$$ and this is a normal fine $\kappa$-complete ultrafilter.

To obtain this extension $\hat j$, use the standard coding fact that for any infinite $\alpha$, there is (uniformly in $\alpha$) a model theoretic interpretation $f_\alpha : V_{\alpha+1}\to H_{|V_\alpha|^+}$. Essentially $f_\alpha(x) = s$ if $x\subseteq V_\alpha$ codes $(E,p)$ where $E$ is a wellfounded extensional relation on a set $A\subseteq V_\alpha$ with $p\in A$, and $\pi(p) = s$ where $\pi : (A,E)\to (M,\in)$ is the transitive collapse of $E$.

Let $f = f_{\kappa+\eta}$ and $f_* = f_{\kappa_*+\eta_*}$. One then sets $\hat j(s) = f_*(j(x))$ for any $x\in f^{-1}(\{s\})$ and checks that $\hat j$ is well-defined and elementary.

Second question. The question is a bit vague: it depends what you mean by "stronger than" and what you mean by "known $\eta$." But anyway the answer is probably no. For example if $\kappa$ is supercompact, then $V_\kappa$ satisfies that for any $\delta$ there is some $\bar \kappa$ that is $\delta$-extendible. Moreover if $\kappa$ is supercompact and $\eta > \kappa$, then there are arbitrarily closed $\kappa$-complete ultrapowers of $V$ in which $\kappa$ is $\eta$-extendible. In particular, if $\varphi(x,y)$ is a $\Sigma_2$ formula with parameters in $V_\kappa$ and there is some $\alpha > \kappa$ such that $\varphi(\kappa,\alpha)$ holds, then $\kappa$ is a limit of cardinals $\bar \kappa$ that are $\bar \alpha$-extendible for some $\bar \alpha < \kappa$ such that $\varphi(\bar \kappa,\bar \alpha)$ holds.

Answer 2

Also, is there some known $\eta$ such that $\eta$-extendibility is (consistency-wise) stronger than full supercompactness?

If $\kappa$ is $\eta$-extendible (or even $\lt\eta$-extendible, that is, $\gamma$-extendible for every $\gamma \lt \eta$) and $\eta$ is the least worldly cardinal greater than $\kappa$ then it follows from Gabe Goldberg's answer that $\kappa$ is $\lt\eta$-supercompact, which implies that $V_\eta \vDash$ ZFC + "$\kappa$ is supercompact" since $\lambda$-supercompactness is downward absolute to $V_{\lambda+2}$. By elementarity of any 1-extendibility embedding, $\kappa$ is a limit of cardinals that are similarly supercompact up to their respective next worldly cardinal.

Added later: For direct implication, suppose $\eta$ is the least $\Sigma_2$-correct cardinal greater than $\kappa$ and $\kappa$ is $\lt\eta$-extendible. Then it is $\lt\eta$-supercompact and by $\Sigma_2$-correctness this is equivalent to full supercompactness.