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



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> 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.
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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:



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> If $\kappa$ is $\eta$-extendible and $\delta+5\leq\eta$, then $\kappa$ is $|V\_{\kappa+\delta}|$-supercompact.
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Extendibility vs supercompactness
$2$-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.


