# Three questions from Kentaro Sato's paper about the n-fold large cardinal hierarchy

In the paper Double helix in large large cardinals and iteration of elementary embeddings there are three things mentioned as unknown which I can answer:

[S]everal results known for ordinary supercompactness do not seem to carry over to $$n$$-fold versions in the same way, e.g., the statement that if $$\lambda$$ is supercompact, then $$\kappa \lt \lambda$$ is supercompact iff $$V_\lambda \vDash \text{“\kappa is supercompact”}$$.

Are “there is an $$n$$-fold extendible cardinal” and “there is an $$n$$-fold supercompact cardinal” equiconsistent, for $$n\ge2$$?

In one direction, theorem 8.3/corollary 8.4 of Sato's paper shows that $$n+1$$-fold extendible cardinals are $$n+1$$-fold supercompact (we can alternatively say $$n$$-fold hyperhuge following Usuba 2017). By this Mathoverflow answer by Gabe Goldberg, theorem 8.3 can be improved to "if $$\kappa$$ is $$n$$-fold $$\gamma+2$$-extendible then it is $$n$$-fold $$\beth_{\kappa+\gamma+1}$$-supercompact". Thus the question is about the converse direction.

Are “there is an $$n+1$$-S-strong cardinal” and “there is an $$n$$-S-huge cardinal” equiconsistent?

By proposition A.3 of Sato's paper, any $$n$$-S-strong/$$n+1$$-fold Shelah cardinal is $$n-1$$-S-huge (following Perlmutter 2013, $$n$$-S-huge cardinals can be called $$n+1$$-fold Shelah for supercompactness). Thus the question is about the converse direction.

• Are you asking a question here, or do you rather just state a result? Sep 3, 2021 at 17:28
• That doesn't explain why you're doing this. Why don't you just post a preprint on the arxiv? Sep 3, 2021 at 19:19
• I don't know how to post preprint at Arxiv. Sep 3, 2021 at 19:50
• @ArvidSamuelsson Arxiv is pretty easy to figure out. See arxiv.org/help/submit Sep 8, 2021 at 12:10

the statement that if $$\lambda$$ is supercompact, then $$\kappa \lt \lambda$$ is supercompact iff $$V_\lambda \vDash \text{“\kappa is supercompact”}$$.

This holds for $$n$$-fold supercompact cardinals too. If $$\lambda$$ is 2-fold supercompact (also known as hyperhuge; Usuba 2017) then it is extendible and thus $$\Sigma_3$$-reflecting and as the $$n+1$$-fold supercompactness (we can alternatively say $$n$$-fold hyperhugeness) of $$\kappa$$ is $$\Pi_3$$-definable, it is absolute between $$V$$ and $$V_\lambda$$.

Are “there is an $$n$$-fold extendible cardinal” and “there is an $$n$$-fold supercompact cardinal” equiconsistent, for $$n\ge2$$?

If $$\kappa$$ is $$n$$-fold $$\beth_{\kappa+\gamma}$$-huperhuge then it is $$n+1$$-fold $$\gamma$$-extendible. Thus $$\kappa$$ is $$n$$-fold huperhuge iff it is $$n+1$$-fold extendible. By the proof of theorem 8.5 of Sato's paper, if $$\kappa$$ is $$n$$-fold $$\beth_{\kappa+\gamma}$$-huperhuge, witnessed by an elementary embedding $$j : V \to M$$, then $$V_{j^{n+2}(\kappa)}^{M^{(n+2)}} \vDash \text{"\kappa is n+1-fold \gamma-extendible"}$$. By elementarity of the $$n+1$$-th iterate of $$j(j)$$, this reflects to $$V_{j(\kappa)}$$. The $$n+1$$-fold $$\gamma$$-extendibility of $$\kappa$$ is upward absolute from $$V_{j(\kappa)}$$ to $$V$$.

Are “there is an $$n+1$$-S-strong cardinal” and “there is an $$n$$-S-huge cardinal” equiconsistent?

A cardinal is a $$n+1$$-S-strong/$$n+2$$-fold Shelah cardinal iff it is $$n$$-S-huge/$$n+1$$-fold Shelah for supercompactness.

Lemma: $$n+2$$-fold Shelah cardinals can also be characterized as $$n+1$$-fold Shelah for extendibility. This follows from the fact that any elementary embedding $$j : V_{\beta+1} \to V_{\gamma+1}$$ can be extended to an embedding $$j : V \to M$$ such that $$V_\gamma \subset M$$, as noted by this Mathoverflow answer by Joel David Hamkins.

Suppose $$\kappa$$ is $$n+1$$-fold Shelah for supercompactness. Then it is $$n+2$$-fold Woodin/$$n+1$$-fold Vopenka so for any function $$f : \kappa \to \kappa$$, $$n+1$$-fold $$f$$-extendible cardinals $$\beta$$ (meaning that there is an extendibility embedding $$i : V_{f(j^{n}(\beta))} \to V_\gamma$$ with critical point $$\beta$$ such that $$\beta$$ is closed under $$f$$ and $$i (f\upharpoonright f(j^{n}(\beta))) = f\upharpoonright \gamma$$). For a given $$f$$ define $$g$$ such that $$g(\alpha)$$ is the least $$\gamma$$ such that there is an $$f$$-extendibility embedding $$i : V_{f(j^{n}(\beta))} \to V_\gamma$$ with critical point $$\beta$$, where $$\beta$$ is the $$\alpha$$-th $$n$$-fold $$f$$-extendible cardinal. Since $$\kappa$$ is $$n+1$$-fold Shelah for supercompactness, there is an elementary embedding $$j : V \to M$$ such that $$M^{\beth_{j(g)(\kappa)}} \subset M$$. Using the argument of theorem 8.5 of Sato's paper, we can see that $$j \upharpoonright V_{j^{n}(j(g)(\kappa))}$$ witnesses that $$M \vDash \text{"\kappa is n-fold j(g)-extendible}$$. Note that $$M^{\beth{j(g)(\kappa)}} \subset M$$ implies $$H_{\beth{j(g)(\kappa)}} \subset M$$. By definition of $$g$$ and elementarity of $$j$$, there is an $$j(f)$$-extendibility embedding $$k : V_{f(j^{n}(\kappa))} \to V_{j(g)(\kappa)}$$ with critical point $$\kappa$$. Since $$f(\alpha) \lt g(\alpha)$$ for every $$\alpha$$, we have $$j(f)(\kappa) \lt (g)(\kappa)$$, so $$H_{\beth{j(g)(\kappa)}}$$ sees that $$k$$ witnesses that $$\kappa$$ is $$n+1$$-fold Shelah for extendibility with respect to $$f$$.