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Suppose $\kappa$ is $\kappa^{+\omega}$-supercompact. If $U$ is a normal measure on $\mathcal P_\kappa(\kappa^{+\omega})$, and $j : V \to M$ is the derived embedding, then it is easy to see that $j[\mathcal P_\kappa(\kappa^{+\omega})] \in M$, and thus $M$ is closed under $\kappa^{+\omega+1}$-sequences by the cardinal arithmetic. We can then derive a normal measure $U'$ on $\mathcal P_\kappa(\kappa^{+\omega+1})$ by putting $X \in U'$ iff $j[\kappa^{+\omega+1}] \in j(X)$. Let $i : V \to N$ be the ultrapower embedding by $U'$. There is an embedding $k : N \to M$ such that $j = k \circ i$. Define $k([f]) = j(f)(j[\kappa^{+\omega+1}])$.

Question 1: Is $k$ the identity?

Now let $\pi : \mathcal P_\kappa(\kappa^{+\omega+1}) \to \mathcal P_\kappa(\kappa^{+\omega})$ be $\pi(z) = z \cap \kappa^{+\omega}$. Let $U'' = \{ X \subseteq \mathcal P_\kappa(\kappa^{+\omega}) : \pi^{-1}[X] \in U' \}$.

Question 2: Does $U'' = U$?

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  • $\begingroup$ I have a deja vú. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 3, 2017 at 17:56
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    $\begingroup$ For the second question, if $X\in U$, then $j''\kappa^{+\omega}\in j(X)$. To show $X\in U''$, we only need to show $\pi^{-1} X \in U'$, which is the same as $j'' \kappa^{+\omega+1}\in j(\pi^{-1}X)=j(\pi)^{-1} j(X)$ so we will be done if $j(\pi)(j''\kappa^{+\omega+1})\in j(X)$ but $j(\pi)(j''\kappa^{+\omega+1})=j''\kappa^{+\omega+1}\cap j(\kappa^{+\omega})=j''\kappa^{+\omega}$. $\endgroup$
    – Jing Zhang
    Commented Nov 3, 2017 at 18:23
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    $\begingroup$ Now that you know that Q2 has a positive answer, I think a positive answer for Q1 follows. $\pi$ is a Rudin-Keisler map witnessing that $U\leq_{\text{RK}} U'$, so, together with your first paragraph, $U$ and $U'$ are isomorphic ultrafilters and it follows that $k$ is the identity. $\endgroup$
    – Miha Habič
    Commented Nov 3, 2017 at 21:51

1 Answer 1

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I think about this a little more abstractly using what I call seed theory. See for example my article,
Hamkins, Joel David, Canonical seeds and Prikry trees, J. Symb. Log. 62, No.2, 373-396 (1997). ZBL0890.03024, which provides an elementary development of this theory. I tried hard in that paper to find a nice, careful treatment of the ideas, which I view as providing a useful framework for answering all these kinds of questions; so I recommend taking a look at the paper, particularly sections 1, 2 and 3. (The paper was adapted from a chapter in my dissertation.)

If you have an elementary embedding $j:V\to M$, then the seed hull of an object $a\in M$ is the class $X_a=\{j(f)(a)\mid f\in V\}$, and this is an elementary substructure of $M$, which is isomorphic to the ultrapower by the measure $\mu_a=\{X\mid a\in j(X)\}$ by the association $[f]_{\mu_a}\mapsto j(f)(a)$. It is a basic fact that if two seeds $a$ and $b$ generate each other, then the measures $\mu_a$ and $\mu_b$ are isomorphic. In particular, any seed $a$ generating all of $M$ will have $j_{\mu_a}=j$.

In your case, for a $\lambda$-supercompactness ultrapower, the seed $j"\lambda$ and the seed $j"\lambda^{<\kappa}$ easily generate each other, for the reasons you gave in your post. Namely, from $j"\lambda$ we can easily construct $j"P_\kappa(\lambda)$, and from $j"\lambda^{<\kappa}$ we can project to $j"\lambda$. It follows that these seeds generate each other and therefore all of $M$, and so the measures they give are isomorphic, the ultrapowers are in each case $j$ itself and the factor embedding $k$ is the identity. The same is true whenever we have two elements $a$ and $b$ in $M$ that can define each other in $M$ using parameters in $\text{ran}(j)$; the corresponding measures $\mu_a$ and $\mu_b$ will be isomorphic and the factor embeddings identical.

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