Elementary embeddings with the same critical point

Question

Question: Is it consistent (relative to the existence of large cardinals) that there is an elementary embedding $j\colon V\to M$ (where $M$ is transitive model) that factors as $j = j_n \circ k_n$ for $n < \omega$, such that for every $n$, $\text{crit }j_n = \kappa$ (the same ordinal), but $j_{n} (\kappa) < j_{n + 1} (\kappa)$ for every $n$?

The motivation for this question is the following observation: It is possible to get from single measurable cardinal many different elementary embeddings $j_n\colon V \to M_n$ with the same critical point (by using iterated ultrapower) and a single model $M$ and elementary embedding $j \colon V \to M$ such that $j = k_n \circ j_n$ for $k_n \colon M_n \to M$. The question is whether the "dual" situation, which seems to be more problematic, is possible.

Answer 1

Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers.

Specifically, suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence.

For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull $$X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\},$$ and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. You can find this and related results in section 3 of my paper, Canonical seeds and Prikry trees, JSL 62, 1997 (adapted from a chapter of my dissertation). It uses the normality of $\mu$, and this particular fact is not necessarily true without that assumption, as shown in the paper.

Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Thus, rather than including all seeds up to $\kappa_n$, which is how you might have proved the situation you mentioned at the end of your question, we instead undertake the dual set, using only seeds from $\kappa_n$ and upward. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$.

Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i<n$, and consequently $X_{S_n}\cap[\kappa,\kappa_n)=\emptyset$, leading to $\pi(\kappa_n)=\kappa$. Thus, $j_n(\kappa)=\kappa_n$. In particular, these all have the same critical point, and they reach higher as $n$ increases. So the situation is just what you requested.

Answer 2

This isn't really an answer, but here's a way you can get the maps as you want with the embeddings defined outside of the model. Assume $0^\sharp$ exists, and let $\langle \alpha_i : i < \omega^2 \rangle$ be the first $\omega^2$ indiscernibles for $L$. Let $k$ send $\alpha_i$ to $\alpha_{\omega+i}$, for $i < \omega^2$ and fix the rest. Let $j_n$ send $\alpha_i$ to $\alpha_{i+n}$ for $i < \omega$ and fix the rest. Then for all $i < \omega^2$, $j_n \circ k(\alpha_i) = k(\alpha_i)$. We can extend the $j_n$'s and $k$ to elementary embeddings from $L$ to $L$. Then putting $j = k = k_n$ for each $n$, we have a collection of maps satisfying the requirements.