The core model and elementary embeddings

Question

Let $K$ be the core model (below a Woodin cardinal). Let $j \colon K \to M$ be an elementary embedding, where $M$ is well founded. Under which conditions can we conclude that $j$ is an iterated ultrapower of extenders in $K$ (possibly a branch in an iteration tree)?

Are there generalizations of this result to larger inner models?

Answer 1

Some remarks:

By Schindler's paper "Iterates of the core model", if $j:V\to N$ is elementary ($N$ transitive) and $N$ is closed under $\omega$-sequences, and $k:K\to K^N$ is the restriction of $j$, then $K^N$ is an iterate of $K$ and $k$ is the iteration map.

Note that if $M$ is a mouse modelling ZFC + ``$\delta$ is Woodin'' and $(\delta^+)^M$ is countable, then letting $j:M\to U$ be a countable stationary tower embedding, $U$ is not an iterate of $M$, since $\omega_1^U=\delta$.

But for example, assuming $M_1^\sharp$ exists and is fully iterable, if $U$ is already known to be a non-dropping iterate of $M_1$ and $j:M_1\to U$ is elementary, then $j=j_1\circ j_0$ where $j_0$ is the iteration map $M_1\to U$ and $j_1:U\to U$ has $\mathrm{crit}(j_1)$ above the Woodin of $U$.

Given a mouse $M$ and $j:M\to N$ definable from parameters over $M$, with $N$ transitive, one can ask the same question, i.e. whether $N$ is an iterate and $j$ is an iteration map. There are various partial results on this (Schindler's mentioned above is relevant), but I think the question is open in general, even assuming that $M$ knows fully how to iterate itself.