# The core model and elementary embeddings

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?

• Have you looked at chapter 8 of Steel's "The Core Model Iterability Problem?" There is a sketch of a negative result showing that even if K is "below two strong cardinals," there may be an embedding of $K$ that is not an iteration map on pages 85-65. Mar 15, 2021 at 16:23
• I meant 85 to 86... Mar 15, 2021 at 20:50
• I was wondering whether the instruction was to read backwards. Mar 15, 2021 at 20:57
• @GabeGoldberg: Thanks. Both the positive results (8.13, 8.14) and the negative result are essentially what I was looking for. Mar 15, 2021 at 21:19

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.

• Thank you. I'm more interested in the last case, assuming that there is no Woodin cardinals, but where we don't assume that N is closed under $\sigma$-sequences. What is known in this case? Mar 15, 2021 at 15:10
• As in, the case that a mouse $M$ satisfies There is no Woodin cardinal and $j:V\to N$ is elementary''? Mar 15, 2021 at 15:59
• Yes, even though based on the comment of @GabeGoldberg above, maybe the right anti-large cardinal is "no overlapping extenders". Mar 15, 2021 at 21:01
• If $M$ has no overlapping extenders on its sequence, then it's much simpler. If $M$ is a mouse modelling ZFC with no overlapping extenders on its sequence, $j,N$ are definable from parameters classes of $M$, $N$ is transitive, and $j:M\to N$ elementary, then $N$ is just an iterate of $M$ and $j$ is the iteration map. Mar 15, 2021 at 21:29
• There are also natural positive results if $j:M\to N$ is an internal ultrapower by a measure, or "extender strong through its length", which apply through superstrong mice. Mar 15, 2021 at 21:39