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?

  • 2
    $\begingroup$ 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. $\endgroup$ Mar 15, 2021 at 16:23
  • $\begingroup$ I meant 85 to 86... $\endgroup$ Mar 15, 2021 at 20:50
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    $\begingroup$ I was wondering whether the instruction was to read backwards. $\endgroup$
    – Farmer S
    Mar 15, 2021 at 20:57
  • $\begingroup$ @GabeGoldberg: Thanks. Both the positive results (8.13, 8.14) and the negative result are essentially what I was looking for. $\endgroup$
    – Yair Hayut
    Mar 15, 2021 at 21:19

1 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.

  • $\begingroup$ 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? $\endgroup$
    – Yair Hayut
    Mar 15, 2021 at 15:10
  • $\begingroup$ As in, the case that a mouse $M$ satisfies ``There is no Woodin cardinal and $j:V\to N$ is elementary''? $\endgroup$
    – Farmer S
    Mar 15, 2021 at 15:59
  • $\begingroup$ Yes, even though based on the comment of @GabeGoldberg above, maybe the right anti-large cardinal is "no overlapping extenders". $\endgroup$
    – Yair Hayut
    Mar 15, 2021 at 21:01
  • $\begingroup$ 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. $\endgroup$
    – Farmer S
    Mar 15, 2021 at 21:29
  • $\begingroup$ 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. $\endgroup$
    – Farmer S
    Mar 15, 2021 at 21:39

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