What is the consistency strength of the following situation?

  1. $j : V \to M$ is an elementary embedding definable from parameters in $V$, with critical point $\kappa$.
  2. $\mathbb P$ is a forcing that collapses all ordinals between $\kappa$ and $j(\kappa)$.
  3. $j$ can be lifted through $\mathbb P$.

One can deduce that $j(\kappa)$ is regular in $V$. The only examples I know of such liftings involve almost-huge cardinals.

  • 1
    $\begingroup$ If the answer is not known, you can define a new notion "pretty huge", and argue that since almost huge is pretty huge, is it always the case that a pretty huge cardinal is almost huge? $\endgroup$
    – Asaf Karagila
    Aug 6, 2020 at 8:29
  • $\begingroup$ You can deduce stronger hypotheses. Let $G$ be $\mathbb P$-generic. Then $\kappa$ is regular in $V[G]$ since it is the critical point of an elementary embedding of $V[G]$ in some outer model. Similarly, $j(\kappa) = (\kappa^+)^{V[G]}$. By assumption, $j(\kappa)\leq (\kappa^+)^{V[G]}$. For the other direction, $j(\kappa)\geq (\kappa^+)^{V[G]}$ since $j(\kappa)$ is the target of the critical point of an elementary embedding of $V[G]$ in some outer model. (The target model $Q$ of this embedding must contain $P^{V[G]}(\kappa)$ and therefore $j(\kappa)\geq\kappa^{+Q}\geq \kappa^{+V[G]}$.) $\endgroup$ Aug 6, 2020 at 22:22

1 Answer 1


$\text{AD}^{L(\mathbb R)}$ suffices. The situation actually holds in the model $H = \text{HOD}^{L(\mathbb R)}$. We will have $\kappa = \omega_1$ and $j : H\to \text{Ult}(H,U)$ equal to the ultrapower of $H$ by the club measure $U$ over $\omega_1$ as computed in $L(\mathbb R)$ (using all functions in $L(\mathbb R)$).

For any number $n$, the $\Sigma_n$-satisfaction predicate of $L(\mathbb R)$ with ordinal parameters is definable over $H$ from its restriction to ordinals less than $\Theta$, so any subclass of $H$ that is ordinal definable over $L(\mathbb R)$ is definable from parameters over $H$. In particular, $j$ is definable from parameters over $H$.

Let $N$ be a $\mathbb P_\text{max}$-extension of $L(\mathbb R)$. Note that $H = \text{HOD}^N$ by the homogeneity and definability of $\mathbb P_\text{max}$. Let $\mathbb P\in H$ be the Vopenka algebra of $N$ for adding a subset of $\omega_2$ to $H$. There is a set $A\subseteq \omega_2$ such that $N= L[A]$, and so $N = H[G_A]$ where $G_A\subseteq \mathbb P$ is the $H$-generic ultrafilter associated to $A$.

In $N$, $\text{NS}_{\omega_1}$ is saturated. Let $G\subseteq P(\omega_1)\setminus\text{NS}_{\omega_1}$ be $N$-generic, and in $N[G]$ let $i : N\to \text{Ult}(N,G)$ be the generic ultrapower embedding associated to $G$ (using functions in $N$).

Now as usual, we cite a theorem due to Woodin: $j = i\restriction H$. This follows from Theorem 4.53 in The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal.

Now in $H$, we have the situation you were looking for with $\kappa = \omega_1.$ Note that $i(\omega_1) = (\omega_2)^N$ by saturation, which means that all $H$-cardinals between $\kappa$ and $j(\kappa)$ are collapsed to $\kappa$ in $N$. Moreover $j$ lifts through the forcing $\mathbb P$ (to $i$) by construction.

  • $\begingroup$ What happens if you require a "super" kind of modification, namely that $j(\kappa)$ can be unbounded when ranging over possible $j$s, how far does this blow up the consistency strength? $\endgroup$
    – Asaf Karagila
    Aug 7, 2020 at 6:48
  • $\begingroup$ Very interesting. How does the extender in $H$ corresponding to $j$ relate to standard large cardinal notions? I wonder how it interacts with other forcings. $\endgroup$ Aug 7, 2020 at 8:19
  • $\begingroup$ The model $H$ can be presented as a fine structure model. It's an open question whether $j$ is the branch extender of an iteration tree by the sequence of this model. The short part of $j$ is just the $\omega_2$-length linear iteration of the unique normal ultrafilter over $\omega_1$ in $H$. $\endgroup$ Aug 7, 2020 at 15:18
  • $\begingroup$ @AsafKaragila I have no idea, I don't see how to get something like that from AD. $\endgroup$ Aug 7, 2020 at 15:48
  • $\begingroup$ Oh, I thought it was obvious you can't. But maybe it's not that obvious... $\endgroup$
    – Asaf Karagila
    Aug 7, 2020 at 15:52

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