So this question probably shows my inner model theoretic ignorance, but:

In "Two remarks on elementary embeddings of the universe" (http://projecteuclid.org/download/pdf_1/euclid.pjm/1102969567), Jech defines - for $j: V\rightarrow M$ a definable-with-parameters elementary embedding - an inner model $L(j)$, and proves that $L(j)$ is the smallest inner model which admits $j$ (in the sense that the inner model thinks $j$ is an elementary embedding as well).

This is a neat, short argument, and the definition of $L(j)$ is very straightforward. Naively, I would suspect that properties of the inner model $L(j)$ would tell us interesting information about the embedding $j$, and similarly for models $L(\overline{j})$ defined for sequences of embeddings $\overline{j}$.

For instance, here's a question which seems natural to me: given an inner model $M\subset V$ and a family of embeddings $j_\eta: V\rightarrow M$ $(\eta\in \kappa)$, it's reasonable to ask for which $I\subset \kappa$ is there an inner model $N$ admitting precisely the $j_\eta$ with $\eta\in I$? In particular, given a left distributive algebra of embeddings, which sub-left distributive algebras can be captured this way? Naively, if I want to capture $\{j_\eta: \eta\in I\}$, my first guess would be to look at $L(\{j_\eta: \eta\in I\})$.

NOTE: this isn't my actual question, I'm just including it to motivate interest in the $L(j)$s.

However, I've never run across these models before, and I can't seem to find any recent reference to them. So, my question is:

Are the $L(j)$s interesting at all from the point of view of modern inner model theory? If not, why not, and if so, why doesn't there seem to be any modern literature about them (maybe they are extremely hard to work with, or maybe there is such literature which I just haven't found yet)?

(For my purposes, "modern" means "since 1990.")

  • $\begingroup$ If the elementary embedding $j$ is looked at as being represented by an extender $E$, then I think it is true that $L[E]$ doesn't give anything more than a measurable cardinal, even if $j$ witnesses that some $\kappa$ is supercompact. But I guess you don't want to look at $L[E]$ but at the smallest inner model of $\text{ZF}$ containing $j$. $\endgroup$ May 6, 2015 at 0:48
  • $\begingroup$ Do such models capture any extra properties of $j$ like degrees of strongness or supercompactness? $\endgroup$ May 6, 2015 at 2:51
  • $\begingroup$ @Monroe: After looking at the way Jech defines $L(j)$, these models shouldn't capture any extra properties of $j$. This is because he defined them as $L[j]$ actually or as $L[E]=L[j_E]$ where $E$ is the extender derived from $j$. If that is the case, then $L[E]$ is just $L[\mu]$ where $\mu$ is a normal measure on some $\kappa$. In this case one would have to use a sequence of extenders, not just one extender, to obtain enough approximations to the elementary embedding, so that one can have large cardinals in "$L(j)$". $\endgroup$ May 6, 2015 at 3:47
  • 2
    $\begingroup$ It sounds like you have a good answer to Noah's question-- the argument for $L(j) = L[\mu]$. $\endgroup$ May 6, 2015 at 3:54
  • $\begingroup$ @CarloVonSchnitzel, if you write your comments up as an answer I will upvote it and probably accept it: although this doesn't convince me that the $L(j)$s are uninteresting (they still seem e.g. a good way to attack the sample problem in my question), it certainly explains why they have no real role in modern inner model theory. $\endgroup$ May 6, 2015 at 19:15

2 Answers 2


I recently noticed that Mitchell returned to $L[j]$ in "Applications of the covering lemma for sequences of measures," where he shows that the model is quite a bit larger than one might expect. The thing is, $L[j]$ isn't $L[E_j]$ where $E_j$ is the short extender of $j$, it's $L[P]$ where $P = \{(x,A) : x\in j(A)\}$ is the entire proper class long extender of $j$. Assuming there is no inner model with a measurable $\kappa$ of order $\kappa^{++}$, Mitchell shows that if $j$ is nontrivial, $L[j]$ contains an iterate of the core model. Moreover, under the same assumption, he proves that if $j$ is the ultrapower embedding associated to a normal ultrafilter, then $L[j]$ satisfies $V = K$. So $L[j]$ contains many measurable cardinals of high order if $K$ does. As far as I know, no one has worked on the question at the level of strong/Woodin cardinals.

  • 2
    $\begingroup$ The paper is from 1987 (written in 1985), so definitely not qualifying for the 90s or newer definition of modern. Still, though. Cool. $\endgroup$
    – Asaf Karagila
    Sep 6, 2021 at 16:35
  • $\begingroup$ Thanks - I hadn't picked up on the $E_j$ vs. $P$ issue. $\endgroup$ Sep 6, 2021 at 22:20

From the point of view of inner model theory, the models $L(j)$ do not seem to be interesting. The main reason is that if $j: V \to M$ is an elementary embedding with critical point $\kappa$ and say $E_j$ is the $(\kappa,j(\kappa))$-extender derived from $j$ (that is $E=\{(a,X): a\in [j(\kappa)]^{<\omega} \wedge a\in j(X)\}$) then the model $L(j)=L[E_j])$ of all sets constructible from $j$ is only $L[\mu]$ where $\mu$ is the normal measure on $\kappa$ derived from $j$. Therefore using a single extender and constructing over it cannot produce any large cardinals above a measurable cardinal. The idea, due to Mitchell, was to build the constructible universe over enough approximations to $E_j$ to be able to obtain large cardinals in $L[\vec{E}]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.