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ِDefinition (1): Call an inner model M "rigid" if there are no non-trivial elementary embeddings $j: M\longrightarrow M$.

It is possible that $L$ is not rigid, while the problem is open for $HOD$. But there is an important difference between these two models; $L$ is "absolute" while $HOD$ is not. By absoluteness I mean:

Definition (2): An inner model $M$ of $ZFC$ is "absolute" iff for any inner model $N$ of $ZFC$ containing $M$ we have "$M$ in the sense of $N$ is equal to $M$".

Also consider the following definition:

Definition (3): An inner model $M$ of $ZFC$ is "forcing absolute" iff for any set forcing extension $W$ of $V$ we have "$M$ in the sense of $W$ is equal to $M$".

Now my questions are about the possible role of absoluteness of an inner model of $ZFC$ in consistency of its "non-rigidity assumption".

Question (1): Suppose $M\neq L$ is a definable inner model of $ZFC$ which is absolute. Is it possible that $M$ is not rigid?

Question (2): Suppose $M\neq L$ is a definable inner model of $ZFC$ which is forcing absolute. Is it possible that $M$ is not rigid?

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  • $\begingroup$ Your definitions (2) and (3) should refer to formulas defining inner models, instead of inner models directly. $\endgroup$ – Noah Schweber Oct 27 '13 at 6:55
  • $\begingroup$ Also, this should all be relative to some fixed background model $V\models ZFC$; as my answer below shows, the answers might be different in different $V$. $\endgroup$ – Noah Schweber Oct 27 '13 at 6:57
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    $\begingroup$ I suppose you assume the existence of large cardinals in V, since if some M is not rigid, then L is not rigid too, so Zero-sharp exists. $\endgroup$ – Mohammad Golshani Oct 27 '13 at 7:17
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    $\begingroup$ To make the question non-trivial I edited your questions by making $M \neq L$ $\endgroup$ – Mohammad Golshani Oct 27 '13 at 7:18
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    $\begingroup$ I think more work is required to make this nontrivial, since e.g., $L[0^\#]$ has the same trivializing properties. $\endgroup$ – Noah Schweber Oct 27 '13 at 7:48
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A. The Dodd-Jensen core model $K$ is absolute and it is non-rigid, assuming $0^{\dagger}$ exists.

I have no idea about larger core models, but maybe some else know the answer w.r.t them.

B. The following result of Jech might be interesting.

Suppose that $j$ is an elementary embedding of the universe. for a submodel N, let $j_N$ be the restriction of $j$ to $N$. Call $N$ admits $j$ if $N \models j_N$ is an elementary embedding of the universe.

Now we have the following:

Theorem. There is a submodel $L(j)$ which is the least submodel which admits $j$.

$L(j)$ is absolute for inner models models which admit $j$

The reference is: Jech, Thomas J. Two remarks on elementary embeddings of the universe. Pacific J. Math. 39 (1971), 395–400.

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  • $\begingroup$ Interesting. Please add the reference paper. $\endgroup$ – user36136 Oct 27 '13 at 7:39
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    $\begingroup$ I've added a link to the paper, I hope that you don't mind. $\endgroup$ – Asaf Karagila Oct 27 '13 at 8:24
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Let me add one thing to the other good answers: If there is no proper class model with a Woodin cardinal then the core model $K$, defined using the background certified construction established in the Core Model Iterability problem (Steel), has no nontrivial elementary embedding into itself. This is a theorem of Steel. $K$ is also universal is some sense: it is the unique universal weasel that elementarily embeds into all other universal weasel (see Schimmerling's handbook article for the definitions) and absolute for set forcing extensions.

In particular, $K^c$, the fine structural countably iterable extender model (using only short extenders) has no sharps, so there is no notrivial elementary embedding of $K^c$ into itself. $K^c$ is constructed by taking a Skolem hull after $K$ is built.

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  • $\begingroup$ No problem Ali. The ideas are developed in "The Core Model Iterability Problem", Steel and in the handbook article of Schimmerling. A recent theorem of Jensen and Steel shows how to remove the "technical assumption" of some $\Omega$ being measurable to define $K$ and prove its absoluteness, maximality, covering and universality properties. $\endgroup$ – Rachid Atmai Oct 27 '13 at 23:15
  • $\begingroup$ Really interesting. $\endgroup$ – user36136 Oct 28 '13 at 0:10
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The answer to both questions is yes, assuming large cardinals, and the example is $L$. Assuming the existence of a measurable, there is a self-embedding of $L$ into $L$; and $L$ as usually defined is absolute and forcing absolute. (Note that there will be, of course, definitions of $L$ in $V$ which are neither absolute nor forcing absolute. For example, if $V$ is a forcing extension of a model $W$ of $CH$, and $V\models\neg CH$, then the formula $$ (CH\wedge x=x)\vee (\neg CH\wedge x\in L)$$ defines $L$ in $V$ but defines $W$ in $W$ and defines $V[G]$ in $V[G]$ if the latter happens to model $CH$. So it's more about the defining formula than the particular model.)

On the other hand, if we take as our background universe the minimal transitive model of $ZFC$, then in that model all inner models are itself; and so (as far as it knows) all inner models whatsoever are rigid. (Although, since we can't directly talk about embeddings of the model itself, within the model, this paragraph is slightly nonsense.)

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The answer is yes to both results. Note that the assertion "$L$ is not rigid" is equivalent to "$0^\#$ exists".

But we can talk about other sharps (for sets of ordinals). Indeed $A^\#$ exists if and only if $L[A]$ is not rigid. Models of the form $L[A]$ are absolute, and so if $A^\#$ exists then $L[A]$ is a definable, forcing absolute, non-rigid inner model.

Now to get slightly more practical, if $\kappa$ is measurable then for every set of ordinals $A$ such that $\sup A<\kappa$, then $A^\#$ exists. One can use that and show that if there exists a supercompact cardinal then every set of ordinals admits a sharp.

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