Remark: This question is based on an open question at the end of a paper by Hamkins, Kirmayer, and Perlmutter: "Generalizations of the Kunen Inconistency".

$HOD$ as an inner model of $ZFC$ lies between $L$ and $V$. But its nature shows a wide degree of flexibility between $L$ -like and $V$-like behaviors up to different problems.

One of the problems about any inner model of $ZFC$ is consistency of existence of a non-trivial self elementary embedding. In this direction the following facts are well known:

(a) There is no non-trivial elementary embedding from $V$ to $V$.

(b) Existence of a non-trivial elementary embedding from $L$ to $L$ is an acceptable large cardinal axiom.

Here the open question is:

Question (1): Is the existence of a non-trivial elementary embedding from $HOD$ to $HOD$ an acceptable large cardinal axiom?

I usually discuss with my colleagues about it. I believe that $HOD$ in this problem has a $L$ - like behavior and existence of a non-trivial elementary embedding from $HOD$ to $HOD$ is an acceptable (not too) large cardinal axiom but one of my friends insists that the facts shows there is no such embedding. Finally we bet on it. Here I want to know about any known partial result around the open problem. So:

Question (2): Please introduce any known fact related to the possible answer of question (1) and explain that which one of the scenarios (a) or (b) for $HOD$ seems more possible by this fact.

  • $\begingroup$ If I understand correctly, then one can get a model of ZFC with an elementary embedding from HOD into HOD assuming that a strong large cardinal axiom related to Reinhardt cardinals is consistent with ZF (see J1 in en.wikipedia.org/wiki/Reinhardt_cardinal). However, currently (I believe) the tide of general opinion is against Reinhardts in ZF; so this might not answer your question. $\endgroup$ – Noah Schweber Oct 26 '13 at 21:55
  • $\begingroup$ Related: the consistency with ZFC of "HOD into HOD" is currently unknown: cantorsattic.info/Kunen_inconsistency. $\endgroup$ – Noah Schweber Oct 26 '13 at 21:57
  • $\begingroup$ @Noah: Thanks. What is your personal idea about it? Is it consistent or inconsistent? $\endgroup$ – user36136 Oct 26 '13 at 22:02
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    $\begingroup$ Woodin conjectures that the answer is no, there is no non-trivial elementary embedding from HOD to itself. This follows from the existence of an extendible cardinal, assuming a separate conjecture known as the HOD conjecture. There is a set of notes titled the HOD Dichotomy that goes into these matters in some detail and the notes are actually pretty accessible. You may also want to look at a paper by Hamkins, Kirmayer, and Perlmutter: "Generalizations of the Kunen Inconistency" $\endgroup$ – Everett Piper Oct 26 '13 at 23:01
  • $\begingroup$ @Everett: So existence of such embedding refutes the existence of some very large cardinals like extendibles. Now it seems I should change my side in the betting! $\endgroup$ – user36136 Oct 26 '13 at 23:06

I would personally bet against the existence of a nontrivial embedding j: HOD --> HOD, but it's just a guess. One elementary point that's mentioned in our Generalizations of the Kunen inconsistency paper, but which I think is worth repeating, is that the Kunen inconsistency has not been proven in the choiceless context. if there exists an elementary embedding j: V --> V, without choice, then restricting this embedding to HOD would give an embedding from HOD to HOD. So contrapositively, if one could prove that there is no nontrivial embedding from HOD to HOD (without using choice in V), this would also show that there is no nontrivial embedding from V to V in the choiceless context.

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  • $\begingroup$ Dear Norman, I am really happy to see you here. It seems that the role of Hajnal-Erdos lemma in classic proof of Kunen inconsistency theorem is very essential. So there is a little hope to prove this theorem without axiom of choice. Danke for your answer. $\endgroup$ – user36136 Oct 28 '13 at 18:42
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    $\begingroup$ Implicit in your belief is somehow the notion that $\sf HOD$ should be very close to $V$, even if choice fails in $V$. Going to the extreme, consider Gitik's model where all the limit ordinals have countable cofinality. How could that model be anywhere close to $\sf HOD$? (While that model certainly doesn't serve as an example to an elementary embedding $j\colon V\to V$, it is still an example of how very far $V$ and $\sf HOD$ can be.) $\endgroup$ – Asaf Karagila Oct 28 '13 at 21:14
  • $\begingroup$ @AsafKaragila: Does this mean that you are in my side at betting? One of the fans of L-like HOD? $\endgroup$ – user36136 Oct 28 '13 at 23:33
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    $\begingroup$ @Ali: I'm still trying to work my intuition on that, but if you'd put a gun to my whisky collection and force me to make a choice, I'd save my Glen Livet and choose that $V$ can be very far from $\sf HOD$. $\endgroup$ – Asaf Karagila Oct 28 '13 at 23:35
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    $\begingroup$ Dear Assaf and Ali, of course HOD can be very far from V, but my belief is that HOD becomes closer to V than L as long as you have larger and larger large cardinals. Woodin's HOD conjecture is in this direction. $\endgroup$ – Mohammad Golshani Oct 29 '13 at 2:49

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