I'm aware that the HOD conjecture is implied by the Ultimate-L conjecture, but I don't know what the evidence is for the Ultimate-L conjecture. On the other hand, I'm aware the evidence against the HOD conjecture is that the consistency of ZF + Berkeley cardinals disproves the HOD conjecture, and the evidence from the theory of Berkeley cardinals seems to indicate that they are consistent with ZF. What is the main evidence for both sides, and what is the expert consensus at the moment on the HOD conjecture?
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11
I believe it is fair to say that the HOD Conjecture has been refuted.
The HOD Conjecture is the statement that the theory ZFC + "There is an extendible cardinal" proves that there is a proper class of regular cardinals that are not measurable in HOD. It is saying that the HOD Dichotomy Theorem is not a dichotomy after all, and the part that says "V is far from HOD" is proven false by the above theory. The HOD Hypothesis (HOD-H) is the statement that that there is a proper class of regular cardinals that are not measurable in HOD. So the Conjecture is that the above theory proves the Hypothesis.
The HOD Conjecture is equivalent to saying that ZFC + "There is an extendible cardinal" is not consistent with the statement that a cobounded class of regular cardinals are measurable in HOD. In general, the way one refutes such a claim of inconsistency is to exhibit a model of the theory in question. Since the negation of the HOD Conjecture is that a theory of high strength is consistent, the method, and the only possible method, to demonstrate it is to use some standard large cardinal hypothesis to construct a model.
As reported by Goldberg, Woodin showed that a model of ZFC + an extendible + not-(HOD-H) can be forced from a model of ZF + a Reinhardt + a proper class of supercompacts. More recently, Aguilera, Bagaria, and Lücke showed that some new large cardinal hypothesis (A), which is consistent with AC, implies not-(HOD-H). They show that the consistency of ZFC+A follows from the consistency of ZF + some standard choiceless large cardinal notions.
Now one could say that the HOD Conjecture has not been refuted because maybe those large cardinal notions are inconsistent. But without a proof of inconsistency, that is just not how this game is played.
answered Nov 19 at 10:07
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4$\begingroup$ I disagree that one can say the HOD conjecture has been refuted. If one can say this, one could have said this when Woodin first posed the conjecture, since it was known then that the consistency of choiceless cardinals refutes the HOD conjecture. $\endgroup$ Commented Nov 20 at 12:51
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5$\begingroup$ There are various large-cardinal-like principles, consistent with AC from $I_0$ or less, that refute the HOD conjecture in the presence of e.g. a strongly compact cardinal. E.g., $\omega$-strongly measurables, or ultraexact cardinals, or $j : \text{HOD}\cap V_{\lambda+2}\to \text{HOD}\cap V_{\lambda+2}$. This has been known since the conjecture was posed in Woodin's Suitable extender models, I; see e.g., Theorem 200 there. A consequence of the HOD conjecture is that the existence of such large cardinals above a strongly compact is inconsistent with ZFC. $\endgroup$ Commented Nov 20 at 12:59
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4$\begingroup$ I think the HOD conjecture is false, because I think Reinhardt cardinals, Berkeley cardinals, whatever are probably consistent. In my opinion, the evidence of this is much weaker than the evidence that $I_0$ is consistent. One can try to argue that the HOD conjecture is false by providing more evidence of the consistency of choiceless cardinals. This is not, however, a mathematical proof that the HOD conjecture is false, just as we cannot mathematically refute those who conjecture that PA is inconsistent. $\endgroup$ Commented Nov 20 at 13:05
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6$\begingroup$ Regarding how the game is played: it could be that the HOD conjecture (that ZFC + supercompact proves the HOD hypothesis) is refutable from Con(ZFC + supercompact) or even from an extendible or $I_0$ cardinal. This would actually refute the HOD conjecture, since the conjecture implicitly takes for granted the consistency of such cardinals. The conjecture explicitly rejects the consistency of ultraexacts and Berkeleys and so on, so it seems the game is to refute it from one of the traditional axioms. $\endgroup$ Commented Nov 20 at 13:22
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5$\begingroup$ It’s true that Reinhardt cardinals have been studied for a long time, but not really intensively in the way supercompact or Woodin cardinals or even I0 have been. It is hard to get started, given the difficulties of working in pure ZF. $\endgroup$ Commented Nov 20 at 17:23