# Large cardinals need not be large in HOD

• Y. Cheng, S. Friedman, and J. D. Hamkins, “Large cardinals need not be large in HOD,” Annals of Pure and Applied Logic, vol. 166, iss. 11, pp. 1186-1198, 2015.
@ARTICLE{ChengFriedmanHamkins2015:LargeCardinalsNeedNotBeLargeInHOD,
title = "Large cardinals need not be large in {HOD} ",
journal = "Annals of Pure and Applied Logic ",
volume = "166",
number = "11",
pages = "1186 - 1198",
year = "2015",
note = "",
issn = "0168-0072",
doi = "10.1016/j.apal.2015.07.004",
eprint = {1407.6335},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/large-cardinals-need-not-be-large-in-hod},
author = "Yong Cheng and Sy-David Friedman and Joel David Hamkins",
keywords = "Large cardinals",
keywords = "HOD",
keywords = "Forcing",
keywords = "Absoluteness ",
abstract = "Abstract We prove that large cardinals need not generally exhibit their large cardinal nature in HOD. For example, a supercompact cardinal κ need not be weakly compact in HOD, and there can be a proper class of supercompact cardinals in V, none of them weakly compact in HOD, with no supercompact cardinals in HOD. Similar results hold for many other types of large cardinals, such as measurable and strong cardinals.",
}

Abstract. We prove that large cardinals need not generally exhibit their large cardinal nature in HOD. For example, a supercompact cardinal $\kappa$ need not be weakly compact in HOD, and there can be a proper class of supercompact cardinals in $V$, none of them weakly compact in HOD, with no supercompact cardinals in HOD. Similar results hold for many other types of large cardinals, such as measurable and strong cardinals.

In this article, we prove that large cardinals need not generally exhibit their large cardinal nature in HOD, the inner model of hereditarily ordinal-definable sets, and there can be a divergence in strength between the large cardinals of the ambient set-theoretic universe $V$ and those of HOD. Our general theme concerns the questions:

Questions.

1. To what extent must a large cardinal in $V$ exhibit its large cardinal properties in HOD?

2. To what extent does the existence of large cardinals in $V$ imply the existence of large cardinals in HOD?

For large cardinal concepts beyond the weakest notions, we prove, the answers are generally negative. In Theorem 4, for example, we construct a model with a supercompact cardinal that is not weakly compact in HOD, and Theorem 9 extends this to a proper class of supercompact cardinals, none of which is weakly compact in HOD, thereby providing some strongly negative instances of (1). The same model has a proper class of supercompact cardinals, but no supercompact cardinals in HOD, providing a negative instance of (2). The natural common strengthening of these situations would be a model with a proper class of supercompact cardinals, but no weakly compact cardinals in HOD. We were not able to arrange that situation, however, and furthermore it would be ruled out by Conjecture 13, an intriguing positive instance of (2) recently proposed by W. Hugh Woodin, namely, that if there is a supercompact cardinal, then there is a measurable cardinal in HOD. Many other natural possibilities, such as a proper class of measurable cardinals with no weakly compact cardinals in HOD, remain as open questions.

## One thought on “Large cardinals need not be large in HOD”

1. I read your very interesting paper and liked it too much. I have some thoughts regarding question 14 of the paper: Can there be a supercompact cardinal in a forcing extension
V [G], if there are no measurable cardinals in V ?

The basic idea would be something like this: Start with $\kappa$ supercompact and do an iteration to add Cohen sets to each inaccessible less or equal $\kappa$ to get the
final extension. Now find some intermediate model, by noting for inaccessible $\alpha, Add(\alpha, 1)$ can be written as two step iteration, first adding an $\alpha$-Souslin tree, and then forcing with that tree. The intermediate model is an iteration with the Souslin parts.

But this idea does not work for trivial reasons. But what if instead of iteration, we have a kind of product. Then we would be able to do the job.

Now, there is a paper by Mack Stanley “Notes on a theorem of Silver” (see http://www.math.sjsu.edu/~stanley/gch.pdf) which gives a proof of Silver’s theorem by a kind of forcing which is not an iteration and can be considered as product.

I don’t see if his forcing adds Cohen subsets to weakly compacts below the supercompact. But maybe a modification of his method can be used to answer the above question and similar ones.