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Is "ZF + $\omega_1$ is supercompact" consistent relative to "ZFC + there is a supercompact cardinal"?

In particular, if $\delta$ is supercompact, does it remain so in $V(\mathbb{R} \cap V[G])$ where $G \subset Col(\omega,<\delta)$ is $V$-generic? This seems to be the case for measurability but I am having trouble proving it for supercompactness. It seems likely that someone else has tried this, so I though I'd ask here.

The appropriate definition of supercompactness in ZF is the one in terms of normal fine measures, where normality is defined using diagonal intersections.

I am aware that $\omega_1$ has some amount of supercompactness under AD. I am interested in a more direct proof using forcing, which I hope will give (full) supercompactness.

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  • $\begingroup$ Related: mathoverflow.net/questions/87430/… $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 3, 2012 at 20:31
  • $\begingroup$ Ah, I thought I remembered a question like that on here but I couldn't find it. Sorry for posting a similar question. $\endgroup$
    – Trevor Wilson
    Commented Aug 3, 2012 at 20:50
  • $\begingroup$ Trevor, I think that my question may be a bit overly broad. It is just like the time I asked about the ability to destroy weak choice principles (or add some of them) via generic extensions (unlike full choice), and later Stefan Geschke asked about a concrete example. It's a reasonable question when trying to tackle a broad question, I think. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 3, 2012 at 20:54
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    $\begingroup$ The Jech construction preserves supercompactness. This is Lemma 1.3 in Apter-Henle, Large cardinal structures below $\aleph_{\omega}$. $\endgroup$
    – user3462
    Commented Aug 13, 2012 at 12:50
  • $\begingroup$ @TanmayInamdar: Great! Could you post that as an answer? (Maybe including the statement of the relevant theorem?) $\endgroup$
    – Trevor Wilson
    Commented Aug 13, 2012 at 15:43

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While might not be a full and satisfactory answer, you might be interested in the following paper by Spector:

Spector, M. Iterated extended ultrapowers and supercompactness without choice., Ann. Pure Appl. Logic 54 (1991), no. 2, 179–194.

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  • $\begingroup$ Thanks, that mentions a way to extend an ultrapower of Ord taken under ZF to an ultrapower of $V$ in a class generic extension. This would be a useful thing to do if $\omega_1$ were supercompact. It looks like the only example given of supercompactness without choice given in the article comes from AD though. $\endgroup$
    – Trevor Wilson
    Commented Aug 3, 2012 at 20:45
  • $\begingroup$ Trevor, I had several chats over a cup of coffee with Magidor on the problem of an inner model for supercompactness. If you think about Solovay's model it has an $L(\mathbb R)$ sort of construction, and in a sense if we make $\aleph_1$ measurable we essentially take some inner model of measurability. However supercompactness does not yet have such canonical inner model, so I don't think that it's that easy to construct a model of $\aleph_1$ being supercompact for all $\lambda$. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 3, 2012 at 20:51
  • $\begingroup$ If you have an idea how to uniformly preserve the supercompact measures on a proper class of cardinals, maybe I can help with the symmetries themselves later on. However if the idea works it seems reasonable that it may solve (or at least advance) the work towards a canonical inner model for supercompactness. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 3, 2012 at 20:52
  • $\begingroup$ I was hoping that the inner model would just be $V(\mathbb{R} \cap V[G])$. Doesn't this work for measurability? Every subset of $\delta$ in this model is added by some proper initial segment of $G$. So we have a measure on $\delta$ given by the union of the small-forcing extensions of the original measure. This is countably complete, again because every countable sequence is added by a proper initial segment of the forcing. But if there is a countably complete measure on $\delta$ then there is a normal measure. $\endgroup$
    – Trevor Wilson
    Commented Aug 3, 2012 at 21:00
  • $\begingroup$ ...This argument fails to adapt to supercompactness in two ways: first, the set being measured is not added by any proper initial segment of the forcing, and second, normality does not come for free with $\mathcal{P}_\kappa(\lambda)$. Could you explain how the question relates to the inner model problem for supercompact cardinals? $\endgroup$
    – Trevor Wilson
    Commented Aug 3, 2012 at 21:03

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