## Luminy – Hugh Woodin: Ultimate L (III)

For the first lecture, see here.

For the second lecture, see here.

III

In the previous lecture we established the Universality Theorem, a version of which is as follows:

Theorem. Suppose $N$ is a weak extender model for $\delta$ is supercompact’. If $\pi:H(\gamma^+)^N\to H(\pi(\gamma)^+)^N$ is elementary, with ${\rm cp}(\pi)\ge\delta$ and $\gamma>\delta$, then $\pi\in N$.

More general versions hold, and even can be obtained directly from the argument from last lecture.

For example, suppose that $\delta$ is supercompact and $\kappa>\delta$ is strongly inaccessible. Let ${\mathcal U}$ be a normal fine measure on ${\mathcal P}_{\delta}(\kappa)$, let $A\in{\mathcal U}$, and consider $N=L[A,{\mathcal U}]\cap V_\kappa$. Then, in $V_\kappa$, $N$ is a weak extender model for $\delta$ is supercompact’. This construction typically “inverts” all forcing constructions one may have previously done, while essentially absorbing all large cardinals in $V_\kappa$. Foreman has studied this construction in some detail.

Question. Let $\delta$ be extendible. Is ${\sf HOD}$ a weak extender model for $\delta$ is supercompact’?

Conjecture. This is indeed the case.

To motivate the conjecture, we argue that refuting it must use techniques completely different from what we currently have at our disposal. (A closely related fact is that if $\delta$ is extendible, then it is ${\sf HOD}$-supercompact (i.e., for all $\lambda$ there is a $\lambda$-supercompactness embedding $j:V\to M$ with $j({\sf HOD}\cap V_\lambda)={\sf HOD}\cap V_\lambda$). Sargsyan has verified that extendible cannot be replaced with supercompact in this case.)

Lemma. Suppose that there is a proper class of Woodin cardinals and every ${\sf OD}$ set $A\subseteq {\mathbb R}$ is universally Baire. Then the $\Omega$-conjecture holds in ${\sf HOD}$. $\Box$

This can be seen as evidence towards the conjecture, since the $\Omega$-conjecture holds in all known extender models. Moreover, the lemma is evidence that, if the conjecture holds, then large cardinals cannot refute the $\Omega$-conjecture.

Definition. Suppose $\gamma>\omega$ is regular. Say that $\gamma$ is $\omega$-strongly measurable in ${\sf HOD}$ iff there is a $\lambda<\gamma$ with $(2^\lambda)^{\sf HOD}<\gamma$ for which there is no partition $\left\in{\sf HOD}$ of $(S^\gamma_\omega)^V$ into sets that are stationary in $V$.

Being $\omega$-strongly measurable in ${\sf HOD}$ is a strong requirement on $\gamma$: In that case, we can perform the following procedure: Start with $S=(S^\gamma_\omega)^V$. Working in ${\sf HOD}$, construct a binary tree of splittings of $S$ as follows: Split $S$ into two $V$-stationary sets, both in ${\sf HOD}$, if possible. Then, consider these two sets and, if possible, split each into two $V$-stationary sets in ${\sf HOD}$, and continue this way, taking intersections along branches (in ${\sf HOD}$) at limit stages. Note that the construction is in ${\sf HOD}$ even if it refers to true stationarity, since this can be represented in ${\sf HOD}$ by making reference at each stage to membership in the ${\sf OD}_A$-filter of $\omega$-club subsets of $A\subseteq S$ (for $A$ the stationary set we are trying to split at a given point in the construction).

Suppose the construction lasts $\lambda$ stages. Since $(2^\lambda)^{\sf HOD}<\gamma$, it cannot be that the construction stops because at limit stages we do not see enough branches. Hence it must be that we stop at a successor stage, and this must happen along each path through the tree. As a consequence, we have split $S^\gamma_\omega$ into a small number of stationary sets, all of which carry, in ${\sf HOD}$, a $\gamma$-complete ultrafilter (namely, the restriction of the $\omega$-club filter). This is a very strong way of witnessing the measurability of $\gamma$ in ${\sf HOD}$, and it is quite difficult to mimic this result with forcing.

${\sf HOD}$-Conjecture. There is a proper class of cardinals $\gamma$ that are regular in $V$ and are not $\omega$-strongly measurable in ${\sf HOD}$.

This is a very plausible conjecture:

• It is not known if there can be more than 3 cardinals that are $\omega$-strongly measurable in ${\sf HOD}$.
• It is not known if the successor of a singular of uncountable cofinality can be $\omega$-strongly measurable in ${\sf HOD}$.
• It is not known whether there can be any cardinals above a supercompact that are $\omega$-strongly measurable in ${\sf HOD}$.

The take-home message is that infinitary combinatorics above a supercompact is hard, since supercompactness is extremely fragile.

Theorem. Suppose that $\delta$ is extendible. Then the following are equivalent:

1. ${\sf HOD}$ is a weak extender model for $\delta$ is supercompact’.
2. There is some $\gamma>\delta$ that is not $\omega$-strongly measurable in ${\sf HOD}$.

Hence if item 2. fails, every regular $\gamma>\delta$ is measurable in ${\sf HOD}$ and, in particular, $(\lambda^+)^{\sf HOD}<\lambda^+$ for any $\lambda\ge\delta$.

As mentioned previously, there is a scenario for the failure of item 2.: It can be forced in ${\sf ZF}$ over $V$ if there is a very strong version of Reinhardt cardinals. But this should really be understood as a scenario towards refuting the existence of Reinhardt cardinals in ${\sf ZF}$, at least in the presence of additional strong large cardinal assumptions.

Proof. $(1.\Rightarrow2.)$ This we already know, since in the corollary shown in the first lecture we saw that item 1. implies that ${\sf HOD}$ computes some successors correctly.

$(2.\Rightarrow1.)$ Here we will need to use extendibility. Let $\gamma_0$ be a cardinal witnessing item 2.

Claim. For all $\alpha>\delta$ there are a $\gamma>\alpha$ and a partition $\vec T=\left\in {\sf HOD}$ of $S^\gamma_\alpha$ into stationary sets.

Proof. Fix $\alpha$. Note that for all $\lambda<\delta$ there is a partition in ${\sf HOD}$ of $S^{\gamma_0}_\omega$ into $\lambda$-many stationary sets. Since $\delta$ is extendible, we can find an embedding

$\pi:V_{\delta+\theta+1}\to V_{\pi(\delta)+\pi(\theta)+1}$ with $\theta$ much larger than $\alpha$, ${\rm cp}(\pi)=\delta$, $\pi(\delta)>\alpha$ and ${\sf HOD}^{V_\theta}={\sf HOD}\cap V_\theta$ (for example, we could pick $\theta$ so that $V_\theta\prec_{\Sigma_3}V$).

Since

$V_\theta\models\gamma_0$ is not $\omega$-strongly measurable in ${\sf HOD}$,

then

$V_{\pi(\theta)}\models\pi(\gamma_0)$ is not $\omega$-strongly measurable in $\pi({\sf HOD})$.

But $\pi({\sf HOD}^{V_{\theta}})=\pi({\sf HOD}\cap V_\theta)=\pi({\sf HOD})\cap V_{\pi(\theta)}={\sf HOD}^{V_{\pi(\theta)}}$ and ${\sf HOD}^{V_{\pi(\theta)}}\subseteq{\sf HOD}\cap V_{\pi(\theta)}$.

This gives us the desired result. $\Box$

Fix $\theta>\delta$ with $V_\theta\prec_{\Sigma_{10000}}V$. Then ${\sf HOD}^{V_\theta}={\sf HOD}\cap V_\theta$. Pick an elementary $\pi:V_\theta\to V_{\pi(\theta)}$ with ${\rm cp}(\pi)=\delta$, $\pi(\delta)>\theta$. Note that $\pi({\sf HOD}\cap V_\theta)={\sf HOD}^{V_{\pi(\theta)}}$.

Claim. For all $\lambda<\theta$, $\pi''\lambda\in {\sf HOD}^{V_{\pi(\theta)}}$.

Since for all $\lambda<\theta$ we have $\pi''{\sf HOD}\cap V_\lambda\in{\sf HOD}^{V_{\pi(\theta)}}$, from this is follows that $V_\theta\models{\sf HOD}$ is a weak extender model for $\delta$ is supercompact’.

Proof. Similar to the proof of Solovay’s lemma in Lecture 1. Fix $\lambda$ and choose a regular $\gamma>\lambda$ with $\gamma<\theta$, and a partition $\left\in{\sf HOD}^{V_\theta}$ of $S^\gamma_\omega$ into stationary sets.

Let $\gamma^*=\sup\pi''\gamma$, and note that $\gamma^*<\pi(\gamma)$, as the latter is regular. Let $\vec E=\left=\pi\left$ and note that $\vec E\in{\sf HOD}^{V_{\pi(\theta)}}$ and $\vec E$ is a partition of $S^{\pi(\gamma)}_\omega$ into stationary sets.

Let $\sigma=\{\eta<\gamma^*\mid E_\eta\cap\gamma^*$ is stationary in $\gamma^*\}$, and note that $\sigma\in {\sf HOD}^{V_{\pi(\theta)}}$.

We can now argue that $\sigma=\pi''\lambda$ just as in the proof of Solovay’s lemma. $\Box$

Since $\pi''{\sf HOD}\cap V_\lambda\in {\sf HOD}^{V_{\pi(\theta)}}$ for all $\lambda<\theta$, if we let ${\mathcal U}_\lambda$ be the measure on ${\mathcal P}_\delta(\lambda)$ derived from $\pi$, we have that ${\mathcal U}_\lambda$ concentrates on ${\sf HOD}$, and its restriction to ${\sf HOD}$ is in ${\sf HOD}$.

This proves that $V_\theta\models{\sf HOD}$ is a weak extender model for $\delta$ is supercompact’. But then we are done, by elementarity. $\Box$

Let us close with some general and sober remarks that Hugh made on how one would go about building extender models. These coarse models use extenders from $V$ (as in the requirement for weak extender models), and typically their analysis suggests how to proceed to their fine-structural counterpart.

When looking at the coarse version for supercompactness, as mentioned before, Magidor’s reformulation is ideally suited to build the models, and this was the original approach of the `suitable extender sequences’ manuscript. Recent results indicate that comparison fails for these models past superstrongs, and in fact, all of $V$ can be coded into these models. This is a serious obstacle to a fine-structural version.

Current results suggest that even if one modifies this approach and directly uses as the extenders in the sequence some measures on ordinals to code supercompactness (which is possible, by Solovay’s lemma), comparison should fail as well around $\kappa^{++}$-supercompactness. This suggests two scenarios, neither particularly appealing: Either iterability (in very general terms) fails, which would force us to completely change the nature of fine-structure theory before we can solve the inner model program for supercompactness, or else the construction of the models collapses quickly, and so a different not yet foreseen approach would be required.

### 7 Responses to Luminy – Hugh Woodin: Ultimate L (III)

1. […] Luminy – Hugh Woodin: Ultimate L (III) « Teaching blog says: October 27, 2010 at 4:28 pm […]

2. Peter Komjath says:

Lovely exposition, well done! I hope it is not just me, who learned a lot from these 3 posts.

3. andso0n says: