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

The XI International Workshop on Set Theory took place October 4-8, 2010. It was hosted by the CIRM, in Luminy, France. I am very glad I was invited, since it was a great experience: The Workshop has a tradition of excellence, and this time was no exception, with several very nice talks. I had the chance to give a talk (available here) and to interact with the other participants. There were two mini-courses, one by Ben Miller and one by Hugh Woodin. Ben has made the slides of his series available at his website.

What follows are my notes on Hugh’s talks. Needless to say, any mistakes are mine. Hugh’s talks took place on October 6, 7, and 8. Though the title of his mini-course was “Long extenders, iteration hypotheses, and ultimate L”, I think that “Ultimate L” reflects most closely the content. The talks were based on a tiny portion of a manuscript Hugh has been writing during the last few years, originally titled “Suitable extender sequences” and more recently, “Suitable extender models” which, unfortunately, is not currently publicly available.

The general theme is that appropriate extender models for supercompactness should provably be an ultimate version of the constructible universe $L$. The results discussed during the talks aim at supporting this idea.

## Ultimate L

I

Let $\delta$ be supercompact. The basic problem that concerns us is whether there is an $L$-like inner model $N\subseteq V$ with $\delta$ supercompact in $N$.

Of course, the shape of the answer depends on what we mean by “$L$-like”. There are several possible ways of making this nontrivial. Here, we only adopt the very general requirement that the supercompactness of $\delta$ in $N$ should “directly trace back” to its supercompactness in $V$.

Recall:

• We use ${\mathcal P}_\delta(X)$ to denote the set $\{a\subseteq X\mid |a|<\delta\}$.
• An ultrafilter (or measure) ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$ is fine iff for all $\alpha<\lambda$ we have $\{a\in{\mathcal P}_\delta(\lambda)\mid \alpha\in a\}\in{\mathcal U}$.
• The ultrafilter ${\mathcal U}$ is normal iff it is $\delta$-complete and for all $F:{\mathcal P}_\delta(\lambda)\to\lambda$, if $F$ is regressive ${\mathcal U}$-ae (i.e., if $\{a\mid F(a)\in a\}\in{\mathcal U}$) then $F$ is constant ${\mathcal U}$-ae, i.e., there is an $\alpha<\lambda$ such that $\{a\mid F(a)=\alpha\}\in{\mathcal U}$.
• $\delta$ is supercompact iff  for all $\lambda$ there is a normal fine measure ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$.

It is a standard result that $\delta$ is supercompact iff for all $\lambda$ there is an elementary embedding $j:V\to M$ with ${\rm cp}(j)=\delta$, $j(\delta)>\lambda$, and $j''\lambda\in M$ (or, equivalently, ${}^\lambda M\subseteq M$).

In fact, given such an embedding $j$, we can define a normal fine ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$ by

$A\in{\mathcal U}$ iff $j''\lambda\in j(A)$.

Conversely, given a normal fine ultrafilter ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$, the ultrapower embedding generated by ${\mathcal U}$ is an example of such an embedding $j$. Moreover, if ${\mathcal U}_j$ is the ultrafilter on ${\mathcal P}_\delta(\lambda)$ derived from $j$ as explained above, then ${\mathcal U}_j={\mathcal U}$.

Another characterization of supercompactness was found by Magidor, and it will play a key role in these lectures; in this reformulation, rather than the critical point, $\delta$ appears as the image of the critical points of the embeddings under consideration. This version seems ideally designed to be used as a guide in the construction of extender models for supercompactness, although recent results suggest that this is, in fact, a red herring.

The key notion we will be studying is the following:

Definition. $N\subseteq V$ is a weak extender model for $\delta$ is supercompact’ iff for all $\lambda>\delta$ there is a normal fine ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$ such that:

1. ${\mathcal P}_\delta(\lambda)\cap N\in {\mathcal U}$, and
2. ${\mathcal U}\cap N\in N$.

This definition couples the supercompactness of $\delta$ in $N$ directly with its supercompactness in $V$. In the manuscript, that $N$ is a weak extender model for $\delta$ is supercompact’ is denoted by $o^N_{\rm long}(\delta)=\infty$. Note that this is a weak notion indeed, in that we are not requiring that $N=L[\vec E]$ for some (long) sequence $\vec E$ of extenders. The idea is to study basic properties of $N$ that follow from this notion, in the hopes of better understanding how such an $L[\vec E]$ model can actually be constructed.

For example, fineness of ${\mathcal U}$ already implies that $N$ satisfies a version of covering: If $A\subseteq\lambda$ and $|A|<\delta$, then there is a $B\in{\mathcal P}_{\delta}(\lambda)\cap N$ with $A\subseteq B$. But in fact a significantly stronger version of covering holds. To prove it, we first need to recall a nice result due to Solovay, who used it to show that ${\sf SCH}$ holds above a supercompact.

Solovay’s Lemma. Let $\lambda>\delta$ be regular. Then there is a set $X$ with the property that the function $f:a\mapsto\sup(a)$ is injective on $X$ and, for any normal fine measure ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$, $X\in{\mathcal U}$.

It follows from Solovay’s lemma that any such ${\mathcal U}$ is equivalent to a measure on ordinals.

Proof. Let $\vec S=\left< S_\alpha\mid\alpha<\lambda\right>$ be a partition of $S^\lambda_\omega$ into stationary sets.

(We could just as well use $S^\lambda_{\le\gamma}$ for any fixed $\gamma<\delta$. Recall that

$S^\lambda_{\le\gamma}=\{\alpha<\lambda\mid{\rm cf}(\alpha)\le\gamma\}$

and similarly for $S^\lambda_\gamma=S^\lambda_{=\gamma}$ and $S^\lambda_{<\gamma}$.)

It is a well-known result of Solovay that such partitions exist.

Hugh actually gave a quick sketch of a crazy proof of this fact: Otherwise, attempting to produce such a partition ought to fail, and we can therefore obtain an easily definable $\lambda$-complete ultrafilter ${\mathcal V}$ on $\lambda$. The definability in fact ensures that ${\mathcal V}\in V^\lambda/{\mathcal V}$, contradiction. We will encounter a similar definable splitting argument in the third lecture.

Let $X$ consist of those $a\in{\mathcal P}_\delta(\lambda)$ such that, letting $\beta=\sup(a)$, we have ${\rm cf}(\beta)>\omega$, and

$a=\{\alpha<\beta\mid S_\alpha\cap\beta$ is stationary in $\beta\}$.

Then $f$ is 1-1 on $X$ since, by definition, any $a\in X$ can be reconstructed from $\vec S$ and $\sup(a)$. All that needs arguing is that $X\in{\mathcal U}$ for any normal fine measure ${\mathcal U}$ on ${\mathcal P}_\delta(\lambda)$. (This shows that to define ${\mathcal U}$-measure 1 sets, we only need a partition $\vec S$ of $S^\lambda_\omega$ into stationary sets.)

Let $j:V\to M$ be the ultrapower embedding generated by ${\mathcal U}$, so

${\mathcal U}=\{A\in{\mathcal P}_\delta(\lambda)\mid j''\lambda\in j(A)\}$.

We need to verify that $j''\lambda\in j(X)$. First, note that $j''\lambda\in M$. Letting $\tau=\sup(j''\lambda)$, we then have that $M\models{\rm cf}(\tau)=\lambda$. Since

$M\models j(\lambda)\ge\tau$ is regular,

it follows that $\tau. Let $\left=j(\left)$. In $M$, the $T_\beta$ partition $S^{j(\lambda)}_\omega$ into stationary sets. Let

$A=\{\beta<\tau\mid M\models T_\beta\cap\tau\mbox{\ is stationary}\}.$

The point is that $A=j''\lambda$.

To prove this, note first that ${}^\lambda M\subseteq M$ and that $j''\lambda$ is an $\omega$-club of $\tau$, since $j$ is continuous at ${\rm cof}(\omega)$ points. Thus, for all $\alpha<\lambda,$ we have $j(S_\alpha)\cap\tau\supseteq j''S_\alpha$ and it follows that $j(S_\alpha)$ is stationary in $\tau$. Hence $A\supseteq j''\lambda$.

Since $j''\lambda\in M$, then $\{T_{j(\alpha)}\mid\alpha<\lambda\}=\{j(S_\alpha)\mid\alpha<\lambda\}\in M$. But $\bigcup\{j(S_\alpha)\colon\alpha<\lambda\}\supseteq j''\lambda\cap{\rm cof}(\omega)$, and this is an $\omega$-club. It follows that no other $T_\beta$ can meet $\tau$ stationarily. So $A=j''\lambda$, and this completes the proof. $\Box$

Solovay’s lemma suggests that perhaps it is possible to build $L[\vec E]$ models for supercompactness in a simpler way than anticipated, by using ultrafilters on ordinals to witness supercompactness.

Our key application of the lemma is the following (which, Hugh points out, could easily have been discovered right after Solovay’s lemma was established):

Corollary. Suppose $N$ is a weak extender model for $\delta$ is supercompact’. Suppose $\gamma>\delta$ is a singular cardinal. Then:

1. $\gamma$ is singular in $N$.
2. $(\gamma^+)^N=\gamma^+$.

Note that item 1. is immediate from covering if ${\rm cf}(\gamma)<\delta$, but a different argument is needed otherwise. Item 2. is a very $L$-like property of $N$. It is not clear to what extent there is a non-negligible (in some sense) class of cardinals for which $N$ computes their cofinality correctly.

Proof. This is immediate from Solovay’s lemma. Both 1. and 2. follow at once from:

$*$ If $\lambda>\delta$ is regular in $N$, then ${\rm cf}^V(\lambda)=|\lambda|^V$.

$(*\Rightarrow 1.)$ If $\gamma>\delta$ is singular but regular in $N$, then ${\rm cf}^V(\gamma)=|\gamma|^V$, but this is impossible since $\gamma$ is singular.

$(*\Rightarrow 2.)$ If $\gamma>\delta$ is singular but $(\gamma^+)^N<\gamma^+$, then ${\rm cf}^V((\gamma^+)^N)=|\gamma|^V$, contradicting that $\gamma$ is singular.

It remains to establish $*$. For this, we use Solovay’s lemma within $N$.

Let ${\mathcal U}$ be a normal fine ultrafilter on ${\mathcal P}_\delta(\lambda)$ such that ${\mathcal U}\cap N\in N$ and $N\cap{\mathcal P}_\delta(\lambda)\in{\mathcal U}$. Note that such ${\mathcal U}$ exists, even if $\lambda$ is not a cardinal in $V$: Just pick a larger regular cardinal in $V$, and project the appropriate measure.

By Solovay’s lemma there is $X\in{\mathcal U}\cap N$ such that $a\mapsto\sup(a)$ is 1-1 on $X$. Suppose that ${\rm cf}^V(\lambda)<|\lambda|^V$. In $V$, let $C\subseteq\lambda$ be club, ${}|C|={\rm cf}^V(\lambda)$. Then $\{a\in X\mid\sup(a)\in C\}\in{\mathcal U}$ since $\sup(j''\lambda)\in j(C)$ for $j$ the ultrapower embedding induced by ${\mathcal U}$. However, if $Y=\{a\in X\mid\sup(a)\in C\}$, then ${}|Y|={\rm cf}^V(\lambda)<|\lambda|^V$ while $\lambda\subseteq\bigcup Y$, by fineness. Contradiction. $\Box$

It follows that if $\delta$ is supercompact in $V$ and in a forcing extension a $V$-regular $\gamma>\delta$ turns into singular while measures on all ${\mathcal P}_\delta(\lambda)$ in $V$ lift (so, in particular, supercompactness of $\delta$ is preserved in the extension), then $\gamma$ is no longer a cardinal in the extension.

We arrive at a key notion. Say that an inner model $M$ is universal iff (sufficiently) large cardinals relativize down to $M$. The corollary seems to suggest that weak extender models for supercompactness ought to be universal, so solving the inner model problem for supercompactness essentially solves the problem for all large cardinals. In fact, we have:

Universality Theorem. Suppose $N$ is a weak extender model for $\delta$ is supercompact’. Suppose $\gamma>\delta$, $\pi:N\cap V_{\gamma+1}\to N\cap V_{\pi(\gamma)+1}$ is elementary, and ${\rm cp}(\pi)\ge\delta$. Then $\pi\in N$.

We will present the proof in the next lecture. In brief: Any extender that coheres with $N$ and has large critical point is in $N$. To see why this is a universality result, notice for example that if in $V$ there is a proper class of $k$-huge cardinals (for all $k<\omega$), then there is such a class in $N$. Contrast this with the traditional situation in inner model theory, where inner models for a large cardinal notion do not capture any larger notions. (Similar results hold for rank into rank embeddings and larger, though some additional ideas are required here.)

In a sense, the universality theorem says that $N$ must be rigid. This is not literally true, but it is in the appropriate sense that there can be no sharps for $N$:

Corollary. Suppose $N$ is an extender model for $\delta$ is supercompact’. Then there is no $j:N\to N$ with ${\rm cp}(j)\ge\delta$.

Proof. Otherwise, $j$ is amenable to $N$, by the universality theorem. But then $N\models\exists i:N\to N$, contradicting Kunen’s theorem. $\Box$

(This is another $K$-like feature that $N$ inherits.) Note the restriction to ${\rm cp}(j)\ge\delta$. This cannot be removed:

Example. Suppose $\delta$ is supercompact and $\delta_0<\delta$ is measurable. Let ${\mathcal U}$ be a normal measure on $\delta_0$, and let $M_\omega$ be the $\omega$-th iterate of the ultrapower embedding $j:V\to M_0\cong V^{\delta_0}/{\mathcal U}$. Then:

1. $M_\omega$ is a weak extender model for $\delta$ is supercompact’.
2. $j(M_\omega)=M_\omega$, so we cannot drop “${\rm cp}(j)\ge\delta$” in the Corollary.
3. Let $N=M_\omega[\left<\delta_i\mid i<\omega\right>]$ where $\left<\delta_i\mid i<\omega\right>$ is the critical sequence ($\delta_{i+1}=j(\delta_i)$ for all $i$). Then $N=\bigcap_i M_i$ where $M_i$ is the $i$-th iterate of $j$. It follows that $N$ is closed under $\delta_0$-sequences. Since $N$ is a forcing extension of $M_\omega$ by small forcing (Prikry forcing), $N$ is also a weak extender model for $\delta$ is supercompact’, and clearly $j(N)=N$ as well. Hence, “${\rm cp}(j)\ge\delta$” cannot be dropped from the Corollary, even if we require some form of strong closure of $N$.

We are now in the position to state a key dichotomy result, the proof of which will occupy us in the third lecture.

Definition. $\delta$ is extendible if for all $\alpha$ there is $j:V_{\delta+\alpha+1}\to V_{j(\delta)+j(\alpha)+1}$ with ${\rm cp}(j)=\delta$ and $j(\delta)>\alpha$.

Lemma. Assume $\delta$ is extendible. The following are equivalent:

1. ${\sf HOD}$ is a weak extender model for $\delta$ is supercompact’.
2. There is a regular $\gamma>\delta$ that is not measurable in ${\sf HOD}$.
3. There is a $\gamma>\delta$ such that $(\gamma^+)^{\sf HOD}=\gamma^+$.

Note that this is indeed a dichotomy result: In the presence of extendible cardinals, either ${\sf HOD}$ is very close to $V$, or else it is very far.

Conjecture. If $\delta$ is extendible, then ${\sf HOD}$ is an extender model for `$\delta$ is supercompact’.

Let us close with a brief description of the proof of the Dichotomy Lemma. Note we already have that items 2. and 3. follow from 1. To prove $(2.\Rightarrow 1.)$, given $\gamma$, we consider the $\omega$-club filter on $\gamma$, and try in ${\sf HOD}$ to split $S^\gamma_\omega$ into stationary sets in $V$. Failure of this will give us that $\gamma$ is measurable in ${\sf HOD}$. Assuming 2., this means we succeed, and we will use the stationary sets to verify that normal fine measures on ${\mathcal P}_\delta(\lambda)$ are absorbed into ${\sf HOD}$. Then extendibility will give us a proper class of such $\gamma$, and item 1. follows.

### 8 Responses to Luminy – Hugh Woodin: Ultimate L (I)

1. […] For the first lecture, see here. […]

2. […] Luminy – Hugh Woodin: Ultimate L (III) For the first lecture, see here. […]

3. […] Miller and Hugh Woodin. Notes taken by Andrés Caicedo of Woodin’s mini course can be found here here on Caicedo’s blog. Below are the available slides and […]

4. […] Well, that is what I wanted to write about – Hugh Woodin‘s theories of infinity, and indeed I found a blog that discussed a lot of this – https://caicedoteaching.wordpress.com/2010/10/19/luminy-hugh-woodin-ultimate-l-i/ […]

5. […] Caicedo also has posted his notes from a workshop given by Woodin in 2010: http://caicedoteaching.wordpress….Here are some slides by Matthew Foreman that explain his thoughts on Woodin's approach: […]

6. […] between (suitable) theories where CH holds and where it fails. That is what Woodin‘s “Ultimate L” theory would really accomplish: […]

7. […] For the first lecture, see here. […]

8. […] For the first lecture, see here. […]