Thursday, February 25, 2016

UCI Summer School, part 7: Sacks forcing (Brent Cody)

This lecture will introduce some basic properties of Sacks forcing for uncountable inaccessible cardinals, and examine an Easton support iteration of such forcing.

The Sacks forcing on $\omega$ adds a real of minimal constructibility degree, and crucially satisfies a fusion property. Although this was reviewed in the summer school, I'm going to omit the discussion for this post.

Instead we will start with Sacks forcing on uncountable cardinals, which traces back to Kanamori (1980), where using $\diamond_\kappa$ it was shown that long products and iterations of $\mathrm{Sacks}(\kappa)$ preserve $\kappa^+$.

Definition: We say $p\subseteq 2^{<\kappa}$ is a perfect $\kappa$-tree if:
  1. If $s\in p$ and $t\subseteq s$ then $t\in p$.
  2. If $\langle s_\alpha:\alpha<\eta\rangle$ is a sequence of nodes in $p$, then $s=\bigcup_{\alpha<\eta} s_\alpha\in p$.
  3. For every $s\in p$ there is $t\supset s$ with $t\frown 0, t\frown 1\in p$.
  4. Let $\mathrm{Split}(p)=\{s\in p: s\frown 0, s\frown 1\in p\}$. Then for some unique club $C(p)\subseteq \kappa$, we have $$\mathrm{Split}(p)=\{s\in p: \mathrm{length}(s)\in C(p)\}.$$
$\mathrm{Sacks}(\kappa)$ is the poset of perfect $\kappa$-trees ordered by inclusion. We think of the generic subset of $\kappa$ added by $\mathrm{Sacks}(\kappa)$ as the intersection of the trees in the generic filter.

The only surprising thing in the generalization is (4): splitting happens for every node on certain levels, which form a club in $\kappa$.

Exercise: $\mathrm{Sacks}(\kappa)$ is $<\kappa$-closed.

Assume $\kappa>\omega$ is inaccessible. Then $\mathrm{Sacks}(\kappa)$ is $\kappa^{++}$-c.c. We will really only consider this case.

Definition: $\mathrm{Split}_\alpha(p)$ is the set of all nodes $s\in p$ with $\mathrm{length}(s)=\beta_\alpha$, where $\langle \beta_\alpha:\alpha<\kappa\rangle$ is an enumeration of $C(p)$, i.e., the level of $p$ at the $\alpha$th member of $C(p)$.

For $p,q\in \mathrm{Sacks}(\kappa)$, write $p\le_\beta q$ iff $p\le q$ and $\mathrm{Split}_\alpha(p)=\mathrm{Split}_\alpha(q)$ for all $\alpha<\beta$.

A descending sequence $\langle p_\alpha:\alpha<\kappa\rangle$ in $\mathrm{Sacks}(\kappa)$ is a fusion sequence if for all $\alpha<\kappa$, $p_\alpha\le_\alpha p_\alpha$.

Lemma (fusion lemma): If $\langle p_\alpha:\alpha<\kappa\rangle$ is a fusion sequence, then $p=\bigcap_{\alpha<\kappa} p_\alpha$ is a lower bound in $\mathrm{Sacks}(\kappa)$.

Proof: exercise. Hint: show that any node $p$ in the intersection is in a cofinal branch of the intersection.

This important lemma affords us a kind of $\kappa^+$ closure, with the catch that we require more of our decreasing sequence. We can see this in action in the next lemma.

Lemma: $\mathrm{Sacks}(\kappa)$ preserves $\kappa^+$.

Proof: If $\dot{f}$ is the name of a function $\kappa\rightarrow \kappa^+$, then we will find $q\le p$ with $q\Vdash \mathrm{ran}(\dot{f})$ bounded. 

Let $p_0=p$. Given $p_\alpha$, for each $s\in \mathrm{Split}_\alpha(p_\alpha)$, let $\bar{r}^s_\alpha\le (p_\alpha)_s$ be such that $\bar{r}^s \Vdash \dot{f}(\alpha)=\eta^s_\alpha$. Here the $(p)_s$ means the subtree of $p$ of nodes compatible with $s$. 

Note $\bigcup\{\bar{r}^s_\alpha:s\in \mathrm{Split}_\alpha(p_\alpha)\}$ might not be a condition by the requirement on splitting levels. Let $C=\bigcap \{C(\bar{r}^s_\alpha:s\in \mathrm{Split}_\alpha(p_\alpha)\}$ and thin each $\bar{r}^s_\alpha$ to some $r^s_\alpha\le \bar{r}^s_\alpha$ with $C(r^s_\alpha)=C$.

At limits $\gamma<\kappa$, let $p_\gamma=\bigcap_{\alpha<\gamma} p_\alpha$ by the fusion lemma. This defines a fusion sequence where the limit forces that the range of $f$ is bounded.

Exercise: Suppose ${}^\kappa M\subseteq M$, for an inner model $M$. Suppose $\mathrm{Sacks}(\kappa)\in M\subseteq V$. If $G$ is $V$-generic for $\mathrm{Sacks}(\kappa)$ then ${}^\kappa M[G]\subseteq M[G]$ in $V[G]$.

Note: This holds for $\kappa^+$-c.c. forcing, but $\mathrm{Sacks}(\kappa)$ is not $\kappa^+$-c.c.

Now we will see what happens when we iterate these Sacks forcings with Easton support below, and at, a measurable cardinal $\kappa$. Think of this like a Sacks forcing version of the Kunen-Paris iteration, where we use the nice fusion property to replace the $\gamma^+$ closure of the factors there.

Theorem (Friedman-Thompson 2008): Assume GCH holds. Suppose $\kappa$ is measurable and let $\mathbb{P}$ be the length $\kappa+1$ Easton support iteration with $\mathbb{Q}_\gamma=\mathrm{Sacks}(\gamma)$ (computed in $V^{\mathbb{P}_\gamma}$) for $\gamma\le \kappa$ inaccessible, and $\mathbb{Q}_\gamma$ is trivial forcing otherwise. Then if $G\ast H$ is $V$-generic for $\mathbb{P}= \mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa$, then every normal ultrapower lifts to $V[G\ast H]$ (and in a particularly interesting way!)

Proof: Let $j:V\rightarrow M$ be a normal ultrapower by $U\in V$. Then $j(\mathbb{P}_\kappa=\mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa\ast \dot{\mathbb{P}}_{\kappa+1,j(\kappa)}$. We get the actual $\dot{\mathbb{Q}}_\kappa$ factor at the $\kappa$ step by using the $\kappa$ closure of the ultrapower.

Using this closure further, and the last exercise, ${}^\kappa M[G\ast H]\subseteq M[G\ast H]$ in $V[G\ast H]$, so $M[G\ast H] \vDash \dot{\mathbb{P}}_{\kappa+1,j(\kappa)}\textrm{ is }\le \kappa-\textrm{closed.}$ So there are $\kappa^+$ maximal antichains of $\mathbb{P}_{\kappa,j(\kappa)}$ in $M[G][H]$. We can now build as usual a generic $G_{\kappa+1,j(\kappa)}\in V[G\ast H]$ for $\mathbb{P}_{\kappa+1,j(\kappa)}$ over $M[G\ast H]$. Lift to $j:V[G]\rightarrow M[j(G)]$. 

Now we have to lift $j$ through $\mathbb{Q}_\kappa=\mathrm{Sacks}(\kappa)$. Using the Silver method, $j``H$ has size $\kappa^+$, but the target model $M[j(G)]$ does not have this much closure.

The crucial point is to just take $t:=\bigcap j``H$. We claim that $t$ is a "tuning fork": by this we mean that $t$ consists of a single branch up to the level $\kappa$, at which point it splits into two branch which are cofinal (and that's everything in $t$).
  1. The function $f:\kappa\rightarrow 2$ determined by $H$ is in $t$, and this is everything in $t$ below $\kappa$.
  2. Every condition in $j``H$ splits at $\kappa$ since for each $p\in H$, $p$ splits at club many levels below $\kappa$, and therefore $j(p)$ splits at level $\kappa$. Therefore, $f\frown 0,f\frown 1\in t$.
  3. Since $H$ is a filter, $t$ is cofinal in $j(\kappa)$.
  4. We will argue that $t$ does not split anywhere else. Given a club $C\subseteq \kappa$, $D_C:=\{p\in \mathrm{Sacks}(\kappa): C(p)\subseteq C\}$ is dense. So there must be $p_C\in H$ so that $C(p_C)\subseteq C$. Now we have:
Claim: $X=\bigcap \{j(C):C\subseteq \kappa \textrm{ club in } V[G]\}=\{\kappa\}$.

Proof of Claim: Clearly $\kappa\in X$. For the other inclusion, suppose $\alpha\in X$, $\alpha>\kappa$. Then choose $f:\kappa\rightarrow \kappa$, $f\in V[G]$ so that $j(f)(\kappa)=\alpha$. Then let $C_f=\{\nu<\kappa: f``\nu\subseteq \nu\}$ is club, but $\alpha\not\in j(C_f)$ since $\alpha$ is not a closure point of $j(f)$ ($\kappa<\alpha$ maps to $\alpha$). This proves the claim.

Let $t_0, t_1$ be the leftmost and rightmost branches through $t$, respectively. Let $K_0=\{p\in j(\mathbb{Q}_\kappa):t_0\subseteq p\}$. Clearly $j``H\subseteq K_0$.

It remains to show that $K_0$ is $M[j(G)]$-generic for $j(\mathbb{Q}_\kappa)$. Let $D$ be a dense open subset of $j(\mathbb{Q}_\kappa$ in $M[j(G)]$. Then there is a sequence $\vec{D}=\langle D_\alpha:\alpha<\kappa\rangle \in V[G]$ such that $j(\vec{D})_\kappa=D$, where each $D_\alpha$ is a dense open subset of $\mathbb{Q}_\kappa$. 

Claim: Every condition $p\in \mathrm{Sacks}(\kappa)$ can be extended to $q_\infty \le p$ so that for every $\alpha<\kappa$ there is $\beta<\kappa$ so that for any node $s\in \mathrm{Split}_\beta(q_\infty)$, the condition $(q_\infty)_s$ meets $D_\alpha$. 

Proof of Claim: exercise, a fusion argument.

Let $q_\infty\in H$ be as in the claim, using genericity of $H$. By elementarity, $j(q_\infty)$ has the property that at some splitting level of $j(q_\infty)$, say $\beta<j(\kappa)$, any node $s\in \mathrm{Split}_\beta(j(q_\infty))$ is such that $(j(q_\infty))_s$ meets $D$. Now we can just take $s$ to be $t_0\upharpoonright \delta_\beta$, where $\delta_\beta$ is the $\beta$th splitting level of $j(q_\infty)$.

Therefore $K_0$ is generic as claimed, and it is in $V[G\ast H]$, so $j$ lifts to
$$j:V[G\ast H]\rightarrow M[j(G)\ast j(H)].$$ 

Friday, February 19, 2016

The rearrangement inequality is everywhere

Recently, I've been talking with John Susice about some elementary Olympiad-style problems. I told him that in high school I was taught that virtually every inequality problem that appeared in this setting follows from the Rearrangement Inequality. This states that if $x_1\le \cdots\le x_n$ and $y_1\le \cdots\le y_n$ are real numbers, then the expression
$$x_1 y_{\sigma(1)}+\cdots+x_n y_{\sigma(n)}$$
for $\sigma$ a permutation on $[n]$ is maximized when $\sigma$ is the identity permutation and minimized when $\sigma$ is the reversing permutation. To me, this neatly isolates a useful and general principle that seems kind of obvious in hindsight.

The proof of the $n=2$ case of the inequality, which is all I'll need below, is just to expand $(x_2-x_1)(y_2-y_1)\ge 0$. This case also implies the AM-GM inequality (taking $x_1=y_1=\sqrt{a}$ and  $x_2=y_2=\sqrt{b}$).

Now sometime later, John told me an interesting problem about factoring numbers. He had a solution using a trick similar to Euler's product, but for me this was a chance to test my thesis about the rearrangement inequality:

Problem: Prove that every natural number $n$ has more factors congruent to 1 (mod 4) than factors congruent to 3 (mod 4).

Solution: By induction. Clearly this holds for 1 and for primes. Now suppose $n$ is composite, so write $n=pq$ where $p,q<n$. For any integer $k$, let $r_k$ denote the number of factors it has which are congruent to 1 (mod 4) and $s_k$ the number of factors congruent to 3 (mod 4).

The product of two 1 (mod 4) numbers is still 1 (mod 4), and the product of two 3 (mod 4) numbers is 3 (mod 4). On the other hand, the product of a 1 (mod 4) and a 3 (mod 4) is a 3 (mod 4), and even factors can never be 1 or 3 (mod 4). This proves that $r_n=r_pr_q+s_ps_q$, which must be greater than $s_n=r_ps_q+r_qs_p$ by the rearrangement inequality!


UCI Summer School, part 6: the number of normal measures (Brent Cody)

Sorry for the delay, loyal readers. Here is the beginning of Brent's part of the summer school.

Assume that is consistent that there is a measurable cardinal.

Question: How many normal measures can a measurable cardinal carry?

Let $\mathrm{NM}(\kappa)$ denote the set of normal measures on $\kappa$. Trivially, $\#\mathrm{NM}(\kappa)\le 2^{2^\kappa}$.

One interesting case happens in a canonical inner model.

Theorem (Kunen '71): In $L[U]$, there is exactly one normal measure (on $\kappa$, the unique measurable cardinal).

We can also realize the other extreme:

Theorem (Kunen-Paris '71): There is a forcing extension in which $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

We will prove this one later today. In the middle, we have

Theorem (Mitchell '74): It is consistent relative to a measurable $\kappa$ of order $\delta$ that $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

And for one case, we can lower the large cardinal assumption used.

Theorem (Apter-Cummings-Hamkins '07): It is consistent relative to a measurable cardinal that that $\#\mathrm{NM}(\kappa)=\kappa^+$.

Finally, we can do it in all cases.

Theorem (Friedman-Magidor '09): Assume GCH. Suppose $\kappa$ is measurable and let $\mu\le \kappa^{++}$ be a cardinal. Then in a cofinality-preserving forcing extension, $\#\mathrm{NM}(\kappa)=\mu$.

The goal eventually will be to show the proof of this result.

Lemma: Suppose $j:V\rightarrow M$ is the ultrapower by a normal measure on $\kappa$. Let $G$ be $V$-generic for $\mathbb{P}$. Assume that in $V[G]$,
$$j_0:V[G]\rightarrow M[j_0(G)]$$
and
$$j_1:V[G]\rightarrow M[j_1(G)]$$
are elementary embeddings extending $j$. Then the following are equivalent:
1) $j_0=j_1$.
2) $j_0(G)=j_1(G)$.
3) The normal measure $U_0$ derived from $j_0$ is equal to the normal measure $U_1$ derived from $j_1$.

Proof: Exercise.

Exercise (Levy-Solovay): Show that every normal measure extends uniquely to a normal measure in any forcing extension by small forcing.

As promised, we will now prove the Kunen-Paris Theorem.

Theorem (Kunen-Paris '71): There is a forcing extension in which $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

Proof: By a preparation forcing if necessary, assume that $2^\kappa=\kappa^+$. Let $\mathbb{P}$ be the length $\kappa+1$ Easton support iteration that forces at cardinal stages $\gamma\le \kappa$ with $\mathbb{Q}_\gamma:=\mathrm{Add}(\gamma^+,1)$ (computed in the extension by $\mathbb{P}_\gamma$), trivial forcing at other stages. This is a standard way of forcing the GCH to hold below $\kappa$.

Let $G\ast H$ be $V$-generic for $\mathbb{P}=\mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa$. Let $j:V\rightarrow M$ be the ultrapower by a normal measure. Factor $j(\mathbb{P}_\kappa)\simeq \mathbb{P}_\kappa \ast \dot{\mathbb{P}}_{\kappa,j(\kappa)}$.

Since ${}^\kappa M[G]\subseteq M[G]$ in $V[G]$ (a name exercise which uses closure under $\kappa$-sequences of $M$ in $V$), we have that in $M[G]$, $\mathbb{P}_{\kappa,j(\kappa)}$ is $\le \kappa$-closed (it is the composition of increasingly closed posets starting with $\mathbb{Q}_\kappa=\mathrm{Add}(\kappa^+,1)$.

Furthermore, computing in $V[G]$, $\mathbb{P}_{\kappa,j(\kappa)}$ has at most $\kappa^+$ maximal antichains inside $M[G]$.  since $|j(\kappa)|=\kappa^+$ by our cardinal arithmetic assumption, and $\mathbb{P}_{\kappa,j(\kappa)}$ is $j(\kappa)$-c.c. of size $j(\kappa)$ in $M[G]$. By enumerating the maximal antichains of $M[G]$ in order-type $\kappa^+$, we can meet them one by one, with the closure of the poset in $M[G]$, noticing that closure of the model $M[G]$ gives that the proper initial segments of this enumeration are in $M[G]$.

Since $j``G\subseteq G\ast G_{\kappa,j(\kappa)}$, we can lift the embedding to
$$j^+:V[G]\rightarrow M[j(G)].$$
Using the Lemma/Exercise above, different choices of $j(G)$ will give rise to different embeddings which will give different normal measures. By passing to finer antichains, we can assume in the enumeration $\langle A_\alpha:\alpha<\kappa^+\rangle$ that if $i<j$, then $A_j$ (strictly) refines $A_i$. Looking at how we built $G_{\kappa,j(\kappa)}$, we can form the tree of attempts to build the generic, noticing that at each level there are incompatible ways to extend the generic so far to meet the maximal antichain. This gives $2^{\kappa^+}=2^{2^\kappa}$ many different generics, and hence different normal measures. $\Box$

So it's not hard to force many normal measures. It is harder to force so that there are few normal measures.

We will need to use Hamkins's Gap Forcing Theorem which gives a sufficient condition for an ultrapower embedding in a generic extension to be the lift of a ground model embedding.

Definition: A forcing $\mathbb{P}$ admits a closure point at $\delta$ if it factors as $\mathbb{P}\simeq \mathbb{Q}\ast \dot{\mathbb{R}}$ where $\mathbb{Q}$ is nontrivial, $|\mathbb{Q}|\le \delta$, and $\Vdash_{\mathbb{Q}} \mathbb{R} \textrm{ is }<\delta-\textrm{closed}$.

Theorem (Hamkins '01, Gap Forcing Theorem): If $V\subseteq V[G]$ admits a closure point at $\delta$ and $j:V[G]\rightarrow M[j(G)]$ is an ultrapower in $V[G]$ with $\mathrm{crit}(j)>\delta$, then $j\upharpoonright V:V\rightarrow M$ is a definable class in $V$.

Finally, we prove one more of the theorems in the introduction.

Theorem (Apter-Cummings-Hamkins '07): It is consistent relative to a measurable cardinal that that $\#\mathrm{NM}(\kappa)=\kappa^+$.

Idea of the proof: Start with at least $\kappa^+$ normal measures on $\kappa$ (e.g., by Kunen-Paris forcing). Force with $\mathrm{Col}(\kappa^+,2^{2^\kappa})$, and show that no new normal measures are added.

Proof: Again we assume $2^\kappa=\kappa^+$. Start with $\#\mathrm{NM}(\kappa)\ge \kappa^+$. Let $\mathbb{P}=\mathrm{Add}(\omega,1)\ast \dot{\mathrm{Col}}(\kappa^+, 2^{2^\kappa})$. The point of the Cohen forcing is to give  $\mathbb{P}$ a closure point below $\kappa$. Suppose $c\ast G\subseteq \mathbb{P}$ is $V$-generic. Then every normal measure in $V$ generates a normal measure in $V[c]$ by Levy-Solovay, and these remain normal measures in $V[c][G]$ since $\mathrm{Col}(\kappa^+, 2^{2^\kappa})$ is $\le \kappa$-closed in $V[c]$.  So there are at least $\kappa^+$ normal measures in $V[c][G]$.

To show the other inequality, suppose $U$ is a normal measure on $\kappa$ in $V[c][G]$. Let
$$j:V[c][G]\rightarrow M[c][j(G)]$$
be the ultrapower. By the gap forcing theorem, $j\upharpoonright V:V\rightarrow M$ is a definable class in $V$. So we can lift to $j\upharpoonright V[c]:V[c]\rightarrow M[c]$. We can now define $U$ in $V[c]$, as the derived measure from this embedding. $\Box$


Tuesday, September 1, 2015

UCI Summer School, part 5 (Monroe Eskew)

This is just a placeholder, for now. My notes for this part are quite rough, so it will be a while before I will try to record it here. The next installment of these notes will cover Brent Cody's lectures on some results about the number of normal measures.

UCI Summer School, part 4: Measure algebras (Monroe Eskew)

Here are some more applications of the ideas we have been considering.

Definition: $\mathcal{B}$ is a measure algebra if it is a complete Boolean algebra equipped with some function $\mu:\mathcal{B}\rightarrow [0,1]$ with $\mu(0)=0, \mu(1)=1, \mu(b)>0$ for $b\neq 0$, and $\mu$ countably additive (i.e., if $\langle b_i:i<\omega\rangle$ is an antichain, then $\mu(\sum b_i)=\sum \mu(b_i)$.

Exercise: All measure algebras are c.c.c.

Example: Let $\kappa$ be a cardinal. We will describe a topology on ${}^\kappa 2$. Fix some $x\in [\kappa]^{<\omega}$ and $s:x\rightarrow 2$ (i.e., $s$ is a finite domain partial function from $\kappa$ to $2$). Then basic open sets are of the form $\mathcal{O}_s=\{r\in {}^\kappa 2: \forall \alpha\in x(r(\alpha)=s(\alpha))\}$. Let $\mathcal{B}_\kappa$ be the $\sigma$-algebra generated by these basic open sets, and define $\mu$ on $\mathcal{B}_\kappa$ by setting $\mu(\mathcal{O}_s=\frac{1}{2^{|s|}}$ (standard theorems from real analysis give that $\mu$ extends uniquely to a countably additive probability measure on $\mathcal{B}_\kappa$. 

Define $\mathrm{Null}=\{A\in\mathcal{B}_\kappa: \mu(A)=0\}$. Then  $\mathcal{R}_\kappa:=\mathcal{B}_\kappa/\mathrm{Null}$ is a measure algebra.

Exercise: Prove that $\mathcal{R}_\kappa$ forces $2^\omega\ge \kappa$.

Exercise: If $\mathcal{A}$ is a measure algebra and $\Vdash_A \dot{\mathcal{B}}$ is a measure algebra, then $\mathrm{r.o.}(\mathcal{A}\ast \dot{\mathcal{B}})$ is a measure algebra. (Note: this is not as easy as it may seem at first since for example $\mathcal{A}$ might even add new reals which can be measures of elements of $\mathcal{B}$! We use r.o. for the Boolean completion here since the letter $\mathcal{B}$ is overloaded).

Exercise: If $\mathcal{A}$ is a complete sublagebra of a measure algebra $\mathcal{B}$, then $\mathcal{A}$ is a measure algebra.

Continuing along this line,

Theorem: If $\mathcal{B}$ is a measure algebra, $\mathcal{A}$ a complete subalgebra of  $\mathcal{B}$, and $G\subseteq \mathcal{A}$ is generic over $V$, then in $V[G]$ we have that $\mathcal{B}/G$ is a measure algebra.

Note that in $V[G]$, $G$ is a filter on $\mathcal{B}$, so $\mathcal{B}/G=\{[b]_G:b\in \mathcal{B}\}$. We use $G^*$ for the dual ideal. It's important to distinguish between the orderings of the two Boolean algebras here, and will be good to see how to translate between them using the forcing relation.

Lemma: If $\mathcal{B}$ is complete and $\mathcal{A}$ is a complete subalgebra and $G\subseteq \mathcal{A}$ is generic, then $\mathcal{B}/G$ is complete in $V[G]$.

Proof of Lemma: Suppose $\langle [b_\alpha]_G:\alpha<\kappa\rangle \in P(\mathcal{B}/G)\cap V[G]$. For each $\alpha<\kappa$, let  $X_\alpha:=\{b:1\Vdash_{\mathcal{A}} [b]_{\dot{G}}\le [\dot{b}_\alpha]_{\dot{G}}\}$. Let $c_\alpha=\sum X_\alpha\in V$ (meet taken in $\mathcal{B}$).

We claim that $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}=[\dot{b}_\alpha]_{\dot{G}}$ for each $\alpha$--this suffices to prove the lemma. The proof of the claim is straightforward but a little tedious. First we show $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\le[\dot{b}_\alpha]_{\dot{G}}$. If this doesn't hold, then there are $d,p$ so that $p\in \mathcal{A}$, $d\wedge p\neq 0$, and 
$$p\Vdash [\check{d}]\le [\check{c}_\alpha] \textrm{ and }[\check{d}]\wedge [\dot{b}_\alpha]=0.$$
The first conjunct implies that $p\wedge d\le c_\alpha$ in $\mathcal{B}$, and since $c_\alpha$ is a lub for $X_\alpha$, there is some $b\in X_\alpha$ so that $p\wedge d\wedge b\neq 0$. So $1\Vdash [p\wedge d\wedge b]\le [b_\alpha]$ by the definition of $b\in X_\alpha$. But the second conjunct gives $p\wedge d\wedge b_\alpha=0$, contradiction.

Now to show $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\ge[\dot{b}_\alpha]_{\dot{G}}$, assume for a contradiction that there are $p,a\in \mathcal{A}$ so that $p\Vdash [\check{a}]\le [\dot{b}_\alpha]$ and $p\Vdash [a\wedge \neg c_\alpha]\neq 0$. Now $p$ forces $[p\wedge a \wedge \neg c_\alpha]\le [\dot{b}_\alpha]$. Trivially, $\neg p$ forces $[p\wedge a \wedge \neg c_\alpha]=0$. So it's just outright forced that $[p\wedge a \wedge \neg c_\alpha]\le [b_\alpha]$ and thus $p\wedge a\wedge \neg c_\alpha \in X_\alpha$, which contradicts $c_\alpha$ is an upper bound for $X_\alpha$, completing the proof of the claim and the lemma. $\Box$

Proof of Theorem: Let $\mu$ be a measure on $\mathcal{B}$, $\mathcal{A}$ a complete subalgebra of $\mathcal{B}$. Define 
$$\mu(b\mid a)=\frac{\mu(a \wedge b)}{\mu(a)}.$$
(We say the measure of $b$ conditioned on $a$).

Definition: For $a\in A,b\in B, \epsilon>0$, say $a$ is $\epsilon$-stable for $b$ if for all $x\le a$ in $\mathcal{A}$, $|\mu(b\mid x)-\mu(b\mid a)|<\epsilon$. 

Lemma: For all $b\in B$ and for all $\epsilon>0$ the set $\{a\in A:a \textrm{ is }\epsilon-\textrm{stable for} b\}$ is dense in $A$.
Proof: Exercise. An interesting one.

In $V[G]$, let $\nu:\mathcal{B}/G\rightarrow [0,1]$ be given by $\nu([b])=r$ if for every $\epsilon>0$ there is some $a\in G$ so that $a$ is $\epsilon$-stable for $b$ and $|\mu(b\mid a)-r|<\epsilon$. The idea is that $G$ could add new reals, so we can only have approximations to the measure of $[b]$ using ground model reals attached to the members of $\mathcal{A}$.

We can check that this is well-defined: if $[b]_G=[c]_G$, then some $a\in G$ forces $b\Delta c\in G^*$. This means that $a\perp (b\Delta c)$, so $\mu(a\wedge (b\Delta c)=0$. Therefore $\mu(b\mid x)=\mu(c\mid x)$ for all $x\le a$. Suppose $r_0\neq r_1$ both satisfy $\nu(b)=r_i$. Take $\epsilon<|r_1-r_0|$. Let $a_0,a_1\in G$ be $\epsilon/4$-stable for $b$ with $|\mu(b\mid a_i)-r_i|<\epsilon/4$. Now take $a\le a_0, a_1$ in $\mathcal{A}$. By a triangle inequality argument, we have $|r_1-r_0|<\epsilon$, a contradiction.

Exercise: Check that $\nu(b)>0$ for all $b\neq_G 0$.

Exercise: Check that $\nu$ is countably additive. First prove that it is finitely additive.
$\Box$.

Now suppose $P(Z)/I$ is a measure algebra. The duality theorem (ccc case) says that for any $\theta$, $\mathcal{R}_\theta\ast P(Z)/\bar{I}\cong P(Z)/I\ast j(\mathcal{R}_\theta)$,
where $j:V\rightarrow M$ is the generic embedding in $V[G]$, $G$ generic for $P(Z)/I$.

The right hand side of this isomorphism is a measure algebra, since $P(Z)/I$ is a measure algebra by assumption, and $j(\mathcal{R}_\theta)$ is a measure algebra of $M$, a model which is closed under countable sequences (and so has all the countable sequences to witness countable additivity and completeness of the Boolean algebra).

We have a map $e:\mathcal{R}_\theta\rightarrow \mathrm{r.o.}(P(Z)/I\ast j(\mathcal{R}_\theta)$, so $\mathcal{R}_\theta$ is isomorphic to a complete subalgebra of the right hand side. Now if $H\subseteq \mathcal{R}_\theta$ is generic, then $B/e''H$ is a measure algebra. Therefore $P(Z)/\bar{I}$ is also a measure algebra.

A real-valued measurable cardinal is a cardinal $\kappa$ which carries a $\kappa$-additive probability measure on all subsets of $\kappa$ which gives measure 0 to singletons. It is atomless if every set of positive measure has a subset of strictly smaller positive measure.

Corollary: If $\langle \kappa_i:i<\theta\rangle$ is a sequence of measurable cardinals, then if $\gamma=\sup \kappa_i$, $\mathcal{R}_\gamma$ forces all $\kappa_i$ to be atomless real-valued measurable cardinals (RVMs).

We note the fact that if $\kappa$ is atomlessly RVM, then $2^\omega\ge \kappa$, so we can't get class many RVMs.

However, if $\kappa$ is strongly compact, then $\mathcal{R}_\kappa$ forces that for all regular $\lambda\ge \kappa$, there is a countably additive real-valued probably measure $\mu_\lambda$ on $\lambda$ giving measure 0 to all subsets of size $<\lambda$.


Friday, August 7, 2015

UCI Summer School, part 3: Applications of Duality Theorem (Monroe Eskew)

We now turn towards applications of the duality theorem. It is recommended that the reader recalls the notation ($I,j,K,J,\hat{H},e,\iota$, etc.) from the previous lecture before proceeding.

The basic idea is that one uses the isomorphism there:
$$\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})$$
to calculate the quotient algebra $P(Z)/J$ as $\mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})/e''H$.

As discussed near the end of the last lecture, under certain assumptions, the statement of the duality theorem becomes somewhat simpler. The first examples will fall into this case.

Special Case: If $I$ is $\kappa$-complete and $\mathbb{P}$ is $\kappa$-c.c., then the hypothesis of the duality theorem holds, $K=\{0\}$ and $J=\bar{I}$, the ideal generated by $I$ in $V^\mathbb{P}$.

Exercise: Show that in the above case, if $p\in \mathbb{P}$ and $A\in (P(Z)/I)\cap V$, then
$$\iota(p,\check{A})=(A,j(\dot{p})).$$

A further simplification will be that we will usually start with a measurable cardinal $\kappa$ and take $I$ to be  the dual to the measure on $\kappa$, so $P(\kappa)/I$ is the trivial Boolean algebra.

A measurable cardinal $\kappa$ has a 2-saturated, $\kappa$-complete ideal, namely the dual to the measure on $\kappa$, and under GCH every cardinal $\kappa$ carries a $\kappa^{++}$-saturated, $\kappa$-complete ideal, namely the ideal of bounded subsets. This motivates the following natural questions, which are the main focus of this lecture:

Question: Suppose $\mu\le \kappa^+$ is a regular cardinal. Is it consistent that there is a cardinal $\kappa$ which is not measurable, but still $\kappa$ carries a $\mu$-saturated, $\kappa$-complete ideal? (Here we want the amount of saturation to be exactly $\mu$).

Digression: does the answer change if we require $\kappa$ to be a successor cardinal?

For the case where $\kappa$ is a successor cardinal, $\kappa^+$-saturation is the strongest we can hope to achieve.

Exercise: Prove using the method of generic ultrapowers that if $\kappa$ is a successor cardinal then there is no $\kappa$-complete, $\kappa$-saturated ideal on $\kappa$.

Kunen showed that if $\kappa$ is a successor cardinal, then getting a $\kappa^+$-saturated ideal on $\kappa$ requires large cardinals much stronger than a measurable, although we won't do this argument here (you can find it in this previous Specinar post. We now return to the original question.

Answer to question 1, if $\mu<\kappa$ ($\mu$ regular): We will use the basic technique of computing the quotient algebra $P(\kappa)/J$ in $V[H]$ using the duality theorem. Start with $\kappa$ measurable in the ground model. Let $\theta\ge \kappa$, and consider $\mathrm{Add}(\mu,\theta)$ which adds $\theta$ Cohen subsets of $\mu$. $\mathrm{Add}(\mu,\theta)$ is $\kappa$-c.c., and under the GCH it is even $\mu^+$-c.c. Let $I=U^*$, where $U$ is a $\kappa$-complete normal ultrafilter on $\kappa$ (here the star means taking the dual ideal). Let $j:V\rightarrow M$ be the ultrapower embedding.

The duality theorem gives the isomorphism:
$$\mathrm{Add}(\mu,\theta)\ast P(\kappa)/\bar{I}\cong P(\kappa)/I\ast \mathrm{Add}(\mu,j(\theta))\cong\mathrm{Add}(\mu,j(\theta)),$$
since $P(\kappa)/I$ is trivial.

If $H$ is generic for $\mathrm{Add}(\mu,\theta)$ over $V$, then
$$e''H=\{(1,j(p)):p\in H\}.$$
So
$$P(\kappa)/\bar{I}\cong \mathrm{Add}(\mu,j(\theta))/e''H\cong \mathrm{Add}(\mu,j(\theta)).$$
In $V[H]$, $P(\kappa)/\bar{I}\cong \mathcal{B}(\mathrm{Add}(\mu,j(\theta))$, so $\bar{I}$ is $(\mu^{<\mu})^+$ saturated. Furthermore, it is easy to check that $\bar{I}$ is $\kappa$-complete, and $2^\mu\ge \kappa$ in $V[H]$, so $\kappa$ is not measurable. This answers question 1 for the case where $\mu<\kappa$. $\Box$

We might ask what large cardinal properties of $\kappa$ are implied by this ideal hypothesis.

Proposition: If $\kappa$ carries a $\kappa$-complete $\mu$-saturated ideal for some $\mu<\kappa$, then:

  1. $\kappa$ is weakly Mahlo
  2. $\kappa$ has the tree property.
Exercise: Prove (1) of the proposition using generic ultrapowers.

Proof of Proposition (2): Suppose $T$ is a $\kappa$-tree. If $G\subseteq P(\kappa)/I$ is generic, then in $V[G]$, $T$ has a branch $b$ given by taking any member of level $\kappa$ of $j(T)$, where $j$ is the generic ultrapower embedding. Now for each $\alpha<\kappa$, $S_\alpha=\{x\in T_\alpha: \exists p(p\Vdash \check{x}\in \dot{b})\}<\mu$ by the saturation. Now $\bigcap_{\alpha<\kappa} S_\alpha$ is a $\kappa$-tree all of whose levels have size $<\mu<\kappa$. It is well-known (or a good exercise) that such trees have cofinal branches. $\Box$

Definition: An ideal $I$ is nowhere prime if there is no $I$-positive set $A$ so that $I\upharpoonright A$ is prime (i.e., dual to an ultrafilter).

Exercise: Show that if there is a nowhere prime, $\kappa$-complete, $\mu^+$-saturated ideal, where $\mu<\kappa$, then $2^\mu\ge \kappa$.

We continue with Question 1 with other arrangements of $\mu$ and $\kappa$.

Answer to question 1, if $\mu=\kappa^+$: Start with $\kappa$ measurable with $2^\kappa=\kappa^+$, $U$ a normal ultrafilter and $j:V\rightarrow M$ the ultrapower embedding. Let $\langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle$ be the Easton support iteration where for regular $\alpha$, $\Vdash_{\mathbb{P}_\alpha} \dot{Q}=\mathrm{Add}(\alpha,1)$. It is straightforward to verify that $\mathbb{P}_\kappa$ has the $\kappa$-c.c., so we are in the special case again.

Note that $j(\mathbb{P}_\kappa)=\mathbb{P}_\kappa\ast \mathbb{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M$ (We use the notation $\mathbb{P}_{\xi,j(\kappa)}$ for $j(\mathbb{P}_\kappa)_{\xi,j(\kappa)}$). The tail part is computed differently in $M$ than in $V$, e.g., the support is on $M$-regular cardinals. 

If $G_\kappa\subseteq \mathbb{P}_\kappa$ is generic over $V$, then the special case of the duality theorem says that  in $V[G_\kappa]$, $P(\kappa)/\bar{I}\cong (\mathbb{P}_{\kappa,j(\kappa)})^M$. However, $P(\kappa)/\bar{I}\cong(\mathbb{P}_{\kappa,j(\kappa)})^M$ does not have the $\kappa^+$-c.c. since there are $M$-regular cardinals between $\kappa^+$ and $j(\kappa)$. So $\bar{I}$ is not $\kappa^+$-saturated.

Now we could also satisfy the hypothesis of the duality theorem of adding a $j(\mathbb{P}_\kappa)=P(\kappa)/J\cong  \mathrm{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M$ generic filter $\hat{H}$ over $M$ in a different way. By a standard technique, a $j(\mathbb{P})$-generic over $M$ exists in $V[G_{\kappa+1}]$ (where $G_{\kappa+1}$ is $\mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1)$-generic). This is because we clearly get a generic for the initial part $\mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1)$, just $G_{\kappa+1}$ itself. For the tail, $(\mathbb{P}_{\kappa+1,j(\kappa)})^{M[G_{\kappa+1}]}$ is is $j(\kappa)$-c.c. of size $j(\kappa)$ in $M[G_{\kappa+1}]$, so $M[G_{\kappa+1}]$ thinks the poset has at most $j(\kappa)$ maximal antichains. In $V[G_{\kappa+1}]$, $|j(\kappa)|=\kappa^+$, and the poset is $\kappa^+$-closed (in $M[G_{\kappa+1}]$, but also in $V[G_{\kappa+1}]$ by the agreement between these models). So we can construct a generic by hand in $V[G_{\kappa+1}]$ by enumerating all of the maximal antichains in $M[G_{\kappa+1}]$. This completes the construction of $\hat{H}$ in the extension by $\mathrm{Add}(\kappa,1)$.

In this construction, we have that $j(\mathbb{P})/K\cong \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1)$, since the Boolean algebra homomorphism $j(\mathbb{P}_\kappa)\rightarrow \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1)$ given by $p\mapsto \|p\in \hat{H}\|$.  has kernel exactly $K$ as defined in the last lecture, and the map is surjective since the codomain completely embeds into the domain. So in the duality theorem calculation, we obtain an ideal $J$ so that $P(\kappa)/J\cong \mathrm{Add}(\kappa,1)$. So $J$ is a $\kappa^+$-saturated ideal on $\kappa$. $\Box$

Note that $\kappa$ is inaccessible.

Exercise: Prove that $\kappa$ is weakly compact in $V[G_\kappa]$. (Hint: use the tree property characterization.)

Exercise: Prove that $\kappa$ is not measurable in $V[G_\kappa]$, but it is measurable in $V[G_{\kappa+1}]$.

Remark: By forcing with $(\mathbb{P}_{\kappa,j(\kappa)})^{M[G_{\kappa+1}]}$ instead of just $\mathrm{Add}(\kappa,1)$ to add the  $j(\mathbb{P})$-generic, we can get a nowhere prime $\kappa$ complete $\kappa^+$-saturated ideal on $\kappa$ in $V[G_{\kappa+1}]$.

Answer to question 1, if $\mu=\kappa$: We will find an example so that $\kappa$ is not weakly compact (compare to earlier results for saturation below $\kappa$), and in fact the quotient algebra is isomorphic to a $\kappa$-Suslin tree.

In the exercises, we will describe how to construct, for $\alpha$ regular with $\alpha^{<\alpha}=\alpha$, a forcing $\mathbb{Q}_\alpha$ which adds an $\alpha$-Suslin tree $\dot{T}_\alpha$ so that $\mathcal{B}(\mathbb{Q}_\alpha\ast \dot{T}_\alpha)\cong\mathrm{Add}(\alpha,1)$. This is due to Kunen.

In the construction for $\mu=\kappa^+$ we got a model (there called $V[G_\kappa]$) where there was an inaccessible $\kappa$ and an ideal $J$ on $\kappa$ so that
$$P(\kappa)/J\cong\mathrm{Add}(\kappa,1)\cong \mathbb{Q}_\kappa\ast \dot{T},$$
where $\dot{T}$ is the $\kappa$-Suslin tree added by $\mathbb{Q}_\kappa$.

Now start with this to be our ground model $V$. Let $H\subseteq \mathbb{Q}_\kappa$ be generic. We want to show that in $V[H]$, there is an ideal $J_1$ on $\kappa$ so that $P(\kappa)/J_1\cong T$.

If $G\subseteq \mathrm{Add}(\kappa,1)$ is generic over $V$, then take in $V[G]$ an embedding
$$j:V\rightarrow M$$
which was constructed before. We want to extend the embedding to $V[H]$.

Now $G\in M$ since $M$ is closed under $\kappa$-sequences in $V[G]$. We can extend $j$ to $V[G]$ by constructing a generic $\hat{G}$ for $\mathrm{Add}(j(\kappa))^M$ over $M$ with $\hat{G}\upharpoonright \kappa=G$ (using the standard method; cf the second exercise following previous construction).

In $V$, by duality theorem there are $J_1$ and $K$ so that
$$\mathbb{Q}_\kappa\ast P(\kappa)/J_1\cong P(\kappa)/J\ast j(\mathbb{Q}_\kappa)/K.$$
In this case, $K$ is a maximal ideal since the $j(\mathbb{Q}_\kappa)$-generic over $M$ is already just added by $P(\kappa)/J$. So in $V[H]$, $P(\kappa)/J_1\cong T$.

Now we turn to Kunen's forcing construction. Conditions in Kunen's forcing $\mathbb{Q}$ are normal trees of successor ordinal height $<\kappa$ which are homogeneous: for all $t\in T$ not on the top level, $T_t\cong T$, where $T_t$ is the tree $\{s\in T: t\le_T s\}$ with the order inherited from $T$.

Exercise:

  1. Show that Kunen's forcing is $\kappa$-strategically closed. Hint: the strategy will go by choosing a particular branch through each of the small trees chosen in a play of the game so far.
  2. Show that $\mathbb{Q}\ast \dot{T}$ has a $\kappa$-closed dense subset, and deduce that $\mathbb{Q}\ast \dot{T}\cong \mathrm{Add}(\kappa)$.
  3. Show that $\dot{T}$ is a Suslin tree.

$\Box$

We will do one last application to construct a precipitous ideal on a cardinal $\kappa$ which is not measurable so that its quotient algebra is $\kappa^+$ closed.

Start with $\kappa$ measurable and $2^\kappa>\kappa^+$. We will use the Easton support iteration $\langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle$, where $\dot{\mathbb{Q}}_\alpha=\dot{\mathrm{Add}(\alpha^+,1)}$ for inaccessible $\alpha<\kappa$ (and is trivial forcing otherwise).

Then $\mathbb{P}_\kappa$ is $\kappa$-c.c., and forces that for all inaccessible $\alpha<\kappa$, $2^\alpha=\alpha^+$ (this is a standard coding trick that was assigned as an exercise in one of Spencer Unger's lectures here). By duality,
$$\mathbb{P}_\kappa\ast P(\kappa)/\bar{I}\equiv j(\mathbb{P}_\kappa).$$
If $G_\kappa\subseteq \mathbb{P}_\kappa$ is generic, then $j(\mathbb{P}_\kappa)/e''G_\kappa\equiv \mathbb{P}_{\kappa,j(\kappa)}$. Since $M[G_\kappa]$ is closed under $\le \kappa$ sequences in $V[G_\kappa]$, this tail is $\kappa^+$-closed forcing over $V[G_\kappa]$.

However, GCH holds at every inaccessible $\alpha<\kappa$ and fails at $\kappa$ in $V[G_\kappa]$. By a reflection argument, $\kappa$ cannot be measurable in $V[G_\kappa]$. $\Box$

Exercise: Show that if $H\subseteq P(\kappa)/\bar{I}$ is generic, then $\kappa$ is measurable in $V[G_\kappa\ast H]$.

Monday, July 27, 2015

UCI Summer School part 2: Duality Theorem (Monroe Eskew)

The Duality Theorem gives a general technique for forcing to make an ideal whose quotient algebra has various properties. It appears in Matthew Foreman's "Calculating quotient algebras of generic embeddings." My version of these notes omits a lot of the dots which indicate that certain objects are just names in a forcing extension. This is for aesthetic reasons, and hopefully does not lead to confusion.

Duality Theorem: Suppose $I$ is a precipitous ideal on $Z$ and $\mathbb{P}$ is any partial order. If: there is a further generic extension of the extension by $P(Z)/I$ so that if $j:V\rightarrow M\subseteq M\subseteq V[G]$ is the ultrapower embedding from $G\subseteq P(Z)/I$, there is $H\subseteq \mathbb{P}$ generic over $V$ and $\hat{H}\subseteq j(\mathbb{P})$ generic over $M$ and some extension of $j$ to $\hat{j}:V[H]\rightarrow M[\hat{H}]$.

Then: there is a $\mathbb{P}$-name for an ideal $J$ on $Z$ and a $P(Z)/I$-name for an ideal $K$ on $j(\mathbb{P})$ and a canonical isomorphism $$\iota:\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K}).$$


So a very general statement of lifting a generic ultrapower map to a forcing extension gives a useful isomorphism for computing $P(Z)/J$ in the generic extension by $\mathbb{P}$. We remark that in some cases, this will be an equivalence.

Note: In what follows, we tacitly identify all of the posets involved with their Boolean completions. Occasionally for emphasis, this identification will be explicit.

Proof: Assume (1). There is some $A\in I^+$ and some $P(Z)/I$-name for a forcing $\dot{\mathbb{Q}}$ so that
$$A\Vdash_{P(Z)/I}(\Vdash_\dot{\mathbb{Q}} \dot{H_0}\subseteq j(\mathbb{P}) \textrm{ is generic over }M \textrm{ and }H:=j^{-1}[H_0]\subseteq \mathbb{P}\textrm{ is generic over }V).$$

Note that the set $\{p\in\mathbb{P}:\quad \Vdash_{P(A)/I\ast \mathbb{Q}} j(p)\not\in H_0\}$ cannot be dense, since it is the complement of the generic $j^{-1}[H_0]$. So there is $p_0\in \mathbb{P}$ so that for all $p\le p_0$, $\| j(p)\in H_0 \|_{P(A)/I\ast \mathbb{Q}}\neq 0$. We will constrain ourselves to work below this $p_0$ in $\mathbb{P}$ and below $A$ in $P(Z)/I$. For simplicity, assume that $A=Z$ and $p_0=1_\mathbb{P}$.

In $V^{P(Z)/I}$, define
$$K=\{p\in j(\mathbb{P}):\quad \Vdash_{\mathbb{\dot{Q}^G}} p\not\in H_0\}.$$
Let $G\ast h$ be generic for $P(Z)/I\ast j(\mathbb{P})/K$. From the $j(\mathbb{P})/K$-generic $h$, we can define a $j(\mathbb{P})$-generic $\hat{H}=\{p:[p]_K\in h\}$.

Claim: The following properties of $H_0$ are also true of $\hat{H}$:

  1. $\Vdash_{P(Z)/I\ast \mathbb{Q}} \hat{H}$ is $j(\mathbb{P})$-generic over $M$.
  2. $\Vdash_{P(Z)/I\ast \mathbb{Q}} j^{-1}[\hat{H}]$ is $\mathbb{P}$-generic over $V$.
  3. For all $p\in \mathbb{P}$, $\not\Vdash_{P(Z)/I\ast \mathbb{Q}} j(p)\not\in \hat{H}$.
Proof of Claim: For (1), suppose $D\in M$ is open dense in $j(\mathbb{P})$. Then $\{[d]_K:d\in D\textrm{ and }d\not\in K\}$ is dense in $j(\mathbb{P})/K$. Otherwise, there would exist $p\in j(\mathbb{P})/K$ so that $p\wedge d\in K$ for all $d\in D$. But this is impossible because we could then force with $\mathbb{Q}$ over $V[G]$ to get a generic $H_0$ containing $p$ (as $p\not\in K$), and then $H_0\cap D=\emptyset$, contradicting genericity of $H_0$ over $M$.

The remaining parts of the claim can be checked similarly, and are left as an exercise. $\Box$.

Now let $e:\mathbb{P}\rightarrow \mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})$ be defined by $e(p)=\|j(p)\in \hat{H}\|$.

By (3) of the claim above, $\mathrm{ker}(e)=0$. Also $e$ preserves Boolean operations simply by the elementarity of $j$. By (2) of the claim, $e$ is a regular embedding (maps maximal antichains pointwise to maximal antichains). 

Exercise: $e:\mathbb{P}\rightarrow \mathbb{Q}$ is a regular embedding iff for every $q\in\mathbb{Q}$ there is $p\in \mathbb{P}$ so that for every $p'\le p$, $e(p')$ is compatible with $q$.

Thus, if $H\subseteq \mathbb{P}$ is generic over $V$, we can force with the quotient $\mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})/e"H$ over $V[H]$ to obtain a generic $G\ast h$ for $P(Z)/I\ast j(\mathbb{P})/K$. By the definition of $e$, we have $j_G"H\subseteq \hat{H}$, where $\hat{H}$ is defined from $h$ as before. So we can extend the embedding $j_G$ to $\hat{j}:V[H]\rightarrow M[\hat{H}]$.

In $V[H]$ we can finally define $J=\{A\subseteq Z:1\Vdash [\mathrm{id}]\not\in \hat{j}(A)\}$, where the forcing is with the quotient $\mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})/e"H$. In $V$, let
$$\iota(p,\dot{A})=e(p)\wedge \|[id]\in\hat{j}(\dot{A})\|.$$

Exercise: $\iota$ is order and incompatibility preserving.

It remains to show that the range of $\iota$ is dense. So take an arbitrary $(B,\dot{q})\in P(Z)/I\ast j(P)/K$. By strengthening this condition, we may assume without loss of generality that there is $f:Z\rightarrow \mathbb{P}$ in $V$ so that $B\Vdash [[f]_M]_K=\dot{q}$.

By regularity of $e$ (using the characterization in the exercise), there is a $p$ so that for all $p'\le p$, $e(p')\wedge (B,\dot{q})\neq 0$. Let $\dot{A}$ be a $\mathbb{P}$-name for a subset of $Z$ such that $p \Vdash \dot{A}=\{z\in B: f(z)\in H\}$ and $\neg p \Vdash \dot{A}\in J^+$.

We check that $(p,\dot{A})$ is actually a condition, which involves checking that $p\Vdash \dot{A}\in J^+$. So take a generic $G\ast \dot{h}$ containing $e(p)\wedge (B,\dot{q})$ (which is nonzero by choice of $p$). Then clearly $B\in G$ and since $[[f]_M]_K=q\in h$, we have $[f]_M=j(f)([\mathrm{id}])\in\hat{H}$. Therefore $[\mathrm{id}]\in \hat{j}(A)$, so this generic $G\ast \dot{h}$ shows that it is not forced by $\mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K}/e"H$ that $[\mathrm{id}]\not\in \hat{j}(A)\}$.

By definition $\iota:=\iota(p,\dot{A})$ forces $j(p)\in \hat{H}$ and $[\mathrm{id}]\in\hat{j}(\dot{A})$. Since $\hat{j}$ extends $j$ and it's forced that $\dot{A}\subseteq B$, $\iota$ must force $B\in G$. And since $j(p)\Vdash_{j(\mathbb{P})} j(\dot{A})=j(\{z:j(f)(z)\in \hat{H}\})$, $\iota$ must force $\dot{q}=[j(f)(\mathrm{id})]_K\in h$. $\Box$

Remark: Suppose $K$ as in the Duality Theorem is forced to be principal, i.e., there is $m$ so that $$\Vdash K=\{p\in j(\mathbb{P}:p\le \neg m\}.$$
Then the Duality Theorem is easily seen to be an equivalence.

We can compute some nice properties of the ideal $J$ arising from the previous theorem.

Proposition: Using the notation of the previous theorem, $J$ is forced to be precipitous, with the same completeness as $I$. If $I$ is normal, then $J$ is also normal. Also, if $\bar{G}\subseteq P(Z)/J$ is generic over $V[H]$ and $G\ast h=\iota[H\ast \bar{G}]$ and $\hat{j}:V[H]\rightarrow M[\hat{H}]$ are as before, then $V[H]^Z/{\bar{G}}=M[\hat{H}]$ and $\hat{j}$ is the ultrapower embedding.

Finally, we relate $J$ to the ideal in $V[H]$ generated by $I$.

Proposition: Suppose $K$ as in the Duality Theorem is forced to be principal, with $m$ so that $\Vdash K=\{p\in j(\mathbb{P}):p\le \neg m\}.$ Suppose further that there exist $f$ and $A$ so that $A\Vdash \dot{m}=[f]_G$ and $\dot{B}$ is a $\mathbb{P}$-name for $\{z\in A: f(z)\in \dot{H}\}$. Then $\bar{I}\upharpoonright B=J\upharpoonright B$, and $A\setminus B\in J$, where $\bar{I}$ is the ideal in $V[H]$ generated by $I$.