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 $I$.

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