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\begin{document}


\setbeamercovered{dynamic}



\title{Stable Homogeneous Trees and Self-Justifying Systems}



\author{S.\ Jackson}




\institute{
Department of Mathematics\\
University of North Texas\\
jackson@unt.edu
}






\date{{\color{red} Second Conference on the Core Model Induction and HOD Mice}\\
August, 2011, Muenster}






\begin{frame}
\titlepage
\end{frame}








\begin{frame}
\section{Introduction}
\frametitle{Homogeneous Trees}






The notion of homogeneous tree was formulated independently by {\color{red}Kunen}
and {\color{red}Martin}, then isolated and 
given in its current form by {\color{red}Kechris}.
\mys


\begin{defn}
A {\color{red}measure} on a set $X$ is a countably additive ultrafilter on $X$.
\end{defn}
\mys


Recall that under $\ad$ every ultrafilter on a set $X$ is a measure.
\mys
\end{frame}



\begin{frame}
\begin{defn}
A tree $T$ is {\color{red}homogeneous} if there are measures $\{ \mu_s\colon s \in \omega^{<\omega}\}$
on $T_s$ satisfying:
\begin{itemize}
\item
If $s \sqsubseteq s'$ then $\mu_{s'}$ projects to $\mu_s$.
\item (homogeneity property)\\ 
For all $x \in \ww$ if $T_x$ is illfounded
then for any $A_{x \res n}$ with $\mu_{x\res n}(A_{x \res n})=1$, there is an 
$f$ such that $\forall n\ f\res n \in A_n$. 
\end{itemize}
\end{defn}
\mys


Note: This is equivalent to saying that the direct limit of the ultrapowers by the measures $\mu_{x \res n}$
is wellfounded.
\mys


Fact ($\zf$, Martin): If $A=p[T]$ where $T$ is homogeneous, then $A$ is determined.
\end{frame}


\begin{frame}
\frametitle{Weakly Homogeneous Trees}
\begin{defn}[Weakly Homogeneous, strong form]
$T$ is {\color{red}weakly homogeneous} if there are measures $\mu_{s,t}$ on $T_s$ defined for 
a set of $(s,t)$ in a tree, and sets $A_{s,t} \subseteq T_s$ satisfying:
\begin{itemize}
\item
$\mu_{s,t}(A_{s,t})=1$.
\item
If $s'$ immediately extends $s$ and $A_{s,t}$ is defined, 
then the $A_{s',t'}$ for $t'$ extending $t$ partition $T_{s'} \cap \{\vec \alpha 
\colon \pi_{s',s}(\vec \alpha) \in A_{s,t}\}$.
\item
If $(s',t') \sqsupseteq (s,t)$ then $\pi_{(s',t'),(s,t)}(\mu_{(s',t')})=\mu_{(s,t)}$. 
\item (homogeneity property)\\ 
For any $x , y \in \ww$, if $\mu_{x \res n, y\res n}$ is defined for all $n$ and 
$\exists f\ \forall n\ (x \res n, f \res n) \in A_{x\res n, y\res n}$, then 
for any $B_{x\res n, y \res n}$ with 
$\mu_{x \res n, y \res n}(B_{x \res n, y\res n})=1$ we have that 
$\exists g \ \forall n\ (x\res n, g\res n) \in B_{x \res n, y \res n}$.
\end{itemize}
\end{defn}
\end{frame}


\begin{frame}
\begin{defn}[Weakly Homogeneous, weak form]
There are measures $\mu_{s,t}$ with $\mu_{s,t}(T_s)=1$
satisfying:
\begin{itemize}
\item
If $(s',t') \sqsupseteq (s,t)$ then $\pi_{(s',t'),(s,t)}(\mu_{s',t'})=\mu_{s,t}$. 
\item
For any $x \in \ww$, if $T_x$ is illfounded then $\exists y \in \ww$ such that 
$\langle \mu_{x \res n, y\res n} \rangle$ has wellfounded direct limit: for any $\mu_{x \res n, y \res n}
(B_{x \res n, y \res n})=1$  there is a $g$ such that $\forall n\ (x \res n, g \res n) \in 
B_{x \res n, y\res n}$. 
\end{itemize}
\end{defn}


{\color{blue}Alternate Version}:
There is a countable set $\sM$ of measures such that whenever $T_x$ is illfounded then there is a 
sequence $\mu_1,\mu_2,\dots$ of measures in $\sM$ with $\mu_n(T_{x \res n})=1$ 
such that $\langle \mu_{x \res n}\rangle$ has wellfounded direct limit.
\end{frame}


\begin{frame}
\begin{fact}[Woodin] In $\zfc$ the strong and weak forms of the definition are equivalent. 
\end{fact}


\begin{fact}
The weak form of the definition is enough to make the Martin-Solovay construction work.
\end{fact}


\begin{fact}
The strong form of the definition s equivalent to saying that 
there is a homogeneous tree $\bar{T}$ on $\omega \times \omega \times \kappa$ which is 
isomorphic to $T$ (under a bijection of $\omega \times \kappa$ with $\kappa$).
\end{fact}
\mys


So, a set $A \subseteq \ww$ is weakly homogeneously Suslin iff it is the projection of 
a strongly homogeneously Suslin set.
\end{frame}



\begin{frame}
From the third (weakest) form of the definition of weakly homogeneous we can get the second
(weak) form easily by restricting to appropriate measure one sets: \medskip


Let $\{ \mu_n\}$ be as in the third definition. We may assume the sequence of measures
is closed under projections. 
For each $s \in \omega^{<\omega}$,
let $\mu_{s,t}$ enumerate those measures in the sequence which satisfy $\mu(T_s)=1$.
\mys


For each $s,t$ let $A_{s,t}$ be such that 
$\mu_{s,t}(A_{s,t})=1$ and $\mu_{s,t'}(A_{s,t})=0$ for all $t' \neq t$. 
We may assume that $A_{s,t} \cap A_{s,t'}=\emptyset$ if $t \neq t'$. 
\medskip

Furthermore, we may assume that $\pi^{s',t'}_{s,t}(A_{s',t'})\subseteq
A_{s,t}$ when $(s',t')$ extends $(s,t)$.
\medskip

Thinning the tree, we may assume that $T_s=\bigcup_t A_{s,t}$.
\end{frame}


\section{A Warm-Up Result}
\begin{frame}
\frametitle{A Warm-Up Result}
We can also show that the strong and weak forms of the definition are equivalent:


\begin{thm}[$\zf+\ad$]
Let $T$ be a tree on $\omega \times \alpha$ for $\alpha <\Theta$ and assume there is a 
countable family of measures $\{ \mu_1,\mu_2,\dots\}$ which witness the weak homogeneity of 
$T$ by the third definition. Then there is a  $T' \subseteq T$ with $p[T]=p[T]$
which is homogeneous by the strong definition ($T'$ is the restriction of $T$ to certain 
measure one sets with respect to the $\mu_n$).
\end{thm}
\medskip

The proof of the fact is a warm-up for the main theorem, so we present it. 


\end{frame}


\begin{frame}
Fix measure one sets $A_{s,t} \subseteq T_s$ as above and measures $\mu_{s,t}$
on $T_{s}$ with $A_{s,t}\cap A_{s,t'}=\emptyset$ if $t \neq t'$ and satisfying the weak form of the
definition  (we assume that $\bigcup_t A_{s,t}=T_s$ w.l.o.g.):
\medskip


If $T_x$ is ill-founded then there is a $y \in \ww$ such that the direct limit of the 
$\mu_{x \res n, y \res n}$ is well-founded. 
\medskip

We must thin the $A_{s,t}$
to measure one sets $B_{s,t}$ such that whenever $T_x$ is ill-founded then 
for all $y$, if $\exists f\  \forall n\ f \res n \in B_{x \res n, y \res n}$ 
then the direct limit of the $\mu_{x \res n, y \res n}$ is well-founded. 
\end{frame}

\begin{frame}
We use from $\ad$ and $\alpha <\Theta$ that there is a fine measure $\nu$ on 
$\sP_{\omega_1}(\bigcup_n (\sP(\alpha^n) ))$.
\medskip

If $\sigma \in \sP_{\omega_1}(\bigcup_n (\sP(\alpha^n) ))$, let 
\[
A^\sigma_{s,t}= A_{s,t} \cap \bigcap \{ A \in \sigma \colon \mu_{s,t}(A)=1\}.
\]

We claim that for $\nu$ almost all $\sigma$ that the sets $A^\sigma_{s, t}$
will work as the $B_{s,t}$.
\medskip


Suppose not. Then for $\nu$ almost all $\sigma$ we define a tree
$V^\sigma$ as follows.
\end{frame}



\begin{frame}
Let $T^\sigma \subseteq T$ consist of the $(s,\vec \alpha) \in T$ 
such that $\forall i < \lh(s)\ (s\res i, \vec\alpha \res i) 
\in A^\sigma_{s\res i, t\res i}$ (for the unique appropriate $t$).
\medskip



$(s,\vec \alpha, \vec \beta) \in V^\sigma \leftrightarrow 
(s,\vec \alpha) \in T^\sigma$ and  for the unique  $t$ such that 
$\lh(t)=\lh(s)$ and   $\vec \alpha \in \bigcap_{i <\lh(s)}
A_{s\res i, t\res i}$   we have for all $i <\lh(s)$ that:
\[
j^{s\res i,t\res i}_{s\res i-1, t\res i-1}(\beta_{i-1})>\beta_{i}.
\]
\medskip


For almost all $\sigma$, let $(x^\sigma, {\vec \alpha}^\sigma,  
{\vec \beta}^\sigma)$ be the left-most branch of $V^\sigma$. Let 
$y^\sigma \in \ww$ be the unique real corresponding to ${\vec \alpha}^\sigma$. 
\end{frame}


\begin{frame}
By countable additivity we may fix $x$, $y$ so $x=x^\sigma$, $y=y^\sigma$
for almost all $\sigma$. 
\medskip


The sequence $\mu_{x \res i, y \res i}$ has ill-founded direct limit (consider 
${\vec \beta}^\sigma$ for almost all $\sigma$).
\medskip

Fix measure one sets $A'_{x\res i, y\res i}\subseteq A_{x\res i, y\res i}$
witnessing the illfoundedness of the direct limit.
\medskip


Fix $\sigma$ such that $x=x^\sigma$, $y=y^\sigma$, and the $A'_{x\res i, y\res i} \in \sigma$.
\medskip

Then ${\vec \alpha}^\sigma$ is a path through the $A^\sigma_{x \res i, y \res i}
\subseteq A'_{x \res i, y\res i}$, a contradiction. 
\end{frame}





\section{Stable Homogeneous Trees}



\begin{frame}
\frametitle{Stable Homogeneous Trees}
Let $T, \{ \mu_s\}$ of $T, \{ \mu_{s,t} \}$ be a homogeneous or weakly homogeneous tree. 
\medskip

If $T_x$ is wellfounded, we let $f_x \colon T_x \to \on$ be the ranking function
(using the Brouwer-Kleene ordering).
\mys


If $\vec A= \{ A_s\}$ (or $\{ A_{s,t} \}$ are measure one sets, we let 
$T^{\vec V}$ be the tree restricted to these measure one sets, that is,

\[
(s,\vec \alpha) \in T^{\vec A} \leftrightarrow 
(s,\vec \alpha)\in T \wedge 
\forall i <\lh(s)\ (s\res i, \vec\alpha\res i)\in A_{s\res i, t\res i}
\]

(for the unique appropriate $t$ determined by $s,\vec \alpha$).
\end{frame}

\begin{frame}
We let $f^{\vec A}_x$ denote the ranking function on $T^{\vec A}_x$.
\medskip



We let $(f^{\vec A}_x)_s$ (or  $(f^{\vec A}_x)_{s,t}$ in the weakly homogeneous case)
denote the sub-function 
on $T_s$ (or $T_{s,t}$) induced by $f^{\vec A}_x$. If $x$ (or $x,y$) are understood,
we let $(f^{\vec A}_x)_n$ denote $(f^{\vec A}_x)_{s\res n}$ (or 
 $(f^{\vec A}_x)_{s\res n, t\res n}$).
\medskip


We let $[f^{\vec A}_x]_{\mu_{x\res n}}$ denote the ordinal represented by $f^{\vec A}_x$
w.r.t.\ the measure $\mu_{x \res n}$ (or $\mu_{x\res n,t}$ in the weakly homogeneous case).
\end{frame}


\begin{frame}
\begin{defn}
The homogeneous tree $(T,\mu_s)$ (or weakly homogeneous tree 
$(T, \mu_{s,t})$) is {\color{red}stable} if there are measure one sets $A_s$
w.r.t.\ the $\mu_s$ (or $A_{s,t}$) such that for any $x \in B=\ww-p[A]$
and any measure one sets $B_{x \res n}$ (w.r.t.\ $\mu_{x \res n}$) we have that
\[
[(f^{\vec A}_{x})_n ]_{\mu_{x \res n}}   \leq  [(f^{\vec B}_{x})_n]_{\mu_{x \res n}}
\]
for all $n$.
\end{defn}
\mys

[For weak homogeneity we have 
\[
[(f^{\vec A}_{x})_{x\res n, t} ]_{\mu_{x \res n,t}}   \leq  
[(f^{\vec B}_{x})_{x\res n,t}]_{\mu_{x \res n,t}}
\]
for all $n,t$].
\end{frame}








\section{Main Theorem}
\begin{frame}
\frametitle{Main theorem}

\begin{thm}[$\zf+\ad+\dc$]
Every homogeneous tree  $(T, \{\mu_s\})$ on $\alpha < \Theta$ is stable. Likewise,
every weakly homogeneous tree $(T,\mu_{s,t})$ on $\alpha <\Theta$ is stable. 
\end{thm}
\mys

Proof uses ideas from the {\color{red}Martin-Woodin} proof that all trees are weakly homogeneous.
\mys



{\color{blue}sketch of proof:}
Let $\nu$ be a fine measure on $\sP_{\omega_1}(\cup_n \sP(\kappa^n))$. 
\mys

For $\sigma \in \sP_{\omega_1}(\cup_n \sP(\kappa^n))$, let $A^\sigma =\{ A^\sigma_s\}_s$
where $A^\sigma_s= \cap \{ A \in \sigma \colon \mu_s(A)=1\}$.
\mys

We show that for $\nu$ almost all $\sigma$ that the $A^\sigma_s$ stabilize $T$. We suppose not.
\end{frame}



\begin{frame}
Uniformly in $\sigma$ we define a tree $U^\sigma$ on $\omega \times \lambda^{<\omega}\sim 
\omega \times \lambda$ where $\lambda= \sup j_{\mu_s}(\kappa)$.
\mys


For $x \in B$ the tree $U^\sigma_x$ attempts to produce ordinals $[f^i_j]_{\mu_{x \res i}}$
which witness that the sets $A^\sigma$ have not yet attained the minimal ranking functions. 
\mys



They do this, roughly speaking, by describing embeddings of $T_x$ (on measure one sets) into 
proper initial segments of $T^{A^\sigma}_x$.
\end{frame}


\begin{frame}




We set $((s(0),\dots,s(n-1)),(\beta_0,\dots, \beta_{n-1})) \in U^\sigma$ iff:
\mys


1.) $\beta_0 \in \omega$.
\mys

2.) Each $\beta_i$ for $i>0$ codes a finite sequence of integers $t_i$ extending $s$ (roughly, a commitment
that later extensions of $s$ must follow $t_i$) along with a finite sequence of ordinals 
$(\beta^i_0,\dots, \beta^i_{|t_i|-1})$. 
\mys


3.) $s$ must be compatible with all $t_i$ for $i<|s|$.
\mys


4.) Let $f^i_j$ represent $\beta^i_j$ with respect to $\mu_{s \res i}$. Then there are 
$\mu_{s \res i}$ measure one sets restricted to which the maps 
$(\eta_0,\dots \eta_{i-1}) \mapsto (f^i_0(\vec \eta),\dots,f^i_{|t_i|-1})$
give an order-preserving from the BK ordering on $T_{s_n}$ to the BK ordering on 
$T^{A^\sigma}$.
\end{frame}


\begin{frame}
5.) For $i<\beta_0$, each $f^i_j$ is almost everywhere the identity function, and for 
$i =\beta_0$ we have that almost everywhere that $(f^i_0(\vec \eta), \dots, f^i_{|t_i|-1}(\vec \eta))$
is a properly less than $\vec \eta$ in $T_{|t_i|}$.
\mys


6.) We weave in the Martin-Solovay tree into $U^\sigma$ as well. Say at every even level of 
$U^\sigma$ we put in ordinals from the Martin-Solovay tree (so a branch through $(U^\sigma)_x$
also gives a branch through $(T')_x$ and so proves $x \in B$).
\end{frame}



\begin{frame}
For any $x$,  $U^\sigma_x$ is illfounded iff $x \in B$ and there are measure one sets 
$B_{x \res n} \subseteq A^\sigma_{x \res n}$ (with respect to $\mu_{x \res n}$) such that 
for some $m$, $[f^{\vec B}_x]_{\mu_{x \res m}} < [f^{A^\sigma}_x]_{\mu_{x \res m}}$.
\mys


So, for $\nu$ almost all $\sigma$ there is a leftmost branch $(x^\sigma, {\vec \beta}^\sigma)$
through $U^\sigma$. By countable additivity we may fix $x=x^\sigma$ on a measure one set. 
\mys


Fix any $\sigma$ where 
$x^\sigma=x$ and also $\sigma$ contains measure one sets $C_{x \res n}$ (w.r.t.\ the 
$\mu_{x \res n}$) which attain the minimal values for $[f^{\vec C}_x]_{\mu_{x \res n}}$
(which we can do by countable choice).
\mys

The branch $\vec \beta^\sigma$ 
gives a proper embedding from $T^{\vec D}_x$ into $T^{A^\sigma}_x\subseteq T^{\vec C}_x$
for some measure one sets $D_{x \res n}$. This contradicts the choice of the $C_{x \res n}$. 



\end{frame}

\begin{frame}
For $(T,\{ \mu_{s,t}\})$ weakly homogeneous the proof is essentially the same.
\medskip

We put now $(s,\vec \beta)\in U^\sigma$ provided the following hold.
\medskip

Now $\beta_0$ codes a sequence $t_0 \in \omega^{<\omega}$. Each $\beta_i$, $i>0$, 
codes a finite sequence of tuples  $(t^i_j, u^i_j, v^i_j, \beta^i_j)$ where $\lh(t^i_j)
=i \leq  n$, $\lh(u^i_j)=\lh(v^i_j)$.
\medskip


$s$ must be consistent with all the $u^i_j$. $\beta^i_j=[f^i_j]_{\mu_{s\res i, t^i_j}}$
($\lh(t^i_j)=i$). The $f^i_j$ must almost everywhere give an order-preserving map from 
$T_s$ to $T^{A^\sigma}$. For $t< t_0$, we have $u^i_j=t^i_j$ and $f^i_j$ is the identity a.e.\

\medskip

Here we use $(s\res i, t^i_j, (\eta_0,\dots,\eta_{i-1}))\mapsto 
((u^i_j,v^i_j, (\cdots  f^i_j(\vec \eta)))$.

\end{frame}


\section{Applications}
\subsection{The Martin-Solovay Tree}
\begin{frame}
\frametitle{Martin-Solovay Tree}
We apply our main theorem to the Martin-Solovay tree. Recall the definition.
\medskip




Let $(T, \{ \mu_{s,t} \})$ be a weakly homogeneous tree with $A=p[T]$. 
The {\color{red}Martin-Solovay tree}
$T'=T'(T,\{ \mu_{s,t}\}, \lambda)$ is a tree with $p[T']=\ww-A=p[T']$ defined as follows.
\mys



$(s,\vec \alpha) \in T'$ iff $\exists f \colon T_s \to \lambda$ order-preserving
w.r.t.\ the BK ordering on $T_s$ such that $\forall i <\lh(s)$
\[ \alpha_i= [f^i]_{\mu_{s_i,t_i}}
\]
where $(s_i,t_i)$ is a reasonable enumeration of $\omega^{<\omega}\times \omega^{<\omega}$
and $f^i=f\res \{ \vec \alpha \colon |\vec \alpha|=i \wedge (s_i,t_i,\vec \alpha)\in T\}$
if $s_i \sqsubseteq s$, and $f=0$ otherwise.
\end{frame}



\begin{frame}
For $x \in A'$, let \[ \varphi_i(x)=[(f_x)_{s_i,t_i}]_{\mu_{s_i,t_i}} \]
(recall $(s_i,t_i)$ enumerates the pairs of equal length).
\medskip

Recall $f_x$ is the ranking function on $T_x$ and $(f_x)_{s_i,t_t}$ 
is the appropriate subfunction of $f_x$. 
\medskip


The Martin-Solovay theorem says that $\{\varphi_i\}$ is a 
{\color{red}semi-scale} on $A'$.
\medskip


We can extract a {\color{red}scale} from the Martin-Solovay tree $T'$ by using left=most branches,
but this is nor good enough for later purposes.
\end{frame}




\begin{frame}
\frametitle{Two Questions About The Martin-Solovay Tree}



\begin{ques}
When is the semi-scale $\{\varphi_i\}$ from the Martin-Solovay tree a scale?
\end{ques}
\medskip


\begin{ques}
When is the Martin-Solovay tree the tree of a scale?
\end{ques}
\end{frame}



\begin{frame}
For the first question, our theorem gives an answer.



\begin{thm}
If $(T,\mu_s)$ is a stable homogeneous (or weakly homogeneous) tree, 
then the semi-scale $\{ \varphi_i\}$ from the
Martin-Solovay tree $T'$ constructed from $T^{\vec A}$ is a scale. Here 
\end{thm}
\medskip


{\color{blue}Proof}
Suppose $x_m \in A'$ and $x_m \to x$. Suppose $\varphi_n(x_m)$ is eventually equal to 
$\lambda_n$ for each $n$. Say $(\varphi_0(x_m),\dots,\varphi_{m}(x_m))=
(\lambda_0,\dots,\lambda_m)$. 
\medskip


For each $m$, the ranking functions $f_{x_m}$ satisfy 
$[(f_{x_m})_{s_i,t_i}]_{\mu_{s_i,t_i}}=\lambda_i$ for all $i\leq m$. 
\medskip
\end{frame}


\begin{frame}
As in the Martin-Solovay proof this gives measure one sets $\vec A=A_{x\res n, t}$
on which the $f_{x_m}$ together are order-preserving on $T_x$. 
\medskip


From the stability property we know that 
\[
\varphi_n(x) 
=[(f_{x})_{s_n,t_n}]_{\mu_{s_n,t_n}} = 
 [(f^{\vec A}_{x})_{s_n,t_n}]_{\mu_{s_n,t_n}} 
\leq  [(f^{\vec A}_{x_n})_{s_n,t_n}]_{\mu_{s_n,t_n}} =\lambda_n.
\]
\qedsymbol
\end{frame}




\begin{frame}
As for the second question: If $T'$ is the tree of a scale $\{ \psi_m\}$ then 
$\{ \psi_n\}$ is given by the left-most branch of $T'$. 
\medskip


Our arguments show that the left-most branch of $T'$ is given by $\{ \varphi_n\}$
(assuming $T$ has been stabilized), so $\vec \psi=\vec \varphi$. 
\medskip

So, this is equivalent to asking if for every $(s,\vec \alpha)\in T'$ there is an 
$x \in A'$ extending $s$ with $\vec \varphi(x) \res \lh(s)=\vec \alpha$. 
\medskip


Not much appears to be known about this.
\end{frame}



\subsection{Scales and Self-Justifying Systems}


\begin{frame}
Recall $\bg$ has the {\color{red}scale property} if every $A \in \bg$ admits a scale 
$\{ \varphi_n\}$ all of whose norms are in $\bg$.
\medskip



\begin{defn}
We say a pointclass $\Lambda$ closed under $\neg$ is {\color{red} self-justifying}
if every $A \in \Lambda$ admits a scale all of whose norms are in $\Lambda$. 
\end{defn}



We recall some of the scale/Suslin cardinal analysis assuming $\ad$.
\end{frame}


\begin{frame}
Let $\kappa$ denote a limit Suslin cardinal (i.e., $\kappa$ is a limit of Suslin cardinals).
\mys



Let $\bd=\{ A \colon o(A) <\kappa\}$, so $\bd$ is closed under $\neg$, $\exists^\ww$, $\forall^\ww$. 
\mys


We have the following facts from $\ad$.
\mys


Type I.) $\cof(\kappa)=\omega$. 

Let $\bs_0=\bigcup_\omega \bd$. Then $\sca(\bs_0)$, 
$\sca(\bp_1)$, $\sca(\bs_2), \dots$. The next Suslin cardinals are $\kappa$, $\kappa^+$, 
$\lambda_3$, $\delta_3=(\lambda_3)^+, \dots$. Also, $S(\kappa)=\bs_1$, $S(\kappa^+)=
\bs_2$, $S(\lambda_3)=\bs_3$, etc.
\end{frame}

\begin{frame}
Types II, III.) $\cof(\kappa)>\omega$, $\bg$ not closed under real quantifiers. 

Let $\bp_{-1}$ be the Steel pointclass ($o(\bg)=\kappa$), so $\bp_{-1}$ is closed under $\forall^\ww$ 
and $\pwo(\bp_{-1})$. Then $\sca(\bp_{-1})$, $\sca(\bs_0)$, $\sca(\bp_1),\dots$. 
The next Suslin cardinals are $\kappa, \lambda_1,\delta1, \lambda_3,\delta_3,\dots$. 
We have $S(\kappa)=\bs_0$, $S(\lambda_1)=\bs_1$, $S(\kappa_1)=\bs_2,\dots$. 
\mys


Type IV.) Let $\bg$ be closed under quantifiers with $o(\bg)=\kappa$ and $\pwo(\bg)$. 
\mys

Then $\kappa$ is regular and $\sca(\bg)$. We need to construct the next Suslin cardinal.
\end{frame}


\begin{frame}
Recall we have several equivalent definitions of the {\color{red}envelope} $\Lambda= \env(\bg,\kappa)$,
which constructs the next Suslin cardinal.
\mys




\begin{defn}[Martin]
Let $\sA=\{ A_\alpha\}_{\alpha<\alpha_0}$, each $A_\alpha \subseteq \ww$. 
Then $\overline{\sA}$ is the set of $A\subseteq \ww$ such that for all countable
$S\subseteq \ww$, there is an $\alpha <\alpha_0$ such that $S\cap A=S\cap A_\alpha$.

We let 
\[\Lambda(\bg,\kappa)=\{ \overline{\sA} \colon  \sA \subseteq \bg \wedge || \sA||\leq \kappa\}.\]
\end{defn}
\end{frame}



\begin{frame}
\frametitle{Facts about the Envelope}
Assume $\bg$ closed under quantifiers, $\kappa=o(\bd)$. Let $\Lambda=\Lambda(\bg,\kappa)$.
\mys

\begin{itemize}
\item
$\Lambda=\Lambda(\bg,\kappa)=\Lambda(\bd,\kappa)$.
\item
There is a single ``universal'' $\sA=\{ A_\alpha\}_{\alpha<\kappa}$ such that every 
$A \in \Lambda$ is Wadge reducible to a set in $\overline{\sA}$.
\item
There is a $\bgd$ set with no uniformization in $\Lambda$. In particular, a $\bgd$-complete set
cannot have a scale all of whose norms are reducible to a fixed set $B\in \Lambda$.
\item
Every measure on $\kappa$ has a code set reducible to a set in $\Lambda$.
\end{itemize}
\end{frame}



\begin{frame}
\frametitle{Other characterizations of ENV}


Steel gave an alternate definition of the envelope. 
\mys


\begin{fact}
Let $\bg$ be type IV and $\kappa=o(\bg)$. Then the following are equivalent:
\begin{itemize}
\item
$A \in \env(\bg)$.
\item
There is a $z_0 \in \ww$ s.t.\ for all countable $\sigma$ containing $z_0$
we have $A\cap \sigma \in L(T,\sigma)$.
\item
There is a $z_0 \in \ww$ s.t.\ for all countable $\sigma$ containing $z_0$
we have $A \cap \sigma$ is ordinal definable in $L(T,\sigma)$ from $T,\sigma$.
\end{itemize}
\end{fact}


\end{frame}





\begin{frame}
Let $A$ be a $\bg$-complete set, and assume $\kappa$ is not the largest Suslin cardinal. 
\mys


Then from $\sca(\bg)$ and the Martin-Woodin theorem (every tree on an $\alpha$ below a Suslin
cardinal is weakly homogeneous) we have that $A=p[T]$ where $T$ is a weakly homogeneous tree
on $\omega \times \kappa$ (in fact, by Martin-Steel we can take $T$ to be homogeneous).
\mys


By our main theorem we may assume that $T$ is stable. We assume w.l.o.g.\ that for $(s,\vec \alpha)
\in T$ that $\alpha_0 \geq \max (\vec \alpha)$. 
\mys



Let $A'=\ww-A$, and $T'$ the Martin-Solovay tree with $A'=p[T']$. Let $\{ \varphi_n\}$
be the semi-scale from the homogeneous tree construction so, as $T$ is stable, 
$\vec \varphi$ is actually a scale on $A'$. 
\end{frame}



\begin{frame}
From the previous facts, each norm $\varphi_n$ is in $\Lambda$:
\mys


\[
x <_n^* y \leftrightarrow T_x \text{ is wellfounded } \wedge 
\forall_{\mu_{s_n,t_n}}^* \vec \alpha\  \exists \beta <\kappa \ 
[|T_x(\vec \alpha) \leq \beta \wedge \neg (|T_y (\vec \alpha) \leq \beta)].
\]
\mys


Lat $\lambda=o(\Lambda)=$ the supremum of the $\Lambda$ prewellorderings.
\mys


So, every $\bgd$ sets admits a scale with all norms of length less than $\lambda$. 
From this and the previous facts it follows that $\lambda$ is the next Suslin cardinal after $\kappa$.
\end{frame}

\begin{frame}
It follows that $\Lambda$ is self-justifying:


\begin{thm}
With $\bg$, $\Lambda$ as above, every $B \in \Lambda$ admits a scale all of whose norms
lie in $\Lambda$.
\end{thm}

{\color{blue}Proof}
Fix a $\bgd$-complete set $A'$ and a scale $\{ \varphi_n\}$ on $A'$ with all norm relations
in $\Lambda$. Fix $B \in \Lambda$, and let $\gamma=o(B)<\lambda$. Fix a non-selfdual pointclass
$\bg_0 \subseteq \Lambda$ which is closed under $\exists^\ww$, $\forall^\omega$, $\pwo(\bg_0)$,
and with $B \in \bd_0$. 
\mys

So, $\bg_0$ is closed under wellordered unions. 
\end{frame}



\begin{frame}
Fix $\delta >$ the supremum of the $\bg_0$ prewellorderings.
Fix $n$ so that $|\varphi_n| > \delta$.
\mys



For each $\alpha <\delta$ let $A^n_{\alpha}=\{ x \in A' \colon \varphi_n(x)\leq \alpha \}$. 
\mys


We can't have that $o(A^n_\alpha)\leq \gamma$ for all $\alpha <\delta$ as this would give 
a $\delta$ increasing sequence of sets in $\bd_0$, and hence a $\bg_0$ prewellodering of length $\delta$,
a contradiction. 
\mys


Fix $\alpha<\delta$ so that $B \leq_W A^n_\alpha$.
\end{frame}



\begin{frame}
Let $\{ \psi_n\}$ be the restriction of $\{ \varphi_n\}$ to $A^n_\alpha$. 
\mys



$\{ \psi_n\}$ is easily a scale on $A^n_\alpha$, and each $\psi_n$ is a $\Lambda$-norm 
(Boolean combination of $A^n_\alpha$ with the norm relations for $\varphi_n$).
\mys


Since $B\leq_W A^n_\alpha$ it follows that $B$ also has a scale all of whose norms lie in $\Lambda$.

\qedsymbol











\end{frame}


























































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