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\begin{document}


\setbeamercovered{dynamic}

\title{Reflections on Dan's Work}



\author{S.\ Jackson}



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


\date{
June 2012\\
North Texas}


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





\section{Steinhaus Problem}
\begin{frame}

The following question was asked by Steinhaus in the 50's.
\mys


\begin{ques}
Does there exist a set $S\subseteq \R^2$ meeting every isometric copy 
$L$ of the integer lattice $\Z^2$ in exactly one point?
\end{ques}
\mys


Note that the one-dimensional version of the problem is trivial.
\mys




{\color{red}Peter Komj\`{a}t} showed that the corresponding question for 
copies of $\Z$ is true.
\end{frame}


\begin{frame}
\frametitle{Variations}

The problem has many variations.
\mys

\begin{itemize}
\item
We can ask it for any lattice $L$ in $\R^n$ ($L$ doesn't have to $n$-dimensional).
\item
More generally, $L$ could be any set, even a finite set.
\item
We can restrict the class of isometries to particular subgroups.
\item
We can restrict the properties of the Steinhaus set $S$, e.g., we can ask for it to be bounded,
measurable, etc.
\item
We can vary the requirements on the intersection of $S$ and $\pi(L)$.
\end{itemize}
\end{frame}




\begin{frame}
The problem has connections to number theory, geometry, set theory.
\mys


For $n \geq 4$ the problem in $\R^n$ for the standard lattice $\Z^n$
has a negative solution ({\color{red}D.\ Goldstein}):  consider the copy $L$ of $\Z^n$
centered at $(\frac{1}{2}, \frac{1}{2},\frac{1}{2},\frac{1}{2},,0, \dots,0)$.
\mys


\begin{thm}[J., Mauldin] ($\zfc$) There is a Steinhaus set $S \subseteq \R^2$
for the standard lattice $\Z^2$. Any Steinhaus set cannot have the Baire property.
\end{thm}
\end{frame}


\begin{frame}
Consider the version of the problem ($n=2$ or $3$) 
where we restrict the group of isometries to translations,
and we require that no two points of $S$ are at a lattice distance apart.
\mys


We call such a set a {\color{blue}partial Steinhaus set}.
\mys


We can specialize further to just the rational translates of $\Z^2$. In fact,
we can consider translates by $\left( \frac{\Z}{p}\right)^n$. This gives a 
finitary version of the problem.
\mys


For $n=2$ this holds for all primes $p$. Only $p \equiv 1 \pmod{4}$ are non-trivial.
\mys

For $n \geq 4$, this fails for $p=2$ by previous argument. 
\end{frame}



\begin{frame}
We have the following number-theoretic problem:

\begin{ques}
For which $n$ and $p$ does there exist a partial Steinhaus set?
\end{ques}

For $n=3$, there is one for $p=2,3$. For $p=5$ the answer is not clear.
\mys



For $n=2,3$, a {\color{blue} root} is a tuple $(\lambda_1,\dots,\lambda_n)$ 
with $\lambda_1^2+\dots+\lambda_n^2=0$. We usually consider the roots up to 
scaling.
\mys

For $n=2$ there are (up to scaling) two roots if $p \equiv 1 \pmod{4}$,
otherwise no roots. For $n=3$ there are (up to scaling) $p+1$ roots. 
\end{frame}


\begin{frame}
Consider the $p^n$ points in $\left( \frac{\Z}{p}\right)^n \cap [0,1)^n$. 
\mys


Having a partial Steinhaus set is equivalent to having functions 
$\ell_1,\dots,\ell_n \colon \{ 0, \dots,p-1\}^n \to \Z$ 
such that for all $x_1=\frac{\vec{i}_1}{p}$, 
$x_2=\frac{\vec{i}_2}{p}$, if 

\begin{equation*}
\begin{split}
z_1& =\frac{\vec{i}_1}{p}+\vec{\ell}(\vec{i}_1)\\
z_2& =\frac{\vec{i}_2}{p}+\vec{\ell}(\vec{i}_2)
\end{split}
\end{equation*}

then $\rho(z_1,z_2)$ is not a lattice distance.
\end{frame}

\begin{frame}
Saying here that $\rho(z_1,z_2)$ is not a lattice distance is equivalent to
saying that $\rho^2(z_1,z_2)\notin \Z$ by the following:
\mys

\begin{fact}
If an integer is the sum of two (or three) squares of rationals, then it is the sum of
two (or three) squares of integers.
\end{fact}
\mys

There is a cute proof of this fact due to {\color{red}Aubrey} which works 
for both $n=2,3$.
\end{frame}

\begin{frame}
The condition $\rho^2(z_1,z_2)\notin \Z$ holds automatically unless
$z_2-z_1$ is a root. 
\mys

Given a root $\vec \lambda$, we say a {\color{blue}thread} 
is the equivalence class of points in $I_p$ which differ by a multiple of 
$\vec \lambda$ (this only depends on $\vec \lambda$ up to scaling).
\mys

Each equivalence class has size $p$.
\mys

Fix a root $\vec \lambda$ and a starting point $\frac{\bar{i}}{p}$. This defines the thread 
$\frac{1}{p} (\bar{i}+e \vec \lambda \pmod{p})$, where $0 \leq e <p$.
\mys

Let $\vec \eta(e)= \frac{\bar{i}+e \vec \lambda -(\bar{i}+e \vec \lambda \pmod{p})}{p}$.
\end{frame}

\begin{frame}
The condition becomes the following.
\mys


Define

\[
\pi^{\vec \lambda}_{\bar{i}}(e)= \vec \lambda \cdot \vec \ell(e) -\lambda\cdot \vec \eta(e) 
+ e \frac{\delta}{2} \pmod{p}
\]


where $\vec \ell(e)=\vec \ell 
( \frac{\bar{i}+e \vec \lambda \pmod{p}}{p})$, and $\delta=\frac{|| \lambda ||}{p}$.
\mys

Then all of the $\pi^{\vec \lambda}_{\bar{i}}$ are permutations.
\mys

{\color{blue}Note:} If we take $\lambda$ to be a root $\pmod{p^2}$ and use
the ``unwrapped points'' to take the $\ell$ values, then we have 
$\pi^{\vec \lambda}_{\bar{i}}(e)= \vec \lambda \cdot \vec \ell(e)$.

\end{frame}

\begin{frame}
\frametitle{Comments on 3D Problem}
\begin{itemize}
\item
{\color{red}Mauldin} with {\color{red}Goldstein} and {\color{red}McLinden} gave the heuristic  estimate 
for the expected number of 3D partial Steinhaus sets for a given $p$:
\[
\frac{{p!}^{p^3+p^2}}{p^{p^4-2p^3}}
\]
\mys

\begin{center}
\begin{tabular}{c|c}
Prime & Expected Number\\
\hline
$3$ & $1.35 \cdot 10^{15}$\\
$5$ & $5.80 \cdot 10^{49}$\\
$7$ & $102.24$\\
$11$ & $1.08 \cdot 10^{-1438}$
\end{tabular}
\end{center}
\end{itemize}
\end{frame}


\begin{frame}
\begin{itemize}
\item
With {\color{red}Henkis} and {\color{red}Lobe} we observed that a certain overdetermined subset of the 
3D system appears to be consistent.

\item
Every ``pseudo'' partial Steinhaus set is actually a partial Steinhaus set.
\end{itemize}
\mys


The 2D proof also uses a geometric lemma: the ``three circles theorem'' which is a result
about 4-bar mechanical linkages. 
\mys

The proof uses an induction over the size of the continuum, using the ``hull'' method. 
This method naturally suggests the above geometric lemma.
\end{frame}


\begin{frame}
It is still open whether there can be a measurable Steinhaus set in $\R^2$, or
whether there can be a bounded one. However ({\color{red}Croft}), there cannot be a bounded
measurable one.
\mys


{\color{red}Kolountzakis} and {\color{red}Wolff} showed that there is no measurable 
Steinhaus set for the standard lattice $\Z^3$ (or `` almost Steinhaus set 
for $\Z^d$, $d \geq 3$). {\color{red}Kolountzakis} and {\color{red}Papadimitrakis} 
gave another, simpler, proof of this.
\mys


{\color{red}Mauldin} and {\color{red}Yingst} extended this to other lattices in $\R^d$, $d \geq 3$. 
\mys


Let $A\in \gl(d,\R)$, and let $L_A=A\cdot \Z^d$ be the corresponding lattice. 
Let $\hat{L}_A=L_{A^{-T}}$ be the dual lattice. 
\end{frame}


\begin{frame}
For $M \in \gl(d,\R)$, let $\sD(M)=\{ || M x ||^2 \colon x \in \Z^d\}$.
\mys


\begin{defn}
Let $A, B \in \gl(d,\R)$. $A \succ B$ if $\sD(A) \supseteq \sD(B)$. $A \succ_w B$
if $A\succ B$ and $\frac{\det(B)}{\det(A)} \notin \Z$. $A \succ_s B$ if 
$A \succ B$ and $\frac{\det(B)}{\det(A)}  \notin \Q$.
\end{defn}
\mys

\begin{thm}[Mauldin, Yingst]
Let $A \in \gl(d,\R)$ and suppose $A^{-T} \succ_w B$ for some $B \in \gl(d,\R)$. 
Then there is no measurable (almost) Steinhaus set for the lattice $L_A$. 
\end{thm}



{\color{red}Mauldin}, {\color{red}Yingst} show that these results apply to the tetrahedral lattice
in $\R^3$. 
\end{frame}


\begin{frame}
They give an example of a matrix in $\gl(d,\R)$ which does not weakly dominate any matrix. 
\mys


Extending this work of Mauldin and Yingst, {\color{red}Chan} and {\color{red}Mauldin}
showed the following.
\mys

Say a lattice $L_A$ is {\color{blue}integral} if the dot product of any two 
vectors (equivalently, columns of $A$ are integers.
\mys



\begin{thm}[Chan, Mauldin]
Let $L_A$ be an integral lattice in $R^d$, $d \geq 3$. Then there is a
$B$ such that $A \prec_s B$. 
\end{thm}
\end{frame}

\begin{frame}
So, they get:


\begin{thm}[Chan, Mauldin]
If $L$ is equivalent to an integral lattice in $\R^d$, $d \geq 3$, then there is no measurable (almost)
Steinhaus set for $L$.
\end{thm}
\end{frame}

\begin{frame}
\frametitle{Finite Version}

We can consider the version of the Steinhaus problem where we use an arbitrary set 
$A \subseteq \R^d$ in place of a lattice. In particular, $A$ can be a finite set.
\mys

Consider the finite version in $\R^2$. 
\mys


\begin{conj}
For every finite $A \subseteq \R^2$ with $|A|>1$, there does not exist 
a set $S \subseteq \R^2$ meeting every isometric copy of $A$ in exactly one point.
\end{conj}
\mys

For $|A|=2,3$ the result is easily proved. 
\end{frame}

\begin{frame}

{\color{red}Gao}, {\color{red}Miller} showed that certain families of four-point sets
satisfy the conjecture. 
\mys

Then Gao's student {\color{red}Mingzhi Xuan} proved the following.
\mys


\begin{thm}[Xuan]
Every four-point set in the plane satisfies the conjecture,
\end{thm}

\end{frame}





\section{Something Old-Something New}

\begin{frame}
In 1991 Mauldin and J proved the following.

\begin{thm}[$\zf$]
Let $X$, $Y$ be uncountable Polish spaces. There is a $\bp^1_1$ set $P\subseteq 
X\times Y$ with all sections $P_x$ countable, such that $P$
is not the union of countably many $\bs^1_2$ (partial) graphs of functions. 
In fact same conclusion for $\sB(\bs^1_2)$.
\end{thm}


Assuming $\bp^1_1$-determinacy the least pointclass $\bg$ was identified such that every 
$\bp^1_1$ with countable sections can be written as a countable union of graphs in $\bg$. 
Namely, 

\[
\bg= \bigcup_n \Game \omega\cdot n-\bp^1_1.
\]
\end{frame}

\begin{frame}
Recently, the method of proof used for this non-uniformization result has been used to 
get a result concerning a question of {\color{red}Maharam} about $\sigma$-finite measures.
\mys


Let $\mu$, $\nu$ be Borel measures on uncountable Polish spaces $X$, $Y$, and let 
$\phi\colon X\to Y$ be Borel. 
\mys

\begin{defn}
A {\color{blue}disintegration} of $\mu$ with respect to $\nu$, $\phi$ is a 
family $\{ \mu_y\}_{y \in Y}$ on Borel measures on $X$ satisfying:

(1) $\forall B \in \sB(X)\ (y \mapsto \mu_y(B))$ is Borel. 

(2) $\forall y \in Y\ (\mu_y(X-\phi^{-1}(y))=0$.

(3) $\forall B\in \sB(X)\ (\mu(X)=\int \mu_y(B) \, d\nu)$
\end{defn}
\end{frame}

\begin{frame}
We say the disintegration $\{ \mu_y\}_{y \in Y}$ is {\color{blue}$\sigma$-finite} 
if for every $y \in Y$, $\mu_y$ is a $\sigma$-finite measure.
\mys


We say the disintegration is {\color{blue}uniformly $\sigma$-finite} if there are Borel sets 
$B_n$ in $X$ satisfying:

(1) $\forall n\ \forall y \in Y \ \mu_y(B_n)<\infty$.

(2) $\forall y \in Y\ (\mu_y(X-\bigcup_n B_n)=0$.
\mys

\begin{ques}[Maharam]
Is every $\sigma$ finite disintegration uniformly $\sigma$-finite?
\end{ques}
\end{frame}

\begin{frame}
\begin{thm}[Backs, J, Mauldin]
Assuming $V=L$, there is a $\sigma$-finite disintegration which is not uniformly $\sigma$-finite. 
\end{thm}
\mys

The proof uses the following improvement to the earlier non-uniformization result.
\mys

\begin{thm}
Assume $V=L$. Then there is a $\bp^1_1$ set $G  \subseteq \ww \times \ww$ 
satisfying
\begin{enumerate}
\item
$\forall x \in X\ |G_x| \leq \omega$
\item
$G$ is not the union of countably many $\bp^1_1$ (or $\bs^1_2$) graphs.
\item
$\forall n \in \omega\ \forall B\in \sB(Y)\ \{ x \in X \colon |G_x|=n
\} \in \sB(X)$
\end{enumerate}
\end{thm}








\end{frame}





\end{document}





















