% file: ASL-talk.tex
% created: 2014-05-19, 01:24:28 PM (-0500 CDT)

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%=========================================================================={{{1
\title[Dualizable, omits $\{ \1, \5 \}$ implies cube]{Naturally dualizable
  algebras omitting types $\1$ and $\5$ have a cube term}
\subtitle{(the talk formerly known as ``Idempotent congruence modular
  algebras that admit a natural duality have cube terms'')}

\author{Matthew Moore}
\institute[VU]{Vanderbilt University}
\date[2014-05-20]{May 20, 2014}

\begin{document}
\begin{frame}
\titlepage
\end{frame}
%==========================================================================}}}1

%==========================================================================={{{
\begin{frame}[t] \frametitle{What?}
\pause
\begin{thm}
Let $\A$ be a finite algebra such that $\V(\A)$ omits types $\1$ and $\5$.
If $\A$ admits a natural duality, then $\A$ has a cube term.
\end{thm}

\pause \vfill

\begin{itemize}
  \item Naturally dualizable
  \vfill
  \pause \item Omitting types $\1$ and $\5$
  \vfill
  \pause \item Cube term
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{Natural dualities}
\pause
Motivational examples to think of:
\begin{itemize}
  \item Stone duality between Boolean algebras and Stone spaces
  \pause \item Priestley duality between distributive lattices and bounded
    Priestley spaces
  \pause \item Pontryagin duality for abelian groups
\end{itemize}

\pause \vfill

Ingredients for a duality:
\begin{itemize}
  \item a finite algebra $\A = \left< A; F\right>$ 
  \pause \item a structured topological space $\sts{A} = \left< A; G, H, R,
    \mathcal{T} \right>$ ($G=\text{total ops}$, $H=\text{partial ops}$,
    $R=\text{relations}$, $\mathcal{T}=\text{discrete topology}$)
  \pause \item If $\m{B}\in \S\P(\A)$ then $\m{B}^\partial =
    \Hom(\m{B},\A)\in \S_c\P_+(\sts{A})$
  \pause \item If $\sts{B}\in \S_c\P_+(\sts{A})$ then $\sts{B}^\partial =
    \Hom(\sts{B},\sts{A})\in \S\P(\A)$
  \pause \item $\sts{A}$ should be algebraic over $\A$
\end{itemize}
\end{frame}
%===========================================================================}}}
%==========================================================================={{{
\begin{frame}[t] \frametitle{Natural dualities}
For $\m{B}\in \S\P(\A)$ there are mappings
\vspace{-2ex}
\[ \begin{array}{rrcl}
  e_{\m{B}}: & \m{B} & \to & \m{B}^{\partial\partial} \\
  & x & \mapsto & 
    \left( e_{\m{B}}(x):\m{B}^{\partial} \to \sts{A}: y \mapsto y(a)
    \right),
\vspace{-2ex} \end{array} \]
\pause $\sts{A}$ \emph{dualizes} $\A$ if $e_{\m{B}}$ are isomorphisms for
all $\m{B}\in \S\P(\A)$.

\pause \vfill

Examples of naturally dualizable algebras:
\begin{itemize}
  \pause \item Groups whose Sylow subgroups are abelian [Nickodemus 2007]
  \pause \item Rings whose Jacobson radical squares to $(0)$ [Clark, Idziak,
    Sabourin, Szabo, Willard 2001]
  \pause \item Algebras with a compatible semilattice operation [Davey,
    Jackson, Pitkethly, Taukder 2007]
  \pause \item Algebras with near unanimity term [Davey, Heindorf, McKenzie
    1995]
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{What?}
\begin{thm}
Let $\A$ be a finite algebra such that $\V(\A)$ omits types $\1$ and $\5$.
If $\A$ admits a natural duality, then $\A$ has a cube term.
\end{thm}

\vfill

\begin{itemize}
  \item \sout{Naturally dualizable}
  \vfill
  \item Omitting types $\1$ and $\5$
  \vfill
  \item Cube term
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{Omitting $\{\1,\5\}$}
TCT type $\1$ is the unary type, $\5$ is the semilattice type

\pause \vfill

If $\A$ finite and $\V(\A)$ is congruence modular then $\V(\A)$ omits $\1$
and $\5$.

\pause \vfill

\begin{thm}[Kearnes, Kiss]
The following are equivalent for a variety $\V$.
\begin{enumerate}
  \item $\V$ omits types $\1$ and $\5$.

  \item $\V$ has a sequence of idempotent terms $f_i(x,y,v,u)$ for $0\leq
  i\leq 2m+1$, such that 
  \begin{enumerate}
    \item $\V\models f_0(x,y,u,v) \approx x$ and $\V\models
      f_{2m+1}(x,y,u,v) \approx v$,
    \item $\V\models f_i(x,y,y,y) \approx f_{i+1}(x,y,y,y)$ for all even $i$,
    \item $\V\models f_i(x,x,y,y) \approx f_{i+1}(x,x,y,y)$ for all odd $i$,
      and
    \item $\V\models f_i(x,y,x,y) \approx f_{i+1}(x,y,x,y)$ for all odd $i$.
  \end{enumerate}
\end{enumerate}
\end{thm}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{What?}
\begin{thm}
Let $\A$ be a finite algebra such that $\V(\A)$ omits types $\1$ and $\5$.
If $\A$ admits a natural duality, then $\A$ has a cube term.
\end{thm}

\vfill

\begin{itemize}
  \item \sout{Naturally dualizable}
  \vfill
  \item \sout{Omitting types $\1$ and $\5$}
  \vfill
  \item Cube term
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{Cube terms}
For a variety $\V$ the term $t(x_1,\ldots,x_n)$ is a \emph{cube term for
$\V$}, if for all $1\leq i\leq n$ there is a choice of $u_1,\ldots,u_n\in
\{x,y\}$ with $u_i = y$ such that
\[
  \V \models t(u_1,\ldots,u_n) \approx x.
\]

\pause \vfill

Two examples:
\begin{itemize}
  \pause \item The ``weakest'' cube term is 
  \[
    t\mat{ y & x & \cdots & x \\
           x & y & \cdots & x \\
           \vdots & & \ddots & \vdots \\
           x & x & \ldots & y } 
    \approx \mat{x \\ x \\ \vdots \\ x}
  \]
  \pause \item The ``strongest'' cube term ``is''
  \[
    t\mat{ y & x & x & y & y & x & y \\
           x & y & x & y & x & y & y \\
           x & x & y & x & y & y & y } 
    \approx \mat{x \\ x \\ x}
  \]
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{What?}
\begin{thm}
Let $\A$ be a finite algebra such that $\V(\A)$ omits types $\1$ and $\5$.
If $\A$ admits a natural duality, then $\A$ has a cube term.
\end{thm}

\vfill

\begin{itemize}
  \item \sout{Naturally dualizable}
  \vfill
  \item \sout{Omitting types $\1$ and $\5$}
  \vfill
  \item \sout{Cube term}
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{Why?}
\pause
\begin{thm}[Davey, Heindorf, McKenzie 1995]
The following are equivalent for a finite algebra $\A$.
\begin{enumerate}
  \item $\A$ is dualizable and $\V(\A)$ is congruence distributive.
  \item $\A$ has a near unanimity term.
\end{enumerate}
\end{thm}

\pause \vfill

\begin{question}
Is there a similar connection between the existence of a cube term for
congruence modular $\V(\A)$ and dualizability of $\A$?
\end{question}

\pause

Intuition:
\begin{itemize}
  \item The cube term is a generalization of the NU term.
  \item A cube term for $\V$ implies CM just like an NU term for $\V$
    implies CD.
\end{itemize}

\pause

Counter-evidence:
\begin{itemize}
  \item The group $\mathbb{S}_3$ is in a CM variety, has a cube term, and is
    dualizable, but the algebra obtained by adding constants is
    non-dualizable.
\end{itemize}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{Inherent non-dualizability}
A finite algebra $\A$ is \emph{inherently non-dualizable} if for all
finite algebras $\m{B}$,
\[
  \A\in \S\P(\m{B}) \qquad \qquad \text{implies} \qquad\qquad \m{B}
  \text{ non-dualizable.}
\]

\pause \vfill

\begin{thm}[Davey, Idziak, Lampe, McNulty 1997]
Let $\A$ be a finite algebra, $\m{B}\leq \A^Z$, $B_0\subseteq B$ be an
infinite subset such that
\begin{enumerate}
  \pause \item there is a function $\phi:\omega\to \omega$ such that for all
  $k\in \omega$ and all $\theta\in \Con(\m{B})$ of index at most $k$,
  $\theta|_{B_0}$ has a unique block of size greater than $\phi(k)$; and

  \pause \item if the element $g\in A^Z$ is defined by $g(z) = \pi(a_z)$ for
  $z\in Z$, where $a_z$ is an element of the unique block of
  $\ker(\pi_z)|_{B_0}$ of size greater than $\phi(|B|)$, then $g\not\in B$.
\end{enumerate}
\pause Then $\A$ is inherently non-dualizable.
\end{thm}
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] \frametitle{Elements of the proof (assuming idempotency)}
Pick $\A$ the smallest idempotent dualizable algebra omitting $\1$ and $\5$
\emph{without} a cube term. \pause $\A$ has a cube term blocker, so there
are $a,b\in A$ such that there is no $t(\cdots)$ with
\[
  t\mat{\\ \text{cube pattern on } (a,b) \\ \ } = \mat{a \\ \vdots \\ a}.
\]

\pause \vfill

Define elements of $A^{\omega}$
\[
  \alpha_{i_1 \cdots i_n} = \begin{cases}
    b & \text{if } i\in \{i_1,\ldots,i_n\} \\
    a & \text{otherwise}.
  \end{cases}
\]

\pause

We will apply the non-dualizability theorem to
\[
  C_0 = \{ \alpha_i \mid i\in \omega \} 
  \qquad \qquad \qquad\qquad
  \m{C} = \Sg^{\A^\omega}(C_0)
\]
\end{frame}
%===========================================================================}}}
%==========================================================================={{{
\begin{frame}[t] \frametitle{Elements of the proof (assuming idempotency)}
To show:
\begin{enumerate}
  \item \small{ there is a function $\phi:\omega\to \omega$ such that for
    all $k\in \omega$ and all $\theta\in \Con(\m{C})$ of index at most $k$,
    $\theta|_{C_0}$ has a unique block of size greater than $\phi(k)$ }
\end{enumerate}
\hrule \pause \vfill

Take $\phi(k) = 1$, $\theta\in \Con(\m{C})$, and assume that there are two
$\theta|_{C_0}$-blocks of size $> 1$:
\vspace{-2ex} \[
  S = \{ \alpha_1, \alpha_3, \ldots \}
  \qquad \qquad \text{and} \qquad \qquad
  T = \{ \alpha_2, \alpha_4, \ldots \}
\]

\pause 

\begin{claim}
  The set $\{ \alpha_{12}, \alpha_{14}, \alpha_{23}, \alpha_{34},
  \alpha_{124}, \alpha_{123}, \alpha_{134}, \alpha_{234} \}$ lies in single
  $\theta$-block.
\end{claim}
\begin{claimproof}
\pause
... (use the WNU and cube term blockers) ...
\end{claimproof}
\end{frame}
%===========================================================================}}}
%==========================================================================={{{
\begin{frame}[t] \frametitle{Elements of the proof (assuming idempotency)}
To show:
\begin{enumerate}
  \item \small{ there is a function $\phi:\omega\to \omega$ such that for
    all $k\in \omega$ and all $\theta\in \Con(\m{C})$ of index at most $k$,
    $\theta|_{C_0}$ has a unique block of size greater than $\phi(k)$ }
\end{enumerate}
\hrule 
\bigskip

\begin{claim}
$\alpha_1 \;\theta\; \alpha_2$.
\end{claim}
\end{frame}
%===========================================================================}}}
%==========================================================================={{{
\begin{frame}[t]
\vspace{-2ex}
\begin{claim}
$\alpha_1 \;\theta\; \alpha_2$.
\end{claim}
\begin{claimproof}
\pause $\A$ omits $\1$ and $\5$, so we have terms
\begin{enumerate}
  \item $\V\models f_0(x,y,u,v) \approx x$ and $\V\models
    f_{2m+1}(x,y,u,v) \approx v$,
  \item $\V\models f_i(x,y,y,y) \approx f_{i+1}(x,y,y,y)$ for all even $i$,
  \item $\V\models f_i(x,x,y,y) \approx f_{i+1}(x,x,y,y)$ for all odd $i$,
    and
  \item $\V\models f_i(x,y,x,y) \approx f_{i+1}(x,y,x,y)$ for all odd $i$.
\end{enumerate}

\pause \vfill

If $i$ is even then $f_i(\alpha_1,\alpha_{12},\alpha_{12},\alpha_{12}) =
f_{i+1}(\alpha_1,\alpha_{12},\alpha_{12},\alpha_{12})$.

\pause If $i$ is odd then by the previous claim,
\pause
\vspace{-2ex} \begin{align*}
  f_i(\alpha_1, \alpha_{12}, \alpha_{12}, \alpha_{12})
  & \uncover<6->{ \;\theta\; f_i(\alpha_1, \alpha_{12}, \alpha_{34}, \alpha_{234}) } \\
  & \uncover<7->{
    = f_i\mat{b      & b & a & a \\
              a      & b & a & b \\
              a      & a & b & b \\
              a      & a & b & b }
    }
  \uncover<8->{
  = f_{i+1}\mat{b      & b & a & a \\
                a      & b & a & b \\
                a      & a & b & b \\
                a      & a & b & b } \\
  }
  & \uncover<9->{ = f_{i+1}(\alpha_1, \alpha_{12}, \alpha_{34}, \alpha_{234}) }
  \uncover<10->{ \;\theta\; f_{i+1}(\alpha_1, \alpha_{12}, \alpha_{12}, \alpha_{12}). }
  \qedhere
\end{align*}
\end{claimproof}
\end{frame}
%===========================================================================}}}
%==========================================================================={{{
\begin{frame}[t] \frametitle{Elements of the proof (assuming idempotency)}
To show:
\begin{enumerate} \setcounter{enumi}{1}
  \item \small{ if the element $g\in A^\omega$ is defined by $g(z) =
  \pi_z(a_z)$ for $z\in \omega$, where $a_z$ is an element of the unique
  block of $\ker(\pi_z)|_{C_0}$ of size greater than $\phi(|C|)$, then
  $g\not\in C$}
\end{enumerate}
\hrule \vfill \pause

Observe:
\begin{itemize}
  \pause \item $\ker(\pi_z)|_{C_0}$ has two blocks: $X_z = \{ \alpha_z \}$
    and $Y_z = \{ \alpha_i \mid i\neq z \}$.
  \vfill \pause \item Therefore $Y_z$ is the unique large block, and
    $\pi_z(\alpha_i) = a$ for $i\neq z$.
  \vfill \pause \item Thus $g(z) = a$ for all $z\in \omega$.
\end{itemize}
\end{frame}
%===========================================================================}}}
%==========================================================================={{{
\begin{frame}[t] \frametitle{Elements of the proof (assuming idempotency)}
\begin{claim}
$g\not\in C$
\end{claim}
\begin{claimproof}
\pause Suppose that $g\in C$. Then there is a term $t(\cdots)$ such that
$t(\alpha_1,\ldots, \alpha_m) = g$. \pause That is,
\[
  t\mat{b & a & \cdots & a \\
        a & b & \cdots & a \\
        \vdots & & \ddots & \vdots \\
        a & a & \cdots & b }
  = \mat{a \\ a \\ \vdots \\ a}
\]
\pause We chose $a$ and $b$ so that this couldn't happen!
\end{claimproof}

\pause By the non-dualizability theorem $\A$ is inherently non-dualizable.
\pause This is a contradiction.
\end{frame}
%===========================================================================}}}

%==========================================================================={{{
\begin{frame}[t] % thanks for all the fish
\begin{thm}
Let $\A$ be a finite algebra such that $\V(\A)$ omits types $\1$ and $\5$.
If $\A$ admits a natural duality, then $\A$ has a cube term.
\end{thm}

\pause \centering \vfill
Thank you!
\vfill
\end{frame}
%===========================================================================}}}
\end{document}


