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\title % (optional, use only with long paper titles)
{Definability in the infinitesimal subgroup of a simple compact Lie group}

\author%[Author, Another] % (optional, use only with lots of authors)
{Martin Bays\inst{1},
Kobi Peterzil\inst{2}
}
% - Use the \inst{?} command only if the authors have different
%   affiliation.

\institute{\inst{1} WWU M\"{u}nster \\
\inst{2} University of Haifa}

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\date{Padova, 2017-09-14}
%\date[07-12-14] % (optional)
%{Friday}

\providecommand{\U}{\mathcal{U}}
\providecommand{\g}{\mathfrak{g}}
\providecommand{\h}{\mathfrak{h}}
\providecommand{\m}{\mathfrak{m}}
\providecommand{\s}{\mathfrak{s}}
\providecommand{\A}{\mathcal{A}}
\providecommand{\B}{\mathcal{B}}
\providecommand{\st}{\operatorname{st}}
\providecommand{\mod}{\operatorname{mod}}
\providecommand{\Ad}{\operatorname{Ad}}
\providecommand{\ad}{\operatorname{ad}}
\providecommand{\SO}{\operatorname{SO}}
\providecommand{\Spin}{\operatorname{Spin}}
\providecommand{\so}{\operatorname{so}}
\providecommand{\su}{\operatorname{su}}
\providecommand{\GL}{\operatorname{GL}}
\providecommand{\pr}{\operatorname{pr}}

\renewcommand{\O}{\mathcal{O}}
\renewcommand{\v}{\operatorname{v}}

\renewcommand{\o}{\circ}

\renewcommand{\O}{\mathcal{O}}
\newcommand{\Ru}{\mathcal{R}}
\newcommand{\Goo}{{G^{00}}}
\newcommand{\Soo}{{S^{00}}}
\newcommand{\Roo}{{\SO_3^{00}}}

\newcommand{\Xbar}{\bar{X}}

\begin{document}
\begin{frame}
  \titlepage
  %  \tableofcontents
  %  % You might wish to add the option [pausesections]
\end{frame}

\begin{frame}
  %2:30
  \frametitle{$\SO_3(\R)$}
  \begin{theorem}[Nesin-Pillay 1991]
    \begin{itemize}\item $X \subseteq  \SO_3(\R)^n$ is definable in the pure group $(\SO_3(\R);*)$ iff 
    it is definable in the field $(\R;+,\cdot)$.
    \item More generally, same for any simple centreless compact linear algebraic 
    group $G \leq  \GL_n(\R)$.
    \end{itemize}
  \end{theorem}
  \begin{example}
    $\{ (A,B) \in \SO_3(\R) : \det(A-B)>0 \}$ is definable in $(\SO_3(\R);*)$.
  \end{example}
  Sketch of proof:
    \begin{itemize}\item Define a copy of $\SO_3(\R)$ in $(G;*)$;
    \item Reconstruct the field from the projective plane of involutions of 
    $\SO_3(\R)$.
    \item See that this yields a bi-interpretation of $(G;*)$ with 
    $(\R;+,\cdot)$.
    \end{itemize}
\end{frame}

\begin{frame}
  %3:00
  \frametitle{$\Roo$}
  \[ (\Ru;+,\cdot) := (\R;+,\cdot)^\U \elemext (\R;+,\cdot) \]

  \[ 0 \rightarrow  \m \rightarrow  \O \xrightarrow{\st} \R \rightarrow  0 \]

  \[ 1 \rightarrow  \Roo \rightarrow  \SO_3(\Ru) \xrightarrow{\st} \SO_3(\R) \rightarrow  1 \]

  \[ \Roo = \SO_3(\Ru) \cap \begin{pmatrix}
    1+\m & \m   & \m   \\
    \m   & 1+\m & \m   \\
    \m   & \m   & 1+\m
    \end{pmatrix} \]

  $(\Roo;*)$ is interpretable in $(\Ru;+,\cdot,\O) \vDash  RCVF$.
  \begin{problem}
    Which $(\Ru;+,\cdot,\O)$-definable subsets of $(\Roo)^n$ are 
    $(\Roo;*)$-definable?

    Same question for $\Goo$ for $G \subseteq  \GL_n(\R)$ compact?
    % Which it turns out can be seen as algebraic linear groups
  \end{problem}
\end{frame}

\begin{frame}
  %6:30
  \frametitle{$(\Roo;*)$}
  \begin{itemize}\item Lie algebra $\g(\Ru) = \so_3(\Ru) =
  \left\{ \begin{pmatrix}
     0 && -z &&  y \\
     z &&  0 && -x \\
    -y &&  x &&  0
  \end{pmatrix} \right\} \cong  \Ru^3 = \{ (x,y,z) \}$.

  \item Infinitesimal Lie algebra:
    $\g_\m := \st^{-1}(0) \cong  \m^3 \leq  \Ru^3$.

  \item Matrix exponentiation yields a homeomorphism
    $\exp_\m : \g_\m \xrightarrow{\sim} \Roo$.

  \item $\exp_\m(X)*\exp_\m(Y) = \exp_\m(X+Y) + \epsilon$ where $v(\|\epsilon\|) 
  \geq  v(\|X\|)+v(\|Y\|)$.

  \item If $X$ and $Y$ are collinear then
    $ \exp_\m(X)*\exp_\m(Y) = \exp_\m(X+Y)$.

  \item For $x \in \Roo$ and $h \in \SO_3(\Ru)$, group conjugation $x \mapsto  x^h := 
  h*x*h^{-1}$ agrees with the matrix action of $\SO_3(\Ru)$ on $\m^3$:
  %(which is also the adjoint action):
    \[ \exp_\m(X)^h = \exp_\m(h X) .\] %\;\; ( = \exp_\m(\Ad(h) X) ) .\]
  \end{itemize}
  % Now draw $\Roo$ as $\m^3$ and indicate conjugation action
  % (giving them time to absorb the above)
  % and recall the problem, maybe try to convince them it could be quite 
  % weak\ldots 
\end{frame}


\begin{frame}
  %2:30
  \frametitle{Main theorem}
  \begin{theorem}
    $G \subseteq  \GL_n(\R)$ a simple compact linear algebraic group.
    \begin{enumerate}[(i)]\item $X \subseteq  (\Goo)^n$ is $(\Goo;*)$-definable iff it is 
    $(\Ru;+,\cdot,\O)$-definable.

    \item
      Moreover, the interpretation of $(\Goo;*)$ in $(\Ru;+,\cdot,\O)$ can be 
      completed to a bi-interpretation.
    \end{enumerate}
  \end{theorem}
    
  \begin{example}
    $\{ (A,B) \in \Roo : v(\det(A-B))>\alpha \}$ is definable in $(\Roo;*)$.
  \end{example}

  Outline of proof:
  \begin{enumerate}[(I)]\item Find an $(\Roo;*)$-definable ordered interval $J$;
  \item apply o-minimal trichotomy to get a field $K$ in $J$;
  \item Find a copy of $\Roo$ in $\Goo$;
  \item use adjoint representation to see the pair $\Goo \leq  G(\Ru)$ in $K$,
    yielding a bi-interpretation.
  %** the characterisation of definable sets follows.
  \end{enumerate}
\end{frame}

\begin{frame}
  %6:00
  \frametitle{(I): Finding an ordered interval}
  \begin{itemize}\item Let $S := \SO_3(\Ru)$ and $\Soo := \Roo$.
  \item Let $b \in \Soo \setminus \{e\}$.
  \item $C_S(b) := \{ h \in S : b^h=b \} \cong  \SO_2(\Ru)$;
  \item $C_{\Soo}(b) := C_S(b) \cap \Soo \cong  \SO_2^{00} \cong  \m$.
  \item $b^Sb^S = \xi(S^2)$ where $\xi(h,h') = b^h*b^{h'}$.
  \item $b^Sb^S = \exp_\m(B)$ where $B \subseteq  \m^3$ is the closed ball of radius 
  $\|b^2\|$.
  \item $b^Sb^S \cap C_{\Soo}(b)$ is the interval $[b^{-2},b^2]$.
  \item By definable choice for the $(\Ru;+,\cdot)$-definable map $\xi$,
  $X := b^{\Soo}b^{\Soo} \cap C_{\Soo}(b)$ contains some interval $(h,b^2]$.
  \item Translating, get $(\Soo;*)$-definable interval $[e,p) \subseteq  C_{\Soo}(b)$,
  hence $J := (p^{-1},p)$ as an ordered interval.
  \item Explicitly: $p := b^2h^{-1}$, then $(e,p) = h^{-1}X \cap b^2X^{-1}$.
  \end{itemize}
\end{frame}

\begin{frame}
  %2:30
  \frametitle{(II): Trichotomy}
  \begin{itemize}\item $T_e(J^\Soo)$ spans $\Ru^3$, so for appropriate $h_1,h_2 \in \Soo$ and
  after shrinking $J$,
  \[ \phi : J^3 \rightarrow  \Soo; \phi(x_0,x_1,x_2) = x_0*x_1^{h_1}*x_2^{h_2} \]
  is a bijection with a neighbourhood of $e \in \Soo$.
  \item $(J;*,<)$ and $\phi$ are definable both in $(\Soo;*)$ and in
  $(\Ru;+,\cdot)$.
  \item Pulling back the $\Soo$ group structure via $\phi$ puts ``non-linear''
  structure on $J$ at $e$.
  \item By the Peterzil-Starchenko o-minimal trichotomy, a real closed field
  $(K;+,\cdot)$ on an interval $e \in K \subseteq  J$ is definable in this structure on $J$.
  \item So $(K;+,\cdot)$ is definable both in $(\Soo;*)$ and in $(\Ru;+,\cdot)$.
  \end{itemize}
\end{frame}

\begin{frame}
  %?:??
  \frametitle{(III): Finding an $\Roo$ in $\Goo$}
  \begin{itemize}\item $\g_0 := L(G)$
  \item $\h_0 \leq  g_0$ Cartan subalgebra (i.e.\ maximal abelian).
  \item $\g := \g_0 \otimes_\R \C$ and $\h := \h_0 \otimes_\R \C$.
  \item $X \in \g$ an $\ad_\h$-eigenvector for a root $\alpha \in \h^* \setminus 
  \{0\}$;\\
  i.e.\ $[H,X]=\alpha(H)X$ for $H \in \h$.
  \item Set $U := X-\Xbar, V := iX + i\Xbar$.
  \item Since $G$ is compact, $U,V \in \g_0$, and $[U,V] \in \h_0$, and $\s := \left<{
  U,V,[U,V]}\right> \cong  \so_3$. Let $\s' := \h_0 + \s$.
  \item Let $S,S' \leq  G$ with $L(S)=\s, L(S')=\s'$.
  \item So $S \cong  \SO(3)$ or $S \cong  \Spin(3)$ and $\Soo \cong  \Roo$.
  \item Considering root space decomposition, calculate:\\
  %$C_{g_0}(\s') = \ker(\alpha\negmedspace\restriction_{\h_0})$, and
  $\s' = C_{g_0}(C_{g_0}(\s'))$, and $\s = [\s',\s']$. Deduce:\\
  $S' = C_G(C_G(S'))$ and $S=(S',S')_1$;\\
  $S'^{00} = C_{\Goo}(C_{\Goo}(S'^{00}))$ and $\Soo=(S'^{00},S'^{00})_1$.
  \item So $S$ is $(\Ru;+,\cdot)$-definable, $\Soo$ is $(\Goo;*)$-definable.
  \item (Trichotomy argument to find $K$ works when $S=\Spin(3)$.)
  \end{itemize}
\end{frame}

\begin{frame}
  %3:00
  \frametitle{(IV): Bi-interpretation}
  \begin{itemize}\item $(K;+,\cdot)$ is definable both in $(\Goo;*)$ and in $(\Ru;+,\cdot)$.
  \item Otero-Peterzil-Pillay: exists $(\Ru;+,\cdot)$-definable isomorphism 
  $\theta : (\Ru;+,\cdot) \xrightarrow{\sim} (K;+,\cdot)$.
  \item $\theta$ induces $\theta_G : G(\Ru) \xrightarrow{\sim} G(K)$.
  \end{itemize}
  \begin{claim}
    $\theta_G\negmedspace\restriction_{\Goo} : G(\Ru)^{00} \xrightarrow{\sim} G(K)^{00}$ is $(\Goo;*)$-definable.
  \end{claim}
  \begin{proof}[Proof of Main Theorem]
    \begin{itemize}\item $\O$ is definable in $(\Ru;+,\cdot,\Goo)$,
    \item so $(\Ru;+,\cdot,\O)$ is interpreted on $K$ in $(\Goo;*)$ via $\theta$, 
    since $G(K)^{00}$ is $(\Goo;*)$-definable by the claim.
    \item $(\Goo;*)$ is interpreted in $(\Ru;+,\cdot,\O)$ tautologically.
    \item The composed interpretations are $\theta$ and $\theta_G\negmedspace\restriction_{\Goo}$, which 
    are definable in $(\Ru;+,\cdot,\Goo)$ resp. $(\Goo;*)$.
    \end{itemize}
  \end{proof}
\end{frame}

\begin{frame}
  %4:00
  \frametitle{Proof of claim}
  \begin{claim}
    $\theta_G\negmedspace\restriction_{\Goo} : G(\Ru)^{00} \xrightarrow{\sim} G(K)^{00}$ is $(\Goo;*)$-definable.
  \end{claim}
  \begin{proof}
    \begin{itemize}\item Quotienting by the discrete centre, we may assume $G$ is centreless.
      Let $\phi$ be a chart for $G$ in $J$ as above.
      (Exists by simplicity of $G$.)
    \item Differentiation in $K$ yields via $\phi$ an adjoint embedding
    $\Ad : G(\Ru) \rightarrow  \GL_d(K)$.
    \item $\Ad$ is $(\Ru;+,\cdot)$-definable.
    \item $\Ad\negmedspace\restriction_{\Goo}$ is $(\Goo;*)$-definable.
    \item $\eta := \Ad \o \theta_G^{-1} : G(K) \rightarrow  \GL_d(K)$ is 
    $(K;+,\cdot)$-definable by purity, hence $(\Goo;*)$-definable.
    \item So $\theta_G\negmedspace\restriction_{\Goo} = \eta^{-1} \o \Ad\negmedspace\restriction_{\Goo}$ is $(\Goo;*)$-definable.
    \end{itemize}
  \end{proof}
\end{frame}

\end{document}
