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\begin{document}

\begin{frame}
\begin{center}
\textbf{The topology of conjugate Berkovich spaces} \\
(in 10 minutes)\\
\bigskip
\bigskip
\bigskip
Pedro A. Castillejo Blasco\\
\bigskip
Advisor: HÃ©lÃ¨ne Esnault \\
\bigskip
\bigskip
\bigskip
\bigskip
2020
\end{center}
\end{frame}

\begin{frame}[fragile]
$\textcolor{blue}{X}/\bbC$ projective algebraic variety \\
$\textcolor{blue}{\sigma} \in \Aut(\bbC)$ 
\[
\begin{tikzcd}
    \textcolor{blue}{X_\sigma} \arrow{r} \arrow{d} &amp; X \arrow{d} \\
    \Spec(\bbC) \arrow{r}{\textcolor{blue}{\sigma}} &amp; \Spec(\bbC) 
\end{tikzcd}
\] \pause
\noindent \textbf{Definition:} Let $X$, $Y$ be varieties over a field $F$. We say that they are {\color{blue}\textit{conjugate}} if   $ \exists \;\; \textcolor{blue}{\sigma} \in \Aut(F)$  such that 
\begin{footnotesize}
\[
\begin{tikzcd}
	Y \arrow{r} \arrow{d} &amp; X \arrow{d} \\
	\Spec F \arrow{r}{\textcolor{blue}{\sigma}} &amp; \Spec F
\end{tikzcd}
\]
\end{footnotesize}
is Cartesian. We will write $Y= X_{\textcolor{blue}{\sigma}}$

\end{frame}

\begin{frame}[fragile]
\begin{textblock*}{30pt}(230pt,0pt)
\begin{tiny}
\[
\begin{tikzcd}
    X_\sigma \arrow{r} \arrow{d} &amp; X \arrow{d} \\
    \Spec(\bbC) \arrow{r}{\sigma} &amp; \Spec(\bbC) 
\end{tikzcd}
\]
\end{tiny}
\end{textblock*}


\textbf{Example}:
$C/\bbC$  sm. proj. curve, genus $g$, \\
$\textcolor{blue}{C^\an}$ is a Riemann surface. \\
$\textcolor{blue}{C(\bbC)}$ is homeomorphic to a torus with $g$ holes. \\ \pause
For $\sigma \in \Aut(\bbC)$, the same is true for $C_\sigma$. \\ \pause
\textit{The analytifications of {\color{blue}conjugate} smooth projective \textbf{complex} curves are {\color{blue}homeomorphic}}!

\pause

\begin{example}
Over $\bbC$, conjugate varieties $X(\bbC)$ and $Y(\bbC)$ have:
\begin{itemize}
\item The same Betti numbers (comparison theorem of Artin, SGA4) \pause
\item The same Ã©tale fundamental group (hence, \textit{profinite completion} of $\pi_1^\top(X(\bbC))$ and $\pi_1^\top(Y(\bbC))$ coincide)
\end{itemize}
\end{example} \pause
\textbf{Question (60's):} Are (the analytifications of) \textit{complex} conjugate varieties homeomorphic?
\end{frame}

\begin{frame}
\textbf{Question (60's):} Are \textit{complex} conjugate varieties homeomorphic? \\
\textbf{Answer (Serre, '64):} No! \\
He constructs conjugate surfaces whose analytifications have different topological fundamental groups.\\ \bigskip
\pause
\textbf{Question:} What happens in the \textit{non-archimedean} setting? \\
Are \textit{non-archimedean} conjugate varieties homeomorphic?
\end{frame}


\begin{frame}

$K$: complete non-archimedean field \\

\begin{wrapfigure}{R}{0.17\textwidth}
\centering
\includegraphics[width=0.17\textwidth]{Sierpinski_triangle.png}
%\caption{\begin{scriptsize}Rigid analytic disk over $3$-adic numbers.\end{scriptsize}}
\end{wrapfigure}
In non-archimedean geometry there are \textcolor{blue}{different analytification functors}.


\textbf{First approach}: using \textcolor{blue}{rigid analytic spaces Ã&nbsp; la Tate}. \\

\textbf{Problem}: the underlying topological space is totally disconnected.\\

\pause

\begin{wrapfigure}{R}{0.17\textwidth}
\centering
\includegraphics[width=0.17\textwidth]{berkovich-disk.png}
\end{wrapfigure}


\textbf{Second approach}: using \textcolor{blue}{Berkovich spaces}.

\textbf{Remark}: this approach will lead to locally path-connected and locally contractible topological spaces.\\
\pause

\bigskip
\textbf{Example}: The Berkovich analytification of smooth \\ curves of genus 0 ($\bbP^1, \bbA^1, \bbG_m$, etc.) is contractible.

\end{frame}


\begin{frame}
$K$ complete, alg. closed, non-archimedean field with non-trivial valuation

\begin{itemize}
\item $C/K$ a smooth curve of genus $g&gt;0$
\end{itemize}
\pause

\begin{center}
\textcolor{blue}{Homotopy type} of its (Berkovich) analytification $\textcolor{blue}{C^\an}$
\[\shortparallel\]
\textcolor{blue}{Homotopy type} of the \textbf{dual intersection graph} of a \textbf{semistable model} of $C$. \pause

\end{center}
\begin{example}[C., idea from Chambert-Loir]
Assume for simplicity $K = \bbC_p$, with $p&gt;3$. \\
Let $E: y^2 = x^3 + px + a$, with $a \in \bbCp$ of valuation equal to 1 $\Rightarrow \textcolor{blue}{E^\an}$ \textit{is contractible}. \pause

If moreover $a$ is transcendental over $\bbQ$, we can construct an automorphism $\sigma \in \Aut(\bbCp)$ such that the analytification of $E_{\textcolor{blue}{\sigma}}: y^2 = x^3 + px + \textcolor{blue}{\sigma(a)}$ \textit{is \textbf{not} contractible}!\pause \\
(Trick: play with the $j$-invariant) \\
$\Rightarrow$ \textcolor{blue}{$E^\an$ and $E_\sigma^\an$ are non-homeomorphic Berkovich elliptic curves}. 
\end{example}

\end{frame}

\begin{frame}

\textbf{Question}: given $\sigma \in \Aut(\bbCp)$, is there a variety $X/\bbCp$ non-homeomorphic to its conjugate $X_\sigma$? \\ \pause


\begin{theorem}[C.]
Given any non-continuous $\sigma \in \Aut(\bbCp)$, we can construct an elliptic curve $E/ \bbCp$ such that $E^\an$ and $E_\sigma^\an$ are non-homeomorphic.
\end{theorem}
\pause
The construction of the elliptic curve is explicit, and uses again the trick of the $j$-invariant.


\end{frame}

\begin{frame}
Let $K = \bbC\{t\} := \left(\overline{\bigcup \bbC(t^{1/n})}\right)^\wedge$ be the field of Puiseux series. 
\begin{theorem}[Nicaise, '20]
Every connected smooth and proper variety $X$ over $\bbC\{t\}$ is conjugate to a smooth and proper variety $Y$ whose Berkovich analytification is contractible.
\end{theorem}
\pause
We don't know an explicit construction of the conjugate variety, this is an existence statement.
\end{frame}

\begin{frame}[fragile]
\textbf{Question}: Can we construct more explicit examples of non-homeomorphic conjugate Berkovich spaces?\\


\textbf{Idea}: Use tropicalizations to achieve this:

\pause
$\iota: X \hookrightarrow \bbA^n_K$ \\
We can define the (extended) tropicalization map
\[
\begin{tikzcd}
\pi_\iota: X^\an \arrow{r} &amp; (\bbR \cup \{ \infty\})^n \\
 &amp; (\val(K))^n \arrow[right hook-&gt;]{u}
\end{tikzcd}
\] \pause

\begin{theorem}[Payne, '09]
Let $X/K$ be an affine variety, $K$ alg. closed non-archimedean field with nontrivial valuation.\\
 $X^\an$ is homeomorphic to the inverse limit of all extended tropicalizations.
\end{theorem}
\end{frame}

\begin{frame}
We fix now the field of Puiseux series $\bbC\{t\}$.
\begin{block}{Construction (C.)}
For any degree $d \geq 3$, we construct explicit smooth planar curves $C \hookrightarrow \bbG_m^2$ of degree $d$, and find an automorphism $\sigma \in \Aut(\bbC\{t\})$ such that:
\begin{itemize}
\item $C$  is Mumford
\item $C_{\sigma}$ has good reduction
\end{itemize} 
In particular, their (Berkovich) analytifications are non-homeomorphic conjugate curves.
\pause
For example, for $d=4$ the curve is given by the following polynomial $f \in \bbC\{t\}[x,y]$:
\[
\begin{array}{ll}
f(x,y)= &amp; y^{4}  \\
&amp;+ t y^{3} + t x y^{3}    \\
&amp; +t^{3} y^{2} + t^{4} x y^{2} + t^{7} x^{2} y^{2}    \\
&amp;+ t^{14} y + t^{24} x y+  t^{46} x^{2} y + t^{127} x^{3} y  \\
&amp;+ \textcolor{blue}{a_{704}}    + t^{2072} x +  t^{4141} x^{2}y  + t^{12353} x^{3}   + \textcolor{blue}{b_{49229}} x^{4}.
\end{array}
\]
The automorphism $\sigma$ fixes $t$ and moves $\textcolor{blue}{a_{704}}$ and $\textcolor{blue}{b_{49229}}$.
\end{block}
\end{frame}


\begin{frame}


\begin{figure}[h]
\centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=.8\linewidth, keepaspectratio = true]{mumford-curve-unimodular.png}
  \caption{Degree $d=4$}
  \label{fig:sub1}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=.8\linewidth, keepaspectratio = true]{mumford-curve-unimodular-6.png}
  \caption{Degree $d=6$}
  \label{fig:sub2}
\end{subfigure}
\caption{Newton polytopes of Mumford curves with their regular subdivisions.}
\label{fig:test}
\end{figure}

\end{frame}


\begin{frame}
\textbf{Conclusions}
\begin{itemize}
\item Berkovich spaces give us a nice setting to discuss about the topology of non-archimedean spaces \pause
\item There are (lots of) non-homeomorphic conjugate Berkovich spaces \pause
\item Tropicalizations give us an effective way to construct explicit examples (smooth hypersurfaces) \pause
\item (Future project? Question posed by Bernd Sturmfels): take 15 integer numbers $a_1, \ldots, a_{15} \in \bbZ$ defining a smooth planar quartic. Can we construct an algorithm describing the semi-stable reduction at all primes $p$? \pause
\end{itemize}

\begin{center}
\textbf{Thank you for your attention!
}
\end{center}

\end{frame}

\end{document}

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