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\title{Peter Scholze awarded the Fields medal}
\author{Ulrich Görtz}
\date{Bonn, October 1, 2018}

\begin{document}
\begin{frame}
\titlepage
\includegraphics[width=5cm]{esaga}
\hfill\includegraphics[width=4cm]{ude}
\end{frame}

\section{$\bullet$}
\begin{frame}
\frametitle{Most important research prize in mathematics}

\vspace{1cm}
\begin{minipage}{4.5cm}
\includegraphics[width=4cm]{fieldsmedal}
\end{minipage}%
\only<1>{\begin{minipage}{6cm}%
John Charles Fields

\bigskip
Since 1936,

59 medals awarded.

\bigskip
Age limit: 40 years
\end{minipage}
}
\only<2>{
\begin{textblock}{10}(3,2)
\includegraphics[width=10cm]{scholzentv}
\end{textblock}
}
\only<3>{
\begin{textblock}{10}(3,2)
\includegraphics[width=10cm]{scholzespiegelonline}
\end{textblock}
}
% \only<4>{
% \begin{textblock}{10}(3,2)
% \includegraphics[width=10cm]{scholzeheise}
% \end{textblock}
% }
\only<4>{
\begin{textblock}{10}(3,2)
\includegraphics[width=10cm]{scholzefaz}
\end{textblock}
}

% https://tex.stackexchange.com/questions/34921/how-to-overlap-images-in-a-beamer-slide
% https://blogs.helsinki.fi/smsiltan/2012/10/12/precise-positioning-in-latex-beamer/
\end{frame}


\begin{frame}
\frametitle{Goal of this talk}

Some impression of the area, provide context for non-experts.

\bigskip
\begin{overprint}
\onslide<1>
\includegraphics[width=10cm]{map1}
\onslide<2>
\includegraphics[width=10cm]{map2}
\onslide<3>
\includegraphics[width=10cm]{map3}
\end{overprint}

{\footnotesize Urbano Monte's map of the earth, 1587\\
David Rumsey Map Collection CC-BY-NC-SA 3.0}
\end{frame}

\begin{frame}
\frametitle{Goal of this talk}

\begin{figure}
\begin{tikzpicture}
\draw (0, 3) ellipse (2.5cm and 1.2cm) node [align=center] {$p$-adic fields,\\e.~g.~$\mathbb Q_p$};
\draw (-3.2, 5.5) ellipse (2.2cm and 1.2cm) node [align=center] {mod $p$ fields,\\e.~g.~$\mathbb F_p((t))$};
\draw (3.2, 5.5) ellipse (2.2cm and 1.2cm) node [align=center] {Archimedean\\fields, e.~g.~$\mathbb R$, $\mathbb C$};
\draw (0, 0) ellipse (2.2cm and 1.2cm) node [align=center] {Algebraic number\\fields, e.~g.~$\mathbb Q$};

\draw (-1.6, 4) -- (-2.2, 4.35);
\draw (0, 1.75) -- (0, 1.25);
\draw (1.6, 4) -- (2.2, 4.35);

\end{tikzpicture}
\end{figure}
\end{frame}


\section{$p$-adic numbers}

\begin{frame}
\frametitle{Solving equations}

Important problem in mathematics:

\bigskip
Understand set of solutions of an equation.

\pause
\bigskip
\begin{itemize}
\item
Do solutions exist?
\item
Are there only finitely many solutions? Can we count them? Can we write them down explicitly?
\item
If there are infinitely many solutions, does the set of solutions have a (geometric) structure?
\end{itemize}

\end{frame}


\begin{frame}
\frametitle{Where are we looking for solutions?}

Natural numbers \qquad $\mathbb N = \{ 0, 1, 2, 3, \dots \}$.

\bigskip
Integers \emph{(add negative numbers)}
\[
\mathbb Z = \{ \dots, -3, -2, -1, 0, 1, 2, 3, \dots \}
\]

\bigskip
\pause
Rational numbers \emph{(add fractions $\frac 12$, $\frac 13$, $\frac 23$, \dots)}
\[
\mathbb Q \pause = \left\{ \frac ab;\ a, b\in\mathbb Z, b\ne 0\right\}
\]

\pause
Real numbers \emph{(add all decimal numbers)}
\[
-1,\ -0.5,\ 0,\ 0.333\ldots,\ 1,\ 2,\ 3.14159265\ldots \in \mathbb R
\]

\end{frame}

\begin{frame}
\frametitle{Can we detect cases without solutions?}

\bigskip
This is often a very hard problem, for instance:

\begin{theorem}[``Fermat's Last Theorem'', A.~Wiles]
Let $n > 2$ be an integer. Then the equation
\[
x^n + y^n = z^n
\]
has no solutions with integers $x, y, z \ge 1$.
\end{theorem}

\bigskip
\pause
For certain equations, however, it is easy to show that there are no solutions in the integers.
\end{frame}

\begin{frame}
\frametitle{Understand set of solutions in real numbers}

Trivially: If no solutions in $\mathbb R$, then no solutions in $\mathbb Z$.

\bigskip
Sometimes: Good understanding of solutions in real numbers $\rightsquigarrow$ understand solutions in integers.


\pause\bigskip
Over $\mathbb R$, can use analytic methods (Differential calculus, derivatives, \dots)

\bigskip
\begin{figure}
\begin{tikzpicture}
\begin{axis}[
        axis y line=center,
        axis x line=middle, 
        axis on top=true,
        xmin=-2,
        xmax=1,
        ymin=-23,
        ymax=10,
        height=5.0cm,
        width=12.0cm
        % grid,
        % xtick={-5,...,5},
        % ytick={-40,-32,...,40},
    ]
\addplot [domain=-1.9:.8, samples=20, mark=none, thick, red] {1.2};
\addplot [domain=-1.9:.8, samples=100, mark=none, ultra thick, blue] {x^5 + 5*x^4 - 20*x^3 - 40*x^2 + 5*x + 1};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{frame}


\begin{frame}
\frametitle{Looking at the final digit \dots}

We see that the equation
\[
x^4 + 17 = 4y^2
\]
has no solutions with integers $x, y\in\mathbb Z$,

\pause
\medskip
because the \alert{final digit} can only be

\medskip
\begin{columns}
\begin{column}{5cm}
left hand side $x^4+17$:

2, 3, 7 or 8,
\end{column}
\begin{column}{5cm}
right hand side $4y^2$:

0, 4, or 6.
\end{column}
\end{columns}

\pause
\vspace{.8cm}
More powerful: Look at more final digits.

\pause
\bigskip
In other words: division with remainder by $10$, $100$, $1000$, \dots, $10^i$.
\end{frame}


\begin{frame}
\frametitle{Looking at the final digit, refined version}

Could also do division with remainder by other numbers $n = 2, 3, 4, \dots$.

For instance consider the equation
\[
x^4 - 17 = 7y^2
\]

This is ``solvable mod $10$'' (both sides can have final digit $3$, for instance).

\pause\bigskip
But division with remainder by $7$ gives remainder $1$, $3$, $4$, or $5$ on the left, and $0$ on the right.
\end{frame}


\begin{frame}
\frametitle{Division with remainder = $n$-adic final digit}

\pause
Binary expression:
\[
23\pause = 16 + 4 + 2 + 1\pause = 2^4 + 2^2 + 2^1 + 2^0\pause = 10111_2.
\]
\pause
$23 \equiv 1 \mod 2$,

$23 \equiv 11_2 = 3 \mod 4$,

$23 \equiv 111_2 = 7 \mod 16$.

\pause\bigskip
$7$-adic expression:
\[
23 = 21 + 2 = 3\cdot 7^1 + 2\cdot 7^0 = 32_7.
\]
$23 \equiv 2 \mod 7$.

\end{frame}


\begin{frame}
\frametitle{Analytic methods}

Key point: passing to limit.

A sequence of real numbers ``coming arbitrarily close to each other'' converges to a limit in $\mathbb R$.

\vspace{1cm}
\begin{overlayarea}{\textwidth}{7cm}
\only<2>{
\begin{itemize}
\item
\dots not interesting in $\mathbb Z$,
\item
\dots does not work in $\mathbb Q$: We can approximate $\sqrt{2}$ by rational numbers, but it is not rational itself.
\end{itemize}}
\only<3->{
Two real numbers are close to each other if the differences lie far to the right of the decimal point:

123.12345 is much closer to 123.12346 than to 123.22345}
\only<4>{

\medskip
A limit always exists because, naively speaking, we allow infinitely many digits to the right of the decimal point.}
\end{overlayarea}
\end{frame}


\begin{frame}
\frametitle{Setting up an analogy}

``Solving'' an equation so that the final 5 digits match is more difficult than having only the final digits match.

\medskip
Having more digits match is a ``better approximation'' of the solution from this point of view.

\pause
\begin{example}[Lind-Reichardt equation: $x^4 - 17 = 2y^2$] % ~ 1940
\begin{itemize}
\item $x=5$, $y=8$:  608 versus 128
\item $x=85$, $y=548$:  52 200 608 versus 600 608
\end{itemize}
\end{example}

\pause
\begin{block}{}
$10$-adic numbers $\mathbb Z_{10}$:\\
Allow infinitely many digits, extending to the left.
\end{block}

\end{frame}


\begin{frame}
\frametitle{Computing with $10$-adic numbers}

$\mathbb Z_{10} = \{ \dots a_2 a_1 a_0;\ a_i \in \{ 0, 1, \dots, 9\}\}$.

\bigskip
All natural numbers are $10$-adic numbers.
\pause
We can add and multiply $10$-adic numbers.

\bigskip
\begin{overprint}
\onslide<3>
Surprising things may happen:
\[\dots 999 + 1 = 0,\quad\text{hence } \dots999 = -1.
\]%
\onslide<4>
\begin{block}{Properties}
\begin{itemize}
\item
all integers are $10$-adic numbers,
\item
$\mathbb Z_{10}$ has operations $+$, $-$, $\cdot$.
\item
Even some fractions are $10$-adic:\qquad $\ldots 6667 \cdot 3 = 1$.
\end{itemize}
\end{block}
\end{overprint}
\end{frame}


\begin{frame}
\frametitle{Variant: $p$-adic numbers}

Although we can compute in the set $\mathbb Z_{10}$ of $10$-adic numbers, it has some less nice features:
\[
\dots 8212890625\ \cdot\ \ldots 1787109376 = 0.
\]
%$x^2 - x = 0$ has $4$ solutions ($0$, $1$ + two others)

\pause\bigskip
Better: $p$-adic numbers $\mathbb Z_p$ for a \alert{prime number $p$}.

That means: use $p$-adic expression, and allow it to extend infinitely to the left.

\pause\bigskip
$\mathbb Z_2 = \{ \ldots a_2a_1a_0;\ a_i\in \{0, 1\} \}$

$\mathbb Z_7 = \{ \ldots a_2a_1a_0;\ a_i\in \{0, 1, \dots, 6 \} \}$
\end{frame}


\begin{frame}
\frametitle{Geometry of the $p$-adic numbers}

\begin{block}{Absolute value on $\mathbb Z_p$}
\begin{align*}
& |x|_p = \frac{1}{p^n},\\
& \text{where $n$ is the number of zeros at the end of $p$-adic expression}
\end{align*}
\end{block}

\only<2>{
\begin{example}
\begin{itemize}
\item
$|48|_2 = |110000_2|_2 = 1 / 2^4 = 1/16$,
\item
$|23|_7 = |32_7|_7 = 1$.
\end{itemize}
\end{example}
}

\only<3>{
We regard $x$ close to $y$, if $|x-y|$ small.

\bigskip
\pause
Some unusual features:
\begin{itemize}
\item
Every triangle is isosceles.
\item
Any two circles are disjoint or concentric.
\end{itemize}
}
\end{frame}


\begin{frame}
\frametitle{The field of $p$-adic numbers}

The field $\mathbb Q_p$: Enlarge $\mathbb Z_p$ by allowing finitely many digits after decimal point.

\pause
\begin{example}[$p=2$]
$0.1_2 = 1/2$,\quad $0.01_2 = 1/4$,\pause\qquad
$\dots 111.1_2 = -1/2$.
\end{example}

\bigskip\pause
$\mathbb Q_p$ a \emph{field}: have $+$, $-$, $\cdot$, $/$.

\bigskip
\only<4>{
\fbox{\includegraphics[width=10cm]{hensel}} % around 1900
}
\only<5>{
\fbox{\includegraphics[width=10cm]{hensel2}}
}
\only<6>{
\begin{align*}
\mathbb Z_p & = \left\{ \sum_{i=0}^\infty a_i p^i;\quad a_i\in \{0, 1, \dots, p-1\} \right\},\\
\mathbb Q_p & = \left\{ \sum_{i=i_0}^\infty a_i p^i;\quad i_0\in\mathbb Z,\quad a_i\in \{0, 1, \dots, p-1\} \right\}.
\end{align*}
}
\end{frame}

\begin{frame}
\frametitle{$p$-adic geometry}

Tate (around 1962): Rigid analytic spaces

\dots

Huber (around 1990): Adic spaces

$\rightsquigarrow$ reasonable notion of $p$-adic manifold/space.

\pause\bigskip
\begin{block}{}
\begin{quote}
Peter Scholze has revolutionized the field\\
of $p$-adic geometry.

\medskip
\hfill M.~Rapoport, Laudatio for P.~Scholze, ICM 2018
\end{quote}
\end{block}
\end{frame}



\begin{frame}
\frametitle{$p$-adic and complex geometry are similar}

\begin{theorem}[Scholze]
Let $C/\mathbb Q_p$ be complete and algebraically closed. Let $X$ be a smooth proper rigid analytic space over $C$.
For all $i\ge 0$, we have
\[
\sum_{j=0}^i \dim_C H^{i-j}(X, \Omega_{X}^j) = \dim_C H^i_{dR}(X/C) = \dim_{\mathbb Q_p} H^i_{et}(X, \mathbb Q_p)
\]
\end{theorem}
\end{frame}



\begin{frame}
\frametitle{The local-global principle}

\begin{theorem}[Hasse-Minkowski]  % Minkowski - 1909; Hasse 1898 - 1979
Let $n \ge 1$ and let $a_{i} \in\mathbb Q$, $1\le i \le n$. Then the equation
\[
a_1 x_1^2 + a_2 x_2^2 + \cdots + a_n x_n^2 = 1
\]
has a solution $x_i \in \mathbb Q$, if and only if it has a solution in $\mathbb R$ and in every field $\mathbb Q_p$.
\end{theorem}

\bigskip\pause
\begin{example}[Hasse-Minkowski for $n=1$]
$ax^2 = 1$ solvable in $\mathbb Q$ $\Leftrightarrow$ $a$ is a square $\ne 0$
$\Leftrightarrow$ $a>0$ and every prime $p$ occurs with even exponent in factorization of $a$
\end{example}

% \bigskip\pause
% \begin{theorem}[Lind-Reichardt]
% The equation
% \[
% x^4  - 17 = 2y^2
% \]
% has a solution in $\mathbb R$ and in every $\mathbb Q_p$, but not in $\mathbb Q$.
% \end{theorem}

% one variable: for irreducible polynomials, local-global principle holds (Hasse)
% 
% in general, not: $(x^2-p)(x^2-q)(x^2-pq)$, $p=5$, $q=29$.
% cf. https://gdz.sub.uni-goettingen.de/id/PPN235181684_0106?tify={%22pages%22:[459],%22view%22:%22info%22}
\end{frame}


\section{Galois representations}


\begin{frame}
\frametitle{When solutions exist \dots}

\hspace{3cm}\dots can we write them down?

\bigskip
Linear:\hspace{1cm} $\begin{aligned}[t] &  2x - 6 = 0,  \qquad x =  \frac 62 = 3.\\ \pause
& ax - b = 0,\ a\ne 0, \qquad x =  \frac ba \in \mathbb Q.
\end{aligned}$

\pause\bigskip
Quadratic:
\begin{align*}
& ax^2 + bx + c = 0,\ a\ne 0,\\[.8cm]
& x = \frac{-b + \sqrt{b^2-4ac}}{2a} \text{ or }
x = \frac{-b - \sqrt{b^2-4ac}}{2a}
\in \mathbb R.
\end{align*}
\end{frame}

\begin{frame}
Formulas for degrees 3, 4

(del Ferro, Tartaglia, Cardano, Ferrari $\approx$ 1500)

\pause\bigskip
\begin{block}{}
Galois: No formula for higher degree! ($\approx$ 1830)
\end{block}

\medskip\pause
Even worse: For example, the zeros of the polynomial
\hfill $x^5 + 5x^4 - 20x^3 - 40x^2 + 5x + 1$\hfill

cannot be expressed in terms of $+$, $-$, $\cdot$, $/$ and $\sqrt[n]{-}$.

\begin{overprint}
\onslide<3>
\begin{figure}
\begin{tikzpicture}
\begin{axis}[
        axis y line=center,
        axis x line=middle, 
        axis on top=true,
        xmin=-8.5,
        xmax=5.5,
        ymin=-500,
        ymax=2200,
        height=5.0cm,
        width=12.0cm
    ]
\addplot [domain=-8:5, samples=100, mark=none, ultra thick, blue] {x^5 + 5*x^4 - 20*x^3 - 40*x^2 + 5*x + 1};
\end{axis}
\end{tikzpicture}
\end{figure}
\onslide<4>
\begin{figure}
\begin{tikzpicture}
\begin{axis}[
        axis y line=center,
        axis x line=middle, 
        axis on top=true,
        xmin=-2,
        xmax=1,
        ymin=-23,
        ymax=10,
        height=5.0cm,
        width=12.0cm
        % grid,
        % xtick={-5,...,5},
        % ytick={-40,-32,...,40},
    ]
\addplot [domain=-1.9:.8, samples=100, mark=none, ultra thick, blue] {x^5 + 5*x^4 - 20*x^3 - 40*x^2 + 5*x + 1};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{overprint}
\end{frame}


\begin{frame}
\frametitle{Why no formula?}

\dots understand symmetries of set of solutions

\pause\bigskip
Distinguish geometric objects by their ``symmetry group''

\begin{figure}
\onslide<2->{%
\begin{tikzpicture}[scale=2]
\draw[ultra thick] (0, 0) -- (1, 0) -- (0.5, 0.866) -- (0,0);
\draw[fill=blue] (0,0) circle (0.1cm);
\draw[fill=blue] (1,0) circle (0.1cm);
\draw[fill=blue] (0.5, 0.866) circle (0.1cm);
\end{tikzpicture}%
}%
\hspace{.8cm}%
\onslide<3->{%
\begin{tikzpicture}[scale=2]
\draw[ultra thick] (0, 0) -- (1.5, 0) -- (0.5, 0.866) -- (0,0);
\draw[fill=red] (0,0) circle (0.1cm);
\draw[fill=red] (1.5,0) circle (0.1cm);
\draw[fill=red] (0.5, 0.866) circle (0.1cm);
\end{tikzpicture}
\hspace{.8cm}}
\onslide<4->{%
\begin{tikzpicture}[scale=2]
\draw[ultra thick] (0, 0) -- (1, 0) -- (1, 1) -- (0, 1) -- (0,0);
\draw[fill=green] (0,0) circle (0.1cm);
\draw[fill=green] (1,0) circle (0.1cm);
\draw[fill=green] (0,1) circle (0.1cm);
\draw[fill=green] (1,1) circle (0.1cm);
\end{tikzpicture}}
\end{figure}


\end{frame}

\begin{frame}
\frametitle{Galois groups}

Distinguish kind of solution of polynomial by their symmetry group

\begin{definition}[informal]
The Galois group of a polynomial is the group of permutations of the zeros of the polynomial that are compatible with $+$, $-$, $\cdot$.
\end{definition}

\pause
\begin{definition}
Let $f$ be a polynomial with coefficients in a field $K$. Let $L$ be the smallest field containing $K$ and all zeros of $f$ (in some algebraically closed extension field).

The Galois group of $f$ is the group of field automorphisms $L\to L$ which fix all elements of $K$.
\end{definition}

\end{frame}

\begin{frame}
\frametitle{Solvability in terms of Galois groups}

\begin{theorem}
If $f$ is a polynomial over $\mathbb Q$ whose solutions can be expressed in terms of $+$, $-$, $\cdot$, $/$ and $\sqrt[n]{-}$ starting from rational numbers, then the Galois group of $f$ is solvable.
\end{theorem}

\begin{example}
The Galois group of the polynomial
\[
x^5 + 5x^4 - 20x^3 - 40x^2 + 5x + 1
\]
is the symmetric group $S_5$ which is not solvable.
\end{example}
\end{frame}

\begin{frame}
\frametitle{Can we understand Galois groups?}

\begin{definition}[Absolute Galois group]
Let $K$ be a field, and let $\overline{K}$ be a separable closure of $K$. We call $G_K = \mathop{\rm Gal}(\overline{K}/K)$ the absolute Galois group of $K$.
\end{definition}

\pause
\begin{example}[$K=\mathbb Q$]
$G_{\mathbb Q}$ is highly mysterious.

Understanding it properly is one of the principal goals of number theory.
\end{example}

\pause
\begin{example}[$K=\mathbb Q_p$]
$G_{\mathbb Q_p}$ is somewhat easier to understand, but still complicated.
\end{example}

\end{frame}


\section{Perfectoid spaces}

\begin{frame}
\frametitle{The absolute Galois group of a finite field}

\begin{definition}[Finite field with $p$ elements]
Let $p$ be a prime number. We let
\[
\mathbb F_p := \{ 0, 1, \dots, p-1 \}
\]
with addition and multiplication ``modulo $p$''.
\end{definition}

In particular: $\underbrace{1 + \dots + 1}_{\text{$p$ summands}} = 0$ in $\mathbb F_p$. (``Characteristic $p$'')

\pause
\begin{block}{Remark}
Let $K$ be a field of characteristic $p$. Then
\[
(x+y)^p = x^p + y^p \quad\text{for all } x, y\in K.
\]
\end{block}
\end{frame}

\begin{frame}
\begin{block}{Remark}
Let $K$ be a field of characteristic $p$. Then
\[
(x+y)^p = x^p + y^p \quad\text{for all } x, y\in K.
\]
In other words: The map $x\mapsto x^p$ is a field homomorphism, the \emph{Frobenius homomorphism}.
\end{block}

\pause
\begin{block}{Consequence}
The absolute Galois group $G_{\mathbb F_p}$ is isomorphic to $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$. It is topologically generated by the Frobenius automorphism.
\end{block}

\end{frame}

\begin{frame}
\frametitle{How far apart are characteristic $0$ and $p$?}

Compare
\[
\mathbb Q_p = \left\{ \sum_{i=i_0}^\infty a_i p^i;\ i_0\in\mathbb Z,\ a_i\in \{0, 1, \dots, p-1\} \right\}.
\]
versus
\[
\mathbb F_p((t)) = \left\{ \sum_{i=i_0}^\infty a_i t^i;\ i_0\in\mathbb Z,\ a_i\in \{0, 1, \dots, p-1\} \right\}.
\]

These descriptions look similar, but addition is very different!
\end{frame}


\begin{frame}
\frametitle{Perfectoid fields and tilting}

\begin{definition}[Scholze]
A \emph{perfectoid field} is a field $K$, complete with respect to a non-discrete non-archimedean valuation, with residue characteristic $p>0$ with ring of integers $\mathcal O_K = \{ x\in K;\ |x|\le 1\}$, such that the map
\[
\mathcal O_K/p \to \mathcal O_K/p,\quad x\mapsto x^p,
\]
is surjective.
\end{definition}

\begin{example}
$\mathbb Q_p(p^{1/p^\infty})^\wedge$,\quad
$\mathbb Q_p(\mu_{p^\infty})^\wedge$,\quad
$\mathbb F_p((t))(t^{1/p^\infty})^\wedge$.  % == tilt of either of the above
\end{example}
\end{frame}


\begin{frame}
\frametitle{Tilting: Switch from char.~$0$ to positive characteristic}

Every perfectoid field $K$ has a tilt $K^\flat$.

\[
K^\flat = \mathop{\rm Frac}(\varprojlim_{x\mapsto x^p} \mathcal O_K/p).
\]
The tilt $K^\flat$ has characteristic $p$: $1 + \cdots + 1 = 0$ in $K^\flat$.

\vspace{1cm}
\begin{theorem}[Fontaine, Wintenberger]
\[
G_K \cong G_{K^\flat}.
\]
\end{theorem}

% explain idea as in [Scholze, CDM] p. 3 (?)
\end{frame}


\begin{frame}
\frametitle{Perfectoid spaces}

\begin{definition}[Scholze]
Let $K$ be a perfectoid field.
A perfectoid space over $K$ is an adic space which is locally isomorphic to an affinoid adic space, i.e., an adic space of the form $\mathop{Spa}(R, R^+)$ where $R$ is a perfectoid $K$-algebra.
\end{definition}

\pause\bigskip
\textbf{Tilting for perfectoid spaces}

Every perfectoid space $X$ has a tilt $X^\flat$, and both have ``the same étale covers'':
\begin{theorem}[Scholze]
\[
\pi_1(X) \cong \pi_1(X^\flat)
\]
\end{theorem}

\end{frame}


\section{The Langlands program}

\begin{frame}
\frametitle{The Langlands program}

\pause

\begin{theorem}[Quadratic Reciprocity Law, Gauß]
Let $p \ne q$ be prime numbers $>2$, $p\equiv 1\mod 4$. The equation
\[
x^2 \equiv q \mod p \quad\text{has a solution}
\]
if and only if the equation
\[
x^2 \equiv p \mod q\quad\text{has a solution.}
\]
\end{theorem}


\pause\bigskip
\begin{example}[$p=5$, $q=67$]
The equation \quad $x^2 \equiv 67 \equiv 2 \mod 5$\quad has no solution.

Hence\quad $x^2 \equiv 5 \mod 67$\quad has no solution.
\end{example}
\end{frame} 

\begin{frame}
\frametitle{Class field theory}

Describe the maximal abelian quotients $G_{\mathbb Q}^{\rm ab}$ and $G_{\mathbb Q_p}^{\rm ab}$.

\vspace{1cm}
\begin{quote}
La théorie du corps de classes a une réputation de  difficulté  qui  est  en  partie  justifiée.   Mais  il faut  faire  une  distinction:  il  n'est  peut-être  pas en  effet  dans  la  science  de  théorie  où  tout à  la fois les démonstrations soient aussi ardues, et les résultats d'une aussi parfaite simplicité et d'une aussi grande puissance.

\hfill
J. Herbrand, 1936
\end{quote}
\end{frame} 

\begin{frame}
\frametitle{Particular instance: Modularity}

\begin{columns}
\begin{column}{.5\textwidth}
\textbf{Elliptic curve}
\vspace{.5cm}

\begin{overprint}
\onslide<1-2>
$y^2 = x^3 + x^2  - x$
% 6 solutions over Z, Q

\includegraphics[width=5cm]{ellc}
\onslide<3->
Numbers of solutions mod $p$,\\
$p$ a prime number:
\begin{align*}
\# \{ & (x, y);\ 0 \le x, y \le p-1,\\
& y^2 \equiv x^3 + x^2  - x \ {\rm mod}\ p \}
\end{align*}

$\rightsquigarrow$ $L$-function $L(E/\mathbb Q, s)$
\end{overprint}
\end{column}
\begin{column}{.45\textwidth}
\textbf{Modular form}
\vspace{.5cm}

\begin{overprint}
\onslide<1>
$f\colon \mathbb H \to \mathbb C$\\
holom., ``highly symmetric''

\includegraphics[width=\textwidth]{mf1}
\onslide<2-3>
$f\colon \mathbb H \to \mathbb C$\\
holom., ``highly symmetric''

\includegraphics[width=\textwidth]{mf2}
\onslide<4->
Fourier expansion,\\
$q = \exp({2\pi i z})$
\[
q - 2q^3 - q^5 + 2q^7 + q^9 + 2q^{13} + \cdots
\]

\vspace{.7cm}
$\rightsquigarrow$ $L$-function $L(f, s)$
\end{overprint}
\end{column}
\end{columns}

\begin{theorem}[Wiles, \dots]
Every elliptic curve $E$ over $\mathbb Q$ is modular.
\end{theorem}
\end{frame} 

\begin{frame}
\begin{corollary}
The $L$-function $L(E/\mathbb Q, s)$ has a holomorphic continuation to $\mathbb C$.
\end{corollary}

\pause\bigskip
\begin{theorem}[Allen, Calegari, Caraiani, Gee, Helm, Le Hung, Newton, Scholze, Taylor, Thorne]
Let $E$ be an elliptic curve over a CM field $K$. Then the $L$-function of $E$ over $K$ has a meromorphic continuation to $\mathbb C$. 
\end{theorem}
\end{frame} 


\begin{frame}
\frametitle{Other direction: automorphic $\rightarrow$ Galois}

\begin{theorem}[Scholze]
Let $F$ be totally real or CM, let $G = \mathop{\rm Res}_{F/\mathbb Q}(GL_n)$, and let $X_K$ be the locally symmetric space
%$X_K = G(\mathbb Q)\backslash (G(\mathbb A)/K_\infty A_\infty^\circ K)$
attached to $G$ and a compact open subgroup $K\subset G(\mathbb A_f)$.
For every system of Hecke eigenvalues occurring in the cohomology $H^i(X_K, \overline{\mathbb F}_p)$, there exists a continuous Galois representation
\[
\rho \colon \mathop{\rm Gal}(\overline{F}/F) \to GL_n(\overline{\mathbb F}_p)
\]
such that Hecke eigenvalues and Frobenius eigenvalues match.
\end{theorem}

% New twist in Scholze's theorem: $X_K$ not usually algebraic
% Find them in the boundary of algebraic varieties.

% torsion classes ...
\end{frame}


% \begin{frame}
% \frametitle{Further topics ...}
% 
% Diamonds
% 
% Fargues, Fargues-Scholze(?): geometric $p$-adic LLC
% 
% [Bhatt-Morrow-Scholze]
%
% [Scholze-Nikolaus] (e.g. Rio.pdf, p. 20)
% \end{frame}

\section{$\bullet$}


\begin{frame}
\frametitle{Goal of this talk}

\begin{figure}
\begin{tikzpicture}
\draw (0, 3) ellipse (2.5cm and 1.2cm) node [align=center] {$p$-adic fields,\\e.~g.~$\mathbb Q_p$};
\draw (-3.2, 5.5) ellipse (2.2cm and 1.2cm) node [align=center] {mod $p$ fields,\\e.~g.~$\mathbb F_p((t))$};
\draw (3.2, 5.5) ellipse (2.2cm and 1.2cm) node [align=center] {Archimedean\\fields, e.~g.~$\mathbb R$, $\mathbb C$};
\draw (0, 0) ellipse (2.2cm and 1.2cm) node [align=center] {Algebraic number\\fields, e.~g.~$\mathbb Q$};

\draw (-1.6, 4) -- (-2.2, 4.35);
\draw (0, 1.75) -- (0, 1.25);
\draw (1.6, 4) -- (2.2, 4.35);

\end{tikzpicture}
\end{figure}
\end{frame}



\begin{frame}
\only<1,3->{\frametitle{What is this good for?}}
\only<2>{\frametitle{Why are we doing this?}}

\only<2>{
Fascinating to

\begin{itemize}
\item
understand problems that have been studied for more than 2000 years,
\item
gain conceptual understanding of surprising patterns,
\item
teach the subject to others.
\end{itemize}
}

\only<3->{
(Polynomial) equations are everywhere:

\begin{itemize}
\item
Elliptic curve cryptography
\item
Theoretical physics
\item
Computer science
\item
Biochemistry
\item
\dots
\end{itemize}
}

\only<4>{
\begin{textblock}{8}(7,6)
\includegraphics[width=6cm]{pa}
\end{textblock}
}
\only<5>{
\begin{textblock}{8}(7,5)
\includegraphics[width=5cm]{cy}
\end{textblock}
}
\only<6>{
\begin{textblock}{8}(3,8.6)
\fbox{\includegraphics[width=10cm]{computervision}}
\end{textblock}
}
\only<7>{
\begin{textblock}{8}(3,8.6)
\fbox{\includegraphics[width=10cm]{molecules}}
\end{textblock}
}
\end{frame}

\begin{frame}

\frametitle{Congratulations, Peter!}

\only<1>{
\fbox{\includegraphics[width=14cm]{seminarbonn}}
}
\only<2>{
\vspace{1cm}
\hfill\includegraphics[width=6cm]{fieldsmedal2}\hfill
}
\end{frame}

\end{document}

