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\begin{document}
\title{Incidence bounds in positive characteristic via valuations and distality}

\author{Martin Bays}
\date{07.04.2021}

\maketitle
%Abstract:
%    Joint work with \textbf{Jean-François Martin}.
%    The Szemerédi-Trotter theorem bounds the number of incidences between
%    finite sets of points and straight lines in the real plane, and
%    generalisations to other algebraic binary relations play an important
%    role in understanding the interaction between (pseudo)finite sets and
%    field structure in zero characteristic. In positive characteristic,
%    these incidence bounds fail drastically without further restrictions,
%    but may be expected to hold under certain conditions. We confirm this in
%    the case of fields admitting a valuation with finite residue field, e.g.\ 
%    finitely generated extensions of $\F_p$, by seeing that the restricted
%    distality provided by the valuation suffices to trigger a result of
%    Chernikov-Galvin-Starchenko.
%    .
This talk is based on joint work with \textbf{Jean-François Martin}.
\tableofcontents

\section{Background}
\subsection{Incidence bounds in characteristic zero}
\begin{itemize}\item
\begin{factn}[Szemerédi-Trotter '83]
    There exists $C \in \R$ such that,
    given $N$ points and $N$ lines in $\R^2$,
    the number of incidences is bounded as
    $$ \left|{\{ (p,l) : p \in l \}}\right| \leq  C(N^{\frac32 - \frac16}) .$$
\end{factn}

\item This has been generalised in various ways. In particular:

\begin{factn}[Elekes-Szabó '12]
    If $(C_b)_{b \in B}$ is an algebraic family of distinct 
    irreducible plane curves over a field $K$ of characteristic 0,
    there are $C,\eps > 0$ such that
    given $N$ points in $K^2$ and $N$ curves in the family,
    $$ \left|{\{ (a,b) : a \in C_b \}}\right| \leq  C(N^{\frac32-\eps}).$$
\end{factn}

\item Hrushovski: this indicates a certain {\em modularity} of the interaction 
between (pseudo)finite sets and field structure.

\item In particular, abelian groups are the only source of relations on which 
finite sets ``maximally accumulate'':

\begin{factnb}[Elekes-Szabó '12 ($m=3$), Raz-Sharir-de Zeeuw '18 ($m=4$), B-Breuillard '20]
    Let $V \subseteq  \A^m$ be an irreducible affine variety over a field $K$ of characteristic 0.

    Exactly one of the following holds:
    \begin{enumerate}[(i)]\item $\exists C,\eps > 0.\;\forall X_1,\ldots ,X_m \subseteq_{\fin}  K.$
    $$| V(K) \cap (X_1\times\ldots \times X_m) | \leq  C\max(|X_i|)^{\dim V - \eps};$$
    \item OR: up to finite correspondences on co-ordinates and taking products,
    \\$V$ is a subgroup of a power of a 1-dimensional algebraic group.
    \end{enumerate}
\end{factnb}
\end{itemize}

\subsection{Incidence bounds in positive characteristic}

\begin{itemize}\item In positive characteristic, these bounds utterly fail:

\begin{remarkn}
    Let $K := \F_p^{\alg}$.
    For any algebraic set $V \subseteq  K^n$,
    there is $r>0$ such that for arbitrarily large $n$,
    $$|V(\mathbb{F}_{p^n})| \geq  r(p^n)^{\dim V}.$$
\end{remarkn}

\item However, Hrushovski conjectures that the Zilber trichotomy applies:
\\infinite pseudofinite fields should be the only obstruction to modularity.
\\(Above it is $\prod_{n \rightarrow  \U} \mathbb{F}_{p^n} \leq  K^\U$.)

\textit{(In the case of the sum-product theorem, this is true,
i.e.\ failures of sum-product bounds are due to finite subfields
(Bourgain-Katz-Tao, Tao-Vu, Tao, Hrushovski, Wagner).)}

\item As an extreme case, this conjecture suggests that for $K$ with 
\textbf{finite} algebraic part $K \cap \F_p^{\alg}$,
e.g.\ $K = \F_p(t)$,
the characteristic 0 results should go through.
\end{itemize}

\subsection{Distality}

\begin{itemize}\item
\begin{definitionn}
  Let $\M$ be an $\Lang$-structure.
  \begin{itemize}\item
  Let $\phi(x;y)$ be an $\Lang$ formula, let $A,B \subseteq  \M$.

  An $\Lang$-formula $\zeta_\phi(x;z)$ is a \defn{uniform strong honest 
  definition (USHD)} for $\phi$ on $A$ over $B$ if
  for any $a \in A^x$ and finite subset $B_0 \subseteq_{\fin}  B$ with $|B_0| \geq  2$,
  there is $d \in B_0^z$
  such that
  $$\tp(a/B_0) \ni \zeta_\phi(x,d) \vdash  \tp_\phi(a/B_0)$$
  (where
  $\tp_\phi(a/B_0) := \{ \phi(x,b)^\eps \in \tp(a/B_0) : b \in B_0^y \}$).

  \item $B \subseteq  \M$ is \defn{distal in $\M$} if every $\Lang$-formula $\phi(x;y)$ 
  has a USHD on $B$ over $B$.
  \end{itemize}
\end{definitionn}

\item Reduction to one variable:

\begin{factn} \label{f:distal1}
    $B \subseteq  \M$ is distal in $\M$ iff any $\Lang$-formula $\phi(x;y)$ with $|x| 
    = 1$ has a USHD on $B$ over $B$.
\end{factn}
\begin{proofidea}
    Given $a,b \in B$ and $\phi(x,y;z)$ and $B_0 \subseteq_{\fin}  B$,
    \\say $\tp(a/bB_0) \ni \zeta(x,b,c) \vdash  \tp_{\phi(x;y,z)}(a/bB_0)$ with $c \in B_0^z$.

    Now say $\tp(b/B_0) \ni \theta(y,c') \vdash  \tp_{\forall x. (\zeta(x,y,w) \rightarrow  
    \phi(x,y,z))}(b/B_0)$.

    Then $\tp(ab/B_0) \ni \zeta(x,y,c) \wedge \theta(y,c') \vdash  \tp^+_{\phi(x,y;z)}(ab/B_0)$.

    Now repeat with $\neg\phi$.
    Uniformity follows through.
\end{proofidea}

\item
\begin{examplen}
    If $B = (b_i)_i \subseteq  \M$ is a $\emptyset $-indiscernible sequence,
    \\and there is an $\Lang$-formula $\theta_<$ with $\M \vDash  
    \theta_<(b_i,b_j) \Leftrightarrow  i<j$,
    \\then $B$ is distal in $\M$.
\end{examplen}
\item
\begin{factn}[Chernikov-Simon]
  $\Th(\M)$ is \defn{distal} iff $\M$ is distal in $\M$.
\end{factn}
%** Let $B \subseteq  \M$. If $(B; (\phi(B))_{\phi \text{ an $\Lang$-formula}})$ 
%admits QE, it is distal iff $B$ is distal in $\M$.

%** If $B \subseteq  \M$ is distal in $\M$, then so is any $B' \subseteq  B$.

\item Distality implies incidence bounds:
\begin{factb}[Chernikov-Galvin-Starchenko, Chernikov-Starchenko '20] \label{f:distInc}
    Suppose $B \subseteq  \M$ is distal in $\M$ and $\phi(x;y)$ is an $\Lang$-formula.

    Suppose $\phi(B;B)$ is $K_{d,s}$-free,
    \\i.e.\ there is no $X_0\times Y_0 \subseteq  \phi(B;B)$ with $|X_0| = d$ and 
    $|Y_0| = s$.

    Let $\zeta_\phi(x,z)$ be a USHD for $\phi$ on $B$,
    and let $t := |z|$.

    Then there exists $C \in \R$ such that for $X_0 \subseteq_{\fin}  B^x$ and $Y_0 \subseteq_{\fin}  B^y$,
    $$|\phi(X_0;Y_0)| \leq  C(|X_0|^{\frac{(t-1)d}{td-1}} 
    |Y_0|^{\frac{td-t}{td-1}} + |X_0| + |Y_0|).$$
%
%    In particular, if $|X_0|,|Y_0| < N$ and $d=2$,
%    there is $\eps = \eps(t) = \frac1{4t-2} > 0$ such that
%    $$|\phi(X_0;Y_0)| \leq  O_\phi(N^{\frac32 - \eps}).$$
\end{factb}
\end{itemize}

\section{The result}
\subsection{Statements}
\begin{itemize}\item
\begin{theoremn}
    Let $k$ be a valued field with finite residue field.
    \\Then $k$ is distal in $k^{\alg} \vDash  \operatorname{ACVF}$.
\end{theoremn}

(Note: a positive characteristic valued field with finite residue field is not distal as a structure, nor even NIP (by Kaplan-Scanlon-Wagner).)

\item
So by Fact~\ref{f:distInc}:

\begin{corollaryn}
  Let $k$ be a valued field with finite residue field.

  Let $E \subseteq  k^n \times k^m$ be quantifier-free definable in $\Ldiv(k)$.

  Suppose $E$ is $K_{d,s}$-free, where $d,s \in \N$.

  Then there exist $t,C>0$ such that
  for $A_0 \subseteq_{\fin}  k^n$ and $B_0 \subseteq_{\fin}  k^m$,
  $$|E \cap (A_0 \times B_0)| \leq  C(|A_0|^{\frac{(t-1)d}{td-1}} 
  |B_0|^{\frac{td-t}{td-1}} + |A_0| + |B_0|).$$
\end{corollaryn}

\item In particular this yields a Szemerédi-Trotter-style result:
\begin{corollaryn}
  Suppose $(C_b)_{b \in B \subseteq  k^m}$ is an algebraic family of distinct 
  irreducible plane curves over a field $k$ which admits a valuation with finite residue field.

  Then $E := \{ (a,b) : a \in C_b(k) \} \subseteq  k^{2+m}$ is $K_{2,s}$-free for some 
  $s \in \N$.

  So there exist $\eps_0,C>0$ such that for $A_0 \subseteq_{\fin}  k^2$ and $B_0 \subseteq_{\fin}  k^m$,
  $$|E \cap (A_0 \times B_0)| 
  \leq  C(|A_0|^{1-\eps_0} |B_0|^{\frac12(1+\eps_0)} + |A_0| + |B_0|)$$
  ($\eps_0 := \frac1{2t-1} > 0$), or in symmetric form:
  $$|E \cap (A_0 \times B_0)| 
  \leq  C'(\max(|A_0|,|B_0|)^{\frac32-\eps}$$
  ($\eps := \frac{\eps_0}2 > 0$, $C' := 3C$).
\end{corollaryn}

%**
%Moreover, it will follow from the proof that $t$ in Fact~\ref{f:distInc} can be 
%bounded in terms of the size of the residue field.
\end{itemize}

\subsection{Fields admitting finite residue field}
\begin{examplen}
    The $t$-adic valuation on $\F_p(t)$ has residue field $\F_p$.

    Similarly, any finitely generated extension of $\F_p$, i.e.\ any function 
    field over a finite field, admits a valuation with finite residue field.
    %(e.g.\ $\F_q(s,t,\sqrt{s+t}) \subseteq  \F_q((s))((t))^{\alg}$.)
\end{examplen}

However:\
\begin{proposition}
  For any prime $p$,
  there exists an algebraic extension $L \geq  \F_p(t)$ such that $L\cap 
  \F_p^\alg = \F_p$ but no valuation on $L$ has finite residue field.
\end{proposition}
Such an $L$ can be built by recursively adjoining Artin-Schreier roots which 
force Artin-Schreier extensions of the residue fields of valuations on 
previously built fields; using the Artin-Schreier version of Kummer theory, 
one can always do this without extending the algebraic part.

\section{Proof of the result}

\begin{itemize}\item Let $L \vDash  \operatorname{ACVF}$ and let $k \subseteq  L$ be a subfield with $\res(k)$ finite.

\item We want to see that $k \subseteq  L$ is distal in $L$.

\item By reduction to 1 variable, it suffices to see:
\\any $\phi(x,y)$ with $|x| = 1$ has a USHD over $k$.

\end{itemize}

\subsection{Compressing balls}
\begin{itemize}\item By QE, for $a \in k$, $\phi(L,a)$ is a boolean combination of open and closed balls 
$$v(x-a') > \alpha \text{ or } v(x-a') \geq  \alpha$$
centred at points $a'$ with $\deg(k(a')/k)$ bounded, say dividing $d$.

\item Let $B_{k,d}$ be the set of balls (closed and open) centred at points of 
degree $|d$ field extensions of $k$ within $L$.

Using the bounded size of the residue field of such extensions, we obtain:

\begin{lemman}
    $x \in y$ has a USHD over $B_{k,d}$.
\end{lemman}
\begin{proof}
    \begin{itemize}\item Let $a \in L$ and $B_0 \subseteq_{\fin}  B_{k,d}$.
    \item Then $x \in (b \setminus \bigcup_{i<s}b_i) \vdash  \tp_{ \in }(a/B_0)$
    where
        \begin{itemize}\item $b$ is smallest in $B_0$ such that $a \in b$ (or $b := L$).
        \item $b_1,\ldots ,b_s$ are the maximal proper subballs of $b$ in $B_0$.
        \end{itemize}
    \item Sufficient to uniformly bound $s$.
    \item By ultrametricity, joins of finitely many balls are joins of two.

        So we may assume $B_0$ is closed under $\vee$
        \\(where $b' \vee b'' :=$ smallest ball containing both).
    \item Assume $s > 1$.
    \item Say $p_i \in b_i$ is of degree $|d$ over $k$, and let $\alpha \in v(L)$ be the radius of $b$.
    \item Now $$i \mapsto  \lambda_i := \res\left({\frac{p_i-p_1}{p_2-p_1}}\right)$$
    is an injection of $\{1,\ldots ,s\}$ into $\res(L)$:
    \begin{itemize}\item If $i\neq j$ then $b_i \vee b_j = b$, so $v(p_i-p_j)=\alpha$.
    \item Now suppose $\lambda_i=\lambda_j$. 
    Then $\res(\frac{p_i-p_j}{p_2-p_1}) = 0$,
    \\so $v(p_i-p_j) > v(p_2-p_1) = \alpha$, so $i=j$.
    \end{itemize}

    \item Say $\res(k) = \F_q$.
    \item Since each $\lambda_i$ is in the residue field of an extension of $k$ of 
    degree $|d^3$, by the valuation inequality
    $$\lambda_i \in \F_{q^{d^3}}.$$
    \item So $s \leq  q^{d^3}$.
    \end{itemize}
\end{proof}

\item But this is not enough on its own.

To show that $\phi(x,y)$ has a USHD over $k$:
given $C \subseteq_{\fin}  k$ we have to determine $\tp_\phi(a/C)$ using only $C$ as 
parameters -- but the balls involved will generally not be defined over $C$!

\item So we need to be more careful with the QE.
\end{itemize}

\subsection{Compressing cheeses}

\begin{itemize}\item $\phi(L,a)$ has a unique-up-to-permutation ``Swiss cheese decomposition'' 
as a finite union of disjoint cheeses,
$$\phi(L,a) = \bigcup_i (b_i \setminus \bigcup_j b_{ij}),$$
where each cheese is a ball $b_i$ minus a finite union of disjoint proper subballs $b_{ij}$,
\\and no $b_i$ is equal to any hole $b_{i'j}$.

\item Moreover:

\begin{factn}[Uniform Swiss Cheese Decomposition]
    There are $N$ and $d$ depending only on $\phi$ such that for all $a \in L^y$,
    $\phi(L,a)$ has Swiss cheese decomposition involving $\leq N$ balls, where each 
    ball contains a point in a degree $|d$ field extension of the subfield 
    generated by $a$.
\end{factn}

\item (Increasing $N$ and allowing the empty ball and its complement, we can 
assume the form of the decomposition is constant.)
\item So we obtain a relatively definable surjection $f_N$ with uniformly finite fibres
$$(B_{k,d})^N \supseteq X \twoheadrightarrow  \{ \ulcorner \phi(x,c) \urcorner : c \in k^y \},$$
$$f_N(b_1,\ldots ,b_N) = \text{[boolean combination of the $b_i$]}.$$
\end{itemize}

\subsection{Collapsing USHDs}

\begin{itemize}\item We conclude by a general elementary model theoretic lemma on USHDs:

\begin{lemman}
    Let $f$ be definable with uniformly finite fibres.
    Let $C \subseteq  \operatorname{im}(f)$.

    If $\psi(x,f(z))$ has a USHD over $f^{-1}(C)$,
    \\then $\psi(x,y)$ has a USHD over $C$.
\end{lemman}
\newcommand{\z}{\overline{z}}
(Explicitly: to conclude we apply this to
$x \in f_N(\z)$,
which has a USHD by ball compression,
then consider $c \mapsto  \ulcorner \phi(x,c) \urcorner$.)

\begin{proofidea}
  Compress $f^{-1}(C_0)$ with $\zeta(x,\widetilde d)$,
  then find $C_1 \subseteq  C_0$ bounded by the fibre size
  such that if $f(\widetilde d') = f(\widetilde d)$ and $\zeta(x,\widetilde d')$ implies the instances 
  for $C_1$, then it implies all.
  Then ``exists such a $\widetilde d'$ over $f(\widetilde d)$'' is a USHD.
%
%    More precisely:\
%    ***) Let $a \in \M^x$ and $C_0 \subseteq_{\fin}  C$.
%    ** We have $\widetilde d \in f^{-1}(C)$ such that $\tp(a/f^{-1}(C_0) \ni \zeta(x,\widetilde d) \vdash  \tp_{\psi(x,f(z))}(a/f^{-1}(C_0))$,
%    and $f(\widetilde d) =: d \in C^{|z|}$.
%    ** There is $C_1 \subseteq  C_0$ with $|C_1| \leq  |f^{-1}(d)|$,
%    such that if $f(\widetilde d') = d$ and 
%    $\zeta(x,\widetilde d') \vdash  \tp_{\psi(x,y)}(a/C_1)$,
%    then already
%    $\zeta(x,\widetilde d') \vdash  \tp_{\psi(x,y)}(a/C_0)$.
%    ** Then $\exists \widetilde d'. (f(\widetilde d') = d \wedge 
%    [\zeta(x,\widetilde d') \vdash  \tp_{\psi(x,y))}(a/C_1)])$ isolates 
%    $\tp_{\psi(x,y)}(a/C_0)$ within $\tp(a/C_0)$.
%    ** By the bound on $|C_1|$ this works uniformly.
%    *e)*
\end{proofidea}

\item
\begin{remarkn}
    Following the proof gives a bound on the exponent in the distal cell 
    decompositions, hence in the incidence bound for $|x| = 1$, of
    $t = 2(q^{d^3}+1)$ where $\res(k) = \F_q$.

    For $|x| > 1$ a bound can in theory be computed, but it involves QE in 
    ACVF.

    For the Szemerédi-Trotter case $\{ ((x,y),(a,b)) : y=ax+b \}$,
    we get $t = 4(q+1)$, giving an exponent in the symmetric form of $\frac32 
    - \frac1{16(q+1)-2}$.

    Question: in the Szemerédi-Trotter case with $\F_p(t)$, could there be an 
    exponent which doesn't depend on $p$? Worst lower bound I know is 
    $\frac43$ (by a similar "rectangular grid" argument as in the char 0 
    case).
\end{remarkn}
\end{itemize}

% One might wonder whether the use of Swiss cheese decomposition, i.e.\ 
% canonical-up-to-finite boolean combinations, was really essential. I expect 
% so. The following question formulates this more precisely; I expect the 
% answer to be negative.
%Question:
%    If $\phi(x,y)$ is a boolean combination of instances of $\psi(x,f_i(y))$ 
%    where $f_i$ are algebraic quasifunctions,
%    and $\psi$ has a USHD over $\bigcup_i f_i(C)$,
%    does it follow that $\phi$ has a USHD over $C$?
%    .

\section{Elekes-Szabó consequences}

As in the characteristic 0 case, these incidence bounds yield ``modularity of 
coherence'', and hence Elekes-Szabó bounds.

\begin{itemize}\item Let $k_0$ be a field admitting a valuation with finite residue field.
\item For $r\geq 1$, let
$$k_r := \{ a \in (k_0)^\alg : \deg(k_0(a)/k_0) \leq  r  \}.$$
\item From the proof for $k_0$, we get that \underline{each $k_r$ is also distal} 
in $(k_0)^\alg \vDash  \ACVF$.

\item This is sufficient for the arguments of one direction of the 1-dimensional case of the main result of B-Breuillard to go through:

\begin{theoremn}
\renewcommand{\a}{\overline{a}}
\newcommand{\trd}{{\operatorname{trd}}}
\newcommand{\ccl}{{\operatorname{ccl}}}
\providecommand{\bdl}{\boldsymbol\delta}
    Let $\U$ be a non-principal ultrafilter on $\omega$.

    Let $k' := ((k_0)^\U)^\alg \leq  ((k_0)^\alg)^\U =: L$,
    so $k' = \bigcup_{r \in \omega} (k_r)^\U$.

    Let $\xi \in \N^\U \setminus \N$.

    $$\bdl(\prod_{i \rightarrow  \U} X_i) := \operatorname{st} \log_\xi \lim_{i \rightarrow  \U} |X_i| \in \R \cup \{\infty\}.$$

    Equip $L$ with the structure generated by countably many internal 
    relations, including each $(k_r)^\U$, such that $\bdl$ is continuous.

    For $\a \in L^{<\omega}$,
    $$\bdl(\a) = \bdl(\tp(\a)) = \inf_{\phi \in \tp(\a)} \bdl(\phi(L)).$$

    $P \subseteq  L$ is \defn{coherent} if $\bdl(\a) = \trd(\a)$ for all $\a \in P^{<\omega}$.

    Write $\acl^0$ for field-theoretic algebraic closure.

    Let $P \subseteq  k'$ be coherent, and let $\ccl(P) := \{a \in \acl^0(P) : 
    \bdl(a) = \trd(a) \} \subseteq  k'$.

    Then $\ccl(P)$ is coherent, and $\acl^0|_{\ccl(P)}$ is a modular 
    pregeometry.
\end{theoremn}

Using the Evans-Hrushovski characterisation of modular subpregeometries of 
$\acl^0$ (proved by Evans-Hrushovski in arbitrary characteristic), one deduces:

\begin{theoremn}
    Let $k_0$ be a field admitting a valuation with finite residue field.
    Let $V \subseteq  \A^m$ be an irreducible affine variety over $k_0$.

    At least one of the following holds:
    \begin{enumerate}[(i)]\item $\exists C,\eps > 0.\;\forall X_1,\ldots ,X_m \subseteq_{\fin}  k_0.$
    $$| V(k_0) \cap (X_1\times\ldots \times X_m) | \leq  C\max(|X_i|)^{\dim V - \eps};$$
    \item Up to finite correspondences on co-ordinates and taking products,
    \\ $V$ is a subgroup of a power of a 1-dimensional algebraic group.
    \end{enumerate}
\end{theoremn}

%Remark:
%    The higher dimensional ("coarse general position") version doesn't 
%    immediately go through, since it needs better bounds (a more uniform 
%    sublinearity in $s$ where we omit $K_{2,s}$, namely not depending on $k$ 
%    where we omit $K_{k,k}$) and a positive characteristic version of higher 
%    dimensional Evans-Hrushovski. (The original ES case $m=3$ assuming general 
%    position (not just cgp) probably goes through directly, but I haven't 
%    checked.)
%
%    Getting a formulation with "exactly one" would also take some more work.
%    .
\end{itemize}

\end{document}
