text:
\section{Equivalent Subobjects have Isomorphic Domains}
Tags: Categories of Subobjects, Subobjects

\begin{theorem}
Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Let $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ be the [[Definition:Category of Subobjects|category of subobjects]] of $C$.
Let $m, m'$ be [[Definition:Equivalent Subobjects|equivalent subobjects]] of $C$.
Then there exists an [[Definition:Isomorphism (Category Theory)|isomorphism]] $f: m \to m'$.
\end{theorem}

\begin{proof}
For brevity, let us write $\operatorname{dom} m = B$ and $\operatorname{dom} m' = B'$.
Since $m$ and $m'$ are [[Definition:Equivalent Subobjects|equivalent subobjects]] of $C$, we have that:
:$m \subseteq m'$ and $m' \subseteq m$
where $\subseteq$ denotes the [[Definition:Inclusion Relation on Subobjects|inclusion relation on subobjects]].
That is, there exist [[Definition:Morphism|morphisms]] $f: m \to m'$ and $f': m' \to m$ of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$.
Hence, we have [[Definition:Morphism|morphisms]]:
:$f \circ f': m' \to m'$
:$f' \circ f: m \to m$
By [[Category of Subobjects is Preorder Category]], these are the [[Definition:Unique|unique]] such [[Definition:Morphism|morphisms]].
It follows that both are [[Definition:Identity Morphism|identity morphisms]]:
:$f \circ f' = \operatorname{id}_{m'}$
:$f' \circ f = \operatorname{id}_m$
That is, $f: m \to m'$ is an [[Definition:Isomorphism (Category Theory)|isomorphism]].
{{qed}}
\end{proof}

config: wiki


TITLE: Determine $\frac{\partial}{\partial r} \int\!\!\!\!\!\!-_{B(x,r)} \frac{r}{n} u_{xx}(y)dy$
QUESTION [0 upvotes]: I'd like to show
$\frac{\partial}{\partial r} \int\!\!\!\!\!\!-_{B(x,r)} \frac{r}{n} u_{xx}(y)dy = \int\!\!\!\!\!\!-_{\partial B(x,r)}u_{xx}dS + (\frac{1}{n}-1)\int\!\!\!\!\!\!-_{B(x,r)} u_{xx} dy$
where $\int\!\!\!\!\!\!-$ denotes the average of the integrale. Here $u = u(x,t) \in C^m(\mathbb{R}^n \times [0,\infty)), m\geq 2, n\geq 2$
Using the product rule I've gotten
$\frac{\partial}{\partial r} \int\!\!\!\!\!\!-_{B(x,r)} \frac{r}{n} u_{xx}(y)dy = \frac{1}{n} \int\!\!\!\!\!\!-_{B(x,r)} u_{xx}(y,t)dy + \frac{r}{n}\frac{\partial}{\partial r}\int\!\!\!\!\!\!-_{B(x,r)} u_{xx}(y)dy$
and perhaps now I should use the differentiation formula for moving regions so:
$\frac{\partial}{\partial r}\int\!\!\!\!\!\!-_{B(x,r)} u_{xx}(y)dy = \int\!\!\!\!\!\!-_{\partial B(x,r)}u_{xx}(y,t)\textbf{v}dS + \int\!\!\!\!\!\!-_{B(x,r)}\frac{\partial}{\partial r} u_{xx}(y,t)dy$
but I'm not sure what $\textbf{v}$ should be or how to continue. If someone could help me out from there I'd appreciate it!

REPLY [0 votes]: Use the substitution $y = rz + x$ to remove the dependence of the region on $r$ (also removing the multiplication by $r^{-n}$ coming from taking the average) - it will come out much tidier than using the differentiation formula for the moving boundary as $r$ will only appear in the integrand and nowhere else.


TITLE: rotated hyperbolic cylinder parameterization
QUESTION [1 upvotes]: A hyperbolic cylinder is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2} = -1$, but thats a hyperbolic cylinder that goes along the Z-axis. How do you parametrize a hyperbolic cylinder that goes instead of symmetric along z=z, symmetric along some direction z=cx+dy+e. (i need to fit data with a hyperbolic cylinder model, but the hyperbolic cylinder is not defined along the z-axis, but along some other direction)

REPLY [1 votes]: $\newcommand{\R}{\mathbf{R}}\newcommand{\e}{\mathbf{e}}$I'm not entirely sure about your terminology. (For example, $z = cx + dy + e$ defines a plane in $\R^{3}$, not a direction; maybe that's your plane of reflection?) However:
If $\{\e_{1}, \e_{2}, \e_{3}\}$ denotes the standard Cartesian basis of $\R^{3}$, then the cylinder $(x/a)^{2} - (y/b)^{2} = -1$ can be parametrized by
$x = a\sinh u$, $y = b\cosh u$, and $z = v$, namely by the vector-valued function
$$
\Phi(u, v) = (a\sinh u)\e_{1} + (b\cosh u)\e_{2} + v\e_{3}.
$$
Replacing the standard basis by a basis adapted to your needs should parametrize the corresponding surface.
For example, if you merely want the plane $z = cx + dy$ as plane of reflection symmetry, you can take
$$
\e_{1} = \frac{(-d, c, 0)}{\sqrt{c^{2} + d^{2}}},\quad
\e_{3} = \frac{(c, d, -1)}{\sqrt{c^{2} + d^{2} + 1}},\quad
\e_{2} = \e_{3} \times \e_{1}.
$$
If the choice of $\e_{1}$ and $\e_{2}$ matters, you can replace the above with the rotated vectors
$$
\e_{1}' = (\cos\theta)\e_{1} + (\sin\theta)\e_{2},\quad
\e_{2}' = -(\sin\theta)\e_{1} + (\cos\theta)\e_{2},\quad
\e_{3}' = \e_{3}.
$$
If you want the "center" of your cylinder somewhere other than the origin, simple translation (i.e., adding constants to the component functions) will now do the job.


TITLE: Check my proof of convergence?
QUESTION [0 upvotes]: I want to show if the following sequence of functions converges pointwise or uniformly on $R.$
$f_n(x) = 0$ if $x<n $ and $1/x$ if $x\geq n $.
Here's what I did, fix $x\in R $. for all $n \geq 1/x$, $f_n(x) \rightarrow1/x$. Furthermore, $f_n(0)=0$. So $f_n$ converges to $f$ s.t. $f(x) =0$ if $x =0$ and $1/x$ if $0<x\leq 1$. So it's pointwise convergent. Is this correct?
For uniform convergent I say take $x$ s.t. $0<x<n$ then $f_n(x)-f(x) = 0-1/x $ which doesn't $\rightarrow 0 $. So it's not uniformly convergent.

REPLY [2 votes]: Observe that: $$0\leq f_n(x)\leq\frac1n$$ for every $x\in\mathbb R$.
This together with $\lim_{n\to\infty}\frac1n=0$ tells us that $f_n$ converges uniformly (hence also pointwise) to the zero function.


text:
\begin{document}
\title{Lattice polygons and the number $\mathbf{2i+7}$}
\author{Christian Haase}
\address{Fachbereich Mathematik \& Informatik \\
  Freie Universit\"at Berlin \\
  14195 Berlin\\
  Germany}
\email{christian.haase@math.fu-berlin.de}
\urladdr{http://ehrhart.math.fu-berlin.de}
\thanks{Work by the first author was supported by NSF grant
  DMS--0200740 and by Emmy Noether fellowship HA 4383/1 of the German
  Research Foundation (DFG).
  Work by the second author was supported by the FWF in the frame
  of the SFB F013 and the project P15551.}
\author{Josef Schicho}
\address{RICAM \\
  Austrian Academy of Sciences \\
  Altenberger Stra\ss e 69 \\
  4040 Linz \\
  Austria}
\email{Josef.Schicho@oeaw.ac.at}
\urladdr{http://www.ricam.oeaw.ac.at/research/symcomp}

\begin{abstract}
  In this note we classify all triples $(a,b,i)$ such that there is a
  convex lattice polygon $P$ with area $a$ which has $b$ and $i$
  lattice points on the boundary and in the interior, respectively.
  The crucial lemma for the classification is the necessity of $b \le
  2 \ i + 7$. We sketch three proofs of this fact: the original one by
  Scott~\cite{Scott}, an elementary one, and one using algebraic
  geometry.

  As a refinement, we introduce an onion skin parameter $\ell$: how
  many nested polygons does $P$ contain? and give sharper bounds.
\end{abstract}

\maketitle

\setcounter{section}{-1}

\section{Introduction}\noindent

\subsection{How it all began}
When the second author translated a result on algebraic surfaces into
the language of lattice polygons using toric geometry, he obtained an
inequality for lattice polygons. This inequality had originally been
discovered by Scott~\cite{Scott}. The first author then found a third
proof.

Subsequently, both authors
went through a phase of polygon addiction. Once you get started to
draw lattice polygons on graph paper and to discover relations between
their numerical invariants, it is not so easy to stop! (The gentle
reader has been warned.)

Thus, it was just unavoidable that the authors came up with new
inequalities: Scott's inequality can be sharpened if one takes another
invariant into account, which is defined by peeling off the skins of
the polygons like an onion (see Section~\ref{sec:onions}).

\subsection{Lattice polygons}
We want to study convex lattice polygons: convex polygons all whose
vertices have integral coordinates. As it turns out, we need to
consider nonconvex polygons as well. Even nonsimple polygons
-- polygons with self intersection -- will prove useful later on.
In what follows, we will abbreviate
\begin{center}
  ``polygon'' := ``convex lattice polygon'',
\end{center}
and we will emphasize when we allow nonintegral or nonsimple
situations.
\begin{figure}[htbp]
  \begin{center}
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(7366,2581)(0,-10)
\put(7283,1283){\blacken\ellipse{150}{150}}
\put(7283,1283){\ellipse{150}{150}}
\put(7283,1883){\blacken\ellipse{150}{150}}
\put(7283,1883){\ellipse{150}{150}}
\put(7283,683){\blacken\ellipse{150}{150}}
\put(7283,683){\ellipse{150}{150}}
\put(6083,1283){\blacken\ellipse{150}{150}}
\put(6083,1283){\ellipse{150}{150}}
\put(6083,1883){\blacken\ellipse{150}{150}}
\put(6083,1883){\ellipse{150}{150}}
\put(6083,683){\blacken\ellipse{150}{150}}
\put(6083,683){\ellipse{150}{150}}
\put(6683,1283){\blacken\ellipse{150}{150}}
\put(6683,1283){\ellipse{150}{150}}
\put(6683,1883){\blacken\ellipse{150}{150}}
\put(6683,1883){\ellipse{150}{150}}
\put(6683,683){\blacken\ellipse{150}{150}}
\put(6683,683){\ellipse{150}{150}}
\thicklines
\path(7283,1283)(6083,1883)
\path(6324.495,1829.334)(6083.000,1883.000)(6270.830,1722.003)
\path(6083,1883)(6683,1883)
\path(6443.000,1823.000)(6683.000,1883.000)(6443.000,1943.000)
\path(6683,1883)(6083,683)
\path(6136.666,924.495)(6083.000,683.000)(6243.997,870.830)
\path(6083,683)(7283,1283)
\path(7095.170,1122.003)(7283.000,1283.000)(7041.505,1229.334)
\thinlines
\put(3083,1283){\blacken\ellipse{150}{150}}
\put(3083,1283){\ellipse{150}{150}}
\put(3083,1883){\blacken\ellipse{150}{150}}
\put(3083,1883){\ellipse{150}{150}}
\put(3083,683){\blacken\ellipse{150}{150}}
\put(3083,683){\ellipse{150}{150}}
\put(3683,1283){\blacken\ellipse{150}{150}}
\put(3683,1283){\ellipse{150}{150}}
\put(3683,1883){\blacken\ellipse{150}{150}}
\put(3683,1883){\ellipse{150}{150}}
\put(3683,683){\blacken\ellipse{150}{150}}
\put(3683,683){\ellipse{150}{150}}
\put(3683,2483){\blacken\ellipse{150}{150}}
\put(3683,2483){\ellipse{150}{150}}
\put(3083,2483){\blacken\ellipse{150}{150}}
\put(3083,2483){\ellipse{150}{150}}
\put(3683,83){\blacken\ellipse{150}{150}}
\put(3683,83){\ellipse{150}{150}}
\put(3083,83){\blacken\ellipse{150}{150}}
\put(3083,83){\ellipse{150}{150}}
\put(83,1283){\blacken\ellipse{150}{150}}
\put(83,1283){\ellipse{150}{150}}
\put(83,1883){\blacken\ellipse{150}{150}}
\put(83,1883){\ellipse{150}{150}}
\put(83,683){\blacken\ellipse{150}{150}}
\put(83,683){\ellipse{150}{150}}
\put(683,1283){\blacken\ellipse{150}{150}}
\put(683,1283){\ellipse{150}{150}}
\put(683,1883){\blacken\ellipse{150}{150}}
\put(683,1883){\ellipse{150}{150}}
\put(683,683){\blacken\ellipse{150}{150}}
\put(683,683){\ellipse{150}{150}}
\put(683,2483){\blacken\ellipse{150}{150}}
\put(683,2483){\ellipse{150}{150}}
\put(83,2483){\blacken\ellipse{150}{150}}
\put(83,2483){\ellipse{150}{150}}
\put(683,83){\blacken\ellipse{150}{150}}
\put(683,83){\ellipse{150}{150}}
\put(83,83){\blacken\ellipse{150}{150}}
\put(83,83){\ellipse{150}{150}}
\put(1283,1283){\blacken\ellipse{150}{150}}
\put(1283,1283){\ellipse{150}{150}}
\put(1283,1883){\blacken\ellipse{150}{150}}
\put(1283,1883){\ellipse{150}{150}}
\put(1283,683){\blacken\ellipse{150}{150}}
\put(1283,683){\ellipse{150}{150}}
\put(1883,1283){\blacken\ellipse{150}{150}}
\put(1883,1283){\ellipse{150}{150}}
\put(1883,1883){\blacken\ellipse{150}{150}}
\put(1883,1883){\ellipse{150}{150}}
\put(1883,683){\blacken\ellipse{150}{150}}
\put(1883,683){\ellipse{150}{150}}
\put(1883,2483){\blacken\ellipse{150}{150}}
\put(1883,2483){\ellipse{150}{150}}
\put(1283,2483){\blacken\ellipse{150}{150}}
\put(1283,2483){\ellipse{150}{150}}
\put(1883,83){\blacken\ellipse{150}{150}}
\put(1883,83){\ellipse{150}{150}}
\put(1283,83){\blacken\ellipse{150}{150}}
\put(1283,83){\ellipse{150}{150}}
\put(4283,1283){\blacken\ellipse{150}{150}}
\put(4283,1283){\ellipse{150}{150}}
\put(4283,1883){\blacken\ellipse{150}{150}}
\put(4283,1883){\ellipse{150}{150}}
\put(4283,683){\blacken\ellipse{150}{150}}
\put(4283,683){\ellipse{150}{150}}
\put(4283,2483){\blacken\ellipse{150}{150}}
\put(4283,2483){\ellipse{150}{150}}
\put(4883,2483){\blacken\ellipse{150}{150}}
\put(4883,2483){\ellipse{150}{150}}
\put(4883,1883){\blacken\ellipse{150}{150}}
\put(4883,1883){\ellipse{150}{150}}
\put(4883,1283){\blacken\ellipse{150}{150}}
\put(4883,1283){\ellipse{150}{150}}
\put(4883,683){\blacken\ellipse{150}{150}}
\put(4883,683){\ellipse{150}{150}}
\put(4883,83){\blacken\ellipse{150}{150}}
\put(4883,83){\ellipse{150}{150}}
\put(4283,83){\blacken\ellipse{150}{150}}
\put(4283,83){\ellipse{150}{150}}
\thicklines
\path(4883,2483)(4283,1883)
\path(4410.279,2095.132)(4283.000,1883.000)(4495.132,2010.279)
\path(4283,1883)(3083,1883)
\path(3323.000,1943.000)(3083.000,1883.000)(3323.000,1823.000)
\path(3083,1883)(4883,83)
\path(4670.868,210.279)(4883.000,83.000)(4755.721,295.132)
\path(4883,83)(4883,2483)
\path(4943.000,2243.000)(4883.000,2483.000)(4823.000,2243.000)
\path(1583,83)(83,83)
\path(323.000,143.000)(83.000,83.000)(323.000,23.000)
\path(383,2258)(1883,1283)
\path(1649.074,1363.491)(1883.000,1283.000)(1714.473,1464.104)
\path(83,83)(373,2187)
\path(399.668,1941.055)(373.000,2187.000)(280.792,1957.440)
\path(1883,1283)(1592,124)
\path(1592.251,371.386)(1592.000,124.000)(1708.639,342.164)
\end{picture}
}
    \caption{Polygons -- convex, lattice, nonsimple.}
    \label{fig:polygons}
  \end{center}
\end{figure}

Denote the area enclosed by a polygon $P$ by $a=a(P)$,
the number of lattice points on the boundary by
$b=b(P)$, and the number of lattice points strictly inside of $P$ by
$i=i(P)$.
A classic result relates these data.
\begin{theorem}[Pick's Formula~\cite{pick}]
  \begin{equation} \label{eq:pick}
    a = i + \frac{b}{2} - 1
  \end{equation}
\end{theorem}
A thorough discussion of this theorem -- including an
application in forest industry! -- can be found in
\cite{Gruenbaum_Shepard:93}.
Pick's theorem is not the only relation between the three parameters
$a$, $b$, $i$ of
polygons.

There is the rather obvious constraint $b\ge 3$.
From Pick's formula we obtain immediately $a \ge i +
\frac{1}{2}$ and $a \ge \frac{b}{2} - 1$.
Are there other constraints?
For the sake of suspense, we do not want to reveal the final
inequalities just yet. We refer the impatient reader to the
conclusion in Section~\ref{sec:conclusion} which contains a
summary of the main results.

\subsection{Lattice equivalence} \label{sec:equivalence}
Clearly, the area $a(P)$ is invariant under rigid motions of the
plane. On the other hand, the numbers $i(P)$, $b(P)$ are not concepts
of Euclidean geometry, because they are not preserved by rigid
motions. But they {\em are} preserved under {\em
  lattice equivalences\/}: affine maps $\Phi \colon \R^2 \rightarrow
\R^2$ of the plane that restrict to isomorphisms of the lattice
$\Z^2$.

Orientation preserving lattice equivalences form a group, the semi
direct product $\slz \ltimes \Z^2$.

\begin{figure}[htbp]
  \centering
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(2566,1981)(0,-10)
\thicklines
\whiten\path(683,83)(83,83)(83,683)
	(683,683)(683,83)
\path(683,83)(83,83)(83,683)
	(683,683)(683,83)
\thinlines
\put(1883,1883){\blacken\ellipse{150}{150}}
\put(1883,1883){\ellipse{150}{150}}
\put(1283,1883){\blacken\ellipse{150}{150}}
\put(1283,1883){\ellipse{150}{150}}
\put(1883,683){\blacken\ellipse{150}{150}}
\put(1883,683){\ellipse{150}{150}}
\put(683,683){\blacken\ellipse{150}{150}}
\put(683,683){\ellipse{150}{150}}
\put(1883,1283){\blacken\ellipse{150}{150}}
\put(1883,1283){\ellipse{150}{150}}
\put(2483,1283){\blacken\ellipse{150}{150}}
\put(2483,1283){\ellipse{150}{150}}
\put(1283,1283){\blacken\ellipse{150}{150}}
\put(1283,1283){\ellipse{150}{150}}
\put(83,683){\blacken\ellipse{150}{150}}
\put(83,683){\ellipse{150}{150}}
\put(683,83){\blacken\ellipse{150}{150}}
\put(683,83){\ellipse{150}{150}}
\put(1283,83){\blacken\ellipse{150}{150}}
\put(1283,83){\ellipse{150}{150}}
\put(683,1283){\blacken\ellipse{150}{150}}
\put(683,1283){\ellipse{150}{150}}
\put(83,83){\blacken\ellipse{150}{150}}
\put(83,83){\ellipse{150}{150}}
\put(2483,1883){\blacken\ellipse{150}{150}}
\put(2483,1883){\ellipse{150}{150}}
\put(1883,83){\blacken\ellipse{150}{150}}
\put(1883,83){\ellipse{150}{150}}
\put(2483,83){\blacken\ellipse{150}{150}}
\put(2483,83){\ellipse{150}{150}}
\put(2483,683){\blacken\ellipse{150}{150}}
\put(2483,683){\ellipse{150}{150}}
\put(83,1283){\blacken\ellipse{150}{150}}
\put(83,1283){\ellipse{150}{150}}
\put(83,1883){\blacken\ellipse{150}{150}}
\put(83,1883){\ellipse{150}{150}}
\put(683,1883){\blacken\ellipse{150}{150}}
\put(683,1883){\ellipse{150}{150}}
\put(1283,683){\blacken\ellipse{150}{150}}
\put(1283,683){\ellipse{150}{150}}
\end{picture}
}
  \begin{minipage}{1.5in}

      \qquad \raisebox{25mm}{$\xymatrix{
        \ar@<1ex>[rr]^{

          \left(\begin{smallmatrix}
            \ 3\ &\ 1\ \\2&1
          \end{smallmatrix}\right)
        }

        &&
        \ar@<1ex>[ll]^{

          \left(\begin{smallmatrix}
            1&-1\\-2&3
          \end{smallmatrix}\right)
        }
      }$}

  \end{minipage}
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(2566,1981)(0,-10)
\thicklines
\whiten\path(1883,1283)(83,83)(683,683)
	(2483,1883)(1883,1283)
\path(1883,1283)(83,83)(683,683)
	(2483,1883)(1883,1283)
\thinlines
\put(1883,1883){\blacken\ellipse{150}{150}}
\put(1883,1883){\ellipse{150}{150}}
\put(1283,1883){\blacken\ellipse{150}{150}}
\put(1283,1883){\ellipse{150}{150}}
\put(1883,683){\blacken\ellipse{150}{150}}
\put(1883,683){\ellipse{150}{150}}
\put(683,683){\blacken\ellipse{150}{150}}
\put(683,683){\ellipse{150}{150}}
\put(1283,683){\blacken\ellipse{150}{150}}
\put(1283,683){\ellipse{150}{150}}
\put(1883,1283){\blacken\ellipse{150}{150}}
\put(1883,1283){\ellipse{150}{150}}
\put(2483,1283){\blacken\ellipse{150}{150}}
\put(2483,1283){\ellipse{150}{150}}
\put(1283,1283){\blacken\ellipse{150}{150}}
\put(1283,1283){\ellipse{150}{150}}
\put(83,683){\blacken\ellipse{150}{150}}
\put(83,683){\ellipse{150}{150}}
\put(683,83){\blacken\ellipse{150}{150}}
\put(683,83){\ellipse{150}{150}}
\put(1283,83){\blacken\ellipse{150}{150}}
\put(1283,83){\ellipse{150}{150}}
\put(683,1283){\blacken\ellipse{150}{150}}
\put(683,1283){\ellipse{150}{150}}
\put(83,83){\blacken\ellipse{150}{150}}
\put(83,83){\ellipse{150}{150}}
\put(2483,1883){\blacken\ellipse{150}{150}}
\put(2483,1883){\ellipse{150}{150}}
\put(1883,83){\blacken\ellipse{150}{150}}
\put(1883,83){\ellipse{150}{150}}
\put(2483,83){\blacken\ellipse{150}{150}}
\put(2483,83){\ellipse{150}{150}}
\put(2483,683){\blacken\ellipse{150}{150}}
\put(2483,683){\ellipse{150}{150}}
\put(83,1283){\blacken\ellipse{150}{150}}
\put(83,1283){\ellipse{150}{150}}
\put(83,1883){\blacken\ellipse{150}{150}}
\put(83,1883){\ellipse{150}{150}}
\put(683,1883){\blacken\ellipse{150}{150}}
\put(683,1883){\ellipse{150}{150}}
\end{picture}
}
  \vspace{-15mm}
  \caption{Two lattice equivalent quadrangles.}
  \label{fig:square}
\end{figure}

So $\Phi$ has the form $\Phi(\pt{x}) = A \pt{x} + \pt{y}$ for a matrix
$A$, and a vector $\pt{y}$. The lattice preservation property
$\Phi(\Z^2)=\Z^2$ implies that both $A$ and $\pt{y}$ have
integral entries, and the same is true for the inverse transformation
$\Phi^{-1}(\pt{x}) = A^{-1} \pt{x} - A^{-1} \pt{y}$. Hence $\det A =
\pm 1$, and $a(P)$ is preserved under $\Phi$ as well.

In all our arguments, we will treat lattice equivalent polygons as
indistinguishable. For example, the quadrangle in
Figure~\ref{fig:square} on the right looks to us like a perfect
square. We see that angles and Euclidian lengths are not preserved.
A lattice geometric substitute for the length of a lattice line
segment is the number of lattice points it contains minus one.
In this sense, $b$ is the perimeter of $P$.

Here is an exercise that helps to get a feeling for what
lattice equivalences can do and cannot do.\footnote{Hint: if you know
  {\em Chinese\/}, not much {\em remains\/} to be done.}
\begin{exercise} \label{ex:angle}
  Given a vertex $\pt{x}$ of a polygon $P$, show that there is a
  unique orientation preserving lattice equivalence $\Phi$ so that
  \begin{itemize}
  \item $\Phi(\pt{x})=(0,0)\transpose$, and
  \item
    there are (necessarily unique) coprime $0 < p \le q$ so that
    the segments $[(1,0)\transpose,(0,0)\transpose]$ and
    $[(0,0)\transpose, (-p,q)\transpose]$ are contained in edges of
    $\Phi(P)$.
  \end{itemize}
\end{exercise}

\subsection{Why algebraic geometry?}
Toric geometry is a powerful link connecting discrete and algebraic
geometry (see e.g. \cite{Sturmfels:99}).
At the heart of this link is the simple correspondence
\begin{eqnarray*}
  \text{ lattice point \qquad } & & \text{ \qquad Laurent monomial } \\
  \pt{p}=(p_1,\ldots,p_m) \in \Z^m & \longleftrightarrow &
  \pt{x}^{\pt{p}} = x_1^{p_1} \cdot\ldots\cdot x_m^{p_m} \in

  \mathbb{C}[x_1^{\pm1},\ldots,x_m^{\pm1}]
\end{eqnarray*}
It was invented by M.~Demazure~\cite{Demazure:70} for a totally
different purpose (to study algebraic subgroups of the Cremona group)
in algebraic geometry. R.~Stanley used it in combinatorics to classify
the possible face numbers of simplicial convex polytopes~\cite{gThmS}.
R.~Krasauskas~\cite{Krasauskas:02} used it in geometric modeling to
construct surfaces with new control structure (see
Figure~\ref{fig:rimas}).
\begin{figure}[htbp]
  \centering
  \includegraphics[height=6cm,width=8cm]{rimas.ps}
  \caption{Toric surface with hexagonal control structure.}
  \label{fig:rimas}
\end{figure}

For any polygon $P$, the Laurent monomials corresponding to its
lattice points define a {\em toric surface\/} $X_P$ in a projective
space of dimension $b+i-1$ as follows.

Number the lattice points $P \cap \Z^2=\{\pt{p}_0,\dots,\pt{p}_n\}$
(where $n=b+i-1$). Then $X_P$ is the closure of the image of the map
$({\mathbb C}^*)^2\to{\mathbb P}^n$ defined by $\pt{x} \mapsto
(\pt{x}^{\pt{p}_0} : \dots : \pt{x}^{\pt{p}_n})$.
Lattice equivalent polygons define the same toric surface.

As to be expected, there is a dictionary translating toric geometry
to lattice geometry:
the degree of the toric surface is equal to twice
the area, and the number of interior points is equal to
the {\em sectional genus\/} of the surface.
For instance, let $\Gamma$ be the triangle with corners
$(0,0)^t,(1,2)^t,(2,1)^t$ (it has one interior point $(1,1)^t$).
Then the toric surface is given by $(1 : x_1x_2^2 : x_1^2x_2 : x_1x_2)
\in \P^3$, its degree is 3 -- this is also reflected by its implicit
equation $y_1y_2y_3-y_4^3=0$, which has also degree~$3$ --
and its sectional genus is $1$, i.e., if we intersect with a generic
hyperplane in $\P^3$, we obtain a genus one Riemann surface.

In the context of toric geometry, Pick's formula appears
as a consequence of the Riemann-Roch Theorem.

\section{Examples}\noindent

Let us approach the question which parameters are possible for

polygons by looking at some examples. Can we bound $i$ or $a$
in terms of $b$\,?   Figure~\ref{fig:bequal3} shows examples with $b=3$
and arbitrarily high $a$ and $i$. So there is no lattice geometric
analogue of the isoperimetric inequality.
\begin{figure}[htbp]
  \centering
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(9766,3181)(0,-10)
\put(1883,983){\blacken\ellipse{150}{150}}
\put(1883,983){\ellipse{150}{150}}
\put(2483,983){\blacken\ellipse{150}{150}}
\put(2483,983){\ellipse{150}{150}}
\put(3083,983){\blacken\ellipse{150}{150}}
\put(3083,983){\ellipse{150}{150}}
\put(683,983){\blacken\ellipse{150}{150}}
\put(683,983){\ellipse{150}{150}}
\put(1283,983){\blacken\ellipse{150}{150}}
\put(1283,983){\ellipse{150}{150}}
\put(83,2183){\blacken\ellipse{150}{150}}
\put(83,2183){\ellipse{150}{150}}
\put(3683,2183){\blacken\ellipse{150}{150}}
\put(3683,2183){\ellipse{150}{150}}
\put(3683,1583){\blacken\ellipse{150}{150}}
\put(3683,1583){\ellipse{150}{150}}
\put(83,983){\blacken\ellipse{150}{150}}
\put(83,983){\ellipse{150}{150}}
\put(83,1583){\blacken\ellipse{150}{150}}
\put(83,1583){\ellipse{150}{150}}
\put(1883,1583){\blacken\ellipse{150}{150}}
\put(1883,1583){\ellipse{150}{150}}
\put(2483,1583){\blacken\ellipse{150}{150}}
\put(2483,1583){\ellipse{150}{150}}
\put(3083,1583){\blacken\ellipse{150}{150}}
\put(3083,1583){\ellipse{150}{150}}
\put(683,1583){\blacken\ellipse{150}{150}}
\put(683,1583){\ellipse{150}{150}}
\put(1283,1583){\blacken\ellipse{150}{150}}
\put(1283,1583){\ellipse{150}{150}}
\put(1883,2183){\blacken\ellipse{150}{150}}
\put(1883,2183){\ellipse{150}{150}}
\put(2483,2183){\blacken\ellipse{150}{150}}
\put(2483,2183){\ellipse{150}{150}}
\put(3083,2183){\blacken\ellipse{150}{150}}
\put(3083,2183){\ellipse{150}{150}}
\put(683,2183){\blacken\ellipse{150}{150}}
\put(683,2183){\ellipse{150}{150}}
\put(1283,2183){\blacken\ellipse{150}{150}}
\put(1283,2183){\ellipse{150}{150}}
\put(3683,983){\blacken\ellipse{150}{150}}
\put(3683,983){\ellipse{150}{150}}
\put(8483,1883){\blacken\ellipse{150}{150}}
\put(8483,1883){\ellipse{150}{150}}
\put(9083,1883){\blacken\ellipse{150}{150}}
\put(9083,1883){\ellipse{150}{150}}
\put(9683,1883){\blacken\ellipse{150}{150}}
\put(9683,1883){\ellipse{150}{150}}
\put(7283,1883){\blacken\ellipse{150}{150}}
\put(7283,1883){\ellipse{150}{150}}
\put(7883,1883){\blacken\ellipse{150}{150}}
\put(7883,1883){\ellipse{150}{150}}
\put(6683,1883){\blacken\ellipse{150}{150}}
\put(6683,1883){\ellipse{150}{150}}
\put(8483,2483){\blacken\ellipse{150}{150}}
\put(8483,2483){\ellipse{150}{150}}
\put(9083,2483){\blacken\ellipse{150}{150}}
\put(9083,2483){\ellipse{150}{150}}
\put(9683,2483){\blacken\ellipse{150}{150}}
\put(9683,2483){\ellipse{150}{150}}
\put(7283,2483){\blacken\ellipse{150}{150}}
\put(7283,2483){\ellipse{150}{150}}
\put(7883,2483){\blacken\ellipse{150}{150}}
\put(7883,2483){\ellipse{150}{150}}
\put(6683,2483){\blacken\ellipse{150}{150}}
\put(6683,2483){\ellipse{150}{150}}
\put(8483,3083){\blacken\ellipse{150}{150}}
\put(8483,3083){\ellipse{150}{150}}
\put(9083,3083){\blacken\ellipse{150}{150}}
\put(9083,3083){\ellipse{150}{150}}
\put(9683,3083){\blacken\ellipse{150}{150}}
\put(9683,3083){\ellipse{150}{150}}
\put(7283,3083){\blacken\ellipse{150}{150}}
\put(7283,3083){\ellipse{150}{150}}
\put(7883,3083){\blacken\ellipse{150}{150}}
\put(7883,3083){\ellipse{150}{150}}
\put(6683,3083){\blacken\ellipse{150}{150}}
\put(6683,3083){\ellipse{150}{150}}
\put(8483,1283){\blacken\ellipse{150}{150}}
\put(8483,1283){\ellipse{150}{150}}
\put(9083,1283){\blacken\ellipse{150}{150}}
\put(9083,1283){\ellipse{150}{150}}
\put(9683,1283){\blacken\ellipse{150}{150}}
\put(9683,1283){\ellipse{150}{150}}
\put(8483,83){\blacken\ellipse{150}{150}}
\put(8483,83){\ellipse{150}{150}}
\put(9083,83){\blacken\ellipse{150}{150}}
\put(9083,83){\ellipse{150}{150}}
\put(9683,83){\blacken\ellipse{150}{150}}
\put(9683,83){\ellipse{150}{150}}
\put(8483,683){\blacken\ellipse{150}{150}}
\put(8483,683){\ellipse{150}{150}}
\put(9083,683){\blacken\ellipse{150}{150}}
\put(9083,683){\ellipse{150}{150}}
\put(9683,683){\blacken\ellipse{150}{150}}
\put(9683,683){\ellipse{150}{150}}
\put(7283,1283){\blacken\ellipse{150}{150}}
\put(7283,1283){\ellipse{150}{150}}
\put(7283,83){\blacken\ellipse{150}{150}}
\put(7283,83){\ellipse{150}{150}}
\put(7283,683){\blacken\ellipse{150}{150}}
\put(7283,683){\ellipse{150}{150}}
\put(7883,1283){\blacken\ellipse{150}{150}}
\put(7883,1283){\ellipse{150}{150}}
\put(7883,683){\blacken\ellipse{150}{150}}
\put(7883,683){\ellipse{150}{150}}
\put(6683,1283){\blacken\ellipse{150}{150}}
\put(6683,1283){\ellipse{150}{150}}
\put(6683,83){\blacken\ellipse{150}{150}}
\put(6683,83){\ellipse{150}{150}}
\put(6683,683){\blacken\ellipse{150}{150}}
\put(6683,683){\ellipse{150}{150}}
\put(7883,83){\blacken\ellipse{150}{150}}
\put(7883,83){\ellipse{150}{150}}
\thicklines
\path(83,983)(683,2183)(3683,1583)(83,983)
\path(7283,3083)(9683,83)(6683,683)(7283,3083)
\end{picture}
}
  \caption{$b=3$ and $a \gg 0$}
  \label{fig:bequal3}
\end{figure}

What about bounds in the opposite direction? Can we bound $b$ in terms
of $i$\,? Well, there is the family of Figure~\ref{fig:iequal0} with
$i=0$ and arbitrary $b$.

\begin{figure}[htbp]
  \centering
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(3766,781)(0,-10)
\put(83,683){\blacken\ellipse{150}{150}}
\put(83,683){\ellipse{150}{150}}
\put(3683,683){\blacken\ellipse{150}{150}}
\put(3683,683){\ellipse{150}{150}}
\put(3683,83){\blacken\ellipse{150}{150}}
\put(3683,83){\ellipse{150}{150}}
\put(83,83){\blacken\ellipse{150}{150}}
\put(83,83){\ellipse{150}{150}}
\put(1883,83){\blacken\ellipse{150}{150}}
\put(1883,83){\ellipse{150}{150}}
\put(2483,83){\blacken\ellipse{150}{150}}
\put(2483,83){\ellipse{150}{150}}
\put(3083,83){\blacken\ellipse{150}{150}}
\put(3083,83){\ellipse{150}{150}}
\put(683,83){\blacken\ellipse{150}{150}}
\put(683,83){\ellipse{150}{150}}
\put(1283,83){\blacken\ellipse{150}{150}}
\put(1283,83){\ellipse{150}{150}}
\put(1883,683){\blacken\ellipse{150}{150}}
\put(1883,683){\ellipse{150}{150}}
\put(2483,683){\blacken\ellipse{150}{150}}
\put(2483,683){\ellipse{150}{150}}
\put(3083,683){\blacken\ellipse{150}{150}}
\put(3083,683){\ellipse{150}{150}}
\put(683,683){\blacken\ellipse{150}{150}}
\put(683,683){\ellipse{150}{150}}
\put(1283,683){\blacken\ellipse{150}{150}}
\put(1283,683){\ellipse{150}{150}}
\thicklines
\path(83,683)(83,83)(3683,83)(83,683)
\end{picture}
}
  \caption{$i=0$ and $b \gg 0$.}
  \label{fig:iequal0}
\end{figure}

Perhaps surprisingly, for $i>0$ no such families exist. For $i=1$,
there are precisely the $16$ lattice equivalence classes depicted in
Figure~\ref{fig:reflexive}. We see that all values $3 \le b \le 9 =
2i+7$ occur. The polygon labeled $3\Delta$ is the $3$-fold dilation of
the standard triangle $\Delta$ which is the convex hull
$\conv[(0,0)\transpose, (1,0)\transpose, (0,1)\transpose]$ of the
origin together with the standard unit vectors. It will play an
important r\^ole later on.
\begin{figure}[htbp]
  \centering
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(10366,4981)(0,-10)
\texture{88555555 55000000 555555 55000000 555555 55000000 555555 55000000
	555555 55000000 555555 55000000 555555 55000000 555555 55000000
	555555 55000000 555555 55000000 555555 55000000 555555 55000000
	555555 55000000 555555 55000000 555555 55000000 555555 55000000 }
\shade\path(83,1883)(83,83)(1883,83)(83,1883)
\path(83,1883)(83,83)(1883,83)(83,1883)
\put(83,4883){\blacken\ellipse{150}{150}}
\put(83,4883){\ellipse{150}{150}}
\put(1283,4883){\blacken\ellipse{150}{150}}
\put(1283,4883){\ellipse{150}{150}}
\put(2483,4883){\blacken\ellipse{150}{150}}
\put(2483,4883){\ellipse{150}{150}}
\put(3683,4883){\blacken\ellipse{150}{150}}
\put(3683,4883){\ellipse{150}{150}}
\put(4883,4883){\blacken\ellipse{150}{150}}
\put(4883,4883){\ellipse{150}{150}}
\put(6083,4883){\blacken\ellipse{150}{150}}
\put(6083,4883){\ellipse{150}{150}}
\put(7283,4883){\blacken\ellipse{150}{150}}
\put(7283,4883){\ellipse{150}{150}}
\put(8483,4883){\blacken\ellipse{150}{150}}
\put(8483,4883){\ellipse{150}{150}}
\put(9683,4883){\blacken\ellipse{150}{150}}
\put(9683,4883){\ellipse{150}{150}}
\put(683,4883){\blacken\ellipse{150}{150}}
\put(683,4883){\ellipse{150}{150}}
\put(1883,4883){\blacken\ellipse{150}{150}}
\put(1883,4883){\ellipse{150}{150}}
\put(3083,4883){\blacken\ellipse{150}{150}}
\put(3083,4883){\ellipse{150}{150}}
\put(4283,4883){\blacken\ellipse{150}{150}}
\put(4283,4883){\ellipse{150}{150}}
\put(5483,4883){\blacken\ellipse{150}{150}}
\put(5483,4883){\ellipse{150}{150}}
\put(6683,4883){\blacken\ellipse{150}{150}}
\put(6683,4883){\ellipse{150}{150}}
\put(7883,4883){\blacken\ellipse{150}{150}}
\put(7883,4883){\ellipse{150}{150}}
\put(9083,4883){\blacken\ellipse{150}{150}}
\put(9083,4883){\ellipse{150}{150}}
\put(10283,4883){\blacken\ellipse{150}{150}}
\put(10283,4883){\ellipse{150}{150}}
\put(83,4283){\blacken\ellipse{150}{150}}
\put(83,4283){\ellipse{150}{150}}
\put(1283,4283){\blacken\ellipse{150}{150}}
\put(1283,4283){\ellipse{150}{150}}
\put(2483,4283){\blacken\ellipse{150}{150}}
\put(2483,4283){\ellipse{150}{150}}
\put(3683,4283){\blacken\ellipse{150}{150}}
\put(3683,4283){\ellipse{150}{150}}
\put(4883,4283){\blacken\ellipse{150}{150}}
\put(4883,4283){\ellipse{150}{150}}
\put(6083,4283){\blacken\ellipse{150}{150}}
\put(6083,4283){\ellipse{150}{150}}
\put(7283,4283){\blacken\ellipse{150}{150}}
\put(7283,4283){\ellipse{150}{150}}
\put(8483,4283){\blacken\ellipse{150}{150}}
\put(8483,4283){\ellipse{150}{150}}
\put(9683,4283){\blacken\ellipse{150}{150}}
\put(9683,4283){\ellipse{150}{150}}
\put(683,4283){\blacken\ellipse{150}{150}}
\put(683,4283){\ellipse{150}{150}}
\put(1883,4283){\blacken\ellipse{150}{150}}
\put(1883,4283){\ellipse{150}{150}}
\put(3083,4283){\blacken\ellipse{150}{150}}
\put(3083,4283){\ellipse{150}{150}}
\put(4283,4283){\blacken\ellipse{150}{150}}
\put(4283,4283){\ellipse{150}{150}}
\put(5483,4283){\blacken\ellipse{150}{150}}
\put(5483,4283){\ellipse{150}{150}}
\put(6683,4283){\blacken\ellipse{150}{150}}
\put(6683,4283){\ellipse{150}{150}}
\put(7883,4283){\blacken\ellipse{150}{150}}
\put(7883,4283){\ellipse{150}{150}}
\put(9083,4283){\blacken\ellipse{150}{150}}
\put(9083,4283){\ellipse{150}{150}}
\put(10283,4283){\blacken\ellipse{150}{150}}
\put(10283,4283){\ellipse{150}{150}}
\put(83,3683){\blacken\ellipse{150}{150}}
\put(83,3683){\ellipse{150}{150}}
\put(1283,3683){\blacken\ellipse{150}{150}}
\put(1283,3683){\ellipse{150}{150}}
\put(2483,3683){\blacken\ellipse{150}{150}}
\put(2483,3683){\ellipse{150}{150}}
\put(3683,3683){\blacken\ellipse{150}{150}}
\put(3683,3683){\ellipse{150}{150}}
\put(4883,3683){\blacken\ellipse{150}{150}}
\put(4883,3683){\ellipse{150}{150}}
\put(6083,3683){\blacken\ellipse{150}{150}}
\put(6083,3683){\ellipse{150}{150}}
\put(7283,3683){\blacken\ellipse{150}{150}}
\put(7283,3683){\ellipse{150}{150}}
\put(8483,3683){\blacken\ellipse{150}{150}}
\put(8483,3683){\ellipse{150}{150}}
\put(9683,3683){\blacken\ellipse{150}{150}}
\put(9683,3683){\ellipse{150}{150}}
\put(683,3683){\blacken\ellipse{150}{150}}
\put(683,3683){\ellipse{150}{150}}
\put(1883,3683){\blacken\ellipse{150}{150}}
\put(1883,3683){\ellipse{150}{150}}
\put(3083,3683){\blacken\ellipse{150}{150}}
\put(3083,3683){\ellipse{150}{150}}
\put(4283,3683){\blacken\ellipse{150}{150}}
\put(4283,3683){\ellipse{150}{150}}
\put(5483,3683){\blacken\ellipse{150}{150}}
\put(5483,3683){\ellipse{150}{150}}
\put(6683,3683){\blacken\ellipse{150}{150}}
\put(6683,3683){\ellipse{150}{150}}
\put(7883,3683){\blacken\ellipse{150}{150}}
\put(7883,3683){\ellipse{150}{150}}
\put(9083,3683){\blacken\ellipse{150}{150}}
\put(9083,3683){\ellipse{150}{150}}
\put(10283,3683){\blacken\ellipse{150}{150}}
\put(10283,3683){\ellipse{150}{150}}
\put(83,3083){\blacken\ellipse{150}{150}}
\put(83,3083){\ellipse{150}{150}}
\put(1283,3083){\blacken\ellipse{150}{150}}
\put(1283,3083){\ellipse{150}{150}}
\put(2483,3083){\blacken\ellipse{150}{150}}
\put(2483,3083){\ellipse{150}{150}}
\put(3683,3083){\blacken\ellipse{150}{150}}
\put(3683,3083){\ellipse{150}{150}}
\put(4883,3083){\blacken\ellipse{150}{150}}
\put(4883,3083){\ellipse{150}{150}}
\put(6083,3083){\blacken\ellipse{150}{150}}
\put(6083,3083){\ellipse{150}{150}}
\put(7283,3083){\blacken\ellipse{150}{150}}
\put(7283,3083){\ellipse{150}{150}}
\put(8483,3083){\blacken\ellipse{150}{150}}
\put(8483,3083){\ellipse{150}{150}}
\put(9683,3083){\blacken\ellipse{150}{150}}
\put(9683,3083){\ellipse{150}{150}}
\put(683,3083){\blacken\ellipse{150}{150}}
\put(683,3083){\ellipse{150}{150}}
\put(1883,3083){\blacken\ellipse{150}{150}}
\put(1883,3083){\ellipse{150}{150}}
\put(3083,3083){\blacken\ellipse{150}{150}}
\put(3083,3083){\ellipse{150}{150}}
\put(4283,3083){\blacken\ellipse{150}{150}}
\put(4283,3083){\ellipse{150}{150}}
\put(5483,3083){\blacken\ellipse{150}{150}}
\put(5483,3083){\ellipse{150}{150}}
\put(6683,3083){\blacken\ellipse{150}{150}}
\put(6683,3083){\ellipse{150}{150}}
\put(7883,3083){\blacken\ellipse{150}{150}}
\put(7883,3083){\ellipse{150}{150}}
\put(9083,3083){\blacken\ellipse{150}{150}}
\put(9083,3083){\ellipse{150}{150}}
\put(10283,3083){\blacken\ellipse{150}{150}}
\put(10283,3083){\ellipse{150}{150}}
\put(83,2483){\blacken\ellipse{150}{150}}
\put(83,2483){\ellipse{150}{150}}
\put(1283,2483){\blacken\ellipse{150}{150}}
\put(1283,2483){\ellipse{150}{150}}
\put(2483,2483){\blacken\ellipse{150}{150}}
\put(2483,2483){\ellipse{150}{150}}
\put(3683,2483){\blacken\ellipse{150}{150}}
\put(3683,2483){\ellipse{150}{150}}
\put(4883,2483){\blacken\ellipse{150}{150}}
\put(4883,2483){\ellipse{150}{150}}
\put(6083,2483){\blacken\ellipse{150}{150}}
\put(6083,2483){\ellipse{150}{150}}
\put(7283,2483){\blacken\ellipse{150}{150}}
\put(7283,2483){\ellipse{150}{150}}
\put(8483,2483){\blacken\ellipse{150}{150}}
\put(8483,2483){\ellipse{150}{150}}
\put(9683,2483){\blacken\ellipse{150}{150}}
\put(9683,2483){\ellipse{150}{150}}
\put(683,2483){\blacken\ellipse{150}{150}}
\put(683,2483){\ellipse{150}{150}}
\put(1883,2483){\blacken\ellipse{150}{150}}
\put(1883,2483){\ellipse{150}{150}}
\put(3083,2483){\blacken\ellipse{150}{150}}
\put(3083,2483){\ellipse{150}{150}}
\put(4283,2483){\blacken\ellipse{150}{150}}
\put(4283,2483){\ellipse{150}{150}}
\put(5483,2483){\blacken\ellipse{150}{150}}
\put(5483,2483){\ellipse{150}{150}}
\put(6683,2483){\blacken\ellipse{150}{150}}
\put(6683,2483){\ellipse{150}{150}}
\put(7883,2483){\blacken\ellipse{150}{150}}
\put(7883,2483){\ellipse{150}{150}}
\put(9083,2483){\blacken\ellipse{150}{150}}
\put(9083,2483){\ellipse{150}{150}}
\put(10283,2483){\blacken\ellipse{150}{150}}
\put(10283,2483){\ellipse{150}{150}}
\put(83,1883){\blacken\ellipse{150}{150}}
\put(83,1883){\ellipse{150}{150}}
\put(1283,1883){\blacken\ellipse{150}{150}}
\put(1283,1883){\ellipse{150}{150}}
\put(2483,1883){\blacken\ellipse{150}{150}}
\put(2483,1883){\ellipse{150}{150}}
\put(3683,1883){\blacken\ellipse{150}{150}}
\put(3683,1883){\ellipse{150}{150}}
\put(4883,1883){\blacken\ellipse{150}{150}}
\put(4883,1883){\ellipse{150}{150}}
\put(6083,1883){\blacken\ellipse{150}{150}}
\put(6083,1883){\ellipse{150}{150}}
\put(7283,1883){\blacken\ellipse{150}{150}}
\put(7283,1883){\ellipse{150}{150}}
\put(8483,1883){\blacken\ellipse{150}{150}}
\put(8483,1883){\ellipse{150}{150}}
\put(9683,1883){\blacken\ellipse{150}{150}}
\put(9683,1883){\ellipse{150}{150}}
\put(683,1883){\blacken\ellipse{150}{150}}
\put(683,1883){\ellipse{150}{150}}
\put(1883,1883){\blacken\ellipse{150}{150}}
\put(1883,1883){\ellipse{150}{150}}
\put(3083,1883){\blacken\ellipse{150}{150}}
\put(3083,1883){\ellipse{150}{150}}
\put(4283,1883){\blacken\ellipse{150}{150}}
\put(4283,1883){\ellipse{150}{150}}
\put(5483,1883){\blacken\ellipse{150}{150}}
\put(5483,1883){\ellipse{150}{150}}
\put(6683,1883){\blacken\ellipse{150}{150}}
\put(6683,1883){\ellipse{150}{150}}
\put(7883,1883){\blacken\ellipse{150}{150}}
\put(7883,1883){\ellipse{150}{150}}
\put(9083,1883){\blacken\ellipse{150}{150}}
\put(9083,1883){\ellipse{150}{150}}
\put(10283,1883){\blacken\ellipse{150}{150}}
\put(10283,1883){\ellipse{150}{150}}
\put(83,1283){\blacken\ellipse{150}{150}}
\put(83,1283){\ellipse{150}{150}}
\put(1283,1283){\blacken\ellipse{150}{150}}
\put(1283,1283){\ellipse{150}{150}}
\put(2483,1283){\blacken\ellipse{150}{150}}
\put(2483,1283){\ellipse{150}{150}}
\put(3683,1283){\blacken\ellipse{150}{150}}
\put(3683,1283){\ellipse{150}{150}}
\put(4883,1283){\blacken\ellipse{150}{150}}
\put(4883,1283){\ellipse{150}{150}}
\put(6083,1283){\blacken\ellipse{150}{150}}
\put(6083,1283){\ellipse{150}{150}}
\put(7283,1283){\blacken\ellipse{150}{150}}
\put(7283,1283){\ellipse{150}{150}}
\put(8483,1283){\blacken\ellipse{150}{150}}
\put(8483,1283){\ellipse{150}{150}}
\put(9683,1283){\blacken\ellipse{150}{150}}
\put(9683,1283){\ellipse{150}{150}}
\put(683,1283){\blacken\ellipse{150}{150}}
\put(683,1283){\ellipse{150}{150}}
\put(1883,1283){\blacken\ellipse{150}{150}}
\put(1883,1283){\ellipse{150}{150}}
\put(3083,1283){\blacken\ellipse{150}{150}}
\put(3083,1283){\ellipse{150}{150}}
\put(4283,1283){\blacken\ellipse{150}{150}}
\put(4283,1283){\ellipse{150}{150}}
\put(5483,1283){\blacken\ellipse{150}{150}}
\put(5483,1283){\ellipse{150}{150}}
\put(6683,1283){\blacken\ellipse{150}{150}}
\put(6683,1283){\ellipse{150}{150}}
\put(7883,1283){\blacken\ellipse{150}{150}}
\put(7883,1283){\ellipse{150}{150}}
\put(9083,1283){\blacken\ellipse{150}{150}}
\put(9083,1283){\ellipse{150}{150}}
\put(10283,1283){\blacken\ellipse{150}{150}}
\put(10283,1283){\ellipse{150}{150}}
\put(83,683){\blacken\ellipse{150}{150}}
\put(83,683){\ellipse{150}{150}}
\put(1283,683){\blacken\ellipse{150}{150}}
\put(1283,683){\ellipse{150}{150}}
\put(2483,683){\blacken\ellipse{150}{150}}
\put(2483,683){\ellipse{150}{150}}
\put(3683,683){\blacken\ellipse{150}{150}}
\put(3683,683){\ellipse{150}{150}}
\put(4883,683){\blacken\ellipse{150}{150}}
\put(4883,683){\ellipse{150}{150}}
\put(6083,683){\blacken\ellipse{150}{150}}
\put(6083,683){\ellipse{150}{150}}
\put(7283,683){\blacken\ellipse{150}{150}}
\put(7283,683){\ellipse{150}{150}}
\put(8483,683){\blacken\ellipse{150}{150}}
\put(8483,683){\ellipse{150}{150}}
\put(9683,683){\blacken\ellipse{150}{150}}
\put(9683,683){\ellipse{150}{150}}
\put(683,683){\blacken\ellipse{150}{150}}
\put(683,683){\ellipse{150}{150}}
\put(1883,683){\blacken\ellipse{150}{150}}
\put(1883,683){\ellipse{150}{150}}
\put(3083,683){\blacken\ellipse{150}{150}}
\put(3083,683){\ellipse{150}{150}}
\put(4283,683){\blacken\ellipse{150}{150}}
\put(4283,683){\ellipse{150}{150}}
\put(5483,683){\blacken\ellipse{150}{150}}
\put(5483,683){\ellipse{150}{150}}
\put(6683,683){\blacken\ellipse{150}{150}}
\put(6683,683){\ellipse{150}{150}}
\put(7883,683){\blacken\ellipse{150}{150}}
\put(7883,683){\ellipse{150}{150}}
\put(9083,683){\blacken\ellipse{150}{150}}
\put(9083,683){\ellipse{150}{150}}
\put(10283,683){\blacken\ellipse{150}{150}}
\put(10283,683){\ellipse{150}{150}}
\put(83,83){\blacken\ellipse{150}{150}}
\put(83,83){\ellipse{150}{150}}
\put(1283,83){\blacken\ellipse{150}{150}}
\put(1283,83){\ellipse{150}{150}}
\put(2483,83){\blacken\ellipse{150}{150}}
\put(2483,83){\ellipse{150}{150}}
\put(3683,83){\blacken\ellipse{150}{150}}
\put(3683,83){\ellipse{150}{150}}
\put(4883,83){\blacken\ellipse{150}{150}}
\put(4883,83){\ellipse{150}{150}}
\put(6083,83){\blacken\ellipse{150}{150}}
\put(6083,83){\ellipse{150}{150}}
\put(7283,83){\blacken\ellipse{150}{150}}
\put(7283,83){\ellipse{150}{150}}
\put(8483,83){\blacken\ellipse{150}{150}}
\put(8483,83){\ellipse{150}{150}}
\put(9683,83){\blacken\ellipse{150}{150}}
\put(9683,83){\ellipse{150}{150}}
\put(683,83){\blacken\ellipse{150}{150}}
\put(683,83){\ellipse{150}{150}}
\put(1883,83){\blacken\ellipse{150}{150}}
\put(1883,83){\ellipse{150}{150}}
\put(3083,83){\blacken\ellipse{150}{150}}
\put(3083,83){\ellipse{150}{150}}
\put(4283,83){\blacken\ellipse{150}{150}}
\put(4283,83){\ellipse{150}{150}}
\put(5483,83){\blacken\ellipse{150}{150}}
\put(5483,83){\ellipse{150}{150}}
\put(6683,83){\blacken\ellipse{150}{150}}
\put(6683,83){\ellipse{150}{150}}
\put(7883,83){\blacken\ellipse{150}{150}}
\put(7883,83){\ellipse{150}{150}}
\put(9083,83){\blacken\ellipse{150}{150}}
\put(9083,83){\ellipse{150}{150}}
\put(10283,83){\blacken\ellipse{150}{150}}
\put(10283,83){\ellipse{150}{150}}
\path(683,4883)(1283,4283)(83,3683)(683,4883)
\path(1883,3683)(2483,4883)(3083,3683)(1883,3683)
\path(3683,4283)(4283,3683)(4883,3683)
	(4283,4883)(3683,4283)
\path(5483,4883)(5483,3683)(7283,3683)
	(6083,4883)(5483,4883)
\path(7883,3683)(9083,4883)(10283,3683)(7883,3683)
\path(83,3083)(83,2483)(683,1883)
	(1283,1883)(1283,2483)(683,3083)(83,3083)
\path(1883,2483)(2483,3083)(3083,2483)
	(3083,1883)(2483,1883)(1883,2483)
\path(3683,1883)(3683,3083)(4883,3083)
	(4883,2483)(4283,1883)(3683,1883)
\path(5483,1883)(6083,3083)(6683,3083)
	(6683,1883)(5483,1883)
\path(7283,1883)(7283,2483)(7883,3083)
	(8483,2483)(8483,1883)(7283,1883)
\path(9083,1883)(9083,3083)(10283,3083)
	(10283,1883)(9083,1883)
\path(2483,83)(2483,683)(3083,1283)
	(4283,83)(2483,83)
\path(4883,83)(5483,1283)(6683,83)(4883,83)
\path(7283,83)(7883,1283)(8483,683)
	(8483,83)(7283,83)
\path(9083,683)(9683,1283)(10283,683)
	(9683,83)(9083,683)
\put(233,308){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$3 \Delta$}}}}
\end{picture}
}
  \caption{All polygons with $i=1$.}
  \label{fig:reflexive}
\end{figure}

What can we do for $i \ge 2$? The family shown in
Figure~\ref{fig:shoe} yields all $4\le b\le 2i+6$. In fact,
Scott~\cite{Scott} showed that $2i+6$ is how far we can get.

\begin{figure}[htbp]
  \centering
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(6766,2211)(0,-10)
\put(1883,1083){\blacken\ellipse{150}{150}}
\put(1883,1083){\ellipse{150}{150}}
\put(2483,1083){\blacken\ellipse{150}{150}}
\put(2483,1083){\ellipse{150}{150}}
\put(3083,1083){\blacken\ellipse{150}{150}}
\put(3083,1083){\ellipse{150}{150}}
\put(683,1083){\blacken\ellipse{150}{150}}
\put(683,1083){\ellipse{150}{150}}
\put(1283,1083){\blacken\ellipse{150}{150}}
\put(1283,1083){\ellipse{150}{150}}
\put(3683,1083){\blacken\ellipse{150}{150}}
\put(3683,1083){\ellipse{150}{150}}
\put(83,1083){\blacken\ellipse{150}{150}}
\put(83,1083){\ellipse{150}{150}}
\put(1883,1683){\blacken\ellipse{150}{150}}
\put(1883,1683){\ellipse{150}{150}}
\put(2483,1683){\blacken\ellipse{150}{150}}
\put(2483,1683){\ellipse{150}{150}}
\put(3083,1683){\blacken\ellipse{150}{150}}
\put(3083,1683){\ellipse{150}{150}}
\put(683,1683){\blacken\ellipse{150}{150}}
\put(683,1683){\ellipse{150}{150}}
\put(1283,1683){\blacken\ellipse{150}{150}}
\put(1283,1683){\ellipse{150}{150}}
\put(3683,1683){\blacken\ellipse{150}{150}}
\put(3683,1683){\ellipse{150}{150}}
\put(83,1683){\blacken\ellipse{150}{150}}
\put(83,1683){\ellipse{150}{150}}
\put(1883,483){\blacken\ellipse{150}{150}}
\put(1883,483){\ellipse{150}{150}}
\put(2483,483){\blacken\ellipse{150}{150}}
\put(2483,483){\ellipse{150}{150}}
\put(3083,483){\blacken\ellipse{150}{150}}
\put(3083,483){\ellipse{150}{150}}
\put(683,483){\blacken\ellipse{150}{150}}
\put(683,483){\ellipse{150}{150}}
\put(1283,483){\blacken\ellipse{150}{150}}
\put(1283,483){\ellipse{150}{150}}
\put(3683,483){\blacken\ellipse{150}{150}}
\put(3683,483){\ellipse{150}{150}}
\put(83,483){\blacken\ellipse{150}{150}}
\put(83,483){\ellipse{150}{150}}
\put(6083,1083){\blacken\ellipse{150}{150}}
\put(6083,1083){\ellipse{150}{150}}
\put(6683,1083){\blacken\ellipse{150}{150}}
\put(6683,1083){\ellipse{150}{150}}
\put(4883,1083){\blacken\ellipse{150}{150}}
\put(4883,1083){\ellipse{150}{150}}
\put(5483,1083){\blacken\ellipse{150}{150}}
\put(5483,1083){\ellipse{150}{150}}
\put(4283,1083){\blacken\ellipse{150}{150}}
\put(4283,1083){\ellipse{150}{150}}
\put(6083,1683){\blacken\ellipse{150}{150}}
\put(6083,1683){\ellipse{150}{150}}
\put(6683,1683){\blacken\ellipse{150}{150}}
\put(6683,1683){\ellipse{150}{150}}
\put(4883,1683){\blacken\ellipse{150}{150}}
\put(4883,1683){\ellipse{150}{150}}
\put(5483,1683){\blacken\ellipse{150}{150}}
\put(5483,1683){\ellipse{150}{150}}
\put(4283,1683){\blacken\ellipse{150}{150}}
\put(4283,1683){\ellipse{150}{150}}
\put(6083,483){\blacken\ellipse{150}{150}}
\put(6083,483){\ellipse{150}{150}}
\put(6683,483){\blacken\ellipse{150}{150}}
\put(6683,483){\ellipse{150}{150}}
\put(4883,483){\blacken\ellipse{150}{150}}
\put(4883,483){\ellipse{150}{150}}
\put(5483,483){\blacken\ellipse{150}{150}}
\put(5483,483){\ellipse{150}{150}}
\put(4283,483){\blacken\ellipse{150}{150}}
\put(4283,483){\ellipse{150}{150}}
\thicklines
\path(83,483)(4883,483)(6683,1083)
	(83,1683)(83,483)
\put(83,108){\makebox(0,0)[b]{\smash{{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$(0,0)$}}}}}
\put(6683,1308){\makebox(0,0)[b]{\smash{{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$(i+1,1)$}}}}}
\put(83,1908){\makebox(0,0)[b]{\smash{{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$(0,2)$}}}}}
\put(4883,108){\makebox(0,0)[b]{\smash{{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$(b-4,0)$}}}}}
\end{picture}
}
  \caption{$4\le b\le 2i+6$}
  \label{fig:shoe}
\end{figure}

Scott's proof is elementary and short enough to be included in this
paper. We give two other proofs for the same result.
One of them uses toric geometry;
it is merely the observation that a well-known
inequality~\cite[Theorem~6]{Schicho:98d} in algebraic geometry
translates to Scott's inequality when applied to toric surfaces.
The third proof is again elementary, and it was the search for this
proof which sparked polygon addiction in the first author.

For the inequality $b\le 2\ i +6$, we have arbitrary large examples
where equality holds (see Figure~\ref{fig:shoe}); but for all these
examples, all interior points are collinear. Under the additional
assumption that the interior points are not collinear, the inequality
can be strengthened to $b \le i + 9$ (see the remark after
Lemma~\ref{lemma:step}).
The coefficient in front of the $i$ can be improved further
by introducing the {\em level} of a
polygon:
roughly speaking, this is the number of times one can pass
to the convex hull of the interior lattice points.

Before we really get going, here is a little caveat. Most of our
considerations break down in dimension $3$. Pick's formula has no
analogue. Already tetrahedra with no boundary or interior lattice
points except the vertices can have arbitrary volume. This was first
pointed out by J. Reeve~\cite{Reeve} (see Figure~\ref{fig:reeve}).
Nevertheless, the phenomenon that for given $i>0$, the volume is
bounded occurs in arbitrary dimension~\cite{lagariasZiegler}.

\begin{figure}[htbp]
  \centering
  \includegraphics{reeve3.eps}
  \caption{Reeve's simplices}
  \label{fig:reeve}
\end{figure}

\section{Three proofs of $\mathbf{b \le 2i+7}$}\noindent
Let $P$ be a
polygon with interior lattice
points. Denote $a$ its surface area, $i$ the number of interior
lattice points, and $b$ the number of lattice points on $P$'s
boundary. In view of Pick's Theorem~\eqref{eq:pick}, the following
three inequalities are equivalent.
\begin{proposition} \label{prop:main}
  If $i>0$, then

  \begin{align}
    b \ &\le \ 2 \ i \ + \ 7 \label{eq:main:bi} \\

    a \ &\le \ 2 \ i \ + \ 5/2 \label{eq:main:ai}

    \\
    b \ &\le \ a \ + \ 9/2 \label{eq:main:ab}
  \end{align}
  with equality only for the triangle $3\Delta$ in
  Figure~\ref{fig:reflexive}.
\end{proposition}

\subsection{Scott's proof} \label{ssec:scott}

Apply lattice equivalences to $P$
so that $P$ fits tightly into a box $[0,p'] \times [0,p]$
with $p$ as small as possible.
Then $2 \le p \le p'$ (remember,
$i>0$). If $P$ intersects the top and the bottom edge of the box in
segments of length $q \ge 0$ and $q' \ge 0$ respectively, then
(See Figure~\ref{fig:box}.)
\begin{align}
  b &\le q+q'+2p \ , \ \text{ and } \label{eq:sc1} \\
  a &\ge p\,(q+q')/2 \ . \label{eq:sc2}
\end{align}
\begin{figure}[htbp]
  \centering
\begin{picture}(0,0)
\includegraphics{box.pstex}
\end{picture}
\setlength{\unitlength}{1973sp}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
\begin{picture}(3383,2489)(518,-6967)
\put(3901,-5761){\makebox(0,0)[lb]{\smash{\SetFigFont{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$p=3$}
}}}
\put(1876,-6211){\makebox(0,0)[lb]{\smash{\SetFigFont{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$q'=3$}
}}}
\put(1726,-6886){\makebox(0,0)[lb]{\smash{\SetFigFont{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$p'=5$}
}}}
\put(1876,-5011){\makebox(0,0)[lb]{\smash{\SetFigFont{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$q=0$}
}}}
\end{picture}
  \caption{$P$ in a box.}
  \label{fig:box}
\end{figure}

We distinguish three cases
\begin{itemize}
\item[($i$)] $p=2$, or $q+q' \ge 4$, or $p=q+q'=3$
\item[($ii$)] $p=3$, and $q+q' \le 2$
\item[($iii$)] $p \ge 4$, and $q+q' \le 3$.
\end{itemize}
The above inequalities~\eqref{eq:sc1}, and \eqref{eq:sc2} are already
sufficient to deal with the first two cases.

\\[.3\baselineskip]
($i$) We have
\begin{align*}
  2b-2a \ &\le \ 2(q+q'+2p) - p\,(q+q') \\
  &= \ (q+q'-4)(2-p) + 8 \quad \le \ 9 ,
\end{align*}
which shows~\eqref{eq:main:ab} in Proposition~\ref{prop:main}.
(With equality if and only if $p=q+q'=3$, $a=9/2$, $b=9$.)
\footnote{It is an exercise to show that the only $P$ with these
  parameters is the triangle $3\Delta$ in Figure~\ref{fig:reflexive}.}

\\[.3\baselineskip]
($ii$) The estimate $b \le q+q'+2p \le 8$ together with $i \ge 1$ show
  that inequality~\eqref{eq:main:bi} in Proposition~\ref{prop:main} is
  strictly satisfied.

\\[.3\baselineskip]
($iii$)
The only case where we have to work a little is case three.

Choose points $\pt{x}=(x_1,p)^t$, $\pt{x}'=(x'_1,0)^t$,
$\pt{y}=(0,y_2)^t$, and $\pt{y}'=(p',y'_2)^t$ in $P$ so that $\delta =
|x_1-x'_1|$ is as small as possible. Then $a \ge p (p'- \delta)/2$
(see Figure~\ref{fig:case3}).

\begin{figure}[htbp]
  \centering
\begin{picture}(0,0)
\includegraphics{case3a.pstex}
\end{picture}
\setlength{\unitlength}{1184sp}

\begingroup\makeatletter\ifx\SetFigFontNFSS\undefined
\gdef\SetFigFontNFSS#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
\begin{picture}(7080,5691)(436,-5329)
\put(7501,-2911){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\pt{y}'$}
}}}}
\put(451,-1111){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\pt{y}$}
}}}}
\put(2776,-61){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\pt{x}$}
}}}}
\put(4051,-5161){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{10}{12.0}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\pt{x}'$}
}}}}
\put(3601,-2536){\makebox(0,0)[b]{\smash{{\SetFigFontNFSS{9}{10.8}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}\makebox[.3in]{$\overset{\displaystyle\delta}{\longleftrightarrow}$}}
}}}}
\end{picture}

\qquad

\begin{picture}(0,0)
\includegraphics{case3b.pstex}
\end{picture}
\setlength{\unitlength}{1184sp}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
\begin{picture}(6024,4911)(1189,-5260)
\put(4051,-5161){\makebox(0,0)[lb]{\smash{\SetFigFont{7}{8.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{1,1,1}\mbox{}}
}}}
\end{picture}
  \caption{\mbox{Case three. Two triangles of total area $p (p'-
      \delta)/2$.}}
  \label{fig:case3}
\end{figure}

Now the task is to apply lattice equivalences so that $\delta$ becomes
small.
\begin{exercise}
  After applying a lattice equivalence of the form $\left[
    \begin{smallmatrix}
      1&k\\0&1
    \end{smallmatrix}
  \right]$
  it is possible to choose $\delta \le (p-q-q')/2$.
\end{exercise}
This lattice equivalence will leave $q$, $q'$, $p$ unchanged, because
it fixes the $x_1$-axis. We still have $p \le p'$ because $p$ was
supposed to be minimal. Thus, we obtain
\begin{equation} \label{eq:sc3}
  a \ge p(p+q+q')/4 ,
\end{equation}
and
\begin{multline*}
  4(b-a) \le 8p+4q+4q' \ - \ p(p+q+q') \\
  = p(8-p) - (p-4)(q+q') \le p(8-p) \le 16
\end{multline*}
because $p\ge4$ in case three. This proves that
inequality~\eqref{eq:main:ab} in Proposition~\ref{prop:main} is
strictly satisfied.
\qed

\subsection{Clipping off vertices}
This proof proceeds by induction on $i$. If $i=1$, we can check the
inequalities on all $16$ lattice equivalence classes of such
$P$. (See Figure~\ref{fig:reflexive}.)

For the induction step, we want to ``chop off a vertex''.
If $i \ge 2$, and $b \le 10$, nothing is to show.
So assume $b \ge 11$.
By applying a lattice equivalence, we may assume without loss of
generality that $\nv$ and $(1,0)\transpose$ lie in the interior of
$P$. Reflect in the $x_1$-axis if necessary in order to assure
that there are $\ge 5$ boundary lattice points with positive second
coordinate.

First, suppose there is a vertex $v$ with positive second coordinate
which is not unimodular. That is, the triangle formed by $v$ together
with its two neighboring lattice points $v'$ and $v''$ on the boundary
has area $> 1/2$. Denote $P'$ the convex hull $\conv(P \cap \Z^2
\setminus \{v\})$. This omission affects our parameters as follows:
$b' = b+k-2$, $i' = i - k + 1$, and, by Pick's formula, $a' = a-k/2$.
Here $k$ is the lattice length of the boundary of $P'$ that is visible
from $v$. Because $v$ was not unimodular, there is an additional
lattice point in the triangle $vv'v''$. Thus, we have $k \ge
2$. Because there are other lattice points with positive second
coordinate, at least one of $\nv$ or $(1,0)\transpose$ remains in the
interior of $P'$, and we can use induction.

\begin{figure}[htbp]
  \centering
  \includegraphics{omit1.eps}
  \caption{Clipping off a nonunimodular vertex.}
  \label{fig:omit1}
\end{figure}

Now, if all vertices with positive second coordinate are unimodular,
similiarly omit one vertex $v$ together with its two boundary
neighbors $v'$ and $v''$: $P'=\conv(P \cap \Z^2 \setminus
\{v,v',v''\})$.
The parameters change as follows:
$b' = b+k-4$, $i' = i - k + 1$, and $a' = a-k/2-1$, where $k$ is
the lattice length of the boundary of $P'$ that is visible from the
removed points.
In order to see that $k \ge 2$, observe that the point $v'''=v'+v''-v$
belongs to the interior of $P$, and the two adjacent segments of $P'$
are both visible from the removed points.
As observed above, there remain lattice points with
positive second coordinate in $P'$ so that at least one of
$\nv$ or $(1,0)\transpose$ stays in the interior of $P'$.
\hfill $\Box$

\begin{figure}[htbp]
  \centering
\begin{picture}(0,0)
\includegraphics{omit2.pstex}
\end{picture}
\setlength{\unitlength}{1973sp}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
\begin{picture}(3174,3099)(814,-4348)
\put(2401,-2011){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$v'$}
}}}}
\put(1201,-2686){\makebox(0,0)[rb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$v''$}
}}}}
\put(1726,-2086){\makebox(0,0)[rb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$v$}
}}}}
\put(1876,-3136){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}{\color[rgb]{1,0,0}$v'''$}
}}}}
\end{picture}
  \caption{Clipping off a unimodular vertex (and its neighbors).}
  \label{fig:omit2}
\end{figure}

\subsection{Algebraic geometry} \noindent
We use the letters $d$ and $p$ to denote the
degree and the sectional genus of an algebraic surface. The inequality
$p\le (d-1)(d-2)/2$ holds for arbitrary algebraic surfaces.
If the surface is rational, i.e. if it has a parametrization by
rational functions, then there are more inequalities.

\begin{theorem} \label{thm:ag} \mbox{}
  \begin{itemize}
  \item If $p=1$, then $d\le 9$.
  \item If $p\ge 2$, then $d\le 4p+4$.
  \end{itemize}
\end{theorem}

Rational surfaces with $p=1$ are called Del Pezzo surfaces. The degree
bound $9$ is due to del Pezzo~\cite{Pezzo:87}. The bound $d\le 4p+4$
was shown by Jung~\cite{Jung:90}, hence this proof is
actually the oldest one!
A modern proof can be found in Schicho~\cite{Schicho:98d}.

Toric surfaces are rational, and Scott's inequality is equivalent
to Theorem~\ref{thm:ag} for toric surfaces.
\hfill $\Box$

\section{Onion skins} \label{sec:onions}
\begin{figure}[htbp]
  \centering
  \includegraphics[height=43mm]{zwiebel3.eps}

\end{figure}

In flatland, take a solid polygon $P$ into your hand

and peel off the shell. You get another convex polygon $P^{(1)}$, the
convex hull of its interior lattice points. Except, of course, if $i=0$
then $P$ was an empty nut, and if all interior lattice points
are collinear then $P^{(1)}$ is a ``degenerate polygon'', namely a line
segment or a single point.

Repeat this process as long as it is possible, peeling off
the skins of the polygon one after the other: $P^{(k+1)} :=
(P^{(k)})^{(1)}$. After $n$ steps you arrive at the nucleus $P^{(n)}$,
which is either a degenerate polygon or an empty nut. We define the
{\em level\/} $\ell=\ell(P)$ in the following way:
\begin{itemize}
  \item $\ell(P)=n$ if the nucleus is a degenerate polygon,
  \item $\ell(P)=n+1/3$ if the nucleus is $\Delta$,
  \item $\ell(P)=n+2/3$ if the nucleus is $2\Delta$, and
  \item $\ell(P)=n+1/2$ if the nucleus is any other empty nut.
\end{itemize}
Here $\Delta$ stands for (a polygon lattice equivalent to) the
standard triangle $\conv[(0,0)\transpose, (1,0)\transpose,
(0,1)\transpose]$. The purpose of this weird definition is
to ensure the second statement in the exercise below.

\begin{exercise}
  Show that $\ell$ is uniquely defined by
  \begin{itemize}
  \item $\ell(P) = \ell(P^{(1)})+1$ if $P^{(1)}$ is $2$-dimensional,
    and
  \item $\ell(kP)=k\ell(P)$ for positive integers $k$.
  \end{itemize}

 \end{exercise}

\begin{figure}[htbp]
  \centering
  \includegraphics{cases.eps}
  \caption{Polygons of levels \hfill \mbox{} \mbox{$\ell=3$, $\ell=2$,
      $\ell=5/2$, $\ell=7/3$, and $\ell=8/3$.}}
  \label{fig:P(ell)}
\end{figure}

The level of a polygon is an analogue of the radius of the in-circle
in Euclidean geometry. There we have the equation $2a=\ell b$.
In lattice geometry, we have an inequality.

\begin{oniontheorem} \stepcounter{theorem} \label{thm:onion}
  Let $P$ be a convex lattice polygon of area $a$ and
  level $\ell \ge 1$ with $b$
  and $i$ lattice points on the boundary and in the interior,
  respectively. Then $(2\ell-1) b \le 2i + 9\ell^2 - 2$, or
  equivalently $2\ell b\le 2a+9\ell^2$, or equivalently
  $(4l-2)a\le 9\ell^2+4l(i-1)$, with equality
  if and only if $P$ is a multiple of $\Delta$.
\end{oniontheorem}
For $\ell>1$, these inequalities really strengthen the old $b \le
2i+7$.

We give two elementary proofs. One is similar to Scott's proof.
The other is a bit longer, but it gives more insight into the process
of peeling onion skins. For instance, it reveals that the set of
all polygons $P$ such that $P^{(1)}=Q$ for some fixed $Q$ is either
empty or has a largest element.
\subsection{Moving out edges}
Using this technique, it is actually possible to sharpen the bound
in various (sub)cases. E.g.,
\begin{itemize}
  \item if $P^{(\ell)}$ is a point, but $P^{(\ell-1)} \neq 3 \Delta$,
  then $(2\ell-1) b \le 2i + 8 \ell^2 - 2$;
  \item if $P^{(\ell)}$ is a segment, then $(2\ell-1) b \le 2i +
  8\ell^2 - 2$ with equality if and only if $P$ is lattice equivalent
  to a polygon with vertices $\nv, (r,0)\transpose,
  (2pq+r,2p)\transpose, (0,2p)\transpose$ for integers $p \ge 1$, $q,r
  \ge 0$ such that $pq+r \ge 3$;
  \item if $P^{(\ell)}$ has no interior lattice points but is not a
  multiple of $\Delta$, then $(2\ell-1) b \le 2i + 8 \ell^2 - 2$.
\end{itemize}
We reduce the proof to the case that $P$ is obtained from $P^{(1)}$ by
``moving out the edges by one''. This is done in the following three
lemmas. Finally, Lemma~\ref{lemma:step} yields the induction step in
the proof of the Onion--Skin Theorem.

We say that an inequality $\langle \pt{a}, \pt{x} \rangle = a_1 x_1 +
a_2 x_2 \le b$ with coprime $(a_1,a_2)$ defines an edge of a
polygon $Q$ if it is satisfied by all points $\pt{x} \in Q$, and there
are two distinct points in $Q$ satisfying equality.
Then, moving out this edge by one means to relax the inequality to
$\langle \pt{a}, \pt{x} \rangle \le b + 1$.
\begin{lemma} \label{lemma:shift}
  Suppose that the inequality $\langle \pt{a}, \pt{x} \rangle \le b$
  defines an edge of $P^{(1)}$. Then $\langle \pt{a}, \pt{x} \rangle
  \le b + 1$ is valid for $P$.
\end{lemma}
That means, if we move all the edges of $Q=P^{(1)}$ out by one, we
obtain a superset $Q^{(-1)}$ of $P$.

\begin{figure}[htbp]
  \centering
  \includegraphics{shift.eps}
  \caption{If $Q=P^{(1)}$ then $P \subseteq Q^{(-1)}$.}
  \label{fig:shift}
\end{figure}

\begin{proof}
  We may apply a lattice equivalence to reduce to the case
  where the edge is defined by $x_2 \le 0$, and that $(0,0)\transpose$
  and $(1,0)\transpose$ are two lattice points of $P^{(1)}$ lying
  on this edge.
  Suppose indirectly that
  $P$ has a vertex $\pt{v}$ with $v_2 > 1$. Then the triangle formed
  by the three points has area $v_2/2 \ge 1$. It must therefore
  contain another lattice point which lies in the interior of $P$, and
  has positive second coordinate.
\end{proof}
For arbitrary $Q$, this $Q^{(-1)}$ does not necessarily have integral
vertices.
But then, not every
polygon arises as $P^{(1)}$ for some $P$. A necessary condition is
that the polygon has good angles.

\begin{figure}[htbp]
  \centering
  \includegraphics{Qminus1.alt.eps}
  \caption{$Q^{(-1)}$ may be nonintegral.}
  \label{fig:badshift}
\end{figure}

\begin{lemma} \label{lemma:angles}
  If $P^{(1)}$ is $2$--dimensional, then for all vertices $\pt{v}$ of
  $P^{(1)}$, the cones generated by $P^{(1)} - \pt{v}$ are
  lattice equivalent to a cone generated by $(1,0)\transpose$ and
  $(-1,k)\transpose$, for some integer $k \ge 1$.
\end{lemma}

\begin{figure}[htbp]
  \centering
  \includegraphics{goodAngles.eps}
  \caption{Good angles and a bad angle.}
  \label{fig:goodAngles}
\end{figure}

\begin{proof}
  Assume, after a lattice equivalence, that $\pt{v} = \nv$, and the
  rays of the cone in question are generated by $(1,0)\transpose$ and
  $(-p,q)\transpose$, with coprime $0 < p \le q$ (see
  Exercise~\ref{ex:angle}). By Lemma~\ref{lemma:shift}, all points of
  $P$ satisfy $x_2 \ge -1$ and $q x_1 + p x_2 \ge -1$. But this
  implies $x_1 + x_2 \ge -1 + \frac{p-1}{q}$. So, if $p > 1$, because
  $P$ has integral vertices, we have $x_1 + x_2 \ge 0$ for all points
  of $P$. This contradicts the fact that $\nv \in P^{(1)}$.
\end{proof}

For a vertex $\pt{v}$ of a
polygon define the shifted
vertex $\pt{v}^{(-1)}$ as follows. Let $\langle \pt{a}, \pt{x} \rangle
\le b$ and \mbox{$\langle \pt{a}', \pt{x} \rangle \le b'$} be the two
edges that intersect in $\pt{v}$.

The unique solution to $\langle \pt{a}, \pt{x} \rangle = b
+ 1$ and $\langle \pt{a}', \pt{x} \rangle = b' + 1$ is denoted
$\pt{v}^{(-1)}$. According to Lemma~\ref{lemma:angles}, when we deal
with $P^{(1)}$ then $\pt{v}^{(-1)}$ is a lattice point. (In the
situation of the lemma, it is $(0,-1)\transpose$.)

We obtain a characterization of when $Q = P^{(1)}$ for some $P$.
\begin{lemma} \label{lemma:Qminus1integral}
  For a
polygon $Q$, the following are equivalent:
  \begin{itemize}
    \item $Q = P^{(1)}$ for some
      polygon $P$.
    \item $Q^{(-1)}$ has integral vertices.
  \end{itemize}
\end{lemma}
Thus, given $Q$, {\em the\/} maximal polygon $P$ with $P^{(1)} = Q$ is
$P = Q^{(-1)}$. We will (and can) restrict to this situation when we
prove the induction step $\ell \leadsto \ell + 1$ for the Onion--Skin
Theorem.
\begin{proof}
  If $Q^{(-1)}$ has integral vertices, then its interior lattice
  points span $Q$.

  For the converse direction, if $Q = P^{(1)}$ then we claim that
  \begin{equation} \label{eq:Qminus1}
    Q^{(-1)} = \conv \{ \pt{v}^{(-1)} \suchthat \pt{v} \text{ vertex
      of } Q \} \ .
  \end{equation}

  To this end,
  denote $\pt{a_1}, \ldots, \pt{a_n}$ and $b_1, \ldots, b_n$ the
  normal vectors respectively right hand sides of the edges of $Q$ in
  cyclic order. Also, denote $\pt{v}_1, \ldots, \pt{v}_n$ the vertices
  of $Q$ so that edge number $k$ is the segment
  $\segment{v_\text{$k$}}{v_\text{$k\negthinspace+\negthinspace1$}}$
  ($k \mod n$).
  \smallskip

  \noindent ``$\subseteq$'': this inclusion holds for
  arbitrary $Q$\/.

  For a point $\pt{y} \in Q^{(-1)}$, let $\langle \pt{a}_k, \pt{x}
  \rangle \le b_k$ be an edge of $Q$ that maximizes $\langle \pt{a},
  \pt{y} \rangle - b$ over all edges. So if $\langle \pt{a}_k, \pt{y}
  \rangle - b_k \le 0$ then $\pt{y} \in Q$. Otherwise we have
  \begin{itemize}
  \item $b_k \le \langle \pt{a}_k, \pt{y} \rangle \le b_k + 1$,
  \item $\langle \pt{a}_k, \pt{y} \rangle - b_k \ge \langle
    \pt{a}_{k-1}, \pt{y} \rangle - b_{k-1}$, and
  \item $\langle \pt{a}_k, \pt{y} \rangle - b_k \ge \langle
    \pt{a}_{k+1}, \pt{y} \rangle - b_{k+1}$,
  \end{itemize}
  which describes (a subset of) the convex hull of $\pt{v}_k,
  \pt{v}_k^{(-1)}, \pt{v}_{k+1}, \pt{v}_{k+1}^{(-1)}$.
  \smallskip

  \noindent ``$\supseteq$'': For this inclusion we actually use $Q =
  P^{(1)}$. Figure~\ref{fig:Qminus1} shows how
  Equation~\eqref{eq:Qminus1} can fail in general.

  \begin{figure}[htb]
    \centering
\begin{picture}(0,0)
\includegraphics{Qminus1.bad.pstex}
\end{picture}
\setlength{\unitlength}{1973sp}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
\begin{picture}(6766,2435)(2918,-7044)
\put(6601,-4936){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$v^{(-1)} \not\in Q^{(-1)}$}}}}
\put(6001,-6136){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefault}{\updefault}$v$}}}}
\end{picture}
    \caption{Equation \eqref{eq:Qminus1} can fail in general.}
    \label{fig:Qminus1}
  \end{figure}

  In our situation, $Q = P^{(1)}$, and we need to show that $v_k^{(-1)}$
  satisfies all inequalities $\langle \pt{a}_j, \cdot \rangle \le b_j +
  1$ for $Q^{(-1)}$.
  Our assumption implies that
  $P$ (and therefore $Q^{(-1)}$ by Lemma~\ref{lemma:shift}) contains
  points $\pt{w}_j$ with $\langle \pt{a}_j, \pt{w}_j \rangle = b_j + 1$.
  None of the other edge normals belongs to the cone
  generated by $\pt{a}_{k-1}$ and $\pt{a}_k$. So for $j \neq k,k-1$,
  \begin{equation*}
    \begin{array}{lcccccc}
      \text{either } & \langle \pt{a}_j, \pt{v}_k^{(-1)} \rangle &\le&
      \langle \pt{a}_j, \pt{w}_{k-1} \rangle &\le& b_m + 1 , \\
      \text{or } & \langle \pt{a}_j, \pt{v}_k^{(-1)} \rangle &\le& \langle
      \pt{a}_j, \pt{w}_k \rangle &\le& b_m + 1 & \text{ (or both).}
    \end{array}
  \end{equation*}
\end{proof}
Finally, we can prove the key lemma for our induction step.
\begin{lemma} \label{lemma:step}
  Let $b^{(1)}$ denote the number of lattice points on the boundary of
  $P^{(1)}$. Then $b \le b^{(1)} + 9$, with equality if and only if
  $P$ is a multiple of $\Delta$.
\end{lemma}
This immediately shows that $b \le b^{(1)} + 9 \le i + 9$ if $P^{(1)}$
is $2$--dimensional.

For the proof, we need a result of B.~Poonen and
F.~Rodriguez-Villegas~\cite{Poonen12}.
Consider a primitive oriented segment $s=\segment{x}{y}$, i.e.,
$\pt{x}$ and $\pt{y}$ are the only lattice points $s$ contains. Call
$s$ admissible if the triangle $\conv(\nv, \pt{x}, \pt{y})$ contains
no other lattice points. Equivalently, $s$ is admissible if the
determinant $\sign(s) = \left| \begin{smallmatrix} x_1&y_1 \\ x_2&y_2
  \end{smallmatrix} \right|$ is equal to $\pm 1$.
The length of a sequence $(s^{(1)}, \ldots, s^{(n)})$ of admissible
segments is $\sum \sign(s^{(k)})$.

The dual of an admissible segment is the unique integral normal vector
$\pt{a}=\pt{a}(s)$ such that $\langle \pt{a}, \pt{x} \rangle = \langle
\pt{a}, \pt{y} \rangle = 1$. For a closed polygon with segments
$(s^{(1)}, \ldots, s^{(n)})$, the dual polygon walks through the
normal vectors $\pt{a}(s^{(k)})$.

\begin{figure}[htbp]
  \begin{center}
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFontNFSS\undefined
\gdef\SetFigFontNFSS#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(5570,2581)(0,-10)
\put(4515,1883){\blacken\ellipse{150}{150}}
\put(4515,1883){\ellipse{150}{150}}
\put(4515,683){\blacken\ellipse{150}{150}}
\put(4515,683){\ellipse{150}{150}}
\put(3315,1283){\blacken\ellipse{150}{150}}
\put(3315,1283){\ellipse{150}{150}}
\put(3315,1883){\blacken\ellipse{150}{150}}
\put(3315,1883){\ellipse{150}{150}}
\put(3315,683){\blacken\ellipse{150}{150}}
\put(3315,683){\ellipse{150}{150}}
\put(3915,1283){\blacken\ellipse{150}{150}}
\put(3915,1283){\ellipse{150}{150}}
\put(3915,1883){\blacken\ellipse{150}{150}}
\put(3915,1883){\ellipse{150}{150}}
\put(3915,683){\blacken\ellipse{150}{150}}
\put(3915,683){\ellipse{150}{150}}
\put(3915,2483){\blacken\ellipse{150}{150}}
\put(3915,2483){\ellipse{150}{150}}
\put(4515,2483){\blacken\ellipse{150}{150}}
\put(4515,2483){\ellipse{150}{150}}
\put(3315,2483){\blacken\ellipse{150}{150}}
\put(3315,2483){\ellipse{150}{150}}
\put(5115,2483){\blacken\ellipse{150}{150}}
\put(5115,2483){\ellipse{150}{150}}
\put(5115,1883){\blacken\ellipse{150}{150}}
\put(5115,1883){\ellipse{150}{150}}
\put(5115,1283){\blacken\ellipse{150}{150}}
\put(5115,1283){\ellipse{150}{150}}
\put(5115,683){\blacken\ellipse{150}{150}}
\put(5115,683){\ellipse{150}{150}}
\put(5115,83){\blacken\ellipse{150}{150}}
\put(5115,83){\ellipse{150}{150}}
\put(4515,83){\blacken\ellipse{150}{150}}
\put(4515,83){\ellipse{150}{150}}
\put(3915,83){\blacken\ellipse{150}{150}}
\put(3915,83){\ellipse{150}{150}}
\put(3315,83){\blacken\ellipse{150}{150}}
\put(3315,83){\ellipse{150}{150}}
\put(1515,1283){\blacken\ellipse{150}{150}}
\put(1515,1283){\ellipse{150}{150}}
\put(1515,1883){\blacken\ellipse{150}{150}}
\put(1515,1883){\ellipse{150}{150}}
\put(1515,683){\blacken\ellipse{150}{150}}
\put(1515,683){\ellipse{150}{150}}
\put(315,1283){\blacken\ellipse{150}{150}}
\put(315,1283){\ellipse{150}{150}}
\put(315,1883){\blacken\ellipse{150}{150}}
\put(315,1883){\ellipse{150}{150}}
\put(315,683){\blacken\ellipse{150}{150}}
\put(315,683){\ellipse{150}{150}}
\put(915,1883){\blacken\ellipse{150}{150}}
\put(915,1883){\ellipse{150}{150}}
\put(915,683){\blacken\ellipse{150}{150}}
\put(915,683){\ellipse{150}{150}}
\put(4515,1283){\whiten\ellipse{150}{150}}
\put(4515,1283){\ellipse{150}{150}}
\put(915,1283){\whiten\ellipse{150}{150}}
\put(915,1283){\ellipse{150}{150}}
\thicklines
\path(1515,1283)(315,1883)
\path(556.495,1829.334)(315.000,1883.000)(502.830,1722.003)
\path(315,1883)(915,1883)
\path(675.000,1823.000)(915.000,1883.000)(675.000,1943.000)
\path(915,1883)(315,683)
\path(368.666,924.495)(315.000,683.000)(475.997,870.830)
\path(315,683)(1515,1283)
\path(1327.170,1122.003)(1515.000,1283.000)(1273.505,1229.334)
\dashline{150.000}(5115,2483)(4515,1883)
\path(4642.279,2095.132)(4515.000,1883.000)(4727.132,2010.279)
\dashline{150.000}(4515,1883)(3315,1883)
\path(3555.000,1943.000)(3315.000,1883.000)(3555.000,1823.000)
\dashline{150.000}(3315,1883)(5115,83)
\path(4902.868,210.279)(5115.000,83.000)(4987.721,295.132)
\dashline{150.000}(5115,83)(5115,2483)
\path(5175.000,2243.000)(5115.000,2483.000)(5055.000,2243.000)
\put(1140,1583){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}1}}}}
\put(540,2108){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}2}}}}
\put(15,833){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}3}}}}
\put(540,308){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}4}}}}
\put(2865,1958){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}3}}}}
\put(5340,158){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}4}}}}
\put(4215,1958){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}2}}}}
\put(5340,2258){\makebox(0,0)[lb]{\smash{{\SetFigFontNFSS{12}{14.4}{\rmdefault}{\mddefault}{\updefault}1}}}}
\end{picture}
}
    \caption{A polygon and its dual. \hfill \mbox{}
      \mbox{Their lengths are $1-1+1+1$ respectively $1+2+3+4$.}
      }
    \label{fig:dual}
  \end{center}
\end{figure}

\begin{theorem}[Poonen and Rodriguez-Villegas~\cite{Poonen12}]
  The sum of the lengths of an admissible polygon and its dual is $12$
  times the winding number of the polygon.
\end{theorem}

Heuristically, the winding number counts how many times a polygon
winds around the origin. Dual polygons will have equal winding
number. In this article, we will only be concerned with polygons of
winding number one.

\begin{figure}[htbp]
  \begin{center}
\setlength{\unitlength}{0.00041667in}

\begingroup\makeatletter\ifx\SetFigFont\undefined
\gdef\SetFigFont#1#2#3#4#5{
  \reset@font\fontsize{#1}{#2pt}
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}
  \selectfont}
\fi\endgroup
{\renewcommand{\dashlinestretch}{30}
\begin{picture}(1366,1381)(0,-10)
\put(1283,683){\blacken\ellipse{150}{150}}
\put(1283,683){\ellipse{150}{150}}
\put(1283,1283){\blacken\ellipse{150}{150}}
\put(1283,1283){\ellipse{150}{150}}
\put(1283,83){\blacken\ellipse{150}{150}}
\put(1283,83){\ellipse{150}{150}}
\put(83,683){\blacken\ellipse{150}{150}}
\put(83,683){\ellipse{150}{150}}
\put(83,1283){\blacken\ellipse{150}{150}}
\put(83,1283){\ellipse{150}{150}}
\put(83,83){\blacken\ellipse{150}{150}}
\put(83,83){\ellipse{150}{150}}
\put(683,683){\blacken\ellipse{150}{150}}
\put(683,683){\ellipse{150}{150}}
\put(683,1283){\blacken\ellipse{150}{150}}
\put(683,1283){\ellipse{150}{150}}
\put(683,83){\blacken\ellipse{150}{150}}
\put(683,83){\ellipse{150}{150}}
\thicklines
\path(83,683)(1283,1283)
\path(1095.170,1122.003)(1283.000,1283.000)(1041.505,1229.334)
\path(1283,1283)(683,83)
\path(736.666,324.495)(683.000,83.000)(843.997,270.830)
\path(683,83)(83,1283)
\path(243.997,1095.170)(83.000,1283.000)(136.666,1041.505)
\path(83,1283)(1283,683)
\path(1041.505,736.666)(1283.000,683.000)(1095.170,843.997)
\path(1283,683)(83,83)
\path(270.830,243.997)(83.000,83.000)(324.495,136.666)
\path(83,83)(83,683)
\path(143.000,443.000)(83.000,683.000)(23.000,443.000)
\end{picture}
}
    \caption{A polygon of winding number $-2$.}
    \label{fig:winding}
  \end{center}
\end{figure}

\begin{proof}[Proof of Lemma~\ref{lemma:step}.]
  Let $Q = P^{(1)}$, and note that by Lemma~\ref{lemma:shift}, $P
  \subseteq Q^{(-1)}$, and by Lemma~\ref{lemma:Qminus1integral},
  $Q^{(-1)}$ has integral vertices. Notice that the number of boundary
  lattice points of $Q$ is $b^{(1)}$, and let $b'$ be the number of
  boundary lattice points of $Q^{(-1)}$. Since $P$ and $Q^{(-1)}$ have
  the same interior lattice points and $P \subseteq Q^{(-1)}$, by
  Pick's Theorem $Q^{(-1)}$ has at least as many boundary lattice
  points as $P$; in other words, $b' \ge b$.

  For each of the vertices $v^{(1)}_1, \ldots, v^{(1)}_n$ of $Q$
  there is a corresponding vertex $v_1, \ldots, v_n$ of $Q^{(-1)}$.
  Consider the (possibly nonconvex, nonsimple) admissible polygon
  with vertices
  $v_1-v^{(1)}_1, \ldots, v_n-v^{(1)}_n$. It is admissible because
  there are no lattice points between $Q$ and $Q^{(-1)}$. One can
  think of it as what remains of $Q^{(-1)}$ when $Q$ shrinks to a
  point. Each segment measures the difference (with the correct sign)
  between the corresponding edges of $Q^{(-1)}$ and $Q$. I.e., the
  length of that polygon is precisely $b'-b^{(1)}$.

  \begin{figure}[htbp]
    \centering
    \includegraphics{shrink.eps}
    \caption{An admissible polygon from $(P,P^{(1)})$, and its dual.}
    \label{fig:shrink}
  \end{figure}

  Now the dual polygon will walk through the normal vectors of
  $Q$. Therefore all segments will count with positive length,
  and there cannot be less than $3$. Also, there is a unique one with
  $3$ segments, which is the dual to $3 \Delta$. Thus $b-b^{(1)} \le
  b'-b^{(1)} \le 12-3$ with equality only for multiples of $\Delta$.
\end{proof}

\begin{proof}[Proof of the Onion--Skin Theorem.]
  Induction on $\ell$.
  \begin{itemize}
    \item For $\ell=1$, the inequality $b \le 2i + 7$ was proved
    earlier.
    \item For $\ell=4/3$, we have $i=3$, and $P \subseteq 4
    \Delta$. So $b \le 12$.
    \item For $\ell=5/3$, we have $i=6$, and $P \subseteq 5
    \Delta$. So $b \le 15$.
    \item For $\ell=3/2$, Lemma~\ref{lemma:step} reads $b \le i + 8$
    which is stronger than what we need.
  \end{itemize}

  If $\ell \ge 2$, we have
  \begin{align*}
    (2\ell-1) b &\le (2\ell-1) b^{(1)} + 9(2\ell-1) \\
    &= 2 b^{(1)} + (2(\ell-1)-1) b^{(1)} + 9(2\ell-1) \\
    &\le 2 b^{(1)} + 2i^{(1)} + 9(\ell-1)^2 - 2 + 9(2\ell-1) \\
    &= 2i + 9\ell^2 - 2
  \end{align*}
\end{proof}

\subsection{Generalizing Scott's proof}
As in Subsection~\ref{ssec:scott}, we tightly fit $P$ into a box
$[0,p'] \times [0,p]$, with $p\le p'$. Let $q$ and $q'$ be the
length of the top and bottom edge (see Figure~\ref{fig:box}).
We again apply lattice equivalence transformations such that $p$ is
as small as possible, and that $P$ has points on the top and bottom
edges with horizontal distance smaller than or equal to $(p-q-q')/2$.
Again, we obtain the following inequalities:
\begin{align*}
b \le q+q'+2p \tag{\ref{eq:sc1}} \\
a \ge p(q+q')/2 \tag{\ref{eq:sc2}} \\
a \ge p(p+q+q')/4 \tag{\ref{eq:sc3}}
\end{align*}
Set $x:=p/\ell$ and $y:=(q+q')/\ell$. Then $x\ge 2$, because
passing to $P^{(1)}$ reduces the height at least by
2.\footnote{We also have $x \le 3$ with equality only for multiples of
  $\Delta$.}
From \eqref{eq:sc1} and \eqref{eq:sc2}, we get
\begin{align*}
\frac{2\ell b-2a-9\ell^2}{\ell^2} \le 2(q+q'+2p)/\ell - p(q+q')/\ell^2-9 \\
        = -xy+4x+2y-9 ,
\end{align*}
and from \eqref{eq:sc1} and \eqref{eq:sc3}, we get
\begin{align*}
\frac{4\ell b-4a-18\ell^2}{\ell^2} \le 4(q+q'+2p)/\ell - p(p+q+q')/\ell^2-18 \\
        = -x^2-xy+8x+4y-18.
\end{align*}

\begin{figure}[htbp]
  \begin{center}

    \includegraphics{plot.ps}
    \caption{At least one of the polynomials is $\le 0$.}
    \label{fig:plot}
  \end{center}
\end{figure}

For $x\ge 2$ and $y\ge 0$, at least one of the two polynomials
$p_1(x,y) = -xy+4x+2y-9$ and $p_2(x,y) = -x^2-xy+8x+4y-18$ is zero or
negative, as it can be seen in Figure~\ref{fig:plot}. (The two shaded
regions are where $p_1$ and $p_2$, respectively, take non-negative
values.) There is only
one point where both upper bounds reach zero, namely $(x,y)=(3,3)$,
and this is the only case where equality can hold in the Onion--Skin
Theorem. It is an exercise to show that equality actually holds only
for multiples of $\Delta$.

\section{Conclusion} \label{sec:conclusion} \noindent

\subsection{Summary of results.}
For a triple $(a,b,i)$ of numbers the following are equivalent.
\begin{itemize}
\item There is a convex lattice polygon $P$ with
  $(a,b,i)=(a(P),b(P),i(P))$.
\item $b \in \Z_{\ge 3}$, $i \in \Z_{\ge 0}, a = i + b/2 -1$, and
  \begin{itemize}
  \item[\scriptsize$\blacktriangleright$] $i=0$ or
  \item[\scriptsize$\blacktriangleright$] $i=1$ and $b \le 9$ or
  \item[\scriptsize$\blacktriangleright$] $i \ge 2$ and $b \le 2i+6$.
  \end{itemize}
\end{itemize}
Furthermore, if $\ell = \ell(P)$, then $(2\ell-1) b \le 2i + 9\ell^2 -
2$.

\subsection{Outlook.}
Is there a proof of the Onion--Skin Theorem using algebraic geometry?
Currently not (yet). The toric dictionary between
polygons and
algebraic varieties also does not (yet) have an algebraic geometry term
for the level of a polygon. A first step in this direction is the
use of the {\em process\/} of peeling off onion skins -- or rather
its algebraic geometry analogue -- for the simplification of
the rational parametrization of an algebraic surface~\cite{Schicho:03a}.

In any case, the Onion--Skin Theorem gives rise to a {\em conjecture\/}
in algebraic geometry, namely the inequality
$(2\ell-1)d\le 9\ell^2+4\ell(p-1)$ for any rational surface
of degree $d$, sectional genus $p$, and level $\ell$. Here the level
of an algebraic surface should be defined via the process of peeling
mentioned above. For toric surfaces, the inequality holds
by the Onion--Skin Theorem, but for nontoric rational surfaces
we do not have a proof (nor a counterexample).

We thank Ricky Pollack for his encouragement to write up this story.
We also thank Daniel J. Velleman and two anonymous referees who helped
us do so in a comprehensible way.

\bibliographystyle{plain}
\bibliography{alles,josef}

\setlength{\parindent}{0pt}

\end{document}

config: arxiv


TITLE: Question about showing bijection when proving an isomorphism.
QUESTION [0 upvotes]: Here is the problem

Prove that the additive group $\mathbb{Z}_6$ is isomorphic to the multiplicative group of nonzero elements in $\mathbb{Z}_7$.

I know the bijection is $f(x) \equiv 3^x \pmod{7}$ Proving it is an injection is easy, but I'm struggling to show how it's surjective given modulo operation. Is just listing out all the elements to show it's bijective considered bad for a proof? The rest of it is easy, as I know I just show that $f(a+b) \equiv f(a)f(b) \pmod{7}$

REPLY [0 votes]: Let $S, T$ are a finite set of same order. Then for a map, $f:S\mapsto T$   $f$ is injective $\Longleftrightarrow f$ is surjective. Now $\mid \Bbb Z_6\mid=6=\phi (7)=\mid \Bbb Z^×_7\mid$


TITLE: CY 3-folds are $T^2 \times \mathbb{R}$ fibrations over the base $\mathbb{R}^3$. What does it mean?
QUESTION [2 upvotes]: In this article at section 2. Toric geometry and Mirror Symmetry there is the statement that CY 3-folds are $T^2 \times \mathbb{R}$ fibrations over the base $\mathbb{R}^3$. Now, my questions refers to pages 3 and 4. Although I have some familiarity with fiber bundles I cannot figure out what this actually means.
Can you "dumb down" this for me? Explain what this fibration actually means (maybe using a familiar example) and tell me how they get this diagram of Fig. 1?

REPLY [2 votes]: There is a map $F:\mathbb C^3 \to \mathbb R^3$ given by
$$F (z_1, z_2, z_3) = (|z_1|^2 - |z_3|^2, |z_2|^2 - |z_3|^2, Im(z_1z_2z_3))$$
Then one can think of $\mathbb C^3$ as a "fibration" of $\mathbb R^3$:
$$\mathbb C^3 = \bigcup_{x\in \mathbb R^3} F^{-1}(x).$$
One can check that when $x$ does not lie in the "degenerate locus" in figure 1, then $F^{-1}(x)$ is diffeomorphic to $T^2 \times \mathbb R$. Note also that all $F^{-1}(x)$ are special Lagrangian. Special Lagrangian fibration are of great interest in mirror symmetry.
It is not true that all CY 3 folds are of this form locally. I do not know if this is true for toric CY 3 folds though.
Note that "fibration" in this case is not the same as a fiber bundle, as some of the fibers are not homotopic to each other.
I am no expert in this field, but to get an introduction to special Lagrangian fibration, you may take a look at "Calabi-Yau Manifolds and Related Geometries"


TITLE: Weaker notion of first-countability for filters?
QUESTION [5 upvotes]: While trying to understand the motivation behind the definition of a filter I've stumbled upon the following notion, let's call it "almost first-countability":
Let $x$ be a point of a set $X$ and $\mathcal{F}$ be a filter on $X$ such that each $F\in\mathcal{F}$ contains $x$. We say $\mathcal{F}$ is almost first-countable if there is a countable collection $\mathcal{C}=\{C_1,C_2,\dots\}$ of nested subsets of $X$ (so $C_1\supset C_2\supset\cdots$) such that $\Sigma(\mathcal{C})=\Sigma(\mathcal{F})$, where for any $\mathcal{X}\in P(P(X))$ the symbol $\Sigma(\mathcal{X})$ denotes the set of sequences $\phi\colon\mathbb{N}\to X$ converging with respect to the collection $\mathcal{X}$, i.e. each element of $\mathcal{X}$ contains a tail of $\phi$.
Clearly if $\mathcal{F}$ is first-countable, i.e. it can be generated by a countable filter base, then it is almost first-countable (just take $\mathcal{C}$ to be $\{C_1,C_1\cap C_2,C_1\cap C_2\cap C_3,\dots\}$ where $\{C_1,C_2,C_3,\dots\}$ is a countable filter base generating $\mathcal{F}$).
Question:
Is this notion strictly weaker than first-countability? In other words, are there a set $X$, a non-first-countable filter $\mathcal{F}$ at $x\in X$ and a countable collection $\mathcal{C}=\{C_1,C_2,\dots\}$ of nested subsets of $X$ such that $\Sigma(\mathcal{C})=\Sigma(\mathcal{F})$?
Observation:
Call $\mathcal{C}'$ the filter generated by the base $\mathcal{C}$ by adding all supersets of elements of $\mathcal{C}$. Then $\Sigma(\mathcal{C}')=\Sigma(\mathcal{C})$. It follows that the above question boils down to finding two filters at $x$ sharing the same set of convergent sequences and such that one is first-countable while the other is not (or proving they do not exist, of course).

As an exercise for myself let me rewrite user126154's brilliant argument in a more detailed fashion. In order to fix the ideas, instead of letting $\alpha$ be any non-$\omega$-cofinal limit ordinal I decided to choose a concrete example by picking up the smallest limit ordinal which is not $\omega$-cofinal. If I'm not mistaken this should be the first uncountable ordinal $\omega_1$.
Consider the first uncountable ordinal $\omega_1$ and the filter $\mathcal{F}_\infty$ on $\omega_1$ generated by the filter base $\mathcal{B}_\infty:=\{(a,\infty)\mid a\in \omega_1\}$ where $(a,\infty):=\{x\in \omega_1\mid x>a\}$. Observe that $\Sigma(\mathcal{F}_\infty)$ is empty: indeed, assume there is a sequence $\phi\colon\mathbb{N}\to \omega_1$ converging with respect to $\mathcal{F}_\infty$; by induction $\phi$ has a strictly increasing subsequence $\phi'$ also converging w.r.t. $\mathcal{F}_\infty$; the limit of this sequence is just $s:=\sup_{n\in\mathbb{N}}\{\phi'(n)\mid n\in\mathbb{N}\}$, i.e. a countable union of countable ordinals which is still countable, hence $s\in \omega_1$  and the interval $(s+1,\infty)$ does not contain any tail of $\phi'$, which is absurd. Further observe that $\mathcal{F}_\infty$ cannot be generated by a countable filter base, otherwise we could construct a sequence converging w.r.t. the base and hence also w.r.t. $\mathcal{F}$.
Now consider the set $X:=(\sqcup_{n\in\mathbb{N}} (\omega_1)_n)\cup\{\infty\}$ consisting of the disjoint union of countably many copies of $\omega_1$ together with a new point $\infty$, and construct the filter $\mathcal{F}$ on $X$ consisting of all subsets $F\subset X$ satisfying all the three conditions below:

$F\cap(\omega_1)_n\in\mathcal{F}_\infty$ for each $n\in\mathbb{N}$,
there is $j\in\mathbb{N}$ such that $F\cap(\omega_1)_k=\omega_1$ for each $k\geq j$,
$\infty\in F$.

Notice that $\mathcal{F}$ cannot be generated by a countable filter base, otherwise by restricting each element of the base to one copy of $\omega_1$ we would obtain a countable filter base for $\mathcal{F}_\infty$. Now consider the collection $\mathcal{C}=\{C_n\mid n\in\mathbb{N}\}$ of subsets of $X$ where $C_n$ is defined by the next three conditions:

$C_n\cap(\omega_1)_j=\emptyset$ for $j<n$,
$C_n\cap(\omega_1)_j=\omega_1$ for each $j\geq n$,
$\infty\in C_n$ for each $n\in\mathbb{N}$.

Clearly $\mathcal{C}$ satisfies the axioms of a filter base, and the filter that it generates is obviously first-countable.
Now we show that $\Sigma(\mathcal{F})=\Sigma(\mathcal{C})$. Let $x_n$ be a sequence in $X$ converging w.r.t. $\mathcal{C}$. Each $F\in\mathcal{F}$ contains some $C_n$ so $x_n$ converges w.r.t. $\mathcal{F}$ as well. Conversely, assume $x_n$ converges w.r.t. $\mathcal{F}$ and define $\sigma(n)$ to be the index of the copy of $\omega_1$ where $x_n$ lies (put $\sigma(n)=\star$ if $x_n=\infty$). If for each $n$ there is $j_n$ such that for each $k\geq j_n$ $\sigma(k)$ is either greater than $n$ or equal to $\star$ then $x_n$ obviously converges w.r.t. $\mathcal{C}$ and we are done. Otherwise $\sigma$ has a constant subsequence at some natural number, that is there is a subsequence $x_{n_k}$ staying in a single copy of $\omega_1$. Since $x_n$ converges w.r.t. $\mathcal{F}$ by assumption, then $x_{n_k}$ must converge w.r.t. $\mathcal{F}_\infty$ and this is impossible since $\Sigma(\mathcal{F}_\infty)=\emptyset$.
Conclusion: we have found a set $X$, two filters at some point $x\in X$ one of which is first-countable while the other is not (the former being the filter generated by $\mathcal{C}$, the latter being $\mathcal{F}$) sharing the same set of converging sequences in $X$. In other words, $\mathcal{F}$ is "almost first-countable" but not first-countable, therefore the notion spelled out above of "almost first-countability" is really different than first-countability (in particular, it is strictly weaker).

In order to avoid ambiguities I add the definitions used. A filter on $X$ is a nonvoid collection of subsets of $X$ such that:

the empty set does not belong to the collection,
any finite intersection of its elements is a superset of some element of the collection,
any superset of any of its elements belongs to the collection.

A filter base is like a filter but we drop the third assumption. By adding all supersets of all elements of a filter base $\mathcal{F}$ we obtain the filter $\mathcal{F}'$ generated by the base $\mathcal{F}$. It follows that a nonvoid collection $\mathcal{B}\in P(P(X))$ is a base of a given filter $\mathcal{F}$ if and only if $\mathcal{B}\subset\mathcal{F}$ and each $F\in\mathcal{F}$ contains some $B\in\mathcal{B}$.

REPLY [3 votes]: Let $X$ be almost-first-countable. Let $\mathcal C=\{C_i\}$ be a nested countable sequence of sets containing $x$ so that $\Sigma(\mathcal C)=\Sigma(\mathcal F)$.
Suppose there is $F\in\mathcal F$ so that $x\in F$ and no $C_i$ is contained in $F$.
In particular there is $x_i\in C_i$ such that $x_i\notin F$. Therefore the sequence $\{x_i\}$ converges to $x$ for $\mathcal C$ but does not converge to $x$ for $\mathcal F$. This contradicts the hypothesis of almost first countability.
We have so proved that for any $F\in\mathcal F$ there is $C_i\in \mathcal C$ such that $C_i\subset F$.
If our filter comes from a topology and the $C_i$'s are open sets, then the converse is true.
Let's now see an example to show that the converse is not necessarily true in general.
Let $\omega$ be the typer of order of $\mathbb N$.
For any ordinal $\alpha$ let $\mathcal F_\alpha$ be the filter given by the order relation. (i.e. the filter of neighborhoods of $\infty$.)
Choose a limit ordinal $\alpha$ which is not $\omega$-cofinal. That is to say, there is no sequence $x_i\stackrel{\mathcal F_\alpha}{\to}\infty$ (i.e. $\mathcal F_\alpha$ is not first-countable.)
Let $X$ be the set obtained by countable disjoint union of copies of $\alpha$, plus the point $\infty$.
$$X=(\bigsqcup_{i\in\mathbb N}\alpha_i)\cup\{\infty\}$$
where $\alpha_i$ is a copy of $\alpha$.
Therefore, any subset $A\subset X$ is of the form $\bigsqcup_{i=0}^\infty A_i$ with $A_i\subset \alpha_i$ for $i\in\mathbb N$ and $A_\infty\subset\{\infty\}$.
Let $\mathcal F$ be the filter defined as follows: A set $F\subset X$ is in $\mathcal F$ if and only if
$\exists n_0=n_0(F)$ s.t.
1) $F_i=\alpha_i$ for any $n>n_0$
2) $\forall n\leq n_0$ we have $\emptyset\neq F_i\in\mathcal F_{\alpha_i}$
moreover,
3) $\infty\in F$
CLAIM: $\mathcal F$ is not first-countable.
PROOF: by conditions 1) and 2), any countable filter base for $F$ would give, by restriction to $\alpha_i$, a countable filter base on $\alpha$, which is assumed not $\omega$-cofinal. $\square$
Now, let's define $\mathcal C$. This is easy, let
$$C_i=(\bigsqcup_{n\geq i} \alpha_i)\cup\{\infty\}$$
Clearly $\mathcal C$ is a countable filter.
CLAIM: for any sequence $(x_i)_{i\in\mathbb N}$ in $X$ we have
$$x_i\stackrel{\mathcal C}{\to}\infty \Leftrightarrow
x_i\stackrel{\mathcal F}{\to}\infty$$
PROOF. Let $x_i\stackrel{\mathcal C}{\to}\infty$. By condition $1)$ for all $F\in\mathcal F$ we have $x_i\in F$ for $i>n_0(F)$. Thus $x_i\stackrel{\mathcal F}{\to}\infty$.
On the other hand, suppose $x_i\stackrel{\mathcal F}{\to}\infty$
The point $x_i$ belongs to a set $\alpha_k$. Define $\sigma(i)$ so that
$$x_i\in\alpha_{\sigma(i)}$$
If $\sigma(i)$ does not converge to infinity, then we can find a subsequence $i_k$ so that $\sigma(i_k)$ is a constant natural number, say $n$.
Therefore the sequence $x_{i_k}$ is a sequence in $\alpha_n$ which converges to infinity, which is impossible because $\alpha$ is not $\omega$-cofinal.
Therefore $\sigma(i)\to \infty$ and thus   $x_i\stackrel{\mathcal C}{\to}\infty$. $\square$