Advanced handouts

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Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>

src/Advanced/Lattices/main.tex

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+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	singlenumbering
+]{../../../lib/tex/ormc_handout}
+\usepackage{../../../lib/tex/macros}
+
+\usepackage{ifthen}
+%\usepackage{lua-visual-debug}
+
+\renewcommand{\arraystretch}{1.2}
+
+\uptitlel{Advanced 2}
+\uptitler{\smallurl{}}
+\title{Lattices}
+\subtitle{Prepared by Mark on \today}
+
+\begin{document}
+
+	\maketitle
+
+	\input{parts/0 intro.tex}
+	\input{parts/1 minkowski.tex}
+	\input{parts/2 orchard.tex}
+
+\end{document}

src/Advanced/Lattices/meta.toml

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+[metadata]
+title = "Lattices"
+
+[publish]
+handout = true
+solutions = true

src/Advanced/Lattices/parts/0 intro.tex

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+\definition{}
+The \textit{integer lattice} $\mathbb{Z}^n \subset \mathbb{R}^n$ is the set of points with integer coordinates.
+
+\problem{}
+Draw $\mathbb{Z}^2$.
+
+\vfill
+
+
+\definition{}
+We say a set of vectors $\{v_1, v_2, ..., v_k\}$ \textit{generates} $\mathbb{Z}^n$ if every lattice point can be written uniquely as
+$$
+	a_1v_1 + a_2v_2 + ... + a_kv_k
+$$
+for integer coefficients $a_i$. \par
+It is fairly easy to show that $k$ must be at least $n$.
+
+\problem{}
+Which of the following generate $\mathbb{Z}^2$?
+\begin{itemize}
+	\item $\{ (1,2), (2,1) \}$
+	\item $\{ (1,0), (0,2) \}$
+	\item $\{ (1,1), (1,0), (0,1) \}$
+\end{itemize}
+
+\begin{solution}
+	Only the last.
+\end{solution}
+
+\vfill
+
+\problem{}
+Find a set of two vectors that generates $\mathbb{Z}^2$. \\
+Don't say $\{ (0, 1), (1, 0) \}$, that's too easy.
+
+\vfill
+
+\problem{}
+Find a set of vectors that generates $\mathbb{Z}^n$.
+
+
+
+\vfill
+\pagebreak
+
+\definition{}
+A \textit{fundamental region} of a lattice is the parallelepiped spanned by a generating set. The exact shape of this region depends on the generating set we use.
+
+\problem{}
+Draw two fundamental regions of $\mathbb{Z}^2$ using two different generating sets. Verify that their volumes are the same.
+
+\vfill
+\pagebreak

src/Advanced/Lattices/parts/1 minkowski.tex

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+\section{Minkowski's Theorem}
+
+\theorem{Blichfeldt's Theorem}
+Let $X$ be a finite connected region. If the volume of $X$ is greater than $1$, $X$ must contain two distinct points that differ by an element of $\mathbb{Z}^n$. In other words, there exist distinct $x, y \in X$ so that $x - y \in \mathbb{Z}^n$.
+
+\vspace{2mm}
+
+Intuitively, this means that you can translate $X$ to cover two lattice points at the same time.
+
+
+\problem{}
+Draw a region in $\mathbb{R}^2$ with volume greater than 1 that contains no lattice points. Find two points in that region which differ by an integer vector.
+\hint{Area is two-dimensional volume.}
+
+\vfill
+
+
+\problem{}
+The following picture gives an idea for the proof of Blichfeldt's theorem in $\mathbb{Z}^2$. Explain the picture and complete the proof.
+
+\begin{center}
+	\includegraphics[angle=90,width=0.5\linewidth]{proof.png}
+\end{center}
+
+\begin{solution}
+	The fundamental region of $\mathbb{Z}^2$ tiles the plane. Translate these tiles by lattice vectors to stack them on the fundamental region. Then since the union of the intersections of X with these tiles has area greater 1 and they are stacked on a region of area 1, there must be an overlap by a generalization of the pigeonhole principle (if there were no overlap then the sum of the areas would be less than or equal to 1). Take points $x, y$ in the overlap. Then $x - y$ is a lattice point corresponding to the difference in translates, which were lattice points. Hence, $x - y \in \mathbb{Z}^2$.
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+%\problem{}
+%Does your proof of Blichfeldt's theorem in $\mathbb{Z}^2$ extend to a proof of Blichfeldt's theorem in $\mathbb{Z}^n$?
+%\vfill
+
+
+\problem{}
+Let $X$ be a region $\in \mathbb{R}^2$ of volume $k$. How many integral points must $X$ contain after a translation?
+
+\begin{solution}
+	$\lceil k \rceil$
+\end{solution}
+
+\vfill
+
+\definition{}
+A region $X$ is \textit{convex} if the line segment connecting any two points in $X$ lies entirely in $X$.
+
+\problem{}
+\begin{itemize}
+	\item Draw a convex region in the plane.
+	\item Draw a region that is not convex.
+\end{itemize}
+
+\vfill
+\pagebreak
+
+
+
+\definition{}
+We say a region $X$ is \textit{symmetric} if for all points $x \in X$, $-x$ is also in $X$.
+
+\problem{}
+\begin{itemize}
+	\item Draw a symmetric region.
+	\item Draw an asymmetric region.
+\end{itemize}
+
+\vfill
+
+\theorem{Minkowski's Theorem}<mink>
+Every convex set in $\mathbb{R}^n$ that is symmetric with respect to the origin and which has a volume greater than $2^n$ contains an integral point that isn't zero.
+
+
+\problem{}
+Draw a few sets that satisfy \ref{mink} in $\mathbb{R}^2$. \par
+What is the simplest region that has the properties listed above?
+
+\vfill
+
+
+\problem{}
+Let $K$ be a region in $\mathbb{R}^2$ satisfying \ref{mink}. \par
+Let $K'$ be this region scaled by $\frac{1}{2}$.
+
+\begin{itemize}
+	\item How does the volume of $K'$ compare to $K$?
+	\item Show that the sum of any two points in $K'$ lies in $K$ \hint{Use convexity.}
+	\item Apply Blichfeldt's theorem to $K'$ to prove Minkowski's theorem in $\mathbb{R}^2$.
+\end{itemize}
+
+\vfill
+
+\problem{}
+Let $K$ be a region in $\mathbb{R}^n$ satisfying \ref{mink}. Scale this region by $\frac{1}{2}$, called $K' = \frac{1}{2}K$.
+
+\begin{itemize}
+	\item How does the volume of $K'$ compare to $K$?
+	\item Show that the sum of any two points in $K'$ lies in $K$
+	\item Apply Blichfeldt's theorem to $K'$ to prove Minkowski's theorem.
+\end{itemize}
+
+\begin{solution}
+	\begin{itemize}
+		\item The volume of $K'$ is $\frac{1}{2^n}$ the volume of $K$.
+		\item Take $x, y \in K'$. It follows that $2x, 2y \in K$. Since $K$ is convex, we have that the midpoint of the line segment between $2x$ and $2y$ is in $K$, and so $\frac{2x + 2y}{2} = x + y \in K$.
+		\item Since the volume of $K$ is greater than $2^n$, we have the volume of $K'$ is greater than one.
+		Applying Blichfeldt's theorem, we can find two distinct points $x, y \in K'$ such that $x - y \in \mathbb{Z}^n$. Since $K'$ is symmetric with respect to the origin, we have that $-y \in K'$. Therefore, $x + (-y) \in K$ by the previous part. $x \neq y, x - y \neq 0$, so we have found a nontrivial integer point in $K$.
+
+	\end{itemize}
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Lattices/parts/2 orchard.tex

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+\section{Polya's Orchard Problem}
+
+You are standing in the center of a circular orchard of integer radius $R$.
+A tree of radius $r$ has been planted at every integer point in the circle.
+If $r$ is small, you will have a clear line of sight through the orchard.
+If $r$ is large, there will be no clear line of sight through in any direction:
+
+\begin{center}
+	\hfill
+	\begin{tikzpicture}[scale=0.4]
+		\draw[color=black, very thick] (0,0) circle (5);
+
+		\draw[fill=black] (0,0) circle (0.2);
+		\draw [->, very thick](0,0) -- (0.7, 0.7);
+		\draw [->, very thick](0,0) -- (-0.7, 0.7);
+
+		\foreach \x in {-4, ..., 4} {
+		\foreach \y in {-4, ..., 4} {
+		\ifthenelse{
+			% Remove center
+			\( \x=0 \AND \y=0 \) \OR
+			%
+			% Remove corners
+			\( \x=4 \AND \y=-4 \) \OR
+			\( \x=4 \AND \y=4 \) \OR
+			\( \x=-4 \AND \y=4 \) \OR
+			\( \x=-4 \AND \y=-4 \) \OR
+			%
+			% Remove corner corners
+			\( \x=4 \AND \y=-3 \) \OR
+			\( \x=4 \AND \y=3 \) \OR
+			\( \x=-4 \AND \y=3 \) \OR
+			\( \x=-4 \AND \y=-3 \) \OR
+			\( \x=3 \AND \y=-4 \) \OR
+			\( \x=3 \AND \y=4 \) \OR
+			\( \x=-3 \AND \y=4 \) \OR
+			\( \x=-3 \AND \y=-4 \)
+		}{}{
+			\draw[fill=teal] (\x,\y) circle (0.4);
+		}}}
+
+		\draw[fill=teal] (0, 5) circle (0.4);
+		\draw[fill=teal] (5, 0) circle (0.4);
+		\draw[fill=teal] (0, -5) circle (0.4);
+		\draw[fill=teal] (-5, 0) circle (0.4);
+	\end{tikzpicture}
+	\hfill
+	\begin{tikzpicture}[scale=0.4]
+		\draw[color=black, very thick] (0,0) circle (5);
+
+		\draw[fill=black] (0,0) circle (0.2);
+
+		\draw [->, very thick](0,0) -- (-5,3.7);
+		\draw [->, very thick](0,0) -- (5, 3.7);
+
+		\foreach \x in {-4, ..., 4} {
+		\foreach \y in {-4, ..., 4} {
+		\ifthenelse{
+			% Remove center
+			\( \x=0 \AND \y=0 \) \OR
+			%
+			% Remove corners
+			\( \x=4 \AND \y=-4 \) \OR
+			\( \x=4 \AND \y=4 \) \OR
+			\( \x=-4 \AND \y=4 \) \OR
+			\( \x=-4 \AND \y=-4 \) \OR
+			%
+			% Remove corner corners
+			\( \x=4 \AND \y=-3 \) \OR
+			\( \x=4 \AND \y=3 \) \OR
+			\( \x=-4 \AND \y=3 \) \OR
+			\( \x=-4 \AND \y=-3 \) \OR
+			\( \x=3 \AND \y=-4 \) \OR
+			\( \x=3 \AND \y=4 \) \OR
+			\( \x=-3 \AND \y=4 \) \OR
+			\( \x=-3 \AND \y=-4 \)
+		}{}{
+			\draw[fill=teal] (\x,\y) circle (0.2);
+		}}}
+
+		\draw[fill=teal] (0, 5) circle (0.2);
+		\draw[fill=teal] (5, 0) circle (0.2);
+		\draw[fill=teal] (0, -5) circle (0.2);
+		\draw[fill=teal] (-5, 0) circle (0.2);
+	\end{tikzpicture}
+	\hfill ~
+\end{center}
+
+\problem{}
+Show that you will have at least one clear line of sight if $r < \frac{1}{\sqrt{R^2 + 1}}$. \par
+\hint{Consider the line segment from $(0, 0)$ to $(R, 1)$. Calculate the distance from the closest integer points to the ray.}
+
+\begin{solution}
+	Consider the ray from the origin to the point $(R, 1)$.
+
+	The two lattice points closest to this ray are $(1, 0)$ and $(R - 1, 1)$. Say the distance from each of these points to the ray is $\delta$. \par
+
+	Now, consider the triangle with vertices $(0, 0)$, $(1, 0)$, and $(R, 1)$. The area of this triangle is $\frac{1}{2}$.
+
+	The area of this triangle is also equal to $\frac{1}{2} \delta \sqrt{R^2 + 1}$. By algebra, $\delta = \frac{1}{\sqrt{R^2+1}}$. \par
+
+	Therefore, if $r < \frac{1}{\sqrt{R^2+1}}$, we will have a clear line of sight given by this ray.
+\end{solution}
+
+
+\vfill
+\pagebreak
+
+\problem{}
+Show that there is no line of sight through the orchard if $r > \frac{1}{R}$. You may want to use the following steps:
+\begin{itemize}
+	\item Show that there is no line of sight if $r \geq 1$.
+	\item Suppose $r < 1$ and $r > \frac{1}{R}$. Then, $R \geq 2$. Choose a potential line of sight passing through an arbitrary point $P$ on the circle. Thicken this line of sight equally on both sides into a rectangle of width $2r$ tangent to $P$ and $-P$. From here, use Minkowski's theorem to get a contradiction. Don't forget
+	to rule out any lattice points that sit outside the orchard but inside the rectangle.
+\end{itemize}
+
+% Any interest in counting the number of trees in the orchard? If so, google the Gauss circle problem. If orchards are not your slice of fruit, maybe rational approximations are your cup of tea
+
+\begin{solution}
+	Suppose $r < 1$ and let $L$ be a potential line of sight. Consider the rectangle of width $2r$ tangent to $P$ and $-P$. Then this is convex and symmetric with respect to the origin. Its area is $(2R)(2r) > 4 \frac{R}{R} = 4$. By Minkowski, we have a nonzero integral point in this rectangle. Suppose first that the integer point is within the orchard. Then this means that there is a tree whose distance to the line is at most $r$. Therefore, this tree blocks the line of sight. Now notice that there is a part of this rectangle that sits outside the orchard. Can the integer point be in this region? This would mean its distance to the origin, $D$, would satisfy $D > R$. Now since this point is within a distance of $r$ of our line $L$, we have that $D < \sqrt{R^2 + r^2} < \sqrt{R^2 + 1}$. So we have that $R < D < \sqrt{R^2 + 1}$. Then $R^2 < D^2 < R^2 + 1$, but $D^2$ is an integer so this is impossible.
+\end{solution}
+
+\vfill
+
+\problem{Challenge}
+Prove that there exists a rational approximation of $\sqrt{3}$ within $10^{-3}$ with denominator at most $501$. Come up with an upper bound for the smallest denominator of a $\epsilon$-close rational approximation of any irrational number $\alpha > 0$. Your bound can have some dependence on $\alpha$ and should get smaller as $\alpha$ gets larger. \\
+\hint{Use the orchard.}
+
+\begin{solution}
+	Take the line through the origin of slope $\sqrt{3}$. We would like an orchard for which $r = 10^{-3}$ gives no line of sight, since this will guarantee an integer point within a distance of $10^{-3}$. Then by our previous problem, we can take $10^{-3} > \frac{1}{R}$, so take $R > 1000$. Now since this line intersects the boundary of the orchard at $( \frac{R}{2}, \frac{\sqrt{3R}}{2} )$, we have that the $x$-coordinate is at most $\frac{R}{2} = 501$. Then we have that our lattice point $(x, y)$ satisfies $\sqrt{3x} - 10^{-3} < y < \sqrt{3x} + 10^{-3}$, so $\sqrt{3} - 10^{-3} < \frac{x}{y} < \sqrt{3} + \frac{10^{-3}}{x}$. Therefore, $\frac{y}{x}$ is a rational approximation that is $10^{-3}$-close to $\sqrt{3}$ and has denominator at most $501$. Notice that we got closer than we need to.
+
+	Repeating this same process, our upper bound for the denominator of an $\epsilon$-close approximation of $\alpha$ is $\frac{\text{cos}({\text{atan}({\alpha})})}{\epsilon} = \frac{1}{\sqrt{\alpha^2 + 1 \epsilon}}$.
+\end{solution}
+
+\vfill
+\pagebreak
+