294 lines
14 KiB
TeX
294 lines
14 KiB
TeX
\documentclass[a4paper,10pt]{amsart}
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\usepackage[utf8]{inputenc}
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\usepackage{amsfonts}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{tikz}
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\usepackage{tikz-cd}
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\usetikzlibrary{positioning}
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\usetikzlibrary{arrows.meta}
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\usetikzlibrary{decorations.pathmorphing}
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\usetikzlibrary{shapes.multipart}
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\usepackage{hyperref}
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\newtheorem{theorem}{Theorem}
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\newtheorem{lemma}{Lemma}
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\newtheorem{definition}{Definition}
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\newcommand{\CC}{\mathbb{C}}
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\newcommand{\RR}{\mathbb{R}}
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\newcommand{\DD}{\mathbb{D}}
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\newcommand{\BB}{\mathbb{B}}
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\newcommand{\PP}{\mathbb{P}}
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\newcommand{\s}{\mathbb{S}}
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\newcommand{\ZZ}{\mathbb{Z}}
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%opening
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\title{Complement of a projective curve}
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\author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz}
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\begin{document}
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\maketitle
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\begin{abstract}
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\end{abstract}
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\section{Definition of the problem}
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Given $C$ a proyective curve in $\PP^2$ over $\CC$, we know that the complement of the curve has the homotopy type of a CW-complex of real dimension $2$, our problem is how to find an effective model for said CW-complex.
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\section{Smooth curves}
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Given a smooth curve of degree $d$ in $\PP^2$, we know it is isomorphic to a curve of the form
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\begin{equation}
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C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\}
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\end{equation}
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Therefore, we will work with $C_d$, from now on.
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We will denote:
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\begin{eqnarray}
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M_d&=&\{(x,y,z)\in\CC | x^d+y^d+z^d=1\} \\
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S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
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K &=&\{[x:y:z]\in\PP^2 | x^d, y^d, z^d\in\RR_{\geq0}\}
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\end{eqnarray}
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We then have a commutative diagram:
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\begin{equation*}
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\begin{array}{ccc}
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S_d & \longrightarrow & M_d \\
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\downarrow & & \downarrow \\
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K & \longrightarrow & \PP^2\setminus C_d \\
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\end{array}
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\end{equation*}
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The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma:
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\begin{theorem}
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$S_d$ is a strong deformation retract of $M_d$.
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\end{theorem}
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\begin{proof}
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We start by getting a retraction $r:M_d\longrightarrow S_d$ for $d=1$, and then we'll see that we can lift it.
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We can construct $r$ in two steps:
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\begin{equation*}
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\begin{array}{cccccc}
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r:&M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
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\end{array}
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\end{equation*}
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Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow S_1$. Then $r_1$ is given by:
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\begin{equation}
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\begin{array}{ccc}
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r_1: M_1\times I & \longrightarrow & M_1\cap\RR^3 \\
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(x_r+ix_i,y_r+iy_i,z_r+iz_i, t) & \longrightarrow & (x_r+(1-t)ix_i,y_r+(1-t)iy_i,z_r+(1-t)iz_i) \\
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\end{array}
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\end{equation}
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$M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)\in\RR_{\geq0}^3|x+y+z=1\}=\{(x,y,z)\in[0,1]^3|x+y+z=1\}$, this is, $S_1$ is a triangular section of the plane defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described below, and we will retract each of them separately:
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
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\draw (3,4.5) to (-6,-9);
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\draw (-3,4.5) to (6,-9);
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\draw (-7,-4.5) to (7,-4.5);
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\node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
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\node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
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\node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
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\node[right] at (0,0) {(0,0,1)};
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\node[right,above] at (3,-4.5) {(0,1,0)};
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\node[left,above] at (-3,-4.5) {(1,0,0)};
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\end{tikzpicture}
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\end{center}
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\end{figure}
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We denote each section as $(\pm,\pm,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ would be $(+,+,+)$.
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The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
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\begin{equation}
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\begin{array}{ccc}
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r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
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(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
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\end{array}
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\end{equation}
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Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
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\begin{equation}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(x,y,z)&(+,+,+) \\
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(1,0,0)&(+,-,-) \\
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(0,1,0)&(-,+,-) \\
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(0,0,1)&(-,-,+) \\
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(\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
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(\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
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(0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
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\end{array}\right.
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\end{equation}
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We need to see now that this point is well defined, for this, we'll see that in the borders of the different parts it coincides. We will see this for the part $(+,+,-)$, for the rest of permutations it is equivalent.
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We start with the border between $(+,+,-)$ and $(+,+,+)$, this is, the line with $z=0$. In this case
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\begin{equation*}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(x,y,0)&(+,+,+) \\
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(\frac{x}{1-0},\frac{y}{1-0},0)=(x,y,0)&(+,+,-) \\
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\end{array}\right.
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\end{equation*}
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For the border between $(+,+,-)$ and $(-,+,-)$, the border is the line $x=0$, so we get
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\begin{equation*}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(\frac{0}{1-z},\frac{y}{1-z},0)=(0,1,0)&(+,+,-) \\
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(0,1,0)&(-,+,-) \\
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\end{array}\right.
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\end{equation*}
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since when $x=0$ the equation of the plane is $y+z=1$ and then $y=1-z$. And lastly, the border between $(+,+,-)$ and $(+,-,-)$, where $y=0$
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\begin{equation*}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(\frac{x}{1-z},\frac{0}{1-z},0)=(1,0,0)&(+,+,-) \\
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(1,0,0)&(+,-,-) \\
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\end{array}\right.
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\end{equation*}
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similarly to the previous case, in this line $x=1-z$.
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Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we will try to lift them in order to get a retraction from $M_d$ to $S_d$.
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We have the following diagram:
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\begin{equation*}
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\begin{array}{ccccc}
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M_d & \longrightarrow & M_d\cap\{x^d, y^d, z^d\in\RR\}&\longrightarrow& S_d \\
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\downarrow & & \downarrow & & \downarrow\\
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M_1 & \longrightarrow & M_1\cap\RR^3 & \longrightarrow & S_1 \\
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\end{array}
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\end{equation*}
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We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, we can write it as $r_1((x,y,z),t)=(\bar{r}(x,t),\bar{r}(y,t),\bar{r}(z,t))$ with $\bar{r}(x,t)+\bar{r}(y,t)+\bar{r}(z,t)=1$ whenever $(x,y,z)\in M_1$. Then, it suffices to find $G(x,t)$ such that $G(x,t)^d=\bar{r}(x^d,t)$.
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%Since $r_1$ works componentwise, we will study the lifting also for each component separetely. Since $r_1$ is the same for each component, we see it for only one of them, the rest will be analogous.
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%For the first component, $r_1(x,t)=x_r+(1-t)ix_i$, and we have a map from $M_d$ to $M_1$ given by $F(x)=x^d$.
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%$r_1$ is the vertical projection over the real axis of the complex plane. In this first component, $M_d$ is a ramified cover over $M_1$, where $0$ is the only point where it ramifies. Therefore, outside of $0$ we have a cover, and then $r_1$ has a unique lifting outside of $0$. Since $r_1$ is constant on $0$, we get a lifting of $r_1$ by defining $\tilde{r}_1(0,t)=0$.
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In order to see this, we can write an explicit expression of $\bar{r}$. For a point $x\in\CC$ the action of $\bar{r}$ is the projection over the real axis, as seen in figure \ref{realprojection}.
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (-5,0) to (5,0);
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\draw (0,5) to (0,-1);
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\draw[fill] (3,2) circle (2pt) node[above]{$x_r+ix_i$};
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\draw[dashed] (3,2) to (3,0);
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\end{tikzpicture}
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\end{center}
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\caption{projection over the real axis}
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\label{realprojection}
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\end{figure}
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we can then write the retraction as $r_1=m(z,t)e^{2i\pi\theta(t)}$ where $m(t)$ is the function indicating how the modulus of $\bar{r}(x,t)$ changes and $\theta(t)$ portrays the change of the argument.
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For a $z=z_r+iz_i\in\CC$, $\bar{r}(z,t)=z_r+(1-t)iz_i$, we can write $m(z,t)=+\sqrt{z_r^2+(1-t)^2z_i^2}$ and $\theta(t)=\arctan\frac{(1-t)z_i}{z_r}$.
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We can divide $\CC$ en $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}.
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (-5,0) to (5,0);
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\draw (5,1.5) to (-5,-1.5);
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\draw (5,3.2) to (-5, -3.2);
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\draw (4.7,5) to (-4.7,-5);
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\draw (2.8,5) to (-2.8,-5);
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\draw (0.9,5) to (-0.9,-5);
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\draw (-0.9,5) to (0.9,-5);
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\draw (-2.8,5) to (2.8,-5);
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\draw (-4.7,5) to (4.7,-5);
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\draw (-5,3.2) to (5, -3.2);
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\draw (-5,1.5) to (5,-1.5);
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\end{tikzpicture}
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\end{center}
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\caption{Division of the complex plane}
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\label{divisionC}
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\end{figure}
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Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$.
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The homeomorphisms are given by
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\begin{equation}
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f_k(|z|\exp^{i\theta})=|z|^\frac{1}{d}\exp^{i\frac{\theta}{d}+(k-1)i\frac{\pi}{d}}\text{ for }k=1,3,5,\dots,2d-1,
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\end{equation}
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\begin{equation}
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f_k(|z|\exp^{i\theta})=|z|^\frac{1}{d}\exp^{i\frac{\theta-\pi}{d}+(k-1)i\frac{\pi}{d}}\text{ for }k=2,4,6,\dots,2d
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\end{equation}
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We will also need the inverses of these homeomorphisms, which are:
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\begin{equation}
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f_k^{-1}(|z|\exp^{i\theta})=|z|^d\exp^{id(\theta-(k-1)\frac{\pi}{d})}\text{ for }k=1,3,5,\dots,2d-1,
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\end{equation}
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\begin{equation}
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f_k^{-1}(|z|\exp^{i\theta})=|z|^d\exp^{id(\theta-\pi-(k-1)i\frac{\pi}{d})}\text{ for }k=2,4,6,\dots,2d
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\end{equation}
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Then, we define $G=f_k\circ\bar{r}\circ f_k^{-1}$, where $k$ depends on the point $z$ to which we will apply it. For $z\in\CC$, given by $z=|z|\exp^{i\theta}$ for $\theta\in[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$, we have:
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\begin{equation}
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\begin{array}{ccc}
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G(z,t)&=&f_k(H(f_k^{-1}(z),t))\\
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&=&f_k(H(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},t))\\
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&=&f_k(+\sqrt{(z^d)_r^2+(1-t)^2(z^d)_i^2}\exp^{i\arctan((1-t)\tan(d(\theta-(k-1)\frac{\pi}{d})))})\\
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&=&+\sqrt[2d]{(z^d)_r^2+(1-t)^2(z^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))}
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\end{array}
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\end{equation}
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Then, we can see that $G(z,t)^d=H(z^d,t)$, since:
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\begin{equation}
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\begin{array}{ccc}
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G(z,t)^d&=&m(z,t)^d\exp^{i\arctan((1-t)\tan d\theta)}\\
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||& & \\
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H(z^d,t)&=&m(z^d,t)\exp^{i\arctan((1-t)\tan d\theta)}\\
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\end{array}
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\end{equation}
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Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d, y^d, z^d\in\RR\}$ into $S_d$.
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For a point $(x,y,z)\in M_d\cap\{x^d, y^d, z^d\in\RR\}$, the corresponding point in $M_1\cap\RR^3$ is $(x^ d,y^d,z^d)$ and therefore the goal points of the retraction $r_2$ will be
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\begin{equation}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(x^d,y^d,z^d)&(+,+,+) \\
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(1,0,0)&(+,-,-) \\
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(0,1,0)&(-,+,-) \\
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(0,0,1)&(-,-,+) \\
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(\frac{x^d}{1-z^d},\frac{y^d}{1-z^d},0)&(+,+,-) \\
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(\frac{x^d}{1-y^d},0,\frac{z^d}{1-y^d})&(+,-,+) \\
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(0,\frac{y^d}{1-x^d},\frac{z^d}{1-x^d})&(-,+,+) \\
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\end{array}\right.
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\end{equation}
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In the $(+,+,+)$ case, $r_2$ is the constant path, and therefore the lift $\tilde{r}_2$ will also be the constant path.
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In the $(-,-,+)$ case, $r_2(x^d,y^d,z^d)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x,y,z)$. Since $x,y$ are negative real numbers multiplied by some roots of unity, the retraction will push them to $0$, while we will push $z$ to a norm 1 complex number in one of the $d$ copies of $\RR$ that we have in the preimage of the complex plane by $F$. We can write this as $\tilde{r}_2(x,y,z)=(\sqrt[d]{1-t}x,\sqrt[d]{1-t}y,\sqrt[d]{(1-t)z^d+t})$. For the permutations of this case it is equivalent.
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Lastly, for the case $(+,+,-)$ and it's permutations, we can write the lifting as $\tilde{r}_2(x,y,z,t)=(\sqrt[d]{1-t+\frac{t}{1-z^d}}x,\sqrt[d]{1-t+\frac{t}{1-z^d}}y,\sqrt[d]{1-t}z)$. We can see by summing the coordinates to the power $d$ that they sum $1$, so the retraction doesn't leave $M_1$.
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All that is left to see is that the formulas for each case coincide on the borders. For the border of $(+,+,-)$ and $(-,+,-)$, we have that:
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\begin{equation*}
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\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}=\sqrt[d]{(1-t+\frac{t}{y^d})y^d}=\sqrt[d]{y^d-ty^d+t}=\sqrt[d]{y^d(1-t)+t}
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\end{equation*}
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The first root corresponds to $\tilde{r}_2$ in $(+,+,-)$ and the last one corresponds to the lifting in $(-,+,-)$,so we see that it coincides. It would be similar in the rest of borders.
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\end{proof}
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We have then proved that $S_d$ is a strong deformation retract of $M_d$. We denote as $\Delta$ the 2-simplex $[v_0, v_1, v_2]$, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
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\begin{equation*}
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\begin{array}{ccccc}
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\Delta & \longleftrightarrow & S_1 & & \\
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\uparrow& & \uparrow & & \\
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\Delta^d & \longleftrightarrow & S_d & \longrightarrow & M_d \\
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& & \downarrow & & \downarrow \\
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& & K & \longrightarrow & \PP^2\setminus C_d \\
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\end{array}
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\end{equation*}
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where $\Delta^d$ is a simplicial complex that we are trying to find and identify with $S_d$, equivalently as what we have for $d=1$.
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Let $\phi: S_d\longleftarrow S_1$ be the map given by $\phi(x,y,z)=(x^d, y^d, z^d)$, then necessarily $\phi^{-1}(\Delta)=\Delta^d$.
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In particular, the $d^3$ vertices of the simplicial complex $\Delta^d$ will be identified to $(\xi_d,0,0)$, $(0,\xi_d,0)$, $(0,0,\xi_d)$ for all $\xi_d$ the $d$-$th$ roots of unity.
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The edges of $\Delta$ can be identified with the lines of $S_1$ parametrized as $(1-t,t,0)$, $(1-t,0,t)$ and $(0,t,1-t)$ with $t\in[0,1]$. Then, the preimages of these lines by $\phi$ are of the form $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$ in the first case and analogous in the rest, where we denote $\sqrt[s]{x}$ the positive real $d$-root for $x$ real, and where $\xi_i$ and $\xi_j$ are two any $d$-$th$ roots of unity. For each of these preimages we can define an edge between the corresponding vertices in $\Delta^d$.
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The $2$-faces of the simplicial complex $\Delta^d$ will then be identified to the areas between the lines $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$, $(\sqrt[d]{1-t}\xi_i,0, \sqrt[d]{t}\xi_k)$ and $(0,\sqrt[d]{t}\xi_j, \sqrt[d]{1-t}\xi_k)$.
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\end{document} |