84 lines
2.5 KiB
TeX
84 lines
2.5 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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\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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\end{proof}
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\end{document} |