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complementarioconica.tex

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+\documentclass[a4paper,10pt]{amsart}
+\usepackage[utf8]{inputenc}
+\usepackage{amsfonts}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage{tikz}
+\usepackage{tikz-cd}
+\usetikzlibrary{positioning}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{decorations.pathmorphing}
+
+\usepackage{hyperref}
+
+\newtheorem{theorem}{Theorem}
+\newtheorem{lemma}{Lemma}
+\newtheorem{definition}{Definition}
+\newcommand{\CC}{\mathbb{C}}
+\newcommand{\RR}{\mathbb{R}}
+\newcommand{\DD}{\mathbb{D}}
+\newcommand{\BB}{\mathbb{B}}
+\newcommand{\PP}{\mathbb{P}}
+\newcommand{\s}{\mathbb{S}}
+
+
+%opening
+\title{Complement of a projective curve}
+\author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz}
+
+\begin{document}
+
+\maketitle
+
+\begin{abstract}
+
+\end{abstract}
+\section{Definition of the problem}
+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.
+
+\section{Complement of a conic projective curve}
+We start with the case of the conic curve of degree $2$. We know that all projective conic curves are topologicaly equivalent to each other, this is, we have automorphisms of the projective plane that take one conic to any other conic.
+
+We take $C=\{x^2+y^2+z^2=0\}$. We have a natural inclusion of $\RR\PP^2$ into $\PP^2\setminus C$, since there are no real solutions of $x^2+y^2+z^2=0$. At the same time, we have $\PP^1\cong \s^2$ is a universal cover of $\RR\PP^2$. we are going to see now that $\PP^1\times\PP^1\setminus\Delta$ is a universal cover of $\PP^2\setminus C_1$, where $C_1=\{4xz=y^2\}$ is an equivalent curve to $C$ and $\Delta$ is the diagonal of $\PP^1\times\PP^1$. Let $\phi:\CC\times\CC\rightarrow\CC^2$ be the map given by $(t,s)\rightarrow(ts,t+s)$, since $\CC\times\CC\in\PP^1\times\PP^1$ and $\CC^2\in\PP^2$, we can homogenize the coordinates and get a map $\Phi:\PP^1\times\PP^1\rightarrow\PP^2$ be the map given by $([t_1:s_1], [t_2:s_2]) \rightarrow[t_1s_1:t_1s_2+t_2s_1:t_2s_2]$.
+Sea $x=([t:1],[t:1])\in\Delta$, then $\Phi(x)=[t^2:t:1]\in C_1$, so we have a covering map from $\PP^1\times\PP^1\setminus\Delta$ to $\PP^2\setminus C_1$. To see that we have a universal cover, we see that $\PP^1\times\PP^1\setminus\Delta$ is simply connected. For this, we consider the proyection $\PP^1\times\PP^1\setminus\Delta\rightarrow\PP^1$, which has fiber $\PP^1\setminus\{p\}$, with $p$ a point. Then, this fiber has the homotopy type of $\s^2\setminus\{p\}\simeq\DD^2$, which is contractible, and therefore $\PP^1\times\PP^1\setminus\Delta$ has the same homotopy type of $\PP^1$, which is simply connected. We have the following diagram:
+\begin{equation*}
+	\begin{array}{ccc}
+		\PP^1\simeq\s^2 & \longrightarrow & \PP^1\times\PP^1\setminus\Delta \\
+		\downarrow & & \downarrow \\
+		\RR\PP^2 & \longrightarrow & \PP^2\setminus C \\
+	\end{array}
+\end{equation*}
+
+We have already seen that the spaces in the upper row are homotopy equivalent
+\end{document}