From 7b64c5cec50b1506c8fa73efdeed6e2d2368a342 Mon Sep 17 00:00:00 2001 From: alarraya Date: Mon, 9 Sep 2024 17:26:05 +0200 Subject: [PATCH] Upload files to "/" --- complementarioconica.tex | 53 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 53 insertions(+) create mode 100644 complementarioconica.tex diff --git a/complementarioconica.tex b/complementarioconica.tex new file mode 100644 index 0000000..67458a1 --- /dev/null +++ b/complementarioconica.tex @@ -0,0 +1,53 @@ +\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} \ No newline at end of file