From fb0e84322562543c949504c6a35bbf818e49781d Mon Sep 17 00:00:00 2001 From: Enrique Artal Date: Mon, 13 Apr 2026 09:22:53 +0200 Subject: [PATCH] cambios --- generaldegree.tex | 44 ++++++++++++++++++++++---------------------- 1 file changed, 22 insertions(+), 22 deletions(-) diff --git a/generaldegree.tex b/generaldegree.tex index 1c1678b..cd573da 100644 --- a/generaldegree.tex +++ b/generaldegree.tex @@ -12,10 +12,11 @@ \usepackage{hyperref} -\newtheorem{theorem}{Theorem} -\newtheorem{example}{Example} -\newtheorem{lemma}{Lemma} -\newtheorem{definition}{Definition} +\newtheorem{theorem}{Theorem}[section] +\newtheorem{lemma}[theorem]{Lemma} +\theoremstyle{definition} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{example}[theorem]{Example} \newcommand{\CC}{\mathbb{C}} \newcommand{\RR}{\mathbb{R}} \newcommand{\DD}{\mathbb{D}} @@ -24,6 +25,7 @@ \newcommand{\s}{\mathbb{S}} \newcommand{\ZZ}{\mathbb{Z}} +\numberwithin{equation}{section} %opening \title{Complement of a projective curve} @@ -39,40 +41,38 @@ \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{Smooth curves} -Given a smooth curve of degree $d$ in $\PP^2$, we know it is isomorphic to a curve of the form +Given a smooth curve of $C$ of degree $d$ in $\PP^2$, we know $(\PP^2,C)$ is homemorphic to $(\PP^2,C_d)$ where \begin{equation} C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\} \end{equation} Therefore, we will work with $C_d$, from now on. - We will denote: \begin{eqnarray} - M_d&=&\{(x,y,z)\in\CC | x^d+y^d+z^d=1\} \\ - S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\ - K &=&\{[x:y:z]\in\PP^2 | x^d, y^d, z^d\in\RR_{\geq0}\} + M_d&=&\{(x,y,z)\in\CC^3 |\, x^d+y^d+z^d=1\} \\ + S_d&=&\{(x,y,z)\in M_d |\, x^d, y^d, z^d\in\RR_{\geq0}\} \\ + K &=&\{[x:y:z]\in\PP^2 |\, x^d, y^d, z^d\in\RR_{\geq0}\} \end{eqnarray} - We then have a commutative diagram: \begin{equation*} - \begin{array}{ccc} - S_d & \longrightarrow & M_d \\ - \downarrow & & \downarrow \\ - K & \longrightarrow & \PP^2\setminus C_d \\ - \end{array} +\begin{tikzcd}[/tikz/column 2/.append style={anchor=base west}] +S_d\rar[hookrightarrow, "i"]\dar&M_d\dar[start anchor={[shift={(0,-.2)}]center}, end anchor={[shift={(-.32,.35)}]center}]\\ +K\rar[hookrightarrow, "i_\PP"]&\PP^2\setminus C_d. +\end{tikzcd} \end{equation*} - -The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma: +The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following result. \begin{theorem} -$S_d$ is a strong deformation retract of $M_d$. +The space $S_d$ is a strong deformation retract of $M_d$. \end{theorem} \begin{proof} -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. +We start by getting a retraction $r:M_d\longrightarrow S_d$ for $d=1$. Note that for $d=1$ the result is trivial since $M_d$ is an affine complex plane (contractible) and $S_d$ is a closed triangle in an affine real plane. The key point is that in this case the retraction is easily constructed and we will lift it +for $d>1$. We can construct $r$ in two steps: \begin{equation*} - \begin{array}{cccccc} - r:&M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\ - \end{array} +\begin{tikzcd} +M_1\rar["r_1"]\ar[dr, "r"]&M_1\cap\RR^3\dar["r_2"]\\ +&S_1. +\end{tikzcd} \end{equation*} 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: \begin{equation}