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Enrique Artal Bartolo 2026-04-13 09:22:53 +02:00
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generaldegree.tex
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@ -12,10 +12,11 @@
\usepackage{hyperref} \usepackage{hyperref}
\newtheorem{theorem}{Theorem} \newtheorem{theorem}{Theorem}[section]
\newtheorem{example}{Example} \newtheorem{lemma}[theorem]{Lemma}
\newtheorem{lemma}{Lemma} \theoremstyle{definition}
\newtheorem{definition}{Definition} \newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newcommand{\CC}{\mathbb{C}} \newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}} \newcommand{\RR}{\mathbb{R}}
\newcommand{\DD}{\mathbb{D}} \newcommand{\DD}{\mathbb{D}}
@ -24,6 +25,7 @@
\newcommand{\s}{\mathbb{S}} \newcommand{\s}{\mathbb{S}}
\newcommand{\ZZ}{\mathbb{Z}} \newcommand{\ZZ}{\mathbb{Z}}
\numberwithin{equation}{section}
%opening %opening
\title{Complement of a projective curve} \title{Complement of a projective curve}
@ -39,40 +41,38 @@
\section{Definition of the problem} \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. 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} \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} \begin{equation}
C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\} C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\}
\end{equation} \end{equation}
Therefore, we will work with $C_d$, from now on. Therefore, we will work with $C_d$, from now on.
We will denote: We will denote:
\begin{eqnarray} \begin{eqnarray}
M_d&=&\{(x,y,z)\in\CC | x^d+y^d+z^d=1\} \\ 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}\} \\ 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}\} K &=&\{[x:y:z]\in\PP^2 |\, x^d, y^d, z^d\in\RR_{\geq0}\}
\end{eqnarray} \end{eqnarray}
We then have a commutative diagram: We then have a commutative diagram:
\begin{equation*} \begin{equation*}
\begin{array}{ccc} \begin{tikzcd}[/tikz/column 2/.append style={anchor=base west}]
S_d & \longrightarrow & M_d \\ S_d\rar[hookrightarrow, "i"]\dar&M_d\dar[start anchor={[shift={(0,-.2)}]center}, end anchor={[shift={(-.32,.35)}]center}]\\
\downarrow & & \downarrow \\ K\rar[hookrightarrow, "i_\PP"]&\PP^2\setminus C_d.
K & \longrightarrow & \PP^2\setminus C_d \\ \end{tikzcd}
\end{array}
\end{equation*} \end{equation*}
The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following result.
The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma:
\begin{theorem} \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} \end{theorem}
\begin{proof} \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: We can construct $r$ in two steps:
\begin{equation*} \begin{equation*}
\begin{array}{cccccc} \begin{tikzcd}
r:&M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\ M_1\rar["r_1"]\ar[dr, "r"]&M_1\cap\RR^3\dar["r_2"]\\
\end{array} &S_1.
\end{tikzcd}
\end{equation*} \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: 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} \begin{equation}