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Plantear los enunciados para d=1 en términos de retracciones concretas

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Miguel Marco 2026-05-25 12:18:57 +02:00
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generaldegree.tex
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@ -25,7 +25,7 @@ decorations.markings, calc}
%opening %opening
\title{Complement of a projective curve} \title{Retraction of the complement of smooth projective curves to a $2$-dimensional $\Delta$-complex.}
\author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz} \author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz}
\begin{document} \begin{document}
@ -34,14 +34,17 @@ decorations.markings, calc}
\begin{abstract} \begin{abstract}
Due to a result by Androtti and Frankel \cite{andreotti19}, we can see that the complement o a complex projective curve has the homotopy type of a $2$-dimensional CW complex. However, no general method has been given to compute explicitly this complex. Here we give a explicit construction of a $2$- dimensional $\Delta$-complex that is a strong deformation retract of the complement of a Fermat curve of degree $d$ in the complex projective space. Due to a result by Andreotti and Frankel \cite{andreotti1959}, it can be seen that the complement of a complex projective curve has the homotopy type of a $2$-dimensional CW complex. However, no general method has been given to compute explicitly this complex. Here we give a explicit construction of a $2$- dimensional $\Delta$-complex that is a strong deformation retract of the complement of a Fermat curve of degree $d$ in the complex projective space.
The retraction is performed in several steps, using the branched cover structure of the Fermat curves to the degree $1$ case. The retraction is performed in several steps, using the branched cover structure of the Fermat curves to the degree $1$ case.
\end{abstract} \end{abstract}
\section{Definition of the problem} \section{Definition of the problem}
Given $C$ a proyective curve in $\CC^2$, Libgober gave in \cite{libgober1986} a construction of a $2-$ dimensional CW-complex that is a deformation retract of its complement. However, it is not clear how to extend that construction to the projective case (that is a curve in $\CC\PP^2$). Given $C$ a projective curve in $\CC^2$, Libgober gave in \cite{libgober1986} a construction of a $2-$ dimensional CW-complex that is a deformation retract of its complement. However, it is not clear how to extend that construction to the projective case (that is a curve in $\CC\PP^2$).
In this work, we give a different construction to produce a specific $\Delta$-complex that is a deformation retract of the complement of a Fermat curve (which is equivalent to the case of general smooth curves). In this work, we give a different construction to produce a specific $\Delta$-complex that is a deformation retract of the complement of a Fermat curve (which is equivalent to the case of general smooth curves).
Our construction can be easily generalized to higher dimensions. We plan to write a more detailed work including that result in the near future.
\section{Smooth curves} \section{Smooth curves}
Given a smooth curve of degree $d$ in $\PP^2$, we know it can be moved to a curve of the form Given a smooth curve of degree $d$ in $\PP^2$, we know it can be moved to a curve of the form
\begin{equation} \begin{equation}
@ -67,11 +70,10 @@ We then have a commutative diagram:
\end{equation*} \end{equation*}
where the vertical maps consist on taking equivalence classes. where the vertical maps consist on taking equivalence classes.
We are going to see that the inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. To do this we first need the following lemma: We are going to see that the inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence.
\begin{lemma}
$M_1\cap\RR^3$ is a strong deformation retract of $M_1$. We will start by doing the construction for the case $d=1$, since the general case will be constructed on top of it.
\end{lemma}
\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 $d=1$, and then we'll see that we can lift it.
%We can construct $r$ in two steps: %We can construct $r$ in two steps:
@ -80,7 +82,7 @@ $M_1\cap\RR^3$ is a strong deformation retract of $M_1$.
% M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\ % M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
% \end{array} % \end{array}
%\end{equation*} %\end{equation*}
We are going to define a strong retraction $r_1:M_1\times I \longrightarrow M_1$ such that $r_1(M_1,0)=M_1$ and $r_1(M_1,1)=M_1\cap\RR^3$ We will now define a strong retraction $r_1:M_1\times I \longrightarrow M_1$ such that $r_1(M_1,0)=M_1$ and $r_1(M_1,1)=M_1\cap\RR^3$
% and $r_2:M_1\cap\RR^3\longrightarrow S_1$. % and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
In particular, we can take $r_1$ that takes the imaginary part to $0$: In particular, we can take $r_1$ that takes the imaginary part to $0$:
\begin{equation} \begin{equation}
@ -99,13 +101,16 @@ Note that for $(x,y,z)\in M_1$, we already have that
\end{array} \end{array}
\end{equation} \end{equation}
and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x,y,z,1)=(x_r,y_r,z_r)\in M_1\cap\RR^3$ for all $(x,y,z)\in M_1$. $r_1$ is then a strong retraction from $M_1$ to $M_1\cap\RR^3$. and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x,y,z,1)=(x_r,y_r,z_r)\in M_1\cap\RR^3$ for all $(x,y,z)\in M_1$. $r_1$ is then a strong retraction from $M_1$ to $M_1\cap\RR^3$.
\end{proof}
This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations.
We have just proven that
\begin{lemma} \begin{lemma}
$S_1$ is a strong deformation retract of $M_1\cap\RR^3$. The map $r_1$ is a strong deformation retraction from $M_1$ to $M_1\cap\RR^3$.
\end{lemma} \end{lemma}
\begin{proof}
This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations. Note that $M_1\cap\RR^3$ Now we will define a retraction from $M_1 \cap \RR^3$ to $S_1$.
Note that $M_1\cap\RR^3$
$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is $M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is
\[ \[
S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\}, S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
@ -211,7 +216,12 @@ P_{x,y,z}=\left\{\begin{array}{lr}
\end{array}\right. \end{array}\right.
\end{equation*} \end{equation*}
similarly to the previous case, in this line $x=1-z$. similarly to the previous case, in this line $x=1-z$.
\end{proof}
We have now proven that:
\begin{lemma}
$r_2$ is a strong deformation retraction from $M_1\cap \RR^3$ to $S_1$.
\end{lemma}
Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we will lift them in order to get a retraction from $M_d$ to $S_d$. Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we will lift them in order to get a retraction from $M_d$ to $S_d$.