From 0e3cc55745953d15059105ad3f85c472ce1a8ebf Mon Sep 17 00:00:00 2001 From: Miguel Marco Date: Mon, 25 May 2026 12:18:57 +0200 Subject: [PATCH] =?UTF-8?q?Plantear=20los=20enunciados=20para=20d=3D1=20en?= =?UTF-8?q?=20t=C3=A9rminos=20de=20retracciones=20concretas?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- generaldegree.tex | 38 ++++++++++++++++++++++++-------------- 1 file changed, 24 insertions(+), 14 deletions(-) diff --git a/generaldegree.tex b/generaldegree.tex index 6afa4fb..293c415 100644 --- a/generaldegree.tex +++ b/generaldegree.tex @@ -25,7 +25,7 @@ decorations.markings, calc} %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} \begin{document} @@ -34,14 +34,17 @@ decorations.markings, calc} \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. \end{abstract} \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). + +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} Given a smooth curve of degree $d$ in $\PP^2$, we know it can be moved to a curve of the form \begin{equation} @@ -67,11 +70,10 @@ We then have a commutative diagram: \end{equation*} 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: -\begin{lemma} -$M_1\cap\RR^3$ is a strong deformation retract of $M_1$. -\end{lemma} -\begin{proof} +We are going to see that the inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. + +We will start by doing the construction for the case $d=1$, since the general case will be constructed on top of 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: @@ -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 \\ % \end{array} %\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 $r_2:M_1\cap\RR^3\longrightarrow S_1$. In particular, we can take $r_1$ that takes the imaginary part to $0$: \begin{equation} @@ -99,13 +101,16 @@ Note that for $(x,y,z)\in M_1$, we already have that \end{array} \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 $M_1\cap\RR^3$. -\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} - $S_1$ 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} -\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 \[ 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{equation*} 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 $S_d$.