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

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 $S_d$.