.

generaldegreeHypersurfaces.tex

@@ -25,6 +25,7 @@ decorations.markings, calc}
 \newcommand{\PP}{\mathbb{P}}
 \newcommand{\s}{\mathbb{S}}
 \newcommand{\ZZ}{\mathbb{Z}}
+\newcommand{\proy}[1]{\pi_{#1}}
 
 
 \numberwithin{equation}{section}
@@ -67,26 +68,27 @@ The retraction is performed in several steps, using the branched cover structure
 
 \maketitle
 \section{Definition of the problem}
-Given $C$ an affine hypersurface 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 $C$ an affine hypersurface in $\CC^2$ (i.e. a curve), 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 hypersurface (which is equivalent to the case of general smooth hypersurface).
+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 hypersurface. Since any smooth hypersurface is isotopic to a Fermat hypersurface, this particular case solves the problemfor a general smooth hypersurface.
 
 
 \section{Smooth hypersurfaces}
 Given a smooth hypersurface of degree $d$ in $\PP^n$, we know it can be moved to a hypersurface of the form
 \begin{equation*}
-	C_d=\{[x_0:\dots:x_{n}]\in\CC\PP^n | \sum x^d_i=0\}
+	C_d=\left\{[x_0:\dots:x_{n}]\in\CC\PP^n\, \middle|\, \sum x^d_i=0\right\}
 \end{equation*}
-by an ambient isotopy.
+by an ambient isotopy; $C_d$ is called a \emph{Fermat hypersurface}.
 Therefore, we will work with $C_d$, from now on.
 
 We will denote:
 \begin{eqnarray*}
 	M_d&=&\{(x_0,\dots,x_n)\in\CC^{n+1} | \sum x^d_i=1\} \\
 	S_d&=&\{(x_0,\dots,x_n)\in M_d | x^d_i\in\RR_{\geq0} \forall i\} \\
-	K &=&\{[x_0:\dots:x_{n}]\in\CC\PP^n \setminus M_2 | x^d_i\in\RR_{\geq0} \forall i\}
+	K &=&\pi(S_d)
+%)\{[x_0:\dots:x_{n}]\in\CC\PP^n \setminus M_2 | x^d_i\in\RR_{\geq0} \forall i\}
 \end{eqnarray*}
-
+where $\proy{}:\CC^{n+1}\setminus\{\mathbf{0}\}\to\PP^n$ is the quotient map.
 We then have a commutative diagram:
 \begin{equation*}
 %	\begin{array}{ccc}
@@ -95,12 +97,11 @@ We then have a commutative diagram:
 %		K & \longhookrightarrow & \PP^n\setminus C_d \\
 %	\end{array}
 \begin{tikzcd}
-S_d\rar[hookrightarrow]\dar& M_d\dar\\ 
-K\rar[hookrightarrow] & \PP^n\setminus C_d
+S_d\rar[hookrightarrow]\dar["\proy{}"]& M_d\dar["\proy{}"]\\ 
+K\rar[hookrightarrow] & \PP^n\setminus C_d.
 \end{tikzcd}
 \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.
 
 We will start by doing the construction for the case $d=1$, since the general case will be constructed on top of it.
@@ -122,10 +123,10 @@ We will now define a strong retraction map
 \begin{equation}
 \begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
 M_1\times{[0,1]}\rar["r_1"]&M_1\\
-(P, t)\rar[mapsto]& t P_r + (1 - t) P_i=:r_{1t}(P),
+(P, t)\rar[mapsto]& t P_r + (1 - t) P_i=:r_{1t}(P);
 \end{tikzcd}
 \end{equation}
-such that $r_{10}$ is the identity and $r_{11}(M_1)=M_1\cap\RR^{n+1}$
+note that $r_{10}$ is the identity,$r_{11}(M_1)=M_1\cap\RR^{n+1}$
 and the restriction of $r_{1t}$ to $M_1\cap\RR^{n+1}$ is the inclusion.
 % and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
 %In particular, we can take $r_1$ that takes the imaginary part to $0$:
@@ -152,13 +153,16 @@ Let us summarize it.
 The map $r_1$ is a strong deformation retraction from $M_1$ to $M_1\cap\RR^{n+1}$.
 \end{lemma}
 
-Now we will define a map $r_2:(M_1 \cap \RR^{n+1})\times[0, 1]\to M_1 \cap \RR^{n+1}$ 
-which will define a strong deformation retraction from $M_1 \cap \RR^{n+1}$ to $S_1$.
+Now we will define a strong deformation retraction
+\[
+r_2:(M_1 \cap \RR^{n+1})\times[0, 1]\to M_1 \cap \RR^{n+1} 
+\]
+from $M_1 \cap \RR^{n+1}$ to $S_1$.
 Note that $M_1\cap\RR^{n+1}$
 is a hypersurface in $\RR^{n+1}$, in Figure~\ref{fig:triangulo} we can see the example $n=2$. 
 
 \begin{lemma}
-$S_1$ is a $n$-simplex.
+The space $S_1$ is a $n$-simplex.
 \end{lemma}
 \begin{proof}
 Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then 
@@ -167,6 +171,7 @@ Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
 S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
 &=&\{x_0e_0+\dots+x_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
 \end{array}
+\qedhere
 \end{equation*}
 \end{proof}