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generaldegreeHypersurfaces.tex
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@ -25,6 +25,7 @@ decorations.markings, calc}
\newcommand{\PP}{\mathbb{P}} \newcommand{\PP}{\mathbb{P}}
\newcommand{\s}{\mathbb{S}} \newcommand{\s}{\mathbb{S}}
\newcommand{\ZZ}{\mathbb{Z}} \newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\proy}[1]{\pi_{#1}}
\numberwithin{equation}{section} \numberwithin{equation}{section}
@ -67,26 +68,27 @@ The retraction is performed in several steps, using the branched cover structure
\maketitle \maketitle
\section{Definition of the problem} \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} \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 Given a smooth hypersurface of degree $d$ in $\PP^n$, we know it can be moved to a hypersurface of the form
\begin{equation*} \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*} \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. Therefore, we will work with $C_d$, from now on.
We will denote: We will denote:
\begin{eqnarray*} \begin{eqnarray*}
M_d&=&\{(x_0,\dots,x_n)\in\CC^{n+1} | \sum x^d_i=1\} \\ 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\} \\ 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*} \end{eqnarray*}
where $\proy{}:\CC^{n+1}\setminus\{\mathbf{0}\}\to\PP^n$ is the quotient map.
We then have a commutative diagram: We then have a commutative diagram:
\begin{equation*} \begin{equation*}
% \begin{array}{ccc} % \begin{array}{ccc}
@ -95,12 +97,11 @@ We then have a commutative diagram:
% K & \longhookrightarrow & \PP^n\setminus C_d \\ % K & \longhookrightarrow & \PP^n\setminus C_d \\
% \end{array} % \end{array}
\begin{tikzcd} \begin{tikzcd}
S_d\rar[hookrightarrow]\dar& M_d\dar\\ S_d\rar[hookrightarrow]\dar["\proy{}"]& M_d\dar["\proy{}"]\\
K\rar[hookrightarrow] & \PP^n\setminus C_d K\rar[hookrightarrow] & \PP^n\setminus C_d.
\end{tikzcd} \end{tikzcd}
\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. 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 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{equation}
\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}] \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\\ 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{tikzcd}
\end{equation} \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 the restriction of $r_{1t}$ to $M_1\cap\RR^{n+1}$ is the inclusion.
% and $r_2:M_1\cap\RR^3\longrightarrow S_1$. % 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 $0$:
@ -152,13 +153,16 @@ Let us summarize it.
The map $r_1$ is a strong deformation retraction from $M_1$ to $M_1\cap\RR^{n+1}$. The map $r_1$ is a strong deformation retraction from $M_1$ to $M_1\cap\RR^{n+1}$.
\end{lemma} \end{lemma}
Now we will define a map $r_2:(M_1 \cap \RR^{n+1})\times[0, 1]\to M_1 \cap \RR^{n+1}$ Now we will define a strong deformation retraction
which will define a strong deformation retraction from $M_1 \cap \RR^{n+1}$ to $S_1$. \[
r_2:(M_1 \cap \RR^{n+1})\times[0, 1]\to M_1 \cap \RR^{n+1}
\]
from $M_1 \cap \RR^{n+1}$ to $S_1$.
Note that $M_1\cap\RR^{n+1}$ 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 $n=2$. is a hypersurface in $\RR^{n+1}$, in Figure~\ref{fig:triangulo} we can see the example $n=2$.
\begin{lemma} \begin{lemma}
$S_1$ is a $n$-simplex. The space $S_1$ is a $n$-simplex.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then 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\}&& \\ 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\\ &=&\{x_0e_0+\dots+x_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
\end{array} \end{array}
\qedhere
\end{equation*} \end{equation*}
\end{proof} \end{proof}