712 lines
37 KiB
TeX
712 lines
37 KiB
TeX
\documentclass[a4paper,10pt]{amsart}
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\usepackage[utf8]{inputenc}
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\usepackage{amsfonts}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{tikz}
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\usepackage[new]{old-arrows}
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\usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart,
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decorations.markings, calc}
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\usepackage{hyperref}
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\newtheorem{theorem}{Theorem}
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\newtheorem{example}{Example}
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\newtheorem{lemma}{Lemma}
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\newtheorem{definition}{Definition}
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\newcommand{\CC}{\mathbb{C}}
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\newcommand{\RR}{\mathbb{R}}
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\newcommand{\DD}{\mathbb{D}}
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\newcommand{\BB}{\mathbb{B}}
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\newcommand{\PP}{\mathbb{P}}
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\newcommand{\s}{\mathbb{S}}
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\newcommand{\ZZ}{\mathbb{Z}}
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%opening
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\title{Complement of a projective curve}
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\author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz}
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\begin{document}
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\maketitle
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\begin{abstract}
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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.
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The retraction is performed in several steps, using the branched cover structure of the Fermat curves to the degree $1$ case.
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\end{abstract}
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\section{Definition of the problem}
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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 $\CC\PP^2$).
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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).
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\section{Smooth curves}
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Given a smooth curve of degree $d$ in $\PP^2$, we know it can be moved to a curve of the form
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\begin{equation}
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C_d=\{[x:y:z]\in\CC\PP^2 | x^d+y^d+z^d=0\}
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\end{equation}
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by an ambient isotopy.
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Therefore, we will work with $C_d$, from now on.
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We will denote:
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\begin{eqnarray}
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M_d&=&\{(x,y,z)\in\CC^3 | x^d+y^d+z^d=1\} \\
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S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
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K &=&\{[x:y:z]\in\CC\PP^2 \setminus M_2 | x^d, y^d, z^d\in\RR_{\geq0}\}
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\end{eqnarray}
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We then have a commutative diagram:
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\begin{equation*}
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\begin{array}{ccc}
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S_d & \longhookrightarrow & M_d \\
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\downarrow & & \downarrow \\
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K & \longhookrightarrow & \PP^2\setminus C_d \\
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\end{array}
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\end{equation*}
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where the vertical maps consist on taking equivalence classes.
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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:
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\begin{lemma}
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$M_1\cap\RR^3$ is a strong deformation retract of $M_1$.
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\end{lemma}
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\begin{proof}
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%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.
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%We can construct $r$ in two steps:
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%\begin{equation*}
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% \begin{array}{ccccc}
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% M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
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% \end{array}
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%\end{equation*}
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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 $r_1(M_1,1)=M_1\cap\RR^3$
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% and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
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In particular, we can take $r_1$ that takes the imaginary part to $0$:
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\begin{equation}
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\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
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M_1\times{[0,1]}\rar["r_1"]&M_1\cap\RR^3\\
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(P, t)\rar[mapsto]& t \Re P + (1 - t) \Im P.
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\end{tikzcd}
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\end{equation}
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where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively.
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Note that for $(x,y,z)\in M_1$, we already have that
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\begin{equation}
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\begin{array}{lll}
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x_r+y_r+z_r & = & 1 \\
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x_i+y_i+z_i & = & 0 \\
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\end{array}
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\end{equation}
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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 $M_1\cap\RR^3$.
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\end{proof}
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\begin{lemma}
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$S_1$ is a strong deformation retract of $M_1\cap\RR^3$.
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\end{lemma}
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\begin{proof}
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This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations. Note that $M_1\cap\RR^3$
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$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is
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\[
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S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
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\]
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i.e, $S_1$ is the triangular section defined by the intersection of this plane with the first octant, delimited by the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described in Figure~\ref{fig:triangulo}. We will retract each of them separately:
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
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% \draw (3,4.5) to (-6,-9);
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% \draw (-3,4.5) to (6,-9);
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% \draw (-7,-4.5) to (7,-4.5);
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% \node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
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% \node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
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% \node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
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% \node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
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% \node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
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% \node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
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% \node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
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% \node[right] at (0,0) {(0,0,1)};
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% \node[right,above] at (3,-4.5) {(0,1,0)};
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% \node[left,above] at (-3,-4.5) {(1,0,0)};
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\coordinate (A) at (-3,-4.5);
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\coordinate (B) at (3,-4.5);
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\coordinate (C) at (0,0);
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\fill[green!25!white] (C) to (B) to (A);
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\draw ($2*(C)-(A)$) to ($2*(A)-(C)$);
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\draw ($2*(B)-(A)$) to ($2*(A)-(B)$);
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\draw ($2*(B)-(C)$) to ($2*(C)-(B)$);
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\node at (0,-2.5) {$(+,+,+)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (5.5,0) {$(-,+,+)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (7.5,-6.5) {$(-,+,-)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (0,-8.5) {$(+,+,-)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (-8,-6.5) {$(+,-,-)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
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\node at (-6,0) {$(+,-,+)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
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\node at (0,4) {$(-,-,+)$};% {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
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\foreach \x[evaluate={\y=1-\x}] in {.125, .25, ..., 1}
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{
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\draw[->=.5,green!75!black] ($1.75*(C) -.75*\x*(A) -.75*\y*(B)$) -- (C);
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\draw[->=.5,green!75!black] ($1.75*(B) -.75*\x*(A) -.75*\y*(C)$) -- (B);
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\draw[->=.5,green!75!black] ($1.75*(A) -.75*\x*(C) -.75*\y*(B)$) -- (A);
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}
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\foreach \x[evaluate={\y=1-\x}] in {0.25, .3125, ..., .75}
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{
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\draw[->=.5,green!75!black] ($1.75*\x*(C) -.75*(A) + 1.75*\y*(B)$) -- ($\x*(C)+ \y*(B)$);
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\draw[->=.5,green!75!black] ($1.75*\x*(A) -.75*(B) + 1.75*\y*(C)$) -- ($\x*(A)+ \y*(C)$);
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\draw[->=.5,green!75!black] ($1.75*\x*(A) -.75*(C) + 1.75*\y*(B)$) -- ($\x*(A)+ \y*(B)$);
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}
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%\fill (6, -4.5) circle[radius=.1cm];
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\node[right=2pt] at (C) {(0,0,1)};
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\node[above right] at (B) {(0,1,0)};
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\node[above left] at (A) {(1,0,0)};
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\end{tikzpicture}
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\caption{Triangle $S_1$}
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\label{fig:triangulo}
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\end{center}
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\end{figure}
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We denote each section as $(\pm,\pm,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ would be $(+,+,+)$.
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The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
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\begin{equation}
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\begin{array}{ccc}
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r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
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(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
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\end{array}
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\end{equation}
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Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ respectively. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
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\begin{equation}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(x,y,z)&(+,+,+) \\
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(1,0,0)&(+,-,-) \\
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(0,1,0)&(-,+,-) \\
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(0,0,1)&(-,-,+) \\
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(\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
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(\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
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(0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
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\end{array}\right.
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\end{equation}
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Since we have defined the retraction by regions, we are now going to show that in the intersections the definitions coincide, and therefore that, by the gluing lemma, $r_2$ is countinous and well defined. We will see this for the part $(+,+,-)$, for the rest of permutations it is equivalent.
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We start with the border between $(+,+,-)$ and $(+,+,+)$, this is, the line with $z=0$. In this case
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\begin{equation*}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(x,y,0)&(+,+,+) \\
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(\frac{x}{1-0},\frac{y}{1-0},0)=(x,y,0)&(+,+,-) \\
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\end{array}\right.
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\end{equation*}
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For the border between $(+,+,-)$ and $(-,+,-)$, the border is the line $x=0$, so we get
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\begin{equation*}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(\frac{0}{1-z},\frac{y}{1-z},0)=(0,1,0)&(+,+,-) \\
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(0,1,0)&(-,+,-) \\
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\end{array}\right.
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\end{equation*}
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since when $x=0$ the equation of the plane is $y+z=1$ and then $y=1-z$. And lastly, the border between $(+,+,-)$ and $(+,-,-)$, where $y=0$
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\begin{equation*}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(\frac{x}{1-z},\frac{0}{1-z},0)=(1,0,0)&(+,+,-) \\
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(1,0,0)&(+,-,-) \\
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\end{array}\right.
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\end{equation*}
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similarly to the previous case, in this line $x=1-z$.
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\end{proof}
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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 $S_d$.
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\begin{theorem}
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$S_d$ is a strong deformation retract of $M_d$.
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\end{theorem}
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\begin{proof}
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We want to produce the following diagram:
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\begin{equation*}
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\begin{array}{ccccc}
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M_d & \rightleftharpoons & M_d\cap\{x^d, y^d, z^d\in\RR\}&\rightleftharpoons& S_d \\
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\downarrow & & \downarrow & & \downarrow\\
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M_1 & \rightleftharpoons & M_1\cap\RR^3 & \rightleftharpoons & S_1 \\
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\end{array}
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\end{equation*}
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where the horizontal arrows represent retractions, and the vertical arrows are the maps $(x,y,z)\mapsto (x^d,y^d,z^d)$. That is, we want to lift the retractions built for $M_1$ and $S_1$ to $M_d$ and $S_d$ via this branched cover.
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We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, we can write it as $r_1((x,y,z),t)=(\bar{r}(x,t),\bar{r}(y,t),\bar{r}(z,t))$ with $\bar{r}(x,t)+\bar{r}(y,t)+\bar{r}(z,t)=1$ whenever $(x,y,z)\in M_1$. Then, it suffices to find $G(x,t)$ such that $G(x,t)^d=\bar{r}(x^d,t)$.
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%Since $r_1$ works componentwise, we will study the lifting also for each component separetely. Since $r_1$ is the same for each component, we see it for only one of them, the rest will be analogous.
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%For the first component, $r_1(x,t)=x_r+(1-t)ix_i$, and we have a map from $M_d$ to $M_1$ given by $F(x)=x^d$.
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%$r_1$ is the vertical projection over the real axis of the complex plane. In this first component, $M_d$ is a ramified cover over $M_1$, where $0$ is the only point where it ramifies. Therefore, outside of $0$ we have a cover, and then $r_1$ has a unique lifting outside of $0$. Since $r_1$ is constant on $0$, we get a lifting of $r_1$ by defining $\tilde{r}_1(0,t)=0$.
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In order to see this, we can write an explicit expression of $\bar{r}$. For a point $x\in\CC$ the action of $\bar{r}$ is the projection over the real axis, as seen in figure \ref{realprojection}.
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.6]
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\draw (-5,0) to (5,0);
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\draw (0,5) to (0,-1);
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\draw[fill] (3,2) circle (2pt) node[above]{$x_r+ix_i$};
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\draw[dashed] (3,2) to (3,0);
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\end{tikzpicture}
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\end{center}
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\caption{projection over the real axis}
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\label{realprojection}
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\end{figure}
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we can then write the retraction as $\bar{r}(x,t)=m(x,t)e^{2i\pi\theta(x,t)}$ where $m(x,t)$ is the function indicating how the modulus of $\bar{r}(x,t)$ changes and $\theta(x,t)$ portrays the change of the argument.
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For a $x=x_r+ix_i\in\CC$, $\bar{r}(x,t)=x_r+(1-t)ix_i$, we can write $m(x,t)=+\sqrt{x_r^2+(1-t)^2x_i^2}$ and $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$.
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We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}, where each of the sections will be homeomorphic to either the upper half complex plane or to the lower half.
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.6]
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\draw (-5,0) to (5,0);
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\draw (5,1.5) to (-5,-1.5);
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\draw (5,3.2) to (-5, -3.2);
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\draw (4.7,5) to (-4.7,-5);
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\draw (2.8,5) to (-2.8,-5);
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\draw (0.9,5) to (-0.9,-5);
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\draw (-0.9,5) to (0.9,-5);
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\draw (-2.8,5) to (2.8,-5);
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\draw (-4.7,5) to (4.7,-5);
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\draw (-5,3.2) to (5, -3.2);
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\draw (-5,1.5) to (5,-1.5);
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\end{tikzpicture}
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\end{center}
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\caption{Division of the complex plane}
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\label{divisionC}
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\end{figure}
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Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$, which we will denote $f_k$. The lifting of $r_1$ will then be given by the $G$ previously described. This $G$ will act by taking the point $(x,y,z)\in M_d$ and mapping it to the upper or lower complex plane via the inverse homeomorphisms $f^{-1}_k$, then taking $r_1$ to take it to the real line, and finally bringing it back to its original sector with the homeomorphism $f_k$.
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The homeomorphisms are given by
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\begin{equation}
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f_k(|z|\exp^{i\theta})=|z|^\frac{1}{d}\exp^{i\frac{\theta}{d}+(k-1)i\frac{\pi}{d}}\text{ for }k=1,3,5,\dots,2d-1,
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\end{equation}
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\begin{equation}
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f_k(|z|\exp^{i\theta})=|z|^\frac{1}{d}\exp^{i\frac{\theta-\pi}{d}+(k-1)i\frac{\pi}{d}}\text{ for }k=2,4,6,\dots,2d
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\end{equation}
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We will also need the inverses of these homeomorphisms, which are:
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\begin{equation}
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f_k^{-1}(|z|\exp^{i\theta})=|z|^d\exp^{id(\theta-(k-1)\frac{\pi}{d})}\text{ for }k=1,3,5,\dots,2d-1,
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\end{equation}
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\begin{equation}
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f_k^{-1}(|z|\exp^{i\theta})=|z|^d\exp^{id(\theta-\pi-(k-1)i\frac{\pi}{d})}\text{ for }k=2,4,6,\dots,2d
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\end{equation}
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Then, we define $G=f_k\circ\bar{r}\circ f_k^{-1}$, where $k$ depends on the point $z$ to which we will apply it. For $z\in\CC$, given by $z=|z|\exp^{i\theta}$ for $\theta\in[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$, we have:
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\begin{equation}
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\begin{array}{ccc}
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G(z,t)&=&f_k(\bar{r}(f_k^{-1}(z),t))\\
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&=&f_k(\bar{r}(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},t))\\
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&=&f_k(+\sqrt{(z^d)_r^2+(1-t)^2(z^d)_i^2}\exp^{i\arctan((1-t)\tan(d(\theta-(k-1)\frac{\pi}{d})))})\\
|
|
&=&+\sqrt[2d]{(z^d)_r^2+(1-t)^2(z^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))}
|
|
\end{array}
|
|
\label{Gliftedr1}
|
|
\end{equation}
|
|
|
|
Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since:
|
|
\begin{equation}
|
|
\begin{array}{ccc}
|
|
G(z,t)^d&=&m(z,t)^d\exp^{i\arctan((1-t)\tan d\theta)}\\
|
|
||& & \\
|
|
\bar{r}(z^d,t)&=&m(z^d,t)\exp^{i\arctan((1-t)\tan d\theta)}\\
|
|
\end{array}
|
|
\end{equation}
|
|
We can the write the lifting of $r_1$ as $\tilde{r}_1((x,y,z),t)=(G(x,t), G(y,t), G(z,t))$.
|
|
|
|
|
|
Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d, y^d, z^d\in\RR\}$ into $S_d$.
|
|
For a point $(x\xi_1,y\xi_2,z\xi_3)\in M_d\cap\{x^d, y^d, z^d\in\RR\}$ (where $(x,y,z)$ are real and $\xi_i$ are $d$-th roots of unity), the corresponding point in $M_1\cap\RR^3$ is $(x^ d,y^d,z^d)$ and therefore the goal points of the retraction $r_2$ will be
|
|
\begin{equation}
|
|
P_{x,y,z}=\left\{\begin{array}{lr}
|
|
(x^d,y^d,z^d)&(+,+,+) \\
|
|
(1,0,0)&(+,-,-) \\
|
|
(0,1,0)&(-,+,-) \\
|
|
(0,0,1)&(-,-,+) \\
|
|
(\frac{x^d}{1-z^d},\frac{y^d}{1-z^d},0)&(+,+,-) \\
|
|
(\frac{x^d}{1-y^d},0,\frac{z^d}{1-y^d})&(+,-,+) \\
|
|
(0,\frac{y^d}{1-x^d},\frac{z^d}{1-x^d})&(-,+,+) \\
|
|
\end{array}\right.
|
|
\end{equation}
|
|
In the $(+,+,+)$ case, $r_2$ is the constant path, and therefore the lift $\tilde{r}_2$ will also be the constant path.
|
|
|
|
In the $(-,-,+)$ case, $r_2((x^d,y^d,z^d), t)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x\xi_1,y\xi_2,z\xi_3)$, for $\xi_1$, $\xi_2$ and $\xi_3$ some $d-$th roots of unity. Since $x,y$ are negative real numbers, the retraction will push them to $0$, while we will push $z\xi_3$ to $\xi_3$. We can write this as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3), t)=(\sqrt[d]{(1-t)x^d}\xi_1,\sqrt[d]{(1-t)y^d}\xi_2,\sqrt[d]{(1-t)z^d+t}\xi_3)$, where the $d$'th roots take values in the corresponding sector of each coordinate. For the cases $(+,-,-)$ and $(-,+,-)$ it is equivalent.
|
|
|
|
Lastly, for the case $(+,+,-)$ and it's permutations, we can write the lifting as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3),t)=(\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1,\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2,\sqrt[d]{(1-t)z^d}\xi_3)$. We can see by summing the coordinates to the power $d$ that they sum $1$, so the retraction doesn't leave $M_1$.
|
|
|
|
All that is left to see is that the formulas for each case coincide on the intersections of the regions. For the intersection of $(+,+,-)$ and $(-,+,-)$, we have that:
|
|
\begin{equation*}
|
|
\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}=\sqrt[d]{(1-t+\frac{t}{y^d})y^d}=\sqrt[d]{y^d-ty^d+t}=\sqrt[d]{y^d(1-t)+t}
|
|
\end{equation*}
|
|
The first root corresponds to $\tilde{r}_2$ in $(+,+,-)$ and the last one corresponds to the lifting in $(-,+,-)$, so we see that it coincides. It would be similar in the rest of intersections.
|
|
|
|
|
|
We have then proved that $S_d$ is a strong deformation retract of $M_d$.
|
|
\end{proof}
|
|
|
|
\begin{theorem}
|
|
The strong retraction from $M_d$ to $S_d$ induces a strong retraction, and therefore an homotopy equivalence, between $\CC\PP^2\setminus C_d$ and $K$.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
The path followed by each point in $\CC\PP^2\setminus C_d$ . in the retraction will be given by the path followed by a representative in $M_d$ as it is retracted to $S_d$. The class representatives we choose in $\CC\PP^2\setminus C_d$ will be such that the coordinates to the $d$ power add up to $1$. Since this choice is not unique, we are now going to see that if we apply the retraction $r=r_2\circ r_1$ to two such representatives, they are sent to the same class by $r$.
|
|
|
|
Let $p=[x:y:z]\in\CC\PP^2\setminus C_d$ and $\lambda p=[\lambda x:\lambda y: \lambda z]$ be two representatives of some point $p$, since both representatives must be such that $x^d+y^d+z^d=1$, necessarily $\lambda$ is a $d-$th root of unity.
|
|
|
|
First, we see how $r_1$ acts on these representatives.
|
|
\begin{equation}
|
|
\begin{array}{lll}
|
|
\tilde{r}_1(p,t) & = & [G(x,t): G(y,t): G(z,t)] \\
|
|
\tilde{r}_1(\lambda p,t) & = & [G(\lambda x,t): G(\lambda y,t): G(\lambda z,t)] \\
|
|
\end{array}
|
|
\end{equation}
|
|
If we define $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$ for $x\in\CC$, we get that
|
|
\begin{equation}
|
|
\theta(\lambda x,t)=\arctan\frac{(1-t)\lambda x_i}{\lambda x_r}=\arctan\frac{(1-t)x_i}{x_r}=\theta(x,t)
|
|
\end{equation}
|
|
By \ref{Gliftedr1}:
|
|
\begin{equation}
|
|
\begin{array}{lll}
|
|
G(x,t) & = & +\sqrt[2d]{(x^d)_r^2+(1-t)^2(x^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))} \\
|
|
G(\lambda x,t) & = & +\sqrt[2d]{(\lambda^d x^d)_r^2+(1-t)^2(\lambda^d x^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))} \\
|
|
\end{array}
|
|
\end{equation}
|
|
Since $\lambda$ is a $d-$th root of unity, we get that $G(x,t)=G(\lambda x,t)$. Therefore, we get that $\tilde{r}_1(p,t)=\tilde{r}_1(\lambda p,t)$. It's just left to check that $r_2$ also sends them to the same class.
|
|
|
|
For a point $(x,y,z)\in M_d\cap\RR^3$, $\tilde{r}_2$ is defined by the region of $M_1\cap\RR^3$ where $(x^d, y^d, z^d)$ is, and therefore, when passing to the projective, it will depend on where the point $p^d=[x^d: y^d: z^d]$ lies. Since we have that $\lambda$ must be a $d-$th root of unity, we get that both points $p^d$ and $(\lambda p)^d=p$ will lie in the same region.
|
|
|
|
In the case that they lie in the regions $(-,-,+)$, $(-,+,-)$ or $(+,-,-)$, the retraction $\tilde{r}_2$ is, in the first case:
|
|
\begin{equation}
|
|
\tilde{r}_2([x\xi_1:y\xi_2:z\xi_3],t)=\left[\sqrt[d]{(1-t)x^d}\xi_1:\sqrt[d]{(1-t)y^d}\xi_2:\sqrt[d]{(1-t)z^d+t}\xi_3\right]
|
|
\end{equation}
|
|
Then if we apply it to $[\lambda x\xi_1:\lambda y\xi_2:\lambda z\xi_3]$, for $\lambda $ another root of unity, we get:
|
|
\begin{equation}
|
|
\tilde{r}_2([\lambda x\xi_1:\lambda y\xi_2:\lambda z\xi_3],t)=\left[\sqrt[d]{(1-t)(\lambda x)^d}\xi_1:\sqrt[d]{(1-t)(\lambda y)^d}\lambda y\xi_2:\sqrt[d]{(1-t)(\lambda z)^d+t}\xi_3\right]
|
|
\end{equation}
|
|
|
|
The rest of these cases would be equivalent.
|
|
|
|
Next, for the cases $(+,+,-)$, $(+,-,+)$ or $(-,+,+)$, the retraction has the form, in the first case:
|
|
\begin{equation}
|
|
\tilde{r}_2([x\xi_1,y\xi_2,z\xi_3],t)=\left[\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1:\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2:\sqrt[d]{(1-t)z^d}\xi_3\right]
|
|
\end{equation}
|
|
Applying it to another representative of the same class, we get:
|
|
\begin{equation}
|
|
\begin{array}{lll}
|
|
\tilde{r}_2([\lambda x\xi_1,\lambda y\xi_2,\lambda z\xi_3],t)& = &\\
|
|
& = & \left[\sqrt[d]{(1-t+\frac{t}{1-(\lambda z)^d})(\lambda x)^d}\xi_1:\sqrt[d]{(1-t+\frac{t}{1-(\lambda z)^d})(\lambda y)^d}\xi_2:\sqrt[d]{(1-t)(\lambda z)^d}\xi_3\right]\\
|
|
& = &\left[\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1:\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2:\sqrt[d]{(1-t)z^d}\xi_3\right]
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
We can then see that both representatives will go to the same point by the retraction.
|
|
|
|
Lastly, if $[x^d: y^d: z^d]$ is in $(+,+,+)$, the retraction is the identity, and therefore both representatives will stay the same and therefore go to the same class.
|
|
\end{proof}
|
|
|
|
We have now seen that there is a homotopy equivalence between $\CC\PP^2\setminus C_d$ and $K$. Therefore, to get a CW-complex structure on $\CC\PP^2\setminus C_d$, it suffices to get it for $K$.
|
|
|
|
|
|
We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertices, $2$ edges and $1$ face, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
|
|
|
|
\begin{equation*}
|
|
\begin{array}{ccccc}
|
|
\Delta & \longleftrightarrow & S_1 & & \\
|
|
\uparrow& & \uparrow & & \\
|
|
\Delta^d & \longleftrightarrow & S_d & \hookrightarrow & M_d \\
|
|
& & \downarrow & & \downarrow \\
|
|
& & K & \hookrightarrow & \PP^2\setminus C_d \\
|
|
\end{array}
|
|
\end{equation*}
|
|
where $\Delta^d$ is a $\Delta-$complex that we are trying to find and identify with $S_d$, equivalently as what we have done for $d=1$. Since $S_d$ is a ramified cover of $S_1$ by the power to the $d$ map, we want $\Delta^d$ to be a ramified cover of $\Delta$ by the same map.
|
|
|
|
%Let $\phi: S_d\longleftarrow S_1$ be the map given by $\phi(x,y,z)=(x^d, y^d, z^d)$, then necessarily $\phi^{-1}(\Delta)=\Delta^d$.
|
|
%In particular, the $d^3$ vertices of the simplicial complex $\Delta^d$ will be identified to $(\xi_d,0,0)$, $(0,\xi_d,0)$, $(0,0,\xi_d)$ for all $\xi_d$ the $d$-$th$ roots of unity.
|
|
|
|
%The edges of $\Delta$ can be identified with the lines of $S_1$ parametrized as $(1-t,t,0)$, $(1-t,0,t)$ and $(0,t,1-t)$ with $t\in[0,1]$. Then, the preimages of these lines by $\phi$ are of the form $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$ in the first case and analogous in the rest, where we denote $\sqrt[s]{x}$ the positive real $d$-root for $x$ real, and where $\xi_i$ and $\xi_j$ are two any $d$-$th$ roots of unity. For each of these preimages we can define an edge between the corresponding vertices in $\Delta^d$.
|
|
|
|
%The $2$-faces of the simplicial complex $\Delta^d$ will then be identified to the areas between the lines $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$, $(\sqrt[d]{1-t}\xi_i,0, \sqrt[d]{t}\xi_k)$ and $(0,\sqrt[d]{t}\xi_j, \sqrt[d]{1-t}\xi_k)$.
|
|
First, we give a detailed description of $\Delta$. We will denote:
|
|
\begin{equation}
|
|
\begin{array}{lllll}
|
|
v_0 & = & (1,0,0) & & \\
|
|
v_1 & = & (0,1,0) & & \\
|
|
v_2 & = & (0,0,1) & & \\
|
|
l_0 & = & [v_1,v2] & = & \{(1-t)v_1+tv_2|t\in[0,1]\}\\
|
|
l_1 & = & [v_0,v2] & = & \{(1-t)v_0+tv_2|t\in[0,1]\}\\
|
|
l_2 & = & [v_0,v1] & = & \{(1-t)v_0+tv_1|t\in[0,1]\}\\
|
|
X & = & [v_0,v_1,v_2]
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
The face maps of $X$, $l_0$, $l_1$ and $l_2$ are:
|
|
\begin{equation}
|
|
\begin{array}{lllll}
|
|
\delta^2_0(X) & = & [v_1,v2] & = & l_0\\
|
|
\delta^2_1(X) & = & [v_0,v2] & = & l_1\\
|
|
\delta^2_2(X) & = & [v_0,v1] & = & l_2\\
|
|
\delta^1_0(l_0) & = & v_1 \\
|
|
\delta^1_1(l_0) & = & v_2\\
|
|
\delta^1_0(l_1) & = & v_0\\
|
|
\delta^1_1(l_1) & = & v_2\\
|
|
\delta^1_0(l_2) & = & v_0\\
|
|
\delta^1_1(l_2) & = & v_1\\
|
|
\end{array}
|
|
\end{equation}
|
|
Lastly, the border maps are given by:
|
|
\begin{equation}
|
|
\begin{array}{lll}
|
|
\partial^2 & = & \delta^2_2-\delta^2_1+\delta^2_0 \\
|
|
\partial^1 & = & \delta^2_1-\delta^2_0 \\
|
|
\end{array}
|
|
\label{bordermaps}
|
|
\end{equation}
|
|
|
|
|
|
$\Delta^d$ will have $3d$ $0-$simplices, $3d^2$ $1-$simplices and $d^3$ $2-$simplices.
|
|
|
|
|
|
Let $\mu_d=\{\xi| \xi^d=1\}$. We can express $S_d$ as an union
|
|
\begin{equation}
|
|
\begin{array}{rll}
|
|
S_d & = & \{(x,y,z)| x^d+y^d+z^d=1 \text{ and } x^d, y^d, z^d\in\RR_{\geq0}\} \\
|
|
& = & \bigcup_{(\xi_1, \xi_2, \xi_3)\in\mu^3_d} X_{(\xi_1, \xi_2, \xi_3)} \\
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
where $X_{(\xi_1, \xi_2, \xi_3)}=\{(x,y,z)| x^d+y^d+z^d=1 \text{ and } \xi_1x, \xi_2y, \xi_3z\geq0\}$.
|
|
|
|
Then, we can see that the faces of the simplex are given by $\mathring{X}_{(\xi_1, \xi_2, \xi_3)}$, while the edges are given by $X_{(\xi_1, \xi_2, 1)}\cap\{z=0\}$, $X_{(\xi_1, 1, \xi_2)}\cap\{y=0\}$, $X_{(1, \xi_1, \xi_2)}\cap\{x=0\}$ and the vertices by $X_{(\xi, 1, 1)}\cap\{y=0,z=0\}$, $X_{(1, 1, \xi)}\cap\{x=0,y=0\}$, $X_{(1, \xi, 1)}\cap\{x=0, z=0\}$.
|
|
We can remark that $X_{(\xi_1, \xi_2, \xi_3)}=(\xi_1, \xi_2, \xi_3)X_{(1,1,1)}$, and therefore that we have a group action of $\mu^3_d$ over $S_d$.
|
|
|
|
Let
|
|
\begin{equation}
|
|
\begin{array}{rll}
|
|
L^z_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{z=0\}\\
|
|
L^y_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{y=0\}\\
|
|
L^x_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{x=0\}\\
|
|
V^z_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{x=0, y=0\}\\
|
|
V^y_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{x=0, z=0\}\\
|
|
V^x_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{y=0, z=0\}\\
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
Then, the $2-$simplices, $1-$simplices and $0-$simplices of $\Delta^d$ are given by
|
|
\begin{equation}
|
|
\begin{array}{rll}
|
|
C_2(\Delta^d) & = & \ZZ[\mu^3_d]<X_{(1,1,1)}>\\
|
|
C_1(\Delta^d) & = & \ZZ[\mu^3_d]/\ZZ[\mu_d]<L^z_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu_d]<L^y_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu_d]<L^x_{(1,1,1)}>\\
|
|
C_0(\Delta^d) & = & \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^z_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^y_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^x_{(1,1,1)}>\\
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
The face maps of $X_{(1,1,1)}$, $L^z_{(1,1,1)}$, $L^y_{(1,1,1)}$ and $L^x_{(1,1,1)}$ are:
|
|
\begin{equation}
|
|
\begin{array}{lll}
|
|
\delta^2_0(X_{(1,1,1)}) & = & L^x_{(1,1,1)}\\
|
|
\delta^2_1(X_{(1,1,1)}) & = & L^y_{(1,1,1)}\\
|
|
\delta^2_2(X_{(1,1,1)}) & = & L^z_{(1,1,1)}\\
|
|
\delta^1_0(L^z_{(1,1,1)}) & = & V^y_{(1,1,1)} \\
|
|
\delta^1_1(L^z_{(1,1,1)}) & = & V^z_{(1,1,1)}\\
|
|
\delta^1_0(L^y_{(1,1,1)}) & = & V^x_{(1,1,1)}\\
|
|
\delta^1_1(L^y_{(1,1,1)}) & = & V^z_{(1,1,1)}\\
|
|
\delta^1_0(L^x_{(1,1,1)}) & = & V^x_{(1,1,1)}\\
|
|
\delta^1_1(L^x_{(1,1,1)}) & = & V^y_{(1,1,1)}\\
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
For the rest of the simplices, due to the group action, the face maps are given by:
|
|
\begin{equation}
|
|
\begin{array}{lll}
|
|
\delta^2_i(X_{(\xi_1, \xi_2, \xi_3)}) & = & (\xi_1, \xi_2, \xi_3)\delta^2_i(X_{(1,1,1)})\\
|
|
\delta^1_i(L^\bullet_{(\xi_1, \xi_2, \xi_3)}) & = & (\xi_1, \xi_2, \xi_3)\delta^1_i(L^\bullet_{(1,1,1)})\\
|
|
|
|
\end{array}
|
|
\end{equation}
|
|
|
|
Then, the border maps are given by (\ref{bordermaps}).
|
|
|
|
We have then defined $\Delta^d$, the simplicial complex with which we can identify $S_d$. Our last step now is to complete the diagram with a simplicial complex associated to the projective space $K$, which we will call $\Delta^d_\text{proj}$.
|
|
|
|
\begin{equation*}
|
|
\begin{array}{ccccc}
|
|
\Delta & \longleftrightarrow & S_1 & & \\
|
|
\uparrow& & \uparrow & & \\
|
|
\Delta^d & \longleftrightarrow & S_d & \longrightarrow & M_d \\
|
|
\downarrow & & \downarrow & & \downarrow \\
|
|
\Delta^d_\text{proj} & \longleftrightarrow & K & \longrightarrow & \PP^2\setminus C_d \\
|
|
\end{array}
|
|
\end{equation*}
|
|
|
|
Where the map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x,y,z)\longrightarrow[x:y:z]$. Since we are passing to the projective plane, two points $(x\xi_1, y\xi_2, z\xi_3)$ and $(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)$ will be identified, and therefore, so will $X_{(\xi_1, \xi_2, \xi_3)}$ and $X_{(\xi_1\xi, \xi_2\xi, \xi_3\xi)}$. As for the points where one coordinate is $0$, say, for example the first one, since the group action is then by $\mu^3_2/\mu_2$, we write the elements of $\mu^3_2/\mu_2$ as $(1,\xi_1,\xi_2)$, where the $1$ is on the coordinate that is $0$. Then, passing to the projective space, we will say that $L^x_{[1:\xi_1,\xi_2]}$ and $L^x_{[1:\xi'_1,\xi'_2]}$ are the same if $[1:\xi'_1,\xi'_2]=[1:\xi_1\xi,\xi_2\xi]$ for some $\xi\in\mu_2$. The rest of $1-$simplices would be identified similarly. Lastly, for the $0-$simplices, since we have $2$ zero coordinates, we have that all $0-$simplices with the same nonzero coordinate will be identified to each other. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices.
|
|
|
|
\begin{equation}
|
|
\begin{array}{rll}
|
|
C_2(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]<X_{[1:1:1]}>\\
|
|
C_1(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu_d]<L^z_{[1:1:1]}> \bigoplus \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu_d]<L^y_{[1:1:1]}> \bigoplus \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu_d]<L^x_{[1:1:1]}>\\
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C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^z_{[1:1:1]}> \bigoplus \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^y_{[1:1:1]}> \bigoplus \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^x_{[1:1:1]}>\\
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\end{array}
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\end{equation}
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\begin{example}{}
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We are going to compute now the simplicial complex for $S_d$ and $K$ when $d=2$, this is, our curve will be $C_D=\{[x:y:z]\in\PP^2|x^2+y^2+z^2=0\}$, $S_2=\{(x,y,z)\in M_2| x^2, y^2, z^2\in\RR_{\geq 0}\}$ and $K=\{[x:y:z]\in\PP^2|x^2,y^2,z^2\in\RR_{\geq0}\}$.
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We start with $S_2$. First, we can observe that, from all points in $\CC$, the only ones whose square power is real are either the real numbers of the pure imaginary numbers. And from these, the later have negative square power. Therefore, we can conclude that $S_2=\{(x,y,z)\in\RR^3| x^2+y^2+z^2=1\}$, that is, $S_2$ is the real three dimensional sphere. We now see the simplicial complex $\Delta^2$ that can be associated to it such that $\phi(\Delta^2)=\Delta$.
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In this case, we have that $\mu_2=\{\xi|\xi^2=1\}=\{1,-1\}$. Then, we have that $X_{(1,1,1)}=\{(x,y,z)\in\RR^3_{\geq0} |x^2+y^2+z^2=1\}$, this is, $X_{(1,1,1)}$ is the set of points given by the positive square roots of the points in $S_1$. And the rest of $2-$simplices will be given by the action of $\mu^3_2$ on $X_{(1,1,1)}$, which will give us the $2-$simplices $X_{(1,1,-1)}$, $X_{(1,-1,1)}$, $X_{(-1,1,1)}$, $X_{(1,-1,-1)}$, $X_{(-1,1,-1)}$, $X_{(-1,-1,1)}$ and $X_{(-1,-1,-1)}$.
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The face maps of $X_{(1,1,1)}$ are:
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\begin{equation}
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\begin{array}{lllll}
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\delta^2_0(X_{(1,1,1)}) & = & L^x_{(1,1,1)} & = & \{(0,\sqrt{1-t},\sqrt{t})|t\in[0,1]\}\\
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\delta^2_1(X_{(1,1,1)}) & = & L^y_{(1,1,1)} & = & \{(\sqrt{1-t},0, \sqrt{t})|t\in[0,1]\}\\
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\delta^2_2(X_{(1,1,1)}) & = & L^z_{(1,1,1)} & = & \{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
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\end{array}
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\end{equation}
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By applying the group action as shown before we get the faces of the rest of $2-$simplices.
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For the $1-$simplices, the group acting would be $\mu^3_2/\mu_2$, since one of the coordinates is $0$. Therefore, the $1-$simplices will be $L^x_{(1,1,1)}$, $L^x_{(1,1,-1)}$, $L^x_{(1,-1,1)}$, $L^x_{(1,-1,-1)}$, $L^y_{(1,1,1)}$, $L^y_{(1,1,-1)}$, $L^y_{(-1,1,1)}$, $L^y_{(-1,1,-1)}$, $L^z_{(1,1,1)}$, $L^z_{(1,-1,1)}$, $L^z_{(-1,1,1)}$ and $L^z_{(-1,-1,1)}$.
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The face maps of $L^z_{(1,1,1)}$, $L^y_{(1,1,1)}$ and $L^x_{(1,1,1)}$ are:
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\begin{equation}
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\begin{array}{lllll}
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\delta^1_0(L^z_{(1,1,1)}) & = & V^y_{(1,1,1)} & = & (1,0,0)\\
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\delta^1_1(L^z_{(1,1,1)}) & = & V^z_{(1,1,1)} & = & (0,1,0)\\
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\delta^1_0(L^y_{(1,1,1)}) & = & V^x_{(1,1,1)} & = & (1,0,0)\\
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\delta^1_1(L^y_{(1,1,1)}) & = & V^z_{(1,1,1)} & = & (0,0,1)\\
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\delta^1_0(L^x_{(1,1,1)}) & = & V^x_{(1,1,1)} & = & (0,1,0)\\
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\delta^1_1(L^x_{(1,1,1)}) & = & V^y_{(1,1,1)} & = & (0,0,1)\\
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\end{array}
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\end{equation}
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Lastly, the $0-$simplices will be $V^x_{(1,1,1)}=(1,0,0)$, $V^x_{(-1,1,1)}=(-1,0,0)$, $V^y_{(1,1,1)}=(0,1,0)$, $V^y_{(1,-1,1)}=(0,-1,0)$, $V^z_{(1,1,1)}=(0,0,1)$ and $V^z_{(1,1,-1)}=(0,0,-1)$. In Figure \ref{fig:S2} we can see the $\Delta-$complex that we associate to $S_2$.
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%The vertices of $\Delta$, $v_0$, $v_1$ and $v_2$, can be identified to $(1,0,0)$, $(0,0,1)$ and $(0,1,0)$ respectively. The preimages of these points are:
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%\begin{equation}
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%\begin{array}{rll}
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%\phi^{-1}((1,0,0))&=&\{(1,0,0),(-1,0,0)\}\\
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%\phi^{-1}((0,1,0))&=&\{(0,1,0),(0,-1,0)\}\\
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%\phi^{-1}((0,0,1))&=&\{(0,0,1),(0,0,-1)\}
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%\end{array}
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%\end{equation}
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%The edges of the simplex, $[v_0v_1]$, $[v_0v_2]$ and $[v_1v_2]$, can be identified to the points of $S_1$ given respectively by $\{(1-t, 0, t)|t\in[0,1]\}$, $\{(1-t,t,0)|t\in[0,1]\}$ and $\{(0,t,1-t)|t\in[0,1]\}$. Therefore, each edge will have four preimages. In the first case, these preimages are:
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%\begin{equation}
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%\phi^{-1}((1-t,t,0))=\left\{\begin{array}{l}
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% \{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
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% \{(\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}\\
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% \{(-\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
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% \{(-\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}
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%\end{array}\right.
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%\end{equation}
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%For the rest of the edges the preimages by $\phi$ would be analougous. We will denote the edges as a triple of $\{+,-,0\}$ indicating in each coordinate the sign of the root taken.
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%Finally, we will have $8$ $2$-faces. For three edges to form a face on the simplex, we need the corresponding vertices of the edges to coincide, that is, the edges with non-zero first coordinate must both take the same sign on said coordinate, and equivalently for the rest of coordinates. Then, the faces are defined by taking the square root of the coordinates of the points in $S_1$ with sign equal to those in the edges.
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\begin{figure}[!hp]
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw[gray, thin] (-5,0) to (5,0);
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\draw[gray, thin] (0,-5) to (0,5);
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\draw[gray, thin] (-4,-4) to (4,4);
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\draw[fill, red] (4,0) circle (2pt);
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\draw[fill, red] (-4,0) circle (2pt);
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\draw[fill, green] (0,4) circle (2pt);
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\draw[fill, green] (0,-4) circle (2pt);
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\draw[fill, blue] (1.41,1.41) circle (2pt);
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\draw[fill, blue] (-1.41,-1.41) circle (2pt);
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\draw (0,0) circle (4cm);
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\draw (-4,0) arc (180:360:4 and 1.5);
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\draw[dashed] (4,0) arc (0:180:4 and 1.5);
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\draw[dashed] (0,4) arc (90:270:-1.5 and 4);
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\draw (0,-4) arc (270:90:1.5 and 4);
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\draw[pink, thick](-4,0.02) arc (180:250:4 and 1.5);
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\draw[pink, thick](-4.02,0.02) arc (180:90:4);
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\draw[pink, thick](0,4) arc (270:90:1.5 and -4);
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\draw[purple, thick](-4,-0.02) arc (180:250:4 and 1.5);
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\draw[purple, thick](-4,-0.02) arc (180: 270:4);
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\draw[purple, thick] (-1.43,-1.41) arc (200:270:1.5 and 3.9);
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\draw[yellow, thick, dashed](-3.98,0.02) arc (180:70:4 and 1.5);
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\draw[yellow, thick, dashed](-3.98,0.02) arc (180: 90:4);
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\draw[yellow, thick, dashed] (1.39,1.41) arc (200:270:-1.5 and -3.9);
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\draw[orange, thick, dashed](-3.98,-0.02) arc (180:70:4 and 1.5);
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\draw[orange, thick, dashed](-3.98,-0.02) arc (180: 270:4);
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\draw[orange, thick, dashed] (1.39,1.41) arc (160:270:-1.5 and 4);
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\draw[teal, thick, dashed](4.02,0.02) arc (180:110:-4 and 1.5);
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\draw[teal, thick, dashed](4.02,0.02) arc (180: 270:-4);
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\draw[teal, thick, dashed] (1.42,1.41) arc (200:270:-1.5 and -3.9);
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\draw[violet, thick](4.02,-0.02) arc (0:110:4 and -1.5);
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\draw[violet, thick](4.02,-0.02) arc (180: 90:-4);
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\draw[violet, thick] (-1.39,-1.42) arc (200:270:1.5 and 3.9);
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\draw[magenta, thick, dashed](3.98,0.02) arc (180:110:-4 and 1.5);
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\draw[magenta, thick, dashed](3.98,-0.02) arc (180: 90:-4);
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\draw[magenta, thick, dashed] (1.43,1.41) arc (160:270:-1.5 and 4);
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\draw[lime, thick, dashed](4.02,0.02) arc (0:110:4 and -1.5);
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\draw[lime, thick, dashed](3.98,0.02) arc (180: 270:-4);
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\draw[lime, thick, dashed] (0.02,3.98) arc (270:160:1.5 and -4);
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\end{tikzpicture}
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\end{center}
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\caption{Division in simplices of $S_2$}
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\label{fig:S2}
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\end{figure}
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Now, we give a description of $\Delta^2_\text{proj}$. First, for the $0-$simplices, we have that $V^x_{[1,1,1]}=V^x_{[-1,1,1]}=[1:0:0]$, $V^y_{[1,1,1]}=V^y_{[1,-1,1]}=[0:1:0]$ and $V^z_{[1,1,1]}=V^z_{[1,1,-1]}=[0:0:1]$.
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We will have $6$ $1-$simplices, obtained from the $12$ of $\Delta^d$ by identifying: $L^x_{[1:1:1]}=L^x{[1:-1:-1]}$, $L^x_{[1:-1:1]}=L^x{[1:1:-1]}$, $L^y_{[1:1:1]}=L^y{[1:-1:-1]}$, $L^y_{[1:-1:1]}=L^y{[1:1:-1]}$, $L^z_{[1:1:1]}=L^z{[1:-1:-1]}$ and $L^z_{[1:-1:1]}=L^z{[1:1:-1]}$.
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Lastly, we will have $4$ $2-$simplices, $X_{[1:1:1]}=X_{[-1:-1:-1]}$, $X_{[1:1:-1]}=X_{[-1:-1:1]}$, $X_{[1:-1:1]}=X_{[-1:1:-1]}$ and $X_{[-1:1:1]}=X_{[1:-1:-1]}$.
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The face maps will be obtained by passing to the projective the face maps of $\Delta^d$.
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\end{example}
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\begin{thebibliography}{99}
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|
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\bibitem{andreotti1959}
|
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A. Andreotti and T. Frankel,
|
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\textit{The Lefschetz theorem on hyperplane sections},
|
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Annals of Mathematics,
|
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Vol. 69, No. 3,
|
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pp. 713--717,
|
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1959.
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\bibitem{libgober1986}
|
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A. Libgober,
|
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\textit{On the homotopy type of the complement to plane algebraic curves},
|
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J. Reine Angew. Math.,
|
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Vol. 367,
|
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pp. 103--114,
|
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1986.
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DOI: 10.1515/crll.1986.367.103
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\end{thebibliography}
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\end{document}
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