From ef9c05010b1e9a26a552dedd62f789f4f058e8d1 Mon Sep 17 00:00:00 2001 From: Enrique Artal Date: Fri, 11 Sep 2026 10:50:34 +0200 Subject: [PATCH] . --- generaldegreeHypersurfaces.tex | 35 +++++++++++++++++++--------------- 1 file changed, 20 insertions(+), 15 deletions(-) diff --git a/generaldegreeHypersurfaces.tex b/generaldegreeHypersurfaces.tex index 330c388..5d0d02a 100644 --- a/generaldegreeHypersurfaces.tex +++ b/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}