alarraya/Curvas
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

pre-arxiv

Browse Source
This commit is contained in:
Enrique Artal Bartolo 2026-05-25 15:50:11 +02:00
parent 0e3cc55745
commit a55111586c
1 changed files with 259 additions and 134 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

393
generaldegree.tex
View File

@ -11,10 +11,13 @@ decorations.markings, calc}
\usepackage{hyperref}
\newtheorem{theorem}{Theorem}
\newtheorem{example}{Example}
\newtheorem{lemma}{Lemma}
\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\theoremstyle{definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{ntc}[theorem]{Notation}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\DD}{\mathbb{D}}
@ -24,13 +27,36 @@ decorations.markings, calc}
\newcommand{\ZZ}{\mathbb{Z}}
\numberwithin{equation}{section}
%opening
\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}
\title[Retraction of the complement of a smooth curve]{Retraction of the complement of smooth projective curves to a $2$-dimensional $\Delta$-complex.}
\author[E. Artal]{Enrique Artal Bartolo}
\address{Departamento de Matem\'{a}ticas, IUMA \\
Universidad de Zaragoza \\
C.~Pedro Cerbuna 12, 50009, Zaragoza, Spain}
\urladdr{\url{http://riemann.unizar.es/~artal}}
\email{\href{mailto:artal@unizar.es}{artal@unizar.es}}
\author[A. Larraya]{Alba Larraya Sancho}
\address{Basque Center for Applied Mathematics\\
Alameda de Mazarredo 14,
48009 Bilbao, Spain}
%\urladdr{\url{http://riemann.unizar.es/~mmarco}}
\email{\href{mailto:alarraya@bcamath.org}{alarraya@bcamath.org}}
\author[M.Á. Marco]{Miguel Á. Marco-Buzunáriz}
\address{Departamento de Matem\'{a}ticas, IUMA \\
Universidad de Zaragoza \\
C.~Pedro Cerbuna 12, 50009, Zaragoza, Spain}
\urladdr{\url{http://riemann.unizar.es/~mmarco}}
\email{\href{mailto:mmarco@unizar.es}{mmarco@unizar.es}}
\thanks{The first and third named authors are partially supported by PID2024-156181NB-C33 funded by
MICIU/AEI/10.13039/501100011033. The second named author is partially supported by PID2024-156181NB-C31 funded by
MICIU/AEI/10.13039/501100011033.}
\begin{document}, Bilbao, Spain
\begin{document}
\maketitle
\begin{abstract}
@ -38,8 +64,10 @@ Due to a result by Andreotti and Frankel \cite{andreotti1959}, it can be seen th
The retraction is performed in several steps, using the branched cover structure of the Fermat curves to the degree $1$ case.
\end{abstract}
\maketitle
\section{Definition of the problem}
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 $\CC\PP^2$).
Given $C$ an affine 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).
@ -47,26 +75,30 @@ Our construction can be easily generalized to higher dimensions. We plan to writ
\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}
\begin{equation*}
C_d=\{[x:y:z]\in\CC\PP^2 | x^d+y^d+z^d=0\}
\end{equation}
\end{equation*}
by an ambient isotopy.
Therefore, we will work with $C_d$, from now on.
We will denote:
\begin{eqnarray}
\begin{eqnarray*}
M_d&=&\{(x,y,z)\in\CC^3 | x^d+y^d+z^d=1\} \\
S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
K &=&\{[x:y:z]\in\CC\PP^2 \setminus M_2 | x^d, y^d, z^d\in\RR_{\geq0}\}
\end{eqnarray}
\end{eqnarray*}
We then have a commutative diagram:
\begin{equation*}
\begin{array}{ccc}
S_d & \longhookrightarrow & M_d \\
\downarrow & & \downarrow \\
K & \longhookrightarrow & \PP^2\setminus C_d \\
\end{array}
% \begin{array}{ccc}
% S_d & \longhookrightarrow & M_d \\
% \downarrow & & \downarrow \\
% K & \longhookrightarrow & \PP^2\setminus C_d \\
% \end{array}
\begin{tikzcd}
S_d\rar[hookrightarrow]\dar& M_d\dar\\
K\rar[hookrightarrow] & \PP^2\setminus C_d
\end{tikzcd}
\end{equation*}
where the vertical maps consist on taking equivalence classes.
@ -74,6 +106,11 @@ We are going to see that the inclusion $i:S_d\longrightarrow M_d$ induces a homo
We will start by doing the construction for the case $d=1$, since the general case will be constructed on top of it.
\begin{ntc}
For a complex number $x\in\CC$, we denote its real and imaginary parts by $x_r,x_i$, respectively,
i.e. $x_ir,x_i\in\RR$ and $x=x_r + i x_i$. This notation also applies to vectors in $\CC^n$.
\end{ntc}
%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:
@ -82,43 +119,51 @@ We will start by doing the construction for the case $d=1$, since the general ca
% M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
% \end{array}
%\end{equation*}
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 $0$:
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\cap\RR^3\\
(P, t)\rar[mapsto]& t \Re P + (1 - t) \Im P.
M_1\times{[0,1]}\rar["r_1"]&M_1\\
(P, t)\rar[mapsto]& t P_r + (1 - t) P_i=:r_{1t}(P),
\end{tikzcd}
\end{equation}
where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively.
such that $r_{10}$ is the identity and $r_{11}(M_1)=M_1\cap\RR^3$
and the restriction of $r_{1t}$ to $M_1\cap\RR^3$ 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 $0$:
%where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively.
Note that for $(x,y,z)\in M_1$, we already have that
\begin{equation}
\begin{equation*}
\begin{array}{lll}
x_r+y_r+z_r & = & 1 \\
x_i+y_i+z_i & = & 0 \\
\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{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 $(x,y,z)\in M_1$.
Let us summarize it.
This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations.
We have just proven that
%$r_1$ is then a strong retraction from $M_1$ to $M_1\cap\RR^3$.
%
%This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations.
%We have just proven that
\begin{lemma}
The map $r_1$ is a strong deformation retraction from $M_1$ to $M_1\cap\RR^3$.
\end{lemma}
Now we will define a retraction from $M_1 \cap \RR^3$ to $S_1$.
Now we will define a map $r_2:(M_1 \cap \RR^3)\times[0, 1]\to M_1 \cap \RR^3$
which will define a strong deformation 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
is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and
\[
S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
\]
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:
\begin{figure}[h]
\begin{figure}[ht]
\begin{center}
\begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
% \draw (3,4.5) to (-6,-9);
@ -171,17 +216,27 @@ i.e, $S_1$ is the triangular section defined by the intersection of this plane w
\end{figure}
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 $(+,+,+)$.
Note that there is no section $(-,-,-)$.
The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
The retraction in $(+,+,+)$ is the identity for any value of $t$. For the rest of the parts, we define a retraction:
\begin{equation}
\begin{array}{ccc}
r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
\end{array}
\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
(M_1\cap\RR^3)\times{[0,1]}\rar["r_2"]&M_1\cap\RR^3\times [0,1]\\
(Q, t)\rar[mapsto]& t P_Q + (1 - t) P,
\end{tikzcd}
\end{equation}
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:
%\begin{equation}
% \begin{array}{ccc}
% r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
% (x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
% \end{array}
%\end{equation}
where $Q_P$ is a point depending on $Q\in M_1\cap\RR^3$.
If $Q$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_Q$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ respectively. If $Q$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_Q$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $Q_P$ is the point of this line that intersects with the corresponding edge of the triangle. One can follow the
green lines in Figure~\ref{fig:triangulo}. Summarizing:
\begin{equation}
P_{x,y,z}=\left\{\begin{array}{lr}
Q_{(x,y,z)}=\left\{\begin{array}{lr}
(x,y,z)&(+,+,+) \\
(1,0,0)&(+,-,-) \\
(0,1,0)&(-,+,-) \\
@ -196,21 +251,21 @@ Since we have defined the retraction by regions, we are now going to show that i
We start with the border between $(+,+,-)$ and $(+,+,+)$, this is, the line with $z=0$. In this case
\begin{equation*}
P_{x,y,z}=\left\{\begin{array}{lr}
Q_{x,y,z}=\left\{\begin{array}{lr}
(x,y,0)&(+,+,+) \\
(\frac{x}{1-0},\frac{y}{1-0},0)=(x,y,0)&(+,+,-) \\
\end{array}\right.
\end{equation*}
For the border between $(+,+,-)$ and $(-,+,-)$, the border is the line $x=0$, so we get
\begin{equation*}
P_{x,y,z}=\left\{\begin{array}{lr}
Q_{x,y,z}=\left\{\begin{array}{lr}
(\frac{0}{1-z},\frac{y}{1-z},0)=(0,1,0)&(+,+,-) \\
(0,1,0)&(-,+,-) \\
\end{array}\right.
\end{equation*}
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 $y=0$
\begin{equation*}
P_{x,y,z}=\left\{\begin{array}{lr}
Q_{x,y,z}=\left\{\begin{array}{lr}
(\frac{x}{1-z},\frac{0}{1-z},0)=(1,0,0)&(+,+,-) \\
(1,0,0)&(+,-,-) \\
\end{array}\right.
@ -219,7 +274,7 @@ similarly to the previous case, in this line $x=1-z$.
We have now proven that:
\begin{lemma}
$r_2$ is a strong deformation retraction from $M_1\cap \RR^3$ to $S_1$.
The map $r_2$ is a strong deformation retraction from $M_1\cap \RR^3$ to $S_1$.
\end{lemma}
@ -345,7 +400,7 @@ We can the write the lifting of $r_1$ as $\tilde{r}_1((x,y,z),t)=(G(x,t), G(y,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}
Q^d_{x,y,z}=\left\{\begin{array}{lr}
(x^d,y^d,z^d)&(+,+,+) \\
(1,0,0)&(+,-,-) \\
(0,1,0)&(-,+,-) \\
@ -357,7 +412,11 @@ P_{x,y,z}=\left\{\begin{array}{lr}
\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.
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 $M_1$.
@ -418,12 +477,17 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
\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}
\begin{aligned}{}
\tilde{r}_2([\lambda x\xi_1,\lambda y\xi_2,\lambda z\xi_3],t)& =
\left[h_d (x, z, t, \lambda) \xi_1: h_d (y, z, t, \lambda)\xi_2:\sqrt[d]{(1-t)(\lambda z)^d}\xi_3\right]\\
& = \left[h_d (x, z,t,1) \xi_1: h_d (y, z,t,1)\xi_2:\sqrt[d]{(1-t)z^d}\xi_3\right],
\end{aligned}
\end{equation}
where
\[
h_d (x, z, t, \lambda):=
\sqrt[d]{\left(1-t+\frac{t}{1-(\lambda z)^d}\right)(\lambda x)^d}
\]
We can then see that both representatives will go to the same point by the retraction.
@ -436,13 +500,18 @@ We have now seen that there is a homotopy equivalence between $\CC\PP^2\setminus
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}
% \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}
\begin{tikzcd}
\Delta \rar[leftrightarrow]& S_1&\\
\Delta^d\uar\rar[leftrightarrow]&S_d\uar\rar[hookrightarrow]\dar& M_d\dar\\
&K \rar[hookrightarrow]& \PP^2\setminus C_d,
\end{tikzcd}
\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.
@ -458,53 +527,49 @@ First, we give a detailed description of $\Delta$. We will denote:
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]\}\\
l_0 & = & [v_1,v_2] & = & \{(1-t)v_1+tv_2|t\in[0,1]\}\\
l_1 & = & [v_0,v_2] & = & \{(1-t)v_0+tv_2|t\in[0,1]\}\\
l_2 & = & [v_0,v_1] & = & \{(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}
\begin{aligned}
\delta^2_0(X) & =[v_1,v_2] = l_0 &&& \delta^1_0(l_0)&= v_1 &&& \delta^1_1(l_0) & = v_2\\
\delta^2_1(X) & =[v_0,v_2] = l_1 &&& \delta^1_0(l_1)&= v_0 &&& \delta^1_1(l_1) & = v_2\\
\delta^2_2(X) & =[v_0,v_1] = l_2 &&& \delta^1_0(l_2)&= v_0 &&& \delta^1_1(l_2) & = v_1,
\end{aligned}
\end{equation}
Lastly, the border maps are given by:
an the boundary 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}
\partial^2 =\delta^2_2-\delta^2_1+\delta^2_0 \qquad
\partial^1 = \delta^2_1-\delta^2_0.
\label{bordermaps}
\end{equation}
We define now the $\Delta$-complex
$\Delta^d$. It has $3d$ $0-$simplices, $3d^2$ $1-$simplices and $d^3$ $2-$simplices.
$\Delta^d$ 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
Let $\mu_d=\{\xi| \xi^d=1\}$. We can express $S_d$ as a 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)} \\
& = & \displaystyle\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 $X_{(\xi, 1, 1)}\cap\{y=0,z=0\}$, $X_{(1, 1, \xi)}\cap\{x=0,y=0\}$, $X_{(1, \xi, 1)}\cap\{x=0, z=0\}$.
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}
@ -516,63 +581,123 @@ Let
V^x_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{y=0, z=0\}\\
\end{array}
\end{equation}
We can describe the chain complexes. They are abelian groups but they are also $R_d$-complexes
where $R_d$ is the group algebra $\ZZ[\mu_d^3]$. For the $2$-chains we have
\[
C_2(\Delta^d) = R_d\langle X_{(1,1,1)}\rangle,
\]
which is a free principal $R_d$-module, i.e. a free abelian grouop of rank~$d^3$. For the $1$-chains we deal with non-free modules.
Let us denote as $t_x, t_y, t_z$ the variables of $R_d$; note that $t_x^d=t_y^d=t_z^d=1$.
Then
\[
C_1(\Delta^d) = \frac{R_d}{(t_x - 1)} \langle L^x_{(1,1,1)}\rangle\
\oplus\ \frac{R_d}{(t_y - 1)} \langle L^y_{(1,1,1)}\rangle
\oplus\ \frac{R_d}{(t_z - 1)} \langle L^z_{(1,1,1)}\rangle,
\]
i.e, a free abelian group of rank $d^2$. Finalyy the $0$-chains are
\[
C_0(\Delta^d) = \frac{R_d}{(t_y - 1, t_z - 1)} \langle V^x_{(1,1,1)}\rangle\
\oplus\ \frac{R_d}{(t_x - 1, t_z - 1 )} \langle V^y_{(1,1,1)}\rangle
\oplus\ \frac{R_d}{(t_x - 1, t_y - 1)} \langle V^z_{(1,1,1)}\rangle,
\]
a free abelian group of rank~$3d$.
%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}
Then, the $2-$simplices, $1-$simplices and $0-$simplices of $\Delta^d$ are given by
The face maps are $R_d$-morphisms defined as
\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}
\begin{aligned}
\delta^2_0(X_{(1,1,1)}) & =L^x_{(1,1,1)} &&& \delta^1_0(L^x_{(1,1,1)})&= V^y_{(1,1,1)} &&& \delta^1_1(L^x_{(1,1,1)}) & = V^z_{(1,1,1)}\\
\delta^2_1(X_{(1,1,1)}) & =L^y_{(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^2_2(X_{(1,1,1)}) & =L^z_{(1,1,1)} &&& \delta^1_0(L^z_{(1,1,1)})&= V^x_{(1,1,1)} &&& \delta^1_1(L^z_{(1,1,1)}) & = V^y_{(1,1,1)}.
\end{aligned}
\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}
The boundary maps are defines as in \eqref{bordermaps}.
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 $\Delta^d_\text{proj}$.
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 $\Delta^d_\textrm{proj}$.
\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}
% \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}
\begin{tikzcd}
\Delta \rar[leftrightarrow]& S_1&\\
\Delta^d\uar\rar[leftrightarrow]\dar&S_d\uar\rar[hookrightarrow]\dar& M_d\dar\\
\Delta^d_\textrm{proj}\rar[leftrightarrow]&K \rar[hookrightarrow]& \PP^2\setminus C_d,
\end{tikzcd}
\end{equation*}
%\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*}
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
The $2$-cells $X_{(\xi_1, \xi_2, \xi_3)}$ and $X_{(\xi_1\xi, \xi_2\xi, \xi_3\xi)}$ of $\Delta^d$, project the same $2$-cell of $\Delta^d_\textrm{proj}$, which will be denoted as $X_{[\xi_1: \xi_2: \xi_3]}$.
The same convention is followed for the edges and the vertices. Note that for $L^x_{[\xi_1: \xi_2: \xi_3]}$
the value of $\zeta_1$ is irrelevant; something similar happens for $L^y_{[\xi_1: \xi_2: \xi_3]}$
(the value of $\zeta_2$ is irrelevant), and for $L^z_{[\xi_1: \xi_2: \xi_3]}$
(the value of $\zeta_3$ is irrelevant).
For the vertices we have only three vertices $V^x, V^y, V^z$.
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]}>\\
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]}>\\
\end{array}
\end{equation}
The group $\mu_d^3$ also acts, but the action of the diagonal vectors is trivial. Let us denote
\[
T_d:=\ZZ\left[\frac{\mu_d^3}{\mu_d\langle(\xi,\xi,\xi)\rangle}\right];
\]
we use variables $t_x,t_y,t_z$ as for $R_d$ and we have to add the relation $t_x t_y t_z=1$.
We have the following chain complexes:
\begin{align*}
C_2(\Delta^d_\text{proj}) &= T_d\langle X_{[1:1:1]}\rangle\\
C_1(\Delta^d_\text{proj}) &= \frac{T_d}{(t_x - 1)} \langle L^x_{[1:1:1]}\rangle\
\oplus\ \frac{T_d}{(t_y - 1)} \langle L^y_{[1:1:1]}\rangle
\oplus\ \frac{T_d}{(t_z - 1)} \langle L^z_{[1:1:1]}\rangle,\\
C_0(\Delta^d) &= \frac{T_d}{(t_y - 1, t_z - 1)} \langle V^x\rangle\
\oplus\ \frac{T_d}{(t_x - 1, t_z - 1 )} \langle V^y\rangle
\oplus\ \frac{T_d}{(t_x - 1, t_y - 1)} \langle V^z\rangle.
\end{align*}
%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]}>\\
%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]}>\\
%\end{array}
%\end{equation}
\begin{example}{}
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 $K=\{[x:y:z]\in\PP^2|x^2,y^2,z^2\in\RR_{\geq0}\}$.