pre-arxiv

generaldegree.tex

@@ -11,10 +11,13 @@ decorations.markings, calc}
 
 \usepackage{hyperref}
 
-\newtheorem{theorem}{Theorem}
+\newtheorem{theorem}{Theorem}[section]
-\newtheorem{example}{Example}
+\newtheorem{lemma}[theorem]{Lemma}
-\newtheorem{lemma}{Lemma}
+\theoremstyle{definition}
-\newtheorem{definition}{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.}
+\title[Retraction of the complement of a smooth curve]{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}
+\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}
+%	\begin{array}{ccc}
-		S_d & \longhookrightarrow & M_d \\
+%		S_d & \longhookrightarrow & M_d \\
-		\downarrow & & \downarrow \\
+%		\downarrow & & \downarrow \\
-		K & \longhookrightarrow & \PP^2\setminus C_d \\
+%		K & \longhookrightarrow & \PP^2\setminus C_d \\
-	\end{array}
+%	\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 $r_1(M_1,1)=M_1\cap\RR^3$
+We will now define a strong retraction map 
-% and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
-In particular, we can take $r_1$ that takes the imaginary part to $0$:
 \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\\
+M_1\times{[0,1]}\rar["r_1"]&M_1\\
-(P, t)\rar[mapsto]& t \Re P + (1 - t) \Im P.
+(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}
+\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 $M_1\cap\RR^3$.
+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}
+\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
-		r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
+(M_1\cap\RR^3)\times{[0,1]}\rar["r_2"]&M_1\cap\RR^3\times [0,1]\\
-		(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
+(Q, t)\rar[mapsto]& t P_Q + (1 - t) P,
-	\end{array}
+\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}
+	\begin{aligned}{}
-		\tilde{r}_2([\lambda x\xi_1,\lambda y\xi_2,\lambda z\xi_3],t)& = &\\
+		\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[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[\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]
+		& = \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{array}
+	\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}
+%	\begin{array}{ccccc}
-		\Delta & \longleftrightarrow & S_1 & & \\
+%		\Delta & \longleftrightarrow & S_1 & & \\
-		\uparrow& & \uparrow & & \\
+%		\uparrow& & \uparrow & & \\
-		\Delta^d & \longleftrightarrow & S_d & \hookrightarrow & M_d \\
+%		\Delta^d & \longleftrightarrow & S_d & \hookrightarrow & M_d \\
-		& & \downarrow & & \downarrow \\
+%		& & \downarrow & & \downarrow \\
-		& & K & \hookrightarrow & \PP^2\setminus C_d \\
+%		& & K & \hookrightarrow & \PP^2\setminus C_d \\
-	\end{array}
+%	\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_0 & = & [v_1,v_2] & = & \{(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_1 & = & [v_0,v_2] & = & \{(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_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}
+\begin{aligned}
-	\delta^2_0(X) & = & [v_1,v2] & = & l_0\\
+	\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,v2] & = & l_1\\
+	\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,v1] & = & l_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,
-	\delta^1_0(l_0) & = & v_1 \\
+\end{aligned}
-	\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:
+an the boundary maps are given by:
 \begin{equation}
-\begin{array}{lll}
+\partial^2 =\delta^2_2-\delta^2_1+\delta^2_0 \qquad
-\partial^2 & = & \delta^2_2-\delta^2_1+\delta^2_0 \\
+\partial^1 = \delta^2_1-\delta^2_0.
-\partial^1 & = & \delta^2_1-\delta^2_0 \\
-\end{array}
 \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 a union
-
-
-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)} \\
+		&  = & \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}
+\begin{aligned}
-C_2(\Delta^d) & = & \ZZ[\mu^3_d]<X_{(1,1,1)}>\\
+	\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)}\\
-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)}>\\
+	\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)}\\
-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)}>\\
+	\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{array}
+\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:
+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}{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}
+%	\begin{array}{ccccc}
-		\Delta & \longleftrightarrow & S_1 & & \\
+%		\Delta & \longleftrightarrow & S_1 & & \\
-		\uparrow& & \uparrow & & \\
+%		\uparrow& & \uparrow & & \\
-		\Delta^d & \longleftrightarrow & S_d & \longrightarrow & M_d \\
+%		\Delta^d & \longleftrightarrow & S_d & \hookrightarrow & M_d \\
-		 \downarrow & & \downarrow & & \downarrow \\
+%		& & \downarrow & & \downarrow \\
-		\Delta^d_\text{proj} & \longleftrightarrow & K & \longrightarrow & \PP^2\setminus C_d \\
+%		& & K & \hookrightarrow & \PP^2\setminus C_d \\
-	\end{array}
+%	\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.
+The group $\mu_d^3$ also acts, but the action of the diagonal vectors is trivial. Let us denote 
-
+\[
-\begin{equation}
+T_d:=\ZZ\left[\frac{\mu_d^3}{\mu_d\langle(\xi,\xi,\xi)\rangle}\right];
-\begin{array}{rll}
+\]
-C_2(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]<X_{[1:1:1]}>\\
+we use variables $t_x,t_y,t_z$ as for $R_d$ and we have to add the relation $t_x t_y t_z=1$.
-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]}>\\
+We have the following chain complexes:
-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]}>\\
+\begin{align*}
-\end{array}
+C_2(\Delta^d_\text{proj}) &= T_d\langle X_{[1:1:1]}\rangle\\
-\end{equation}
+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 $K=\{[x:y:z]\in\PP^2|x^2,y^2,z^2\in\RR_{\geq0}\}$.