From 6498e97d2f809a0fd91ee47a82b95ad8cb948a1b Mon Sep 17 00:00:00 2001 From: Enrique Artal Date: Thu, 24 Sep 2026 13:30:36 +0200 Subject: [PATCH 1/2] mitad de pagina 7 --- generaldegreeHypersurfaces.tex | 224 +++++++++++++++++---------------- 1 file changed, 113 insertions(+), 111 deletions(-) diff --git a/generaldegreeHypersurfaces.tex b/generaldegreeHypersurfaces.tex index d8fe6c6..431098d 100644 --- a/generaldegreeHypersurfaces.tex +++ b/generaldegreeHypersurfaces.tex @@ -1,5 +1,6 @@ \documentclass[a4paper,10pt]{amsart} \usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amsmath} @@ -28,6 +29,9 @@ decorations.markings, calc} \newcommand{\proy}[1]{\pi_{#1}} +\addtolength{\hoffset}{-1cm} +\addtolength{\textwidth}{2cm} +\linespread{1.2} \numberwithin{equation}{section} %opening @@ -44,7 +48,7 @@ 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} +\author[M.Á. Marco]{Miguel \'A. Marco-Buzunáriz} \address{Departamento de Matem\'{a}ticas, IUMA \\ Universidad de Zaragoza \\ C.~Pedro Cerbuna 12, 50009, Zaragoza, Spain} @@ -55,6 +59,13 @@ C.~Pedro Cerbuna 12, 50009, Zaragoza, Spain} MICIU/AEI/10.13039/501100011033. The second named author is partially supported by PID2024-156181NB-C31 funded by MICIU/AEI/10.13039/501100011033 and CEX2021-001142-S-20-9.} + +\tikzcdset{% + harpoonlurd/.code={\pgfsetarrows{tikzcd right to-tikzcd right to}}, + harpoonldru/.code={\pgfsetarrows{tikzcd left to-tikzcd left to}}, + harpoonluru/.code={\pgfsetarrows{tikzcd right to-tikzcd left to}}, + harpoonldrd/.code={\pgfsetarrows{tikzcd left to-tikzcd right to}}, +} \begin{document} @@ -134,12 +145,10 @@ and the restriction of $r_{1t}$ to $M_1\cap\RR^{n+1}$ is the inclusion. %where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively. Note that for $(x_0,\dots,x_n)\in M_1$, we already have that -\begin{equation*} -\begin{array}{lll} - \sum x^r_j & = & 1 \\ - \sum x^i_j & = & 0 \\ -\end{array} -\end{equation*} +\begin{align*} + \sum_{j=0}^n x^r_j & = 1 \\ + \sum_{j=0}^n x^i_j & = 0 \\ +\end{align*} and therefore $r_1(x_0,\dots,x_n,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x_0,\dots,x_n,1)=(x^r_0,\dots,x^r_n)\in M_1\cap\RR^{n+1}$ for all $(x_0,\dots,x_n)\in M_1$. Let us summarize it. @@ -166,13 +175,11 @@ 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 -\begin{equation*} -\begin{array}{lllll} -S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\ -&=&\{x_0e_0+\dots+x_ne_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\ -\end{array} +\begin{align*} +S_1&=\left\{(x_0,\dots,x_n)\in\RR^{n+1}\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\} \\ +&=\left\{x_0e_0+\dots+x_ne_n\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\}=\Delta_n \qedhere -\end{equation*} +\end{align*} \end{proof} \begin{figure}[ht] @@ -231,12 +238,12 @@ As we can see in Figure \ref{fig:triangulo}, the hyperplanes defined by $\{x_i=0 Note that there is no region $(-,\dots,-)$. The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the rest of the parts, we define a retraction: -\begin{equation} +\begin{equation*} \begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}] (M_1\cap\RR^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\\ (Q, t)\rar[mapsto]& t P_Q + (1 - t) Q, \end{tikzcd} -\end{equation} +\end{equation*} %\begin{equation} % \begin{array}{ccc} % r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\ @@ -246,14 +253,15 @@ The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the re where $P_Q$ is a point depending on $Q\in M_1\cap\RR^{n+1}$. In general, for a point $Q$ in a region $(\pm,\dots,\pm)$, we want the retraction to take the negative coordinates to $0$ and to change the positive ones such that at the end the sum is $1$. Therefore, we can write the point $P_Q$ as: -\begin{equation} -P_Q=\left(\frac{x_0\chi_0}{1-\sum x_j(1-\chi_j)}, \dots,\frac{x_n\chi_n}{1-\sum x_j(1-\chi_j)} \right) -\end{equation} - +\begin{equation*} +P_Q=\left(\frac{x_0\chi_0}{\displaystyle 1-\sum_{j=0}^n x_j(1-\chi_j)}, \dots, +\frac{x_n\chi_n}{\displaystyle 1-\sum_{j=0}^n x_j(1-\chi_j)} \right) +\end{equation*} where $\chi_j=1$ when $x_j\geq0$ and $\chi_j=0$ otherwise. It is easy to check that the points $tP_Q+(1-t)Q$ are in $M_1$ for all $t$. To see this, we remark that $\sum x_i=1$ implies that $\sum x_i\chi_i+\sum x_i(1-\chi_i)=1$ and therefore $\sum x_i\chi_i=1-\sum x_i(1-\chi_i)$. Then -\begin{equation} -\sum \left((1-t)x_i+\frac{tx_i\chi_i}{1-\sum x_j(1-\chi_j)}\right)=(1-t)\sum x_i+\frac{t}{1-\sum x_j(1-\chi_j)}\sum x_i\chi_i=1 -\end{equation} +\begin{equation*} +\sum_{i=0}^n \left((1-t)x_i+\frac{tx_i\chi_i}{\displaystyle 1-\sum_{i=0}^n x_j(1-\chi_j)}\right)= +(1-t)\sum_{i=0}^n x_i+\frac{t}{\displaystyle 1-\sum_{i=0}^n x_j(1-\chi_j)}\sum_{i=0}^n x_i\chi_i=1. +\end{equation*} Then, we have proved the following result: \begin{lemma} @@ -265,18 +273,17 @@ The map $r_2$ is a strong deformation retraction from $M_1\cap \RR^{n+1}$ to $S_ 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 $S_d$. \begin{theorem} - $S_d$ is a strong deformation retract of $M_d$. + The space $S_d$ is a strong deformation retract of $M_d$. \end{theorem} \begin{proof} We want to produce the following diagram: \begin{equation*} - \begin{array}{ccccc} - M_d & \rightleftharpoons & M_d\cap\{x^d_i\in\RR\forall i\}&\rightleftharpoons& S_d \\ - \downarrow & & \downarrow & & \downarrow\\ - M_1 & \rightleftharpoons & M_1\cap\RR^{n+1} & \rightleftharpoons & S_1 \\ - \end{array} +\begin{tikzcd} + (x_0,\dots,x_n)\dar[mapsto] & [-30pt] M_d \rar[shift left=2pt,harpoon] \dar & \lar[shift left=2pt,harpoon] M_d\cap\{x^d_i\in\RR,\,\forall i\}\rar[shift left=2pt,harpoon]\dar & \lar[shift left=2pt,harpoon]S_d \dar\\ + (x^d_0,\dots,x^d_n) &M_1 \rar[shift left=2pt,harpoon] & \lar[shift left=2pt,harpoon] M_1\cap\RR^{n + 1}\rar[shift left=2pt,harpoon] & \lar[shift left=2pt,harpoon]S_1 +\end{tikzcd} \end{equation*} -where the horizontal arrows represent retractions, and the vertical arrows are the maps $(x_0,\dots,x_n)\mapsto (x^d_0,\dots,x^d_n)$. 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. +where the horizontal arrows represent retractions. 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. We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, we can write it as $r_1((x_0,\dots,x_n),t)=(\bar{r}(x_0,t),\dots,\bar{r}(x_n,t))$ with $\sum\bar{r}(x_i,t)=1$ whenever $(x_0,\dots,x_n)\in M_1$. Then, it suffices to find $G(x,t)$ such that $G(x,t)^d=\bar{r}(x^d,t)$. @@ -292,47 +299,55 @@ We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, %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 $F(x)=x^d$. %$r_1$ 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 $\tilde{r}_1(0,t)=0$. -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}. +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}. \begin{figure}[h] \begin{center} \begin{tikzpicture}[scale=0.6] \draw (-5,0) to (5,0); \draw (0,5) to (0,-1); \draw[fill] (3,2) circle (2pt) node[above]{$x_r+ix_i$}; - \draw[dashed] (3,2) to (3,0); + \draw[dashed] (3,2) to (3,0) node[below] {$x_r$}; \end{tikzpicture} \end{center} -\caption{projection over the real axis} +\caption{Projection over the real axis} \label{realprojection} \end{figure} +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. Let us give explicit formulas. -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. - -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 $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$. - -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. +For a $x=x_r+ix_i\in\CC$, $\bar{r}(x,t)=x_r+(1-t)ix_i$, we can write +\[ +m(x,t)=+\sqrt{x_r^2+(1-t)^2x_i^2} \text{ and } \theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}. +\] +We can divide $\CC$ in $2d$ sectors of angle $\frac{\pi}{d}$ 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 one. \begin{figure}[h] \begin{center} -\begin{tikzpicture}[scale=0.6] - \draw (-5,0) to (5,0); - \draw (5,1.5) to (-5,-1.5); - \draw (5,3.2) to (-5, -3.2); - \draw (4.7,5) to (-4.7,-5); - \draw (2.8,5) to (-2.8,-5); - \draw (0.9,5) to (-0.9,-5); - \draw (-0.9,5) to (0.9,-5); - \draw (-2.8,5) to (2.8,-5); - \draw (-4.7,5) to (4.7,-5); - \draw (-5,3.2) to (5, -3.2); - \draw (-5,1.5) to (5,-1.5); +%\begin{tikzpicture}[scale=0.6] +% \draw (-5,0) to (5,0); +% \draw (5,1.5) to (-5,-1.5); +% \draw (5,3.2) to (-5, -3.2); +% \draw (4.7,5) to (-4.7,-5); +% \draw (2.8,5) to (-2.8,-5); +% \draw (0.9,5) to (-0.9,-5); +% \draw (-0.9,5) to (0.9,-5); +% \draw (-2.8,5) to (2.8,-5); +% \draw (-4.7,5) to (4.7,-5); +% \draw (-5,3.2) to (5, -3.2); +% \draw (-5,1.5) to (5,-1.5); +%\end{tikzpicture} +% +\begin{tikzpicture} +\foreach \x in {0, ..., 16}{ + \draw (0,0) to (360/17*\x:3);} \end{tikzpicture} + + \end{center} \caption{Division of the complex plane} \label{divisionC} \end{figure} -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 $f_k$. +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 $(x_0,\dots,x_n)\in M_d$ 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~$f_k$. @@ -344,40 +359,34 @@ Then we can get homeomorphisms from the upper complex plane to the sectors of an The homeomorphisms are given by -\begin{equation} - 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, -\end{equation} -\begin{equation} - 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 -\end{equation} +\begin{align*} + f_k(|z|\exp^{i\theta})&=|z|^\frac{1}{d}\exp{\left(i\frac{\theta}{d}+(k-1)i\frac{\pi}{d}\right)}\text{ for }k=1,3,5,\dots,2d-1,\\ + f_k(|z|\exp^{i\theta})&=|z|^\frac{1}{d}\exp{\left(i\frac{\theta-\pi}{d}+(k-1)i\frac{\pi}{d}\right)}\text{ for }k=2,4,6,\dots,2d. +\end{align*} We will also need the inverses of these homeomorphisms, which are: -\begin{equation} - 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, -\end{equation} -\begin{equation} - 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 -\end{equation} - +\begin{align*} + f_k^{-1}(|z|\exp^{i\theta})&=|z|^d\exp{\left(id\left(\theta-(k-1)\frac{\pi}{d}\right)\right)}\text{ for }k=1,3,5,\dots,2d-1,\\ + f_k^{-1}(|z|\exp^{i\theta})&=|z|^d\exp{\left(id\left(\theta-\pi-(k-1)i\frac{\pi}{d}\right)\right)}\text{ for }k=2,4,6,\dots,2d. +\end{align*} 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: \begin{equation} -\begin{array}{ccc} - G(z,t)&=&f_k(\bar{r}(f_k^{-1}(z),t))\\ - &=&f_k(\bar{r}(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},t))\\ - &=&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} +\begin{aligned} + G(z,t)&=f_k\left(\bar{r}(f_k^{-1}(z),t)\right)\\ + &=f_k\left(\bar{r}(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},t)\right)\\ + &=f_k\left(\sqrt{(z^d)_r^2+(1-t)^2(z^d)_i^2}\, \exp{\left(i\arctan\left((1-t)\tan\left(d\left(\theta-(k-1)\frac{\pi}{d}\right)\right)\right)\right)}\right)\\ + &=\sqrt[2d]{(z^d)_r^2+(1-t)^2(z^d)_i^2}\, + \exp{\left(\frac{i}{d}\arctan\left((1-t)\tan\left(d\theta\right)\right)\right)} +\end{aligned} \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 $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$. +\begin{equation*} +\begin{tikzcd}[column sep=0pt, row sep=5pt] +G(z,t)^d\dar[equal, equal]&=&m(z,t)^d\exp{\left(i\arctan((1-t)\tan d\theta)\right)}\\ +\bar{r}(z^d,t)&=&m(z^d,t)\exp{(i\arctan((1-t)\tan d\theta))}\ +\end{tikzcd} +\end{equation*} +We can then write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$. Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d_0, \ldots , x^d_n\in\RR\}$ into $S_d$. @@ -385,11 +394,11 @@ For a point $P=(x_0\xi_0,\dots,x_n\xi_n)\in M_d\cap\{x^d_0, \ldots , x^d_n\in\RR \begin{equation} P^d_Q=\left(\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}, \dots,\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)} \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 $(+,\dots,+)$ case, $r_2$ is the constant path, and therefore the lift $\tilde{r}_2$ will also be the constant path. -In the rest of the cases, the lift will be given by: +In the rest of the cases, the lift will be given by $\tilde{r}_2((x_0\xi_0,\dots,x_n\xi_n), t)$ equals \[ -\tilde{r}_2((x_0\xi_0,\dots,x_n\xi_n), t)=\left(\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0,\dots,\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right) +\left(\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0,\dots,\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right). \] @@ -399,7 +408,7 @@ 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 $\CC\PP^n\setminus C_d$ and $K$. + The strong retraction from $M_d$ to $S_d$ induces a strong retraction, and therefore, a homotopy equivalence between $\CC\PP^n\setminus C_d$ and $K$. \end{theorem} \begin{proof} The path followed by each point in $\CC\PP^n\setminus C_d$ 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 $\CC\PP^n\setminus C_d$ 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 $r$. @@ -407,43 +416,36 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$. Let $p=[x_0:\dots:x_n]\in\CC\PP^n\setminus C_d$ and $\lambda p=[\lambda x_0:\dots: \lambda x_n]$ be two representatives of some point $p$, since both representatives must be such that $\sum x^d_i=1$, 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_0,t): \dots: G(x_n,t)] \\ - \tilde{r}_1(\lambda p,t) & = & [G(\lambda x_0,t): \dots: G(\lambda x_n,t)] \\ - \end{array} - \end{equation} + \begin{align*} + \tilde{r}_1(p,t) & = [G(x_0,t): \dots: G(x_n,t)], \\ + \tilde{r}_1(\lambda p,t) & = [G(\lambda x_0,t): \dots: G(\lambda x_n,t)]. + \end{align*} If we define $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$ for $x\in\CC$, we get that - \begin{equation} + \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} + \end{equation*} + By \eqref{Gliftedr1}: + \begin{align*} + G(x,t) & = \sqrt[2d]{(x^d)_r^2+(1-t)^2(x^d)_i^2}\exp{\left(\frac{i}{d}\arctan((1-t)\tan(d\theta)))\right)} \\ + G(\lambda x,t) & = \sqrt[2d]{(\lambda^d x^d)_r^2+(1-t)^2(\lambda^d x^d)_i^2}\exp{\left(\frac{i}{d}\arctan((1-t)\tan(d\theta)))\right)} \\ + \end{align*} 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. If $[x^d_0: \dots: x^d_n]$ is in $(+,\dots,+)$, the retraction is the identity, and therefore both representatives will stay the same and therefore go to the same class. For the rest of the regions, we have -\begin{equation} -\begin{array}{lll} -\tilde{r}_2([x_0\xi_0:\dots:x_n\xi_n], t)& =& \left[\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right] -\end{array} -\end{equation} - - +\begin{equation*} +\tilde{r}_2([x_0\xi_0:\dots:x_n\xi_n], t) =\left[H(x_0,\chi_0)\xi_0:\dots:H(x_n,\chi_n) \xi_n\right] +\end{equation*} +where +\[ +H(x_i, \chi_i):=\sqrt[d]{(1-t)x^d_i+t\frac{x^d_i\chi_i}{1-\sum x^d_j(1-\chi_j)}},\quad i=0,\dots,n. +\] Then, for the point $\lambda p$ we get -\begin{equation} -\begin{array}{lll} -\tilde{r}_2([\lambda x_0\xi_0:\dots:\lambda x_n\xi_n], t)&=&\left[\sqrt[d]{(1-t)\lambda^d x^d_0+t\frac{\lambda^d x^d_0\chi_0}{1-\sum \lambda^d x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)\lambda^d x^d_n+t\frac{\lambda^d x^d_n\chi_n}{1-\sum \lambda^d x^d_j(1-\chi_j)}}\xi_n\right]\\ -&=&\left[\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right]\\ -\end{array} -\end{equation} - +\begin{align*} +\tilde{r}_2([\lambda x_0\xi_0:\dots:\lambda x_n\xi_n], t)&=\left[H(\lambda x_0, \chi_0)\xi_0:\dots:H(\lambda x_n, \chi_n)\xi_n\right]\\ +&=\left[H( x_0, \chi_0)\xi_0:\dots:H(x_n, \chi_n)\xi_n\right]. +\end{align*} So we see that $\tilde{r}_2$ also takes both representatives will end to the same point. \end{proof} From 2a87fee69dfb1e6fdd6e608f9b4ee957cc3f5a74 Mon Sep 17 00:00:00 2001 From: Enrique Artal Date: Thu, 24 Sep 2026 14:49:16 +0200 Subject: [PATCH 2/2] falta abstract e introduccion --- .gitignore | 303 +++++++++++++++++++++++++++++++++ generaldegreeHypersurfaces.tex | 89 +++++----- 2 files changed, 344 insertions(+), 48 deletions(-) create mode 100644 .gitignore diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..4579efc --- /dev/null +++ b/.gitignore @@ -0,0 +1,303 @@ +# ---> TeX +## Core latex/pdflatex auxiliary files: +*.aux +*.lof +*.log +*.lot +*.fls +*.out +*.toc +*.fmt +*.fot +*.cb +*.cb2 +.*.lb + +## Intermediate documents: +*.dvi +*.xdv +*-converted-to.* +# these rules might exclude image files for figures etc. +# *.ps +# *.eps +*.pdf + +## Generated if empty string is given at "Please type another file name for output:" +.pdf + +## Bibliography auxiliary files (bibtex/biblatex/biber): +*.bbl +*.bcf +*.blg +*-blx.aux +*-blx.bib +*.run.xml + +## Build tool auxiliary files: +*.fdb_latexmk +*.synctex +*.synctex(busy) +*.synctex.gz +*.synctex.gz(busy) +*.pdfsync + +## Build tool directories for auxiliary files +# latexrun +latex.out/ + +## Auxiliary and intermediate files from other packages: +# algorithms +*.alg +*.loa + +# achemso +acs-*.bib + +# amsthm +*.thm + +# beamer +*.nav +*.pre +*.snm +*.vrb + +# changes +*.soc + +# comment +*.cut + +# cprotect +*.cpt + +# elsarticle (documentclass of Elsevier journals) +*.spl + +# endnotes +*.ent + +# fixme +*.lox + +# feynmf/feynmp +*.mf +*.mp +*.t[1-9] +*.t[1-9][0-9] +*.tfm + +#(r)(e)ledmac/(r)(e)ledpar +*.end +*.?end +*.[1-9] +*.[1-9][0-9] +*.[1-9][0-9][0-9] +*.[1-9]R +*.[1-9][0-9]R +*.[1-9][0-9][0-9]R +*.eledsec[1-9] +*.eledsec[1-9]R +*.eledsec[1-9][0-9] +*.eledsec[1-9][0-9]R +*.eledsec[1-9][0-9][0-9] +*.eledsec[1-9][0-9][0-9]R + +# glossaries +*.acn +*.acr +*.glg +*.glo +*.gls +*.glsdefs +*.lzo +*.lzs +*.slg +*.slo +*.sls + +# uncomment this for glossaries-extra (will ignore makeindex's style files!) +# *.ist + +# gnuplot +*.gnuplot +*.table + +# gnuplottex +*-gnuplottex-* + +# gregoriotex +*.gaux +*.glog +*.gtex + +# htlatex +*.4ct +*.4tc +*.idv +*.lg +*.trc +*.xref + +# hyperref +*.brf + +# knitr +*-concordance.tex +# TODO Uncomment the next line if you use knitr and want to ignore its generated tikz files +# *.tikz +*-tikzDictionary + +# listings +*.lol + +# luatexja-ruby +*.ltjruby + +# makeidx +*.idx +*.ilg +*.ind + +# minitoc +*.maf +*.mlf +*.mlt +*.mtc[0-9]* +*.slf[0-9]* +*.slt[0-9]* +*.stc[0-9]* + +# minted +_minted* +*.pyg + +# morewrites +*.mw + +# newpax +*.newpax + +# nomencl +*.nlg +*.nlo +*.nls + +# pax +*.pax + +# pdfpcnotes +*.pdfpc + +# sagetex +*.sagetex.sage +*.sagetex.py +*.sagetex.scmd + +# scrwfile +*.wrt + +# svg +svg-inkscape/ + +# sympy +*.sout +*.sympy +sympy-plots-for-*.tex/ + +# pdfcomment +*.upa +*.upb + +# pythontex +*.pytxcode +pythontex-files-*/ + +# tcolorbox +*.listing + +# thmtools +*.loe + +# TikZ & PGF +*.dpth +*.md5 +*.auxlock + +# titletoc +*.ptc + +# todonotes +*.tdo + +# vhistory +*.hst +*.ver + +# easy-todo +*.lod + +# xcolor +*.xcp + +# xmpincl +*.xmpi + +# xindy +*.xdy + +# xypic precompiled matrices and outlines +*.xyc +*.xyd + +# endfloat +*.ttt +*.fff + +# Latexian +TSWLatexianTemp* + +## Editors: +# WinEdt +*.bak +*.sav + +# Texpad +.texpadtmp + +# LyX +*.lyx~ + +# Kile +*.backup + +# gummi +.*.swp + +# KBibTeX +*~[0-9]* + +# TeXnicCenter +*.tps + +# auto folder when using emacs and auctex +./auto/* +*.el + +# expex forward references with \gathertags +*-tags.tex + +# standalone packages +*.sta + +# Makeindex log files +*.lpz + +# xwatermark package +*.xwm + +# REVTeX puts footnotes in the bibliography by default, unless the nofootinbib +# option is specified. Footnotes are the stored in a file with suffix Notes.bib. +# Uncomment the next line to have this generated file ignored. +#*Notes.bib + diff --git a/generaldegreeHypersurfaces.tex b/generaldegreeHypersurfaces.tex index 431098d..0ab0f72 100644 --- a/generaldegreeHypersurfaces.tex +++ b/generaldegreeHypersurfaces.tex @@ -452,7 +452,7 @@ So we see that $\tilde{r}_2$ also takes both representatives will end to the sam We have now seen that there is a homotopy equivalence between $\CC\PP^n\setminus C_d$ and $K$. Therefore, to get a CW-complex structure on $\CC\PP^n\setminus C_d$, it suffices to get it for $K$. -We denote as $\Delta$ the $\Delta-$complex $[v_0, \dots, v_n]$ that has $n+1$ vertices, and $n+1-j$ $j$-simplices then, we can identify $S_1$ with $\Delta$, and we get a diagram: +We denote as $\Delta$ the $\Delta-$complex $[v_0, \dots, v_n]$ that has $n+1$ vertices, and $n+1-j$ $j$-simplices. Then, we can identify $S_1$ with $\Delta$, and we get a diagram: \begin{equation*} % \begin{array}{ccccc} @@ -477,24 +477,21 @@ where $\Delta^d$ is a $\Delta-$complex that we are trying to find and identify w %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 the $0$-simplices, the $i$-simplices and the $n$-simplex as: -\begin{equation} -\begin{array}{lllll} - X^j & = & v_j & = & e_j \\ - X^{(j_0,\dots,j_i)} & = & [v_{j_0},\dots,v_{j_i}]\\ - X & = & [v_0,\dots,v_n] -\end{array} -\end{equation} - +\begin{align*} + X^j & = v_j = e_j \\ + X^{(j_0,\dots,j_i)} & = [v_{j_0},\dots,v_{j_i}]\\ + X & = [v_0,\dots,v_n] +\end{align*} The face maps of $X$ and $X^{(j_0,\dots,j_i)}$ are: -\begin{equation} +\begin{equation*} \delta^i_k(X^{(j_0,\dots,j_i)}) =[v_{j_0},\dots,\hat{v},\dots,v_{j_i}] -\end{equation} -where $\hat{v}$ indicates that we take out the vertix in position $k$, for $k\leq i$. +\end{equation*} +where $\hat{v}$ indicates that we take out the vertex in position $k$, for $k\leq i$. an the boundary maps are given by: -\begin{equation} +\begin{equation*} \partial^i =\sum (-1)^{k}\delta^i_k \label{bordermaps} -\end{equation} +\end{equation*} We define now the $\Delta$-complex $\Delta^d$. It has $(n+1)d$ $0-$simplices, ${n+1}\choose{2}$ $d^2$ $1-$simplices, ${n+1}\choose{i}$ $d^{i+1}$ $i-$simplices and $d^{n+1}$ $n-$simplices. @@ -514,17 +511,14 @@ X_{(\xi_0, \dots,\xi_n)}\cap\{x_{i_1},\dots,x_{i_j}=0\}, \] where $\xi_{i_1},\dots,\xi_{i_j}=1$, and the vertices by \[ -X_{(1,\dots,\xi_i, 1, 1)}\cap\{x_j=1\forall j\neq i\}. +X_{(1,\dots,\xi_i, 1, 1)}\cap\{x_j=1,\,\forall j\neq i\}. \] -We can remark that $X_{(\xi_0, \dots, \xi_n)}=(\xi_0, \dots, \xi_n)X_{(1,\dots,1)}$, and therefore that we have a group action of $\mu^{n+1}_d$ over $S_d$. +We can remark that $X_{(\xi_0, \dots, \xi_n)}=(\xi_0, \dots, \xi_n)\cdot X_{(1,\dots,1)}$, and therefore that we have a group action of $\mu^{n+1}_d$ over $S_d$. Let -\begin{equation} -\begin{array}{rll} - X^{(j_0,\dots,j_i)}_{(1,\dots,1)} & = & X_{(1, \dots, 1)}\cap\{x_{j_k}\neq0 \forall k=0,\dots i\text{ and }x_{i}=0\text{ for the rest}\}\\ - X^j_{(1,\dots,1)} & = & X_{(1, \dots, 1)}\cap\{x_i=0\forall i\neq j\}\\ - -\end{array} -\end{equation} +\begin{align*} + X^{(j_0,\dots,j_i)}_{(1,\dots,1)} & = X_{(1, \dots, 1)}\cap\{x_{j_k}\neq0,\, \forall k=0,\dots i\text{ and }x_{i}=0\text{ for the rest}\}\\ + X^j_{(1,\dots,1)} & = X_{(1, \dots, 1)}\cap\{x_i=0,\,\forall i\neq j\}\\ +\end{align*} We can describe the chain complexes. They are abelian groups but they are also $R_d$-complexes where $R_d$ is the group algebra $\ZZ[\mu_d^{n+1}]$. For the $n$-chains we have \[ @@ -536,7 +530,7 @@ Then \[ C_i(\Delta^d) = \bigoplus_{(t_{j_1},\dots,t_{j_{n-i}})}\frac{R_d}{(t_{j_1} - 1,\dots,t_{j_{n-i}} - 1)} \langle X^{(k_0,\dots,k_i)}_{(1,\dots,1)}\rangle, \] -,where $t_{j_s}\neq t_{k_r}\forall s,r$ i.e, a free abelian group of rank $d^i$. Finalyy the $0$-chains are +where $t_{j_s}\neq t_{k_r}\forall s,r$ i.e, a free abelian group of rank $d^i$. Finally the $0$-chains are \[ C_0(\Delta^d) = \bigoplus_{t_j}\frac{R_d}{(t_0 - 1,\dots,t_{j-1}-1,t_{j+1}-1,\dots, t_n - 1)} \langle X^j_{(1,\dots,1)}\rangle \] @@ -551,11 +545,11 @@ a free abelian group of rank~$(n+1)d$. %\end{equation} The face maps are $R_d$-morphisms defined as -\begin{equation} +\begin{equation*} \begin{aligned} \delta^n_i(X_{(1,1,1)}) & =X^{(0,\dots,i-1,i+1,\dots,n)}_{(1,\dots,1)} &&& \delta^j_i(X^{(k_0,\dots,k_j)}_{(1,\dots,1)})&= X^{(k_0,\dots,k_{i-1},k_{i+1},\dots,k_n)}_{(1,\dots,1)} . \end{aligned} -\end{equation} +\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} @@ -604,7 +598,7 @@ We have then defined $\Delta^d$, the simplicial complex with which we can identi % \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 $(x_0,\dots,x_n)\longrightarrow[x_0:\dots:x_n]$. Since we are passing to the projective plane, two points $(x_0\xi_0, \dots, x_n\xi_n)$ and $(x_0\xi_0\xi, \dots, x_n\xi_n\xi)$ will have the same image. +The map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x_0,\dots,x_n)\longmapsto[x_0:\dots:x_n]$. Since we are passing to the projective plane, two points $(x_0\xi_0, \dots, x_n\xi_n)$ and $(x_0\xi_0\xi, \dots, x_n\xi_n\xi)$ will have the same image. The $n$-cells $X_{(\xi_0, \dots, \xi_n)}$ and $X_{(\xi_0\xi, \dots, \xi_n\xi)}$ of $\Delta^d$, project the same $n$-cell of $\Delta^d_\textrm{proj}$, which will be denoted as $X_{[\xi_0: \dots: \xi_n]}$. The same convention is followed for the rest of $i-$simplices. Note that for the vertices, since all coordinates but one are $0$, we have only $n+1$ vertices $X^0, \dots, X^n$. @@ -614,6 +608,7 @@ The group $\mu_d^{n+1}$ also acts, but the action of the diagonal vectors is tri T_d:=\ZZ\left[\frac{\mu_d^{n+1}}{\mu_d\langle(\xi,\dots,\xi)\rangle}\right]; \] we use variables $t_0,\dots,t_n$ as for $R_d$ and we have to add the relation $t_0 \cdots t_n=1$. +This group is actualy isomorphic to $\mu_n^d$ but it is more convenient to use this presentation. We have the following chain complexes: \begin{align*} C_n(\Delta^d_\text{proj}) &= T_d\langle X_{[1:\dots:1]}\rangle,\\ @@ -631,38 +626,36 @@ C_0(\Delta^d) &= \bigoplus_{t_j}\frac{T_d}{(t_0 - 1,\dots,t_{j-1}-1,t_{j+1}-1,\d %\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 $K=\{[x:y:z]\in\PP^2|x^2,y^2,z^2\in\RR_{\geq0}\}$. + We are going to compute now the simplicial complex for $S_d$ and $K$ when $d=n=2$, 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 $K=\{[x:y:z]\in\PP^2|x^2,y^2,z^2\in\RR_{\geq0}\}$. 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 $\phi(\Delta^2)=\Delta$. 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 $X_{(-1,-1,-1)}$. The face maps of $X_{(1,1,1)}$ are: - \begin{equation} - \begin{array}{lllll} - \delta^2_0(X_{(1,1,1)}) & = & L^x_{(1,1,1)} & = & \{(0,\sqrt{1-t},\sqrt{t})|t\in[0,1]\}\\ - \delta^2_1(X_{(1,1,1)}) & = & L^y_{(1,1,1)} & = & \{(\sqrt{1-t},0, \sqrt{t})|t\in[0,1]\}\\ - \delta^2_2(X_{(1,1,1)}) & = & L^z_{(1,1,1)} & = & \{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\ - \end{array} - \end{equation} - + \begin{equation*} +\begin{tikzcd}[row sep=0, column sep=8pt] + \delta^2_0(X_{(1,1,1)}) \rar[equal] & L^x_{(1,1,1)} \rar[equal] & \{(0,\sqrt{1-t},\sqrt{t})|t\in[0,1]\}\\ + \delta^2_1(X_{(1,1,1)})\rar[equal]& L^y_{(1,1,1)} \rar[equal] & \{(\sqrt{1-t},0, \sqrt{t})|t\in[0,1]\}\\ + \delta^2_2(X_{(1,1,1)}) \rar[equal] & L^z_{(1,1,1)} \rar[equal] & \{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\} +\end{tikzcd} + \end{equation*} By applying the group action as shown before we get the faces of the rest of $2-$simplices. 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 $L^z_{(-1,-1,1)}$. The face maps of $L^z_{(1,1,1)}$, $L^y_{(1,1,1)}$ and $L^x_{(1,1,1)}$ are: -\begin{equation} -\begin{array}{lllll} - \delta^1_0(L^z_{(1,1,1)}) & = & V^y_{(1,1,1)} & = & (1,0,0)\\ - \delta^1_1(L^z_{(1,1,1)}) & = & V^z_{(1,1,1)} & = & (0,1,0)\\ - \delta^1_0(L^y_{(1,1,1)}) & = & V^x_{(1,1,1)} & = & (1,0,0)\\ - \delta^1_1(L^y_{(1,1,1)}) & = & V^z_{(1,1,1)} & = & (0,0,1)\\ - \delta^1_0(L^x_{(1,1,1)}) & = & V^x_{(1,1,1)} & = & (0,1,0)\\ - \delta^1_1(L^x_{(1,1,1)}) & = & V^y_{(1,1,1)} & = & (0,0,1)\\ -\end{array} -\end{equation} - -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 $S_2$. +\begin{equation*} +\begin{tikzcd}[row sep=0, column sep=8pt] + \delta^1_0(L^z_{(1,1,1)}) \rar[equal] & V^y_{(1,1,1)} \rar[equal] & (1,0,0)\\ + \delta^1_1(L^z_{(1,1,1)}) \rar[equal] & V^z_{(1,1,1)} \rar[equal] & (0,1,0)\\ + \delta^1_0(L^y_{(1,1,1)}) \rar[equal] & V^x_{(1,1,1)} \rar[equal] & (1,0,0)\\ + \delta^1_1(L^y_{(1,1,1)}) \rar[equal] & V^z_{(1,1,1)} \rar[equal] & (0,0,1)\\ + \delta^1_0(L^x_{(1,1,1)}) \rar[equal] & V^x_{(1,1,1)} \rar[equal] & (0,1,0)\\ + \delta^1_1(L^x_{(1,1,1)}) \rar[equal] & V^y_{(1,1,1)} \rar[equal] & (0,0,1)\\ +\end{tikzcd} +\end{equation*} +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 $S_2$ which corresponds to the partition of the $2$-spheres in triangles with right angles. %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: %\begin{equation} %\begin{array}{rll}