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}