reescribir y completar detalles

generaldegree.tex

@@ -4,11 +4,9 @@
 \usepackage{amsthm}
 \usepackage{amsmath}
 \usepackage{tikz}
-\usepackage{tikz-cd}
+
-\usetikzlibrary{positioning}
+\usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart,
-\usetikzlibrary{arrows.meta}
+decorations.markings, calc}
-\usetikzlibrary{decorations.pathmorphing}
-\usetikzlibrary{shapes.multipart}
 
 \usepackage{hyperref}
 
@@ -62,45 +60,93 @@ We then have a commutative diagram:
 \end{equation*}
 
 The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma:
-\begin{theorem}
+\begin{lemma}
-$S_d$ is a strong deformation retract of $M_d$.
+$M_1\cap\RR^3$ is a strong deformation retract of $M_1$.
-\end{theorem}
+\end{lemma}
 \begin{proof}
-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 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:
+%We can construct $r$ in two steps:
-\begin{equation*}
+%\begin{equation*}
-	\begin{array}{cccccc}
+%	\begin{array}{ccccc}
-		r:&M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
+%		M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
-	\end{array}
+%	\end{array}
-\end{equation*}
+%\end{equation*}
-Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow S_1$. Then $r_1$ is given by:
+We are going to 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 $0$:
 \begin{equation}
 	\begin{array}{ccc}
-		r_1: M_1\times I & \longrightarrow & M_1\cap\RR^3 \\
+		r_1: M_1\times I & \longrightarrow & M_1 \\
 		(x_r+ix_i,y_r+iy_i,z_r+iz_i, t) & \longrightarrow & (x_r+(1-t)ix_i,y_r+(1-t)iy_i,z_r+(1-t)iz_i) \\
 	\end{array}
 \end{equation}
-$M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)\in\RR_{\geq0}^3|x+y+z=1\}=\{(x,y,z)\in[0,1]^3|x+y+z=1\}$, this is, $S_1$ is a triangular section of the plane defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described below, and we will retract each of them separately:
+
+Note that for $(x,y,z)\in M_1$, we already have that
+\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 of $M_1$ to $M_1\cap\RR^3$.
+\end{proof}
+
+\begin{lemma}
+	$M_1\cap\RR^3$ is a strong deformation retract of $S_1$.
+\end{lemma}
+\begin{proof}
+	Note that $M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)\in\RR_{\geq0}^3|x+y+z=1\}=\{(x,y,z)\in[0,1]^3|x+y+z=1\}$, this is, $S_1$ is a triangular region of the plane defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described below, and we will retract each of them separately:
 \begin{figure}[h]
 \begin{center}
 \begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
-	\draw (3,4.5) to (-6,-9);
+%	\draw (3,4.5) to (-6,-9);
-	\draw (-3,4.5) to (6,-9);
+%	\draw (-3,4.5) to (6,-9);
-	\draw (-7,-4.5) to (7,-4.5);
+%	\draw (-7,-4.5) to (7,-4.5);
-	\node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
+%	\node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
-	\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
+%	\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
-	\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
+%	\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
-	\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
+%	\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
-	\node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
+%	\node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
-	\node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
+%	\node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
-	\node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
+%	\node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
-	\node[right] at (0,0) {(0,0,1)};
+%	\node[right] at (0,0) {(0,0,1)};
-	\node[right,above] at (3,-4.5) {(0,1,0)};
+%	\node[right,above] at (3,-4.5) {(0,1,0)};
-	\node[left,above] at (-3,-4.5) {(1,0,0)};
+%	\node[left,above] at (-3,-4.5) {(1,0,0)};
+	\coordinate (A) at (-3,-4.5);
+\coordinate (B) at (3,-4.5);
+\coordinate (C) at (0,0);
+\fill[green!25!white] (C) to (B) to (A);
+	\draw ($2*(C)-(A)$) to ($2*(A)-(C)$);
+	\draw ($2*(B)-(A)$) to ($2*(A)-(B)$);
+	\draw ($2*(B)-(C)$) to ($2*(C)-(B)$);
+	\node at (0,-2.5) {$(+,+,+)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
+	\node at (5.5,0) {$(-,+,+)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
+	\node at (7.5,-6.5) {$(-,+,-)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
+	\node at (0,-8.5) {$(+.+,-)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
+	\node at (-8,-6.5)  {$(+.-,-)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
+	\node at (-6,0) {$(+.-,+)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
+	\node at (0,4) {$(-,-,+)$};% {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
+	\foreach \x[evaluate={\y=1-\x}] in {.125, .25, ..., 1}
+		{
+\draw[->=.5,green!75!black] ($1.75*(C) -.75*\x*(A) -.75*\y*(B)$) -- (C);
+\draw[->=.5,green!75!black] ($1.75*(B) -.75*\x*(A) -.75*\y*(C)$) -- (B);
+\draw[->=.5,green!75!black] ($1.75*(A) -.75*\x*(C) -.75*\y*(B)$) -- (A);
+}
+	\foreach \x[evaluate={\y=1-\x}] in {0.25, .3125, ..., .75}
+		{
+\draw[->=.5,green!75!black] ($1.75*\x*(C) -.75*(A) + 1.75*\y*(B)$) -- ($\x*(C)+ \y*(B)$);
+\draw[->=.5,green!75!black] ($1.75*\x*(A) -.75*(B) + 1.75*\y*(C)$) -- ($\x*(A)+ \y*(C)$);
+\draw[->=.5,green!75!black] ($1.75*\x*(A) -.75*(C) + 1.75*\y*(B)$) -- ($\x*(A)+ \y*(B)$);
+}
+%\fill (6, -4.5) circle[radius=.1cm];
+	\node[right=2pt] at (C) {(0,0,1)};
+	\node[above right] at (B) {(0,1,0)};
+	\node[above left] at (A) {(1,0,0)};
 \end{tikzpicture}
 \end{center}
 \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 $(+,+,+)$.
 
 The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
@@ -122,7 +168,8 @@ P_{x,y,z}=\left\{\begin{array}{lr}
 	(0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
 \end{array}\right.
 \end{equation}
-We need to see now that this point is well defined, for this, we'll see that in the borders of the different parts it coincides. We will see this for the part $(+,+,-)$, for the rest of permutations it is equivalent.
+
+Since we have defined the retraction by regions, we are now going to show that in the intersections the definitions coincide, and therefore that, by the gluing lemma, $r_2$ is countinous and well defined. We will see this for the part $(+,+,-)$, for the rest of permutations it is equivalent.
 
 We start with the border between $(+,+,-)$ and $(+,+,+)$, this is, the line with $z=0$. In this case
 \begin{equation*}
@@ -146,9 +193,15 @@ P_{x,y,z}=\left\{\begin{array}{lr}
 \end{array}\right.
 \end{equation*}
 similarly to the previous case, in this line $x=1-z$.
+\end{proof}
+
 
 Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we will try to lift them in order to get a retraction from $M_d$ to $S_d$.
 
+\begin{theorem}
+	$S_d$ is a strong deformation retract of $M_d$.
+\end{theorem}
+\begin{proof}
 We have the following diagram:
 \begin{equation*}
 	\begin{array}{ccccc}
@@ -184,11 +237,11 @@ In order to see this, we can write an explicit expression of $\bar{r}$. For a po
 \label{realprojection}
 \end{figure}
 
-we can then write the retraction as $r_1=m(z,t)e^{2i\pi\theta(t)}$ where $m(t)$ is the function indicating how the modulus of $\bar{r}(x,t)$ changes and $\theta(t)$ portrays the change of the argument.
+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 $z=z_r+iz_i\in\CC$, $\bar{r}(z,t)=z_r+(1-t)iz_i$, we can write $m(z,t)=+\sqrt{z_r^2+(1-t)^2z_i^2}$ and $\theta(t)=\arctan\frac{(1-t)z_i}{z_r}$.
+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 $\CC$ en $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}.
+We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}.
 
 \begin{figure}[h]
 \begin{center}
@@ -210,7 +263,17 @@ We can divide $\CC$ en $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC
 \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 $k=2,4,...,2d$.
+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 $f_k$. 
+
+
+
+
+
+
+
+
+
+
 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,
@@ -229,19 +292,19 @@ We will also need the inverses of these homeomorphisms, which are:
 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(H(f_k^{-1}(z),t))\\
+	G(z,t)&=&f_k(\bar{r}(f_k^{-1}(z),t))\\
-	&=&f_k(H(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},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}
 \end{equation}
 
-Then, we can see that $G(z,t)^d=H(z^d,t)$, since:
+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)}\\
 	||& & \\
-	H(z^d,t)&=&m(z^d,t)\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}
 
@@ -262,18 +325,22 @@ 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, $r_2(x^d,y^d,z^d)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x,y,z)$. Since $x,y$ are negative real numbers multiplied by some roots of unity, the retraction will push them to $0$, while we will push $z$ to a norm 1 complex number in one of the $d$ copies of $\RR$ that we have in the preimage of the complex plane by $F$. We can write this as $\tilde{r}_2(x,y,z)=(\sqrt[d]{1-t}x,\sqrt[d]{1-t}y,\sqrt[d]{(1-t)z^d+t})$. For the permutations of this case it is equivalent.
+In the $(-,-,+)$ case, $r_2(x^d,y^d,z^d)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x,y,z)$. Since $x,y$ are negative real numbers multiplied by some roots of unity, the retraction will push them to $0$, while we will push $z$ to a norm 1 complex number in one of the $d$ copies of $\RR$ that we have in the preimage of the complex plane by $F$. We can write this as $\tilde{r}_2(x,y,z)=(\sqrt[d]{1-t}x,\sqrt[d]{1-t}y,\sqrt[d]{(1-t)z^d+t})$. For the cases $(+,-,-)$ and $(-,+,-)$ it is equivalent.
 
 Lastly, for the case $(+,+,-)$ and it's permutations, we can write the lifting as $\tilde{r}_2(x,y,z,t)=(\sqrt[d]{1-t+\frac{t}{1-z^d}}x,\sqrt[d]{1-t+\frac{t}{1-z^d}}y,\sqrt[d]{1-t}z)$. We can see by summing the coordinates to the power $d$ that they sum $1$, so the retraction doesn't leave $M_1$.
 
-All that is left to see is that the formulas for each case coincide on the borders. For the border of $(+,+,-)$ and $(-,+,-)$, we have that:
+All that is left to see is that the formulas for each case coincide on the intersections of the regions. For the intersection of $(+,+,-)$ and $(-,+,-)$, we have that:
 \begin{equation*}
 	\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}=\sqrt[d]{(1-t+\frac{t}{y^d})y^d}=\sqrt[d]{y^d-ty^d+t}=\sqrt[d]{y^d(1-t)+t}
 \end{equation*} 
-The first root corresponds to $\tilde{r}_2$ in $(+,+,-)$ and the last one corresponds to the lifting in $(-,+,-)$,so we see that it coincides. It would be similar in the rest of borders.
+The first root corresponds to $\tilde{r}_2$ in $(+,+,-)$ and the last one corresponds to the lifting in $(-,+,-)$, so we see that it coincides. It would be similar in the rest of intersections.
+
+
+We have then proved that $S_d$ is a strong deformation retract of $M_d$. 
 \end{proof}
 
-We have then proved that $S_d$ is a strong deformation retract of $M_d$. We denote as $\Delta$ the 2-simplex $[v_0, v_1, v_2]$, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
+
+We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertices, $2$ eedges and $1$ face, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
 
 \begin{equation*}
 	\begin{array}{ccccc}
@@ -284,7 +351,7 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$. We deno
 		& & K & \longrightarrow & \PP^2\setminus C_d \\
 	\end{array}
 \end{equation*}
-where $\Delta^d$ is a simplicial complex that we are trying to find and identify with $S_d$, equivalently as what we have for $d=1$.
+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.
 
 %Let $\phi: S_d\longleftarrow S_1$ be the map given by $\phi(x,y,z)=(x^d, y^d, z^d)$, then necessarily $\phi^{-1}(\Delta)=\Delta^d$.
 %In particular, the $d^3$ vertices of the simplicial complex $\Delta^d$ will be identified to $(\xi_d,0,0)$, $(0,\xi_d,0)$, $(0,0,\xi_d)$ for all $\xi_d$ the $d$-$th$ roots of unity.
@@ -363,32 +430,65 @@ We have then defined $\Delta^d$, the simplicial complex with which we can identi
 
 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 $X_{(x\xi_1, y\xi_2, z\xi_3)}$ and $X_{(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)}$. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices.
 
-\begin{example}{}
+\begin{equation}
-	We are going to compute now the simplicial complex for $d=2$,  this is, $S_2=\{(x,y,z)\in M_2| x^2, y^2, z^2\in\RR_{\geq 0}\}$.
+\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}
 
-	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 $\phi(\Delta^2)=\Delta$. 
+\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 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 $\phi(\Delta^2)=\Delta$. 
 	
-	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:
+	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 $S_1$.	
-	\begin{equation}
-		\begin{array}{rll}
-			\phi^{-1}((1,0,0))&=&\{(1,0,0),(-1,0,0)\}\\
-			\phi^{-1}((0,1,0))&=&\{(0,1,0),(0,-1,0)\}\\
-			\phi^{-1}((0,0,1))&=&\{(0,0,1),(0,0,-1)\}
-		\end{array}
-	\end{equation}
 	
-	The edges of the simplex, $[v_0v_1]$, $[v_0v_2]$ and $[v_1v_2]$, can be identified to the points of $S_1$ given respectively by $\{(1-t, 0, t)|t\in[0,1]\}$, $\{(1-t,t,0)|t\in[0,1]\}$ and $\{(0,t,1-t)|t\in[0,1]\}$. Therefore, each edge will have four preimages. In the first case, these preimages are:
-	\begin{equation}
-		\phi^{-1}((1-t,t,0))=\left\{\begin{array}{l}
-			\{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
-			\{(\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}\\
-			\{(-\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
-			\{(-\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}
-		\end{array}\right.
-	\end{equation}	
-	For the rest of the edges the preimages by $\phi$ would be analougous. We will denote the edges as a triple of $\{+,-,0\}$ indicating in each coordinate the sign of the root taken. 
 	
-	Finally, we will have $8$ $2$-faces. For three edges to form a face on the simplex, we need the corresponding vertices of the edges to coincide, that is, the edges with non-zero first coordinate must both take the same sign on said coordinate, and equivalently for the rest of coordinates. Then, the faces are defined by taking the square root of the coordinates of the points in $S_1$ with sign equal to those in the edges.
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	
+	%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}
+			%\phi^{-1}((1,0,0))&=&\{(1,0,0),(-1,0,0)\}\\
+			%\phi^{-1}((0,1,0))&=&\{(0,1,0),(0,-1,0)\}\\
+			%\phi^{-1}((0,0,1))&=&\{(0,0,1),(0,0,-1)\}
+		%\end{array}
+	%\end{equation}
+	
+	T%he edges of the simplex, $[v_0v_1]$, $[v_0v_2]$ and $[v_1v_2]$, can be identified to the points of $S_1$ given respectively by $\{(1-t, 0, t)|t\in[0,1]\}$, $\{(1-t,t,0)|t\in[0,1]\}$ and $\{(0,t,1-t)|t\in[0,1]\}$. Therefore, each edge will have four preimages. In the first case, these preimages are:
+	%\begin{equation}
+		%\phi^{-1}((1-t,t,0))=\left\{\begin{array}{l}
+		%	\{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
+		%	\{(\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}\\
+		%	\{(-\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
+		%	\{(-\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}
+		%\end{array}\right.
+	%\end{equation}	
+	%For the rest of the edges the preimages by $\phi$ would be analougous. We will denote the edges as a triple of $\{+,-,0\}$ indicating in each coordinate the sign of the root taken. 
+	
+	%Finally, we will have $8$ $2$-faces. For three edges to form a face on the simplex, we need the corresponding vertices of the edges to coincide, that is, the edges with non-zero first coordinate must both take the same sign on said coordinate, and equivalently for the rest of coordinates. Then, the faces are defined by taking the square root of the coordinates of the points in $S_1$ with sign equal to those in the edges.
 	  
 	  \begin{figure}[h]
 		\begin{center}