redaccion de la retraccion

generaldegree.tex

@@ -47,8 +47,8 @@ Therefore, we will work with $C_d$, from now on.
 We will denote:
 \begin{eqnarray}
 	M_d&=&\{(x,y,z)\in\CC | 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\PP^2 | x^d, y^d, z^d\in\RR^{\geq0}\} 
+	S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
+	K &=&\{[x:y:z]\in\PP^2 | x^d, y^d, z^d\in\RR_{\geq0}\} 
 \end{eqnarray}
 
 We then have a commutative diagram:
@@ -80,7 +80,7 @@ Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow
 		(x_r+ix_i,y_r+iy_i,z_r+iz_i, t) & \longrightarrow & (x_r+(1-t)ix_i,y_r+(1-t)iy_i,z_r+(1-t)iz_i) \\
 	\end{array}
 \end{equation}
-$M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)\in\RR^3|x+y+z=1\}=\{(x,y,z)\in[0,1]^3|x+y+z=1\}$, 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:
+$M_1\cap\RR^3$ 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:
 \begin{figure}[h]
 \begin{center}
 \begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
@@ -91,7 +91,7 @@ $M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)
 	\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
 	\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
 	\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
-	\node at (-5.5,-6) {$x\geq0$,\\ $y\leq$,\\ $z\leq0$};
+	\node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
 	\node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
 	\node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
 	\node[right] at (0,0) {(0,0,1)};
@@ -156,24 +156,95 @@ We have the following diagram:
 		M_1 & \longrightarrow & M_1\cap\RR^3 & \longrightarrow & S_1 \\
 	\end{array}
 \end{equation*}
-We start seeing how the lifting of $r_1$ works. Since $r_1$ works componentwise, we will study the lifting also for each component separetely. Since $r_1$ is the same for each component, we see it for only one of them, the rest will be analogous.
-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 $F(x)=x\longrightarrow x^d$.
-$r_1$ 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 $\tilde{r}_1(0,t)=0$.
+We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, we can write it as $r_1((x,y,z),t)=(\bar{r}(x,t),\bar{r}(y,t),\bar{r}(z,t))$ with $\bar{r}(x,t)+\bar{r}(y,t)+\bar{r}(z,t)=1$ whenever $(x,y,z)\in M_1$. Then, it suffices to find $G(x,t)$ such that $G(x,t)^d=\bar{r}(x^d,t)$. 
 
-In order to see this $\tilde{r}_1$  is continuous, we can write an explicit expression. To get this explicit expression, in $M_1$, the retraction acts as:
+
+
+
+
+
+
+
+%Since $r_1$ works componentwise, we will study the lifting also for each component separetely. Since $r_1$ is the same for each component, we see it for only one of them, the rest will be analogous.
+%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}.
 \begin{figure}[h]
 \begin{center}
 \begin{tikzpicture}[scale=0.7]
 	\draw (-5,0) to (5,0);
-	\draw (0,5) to (0,-5);
+	\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);
 \end{tikzpicture}
 \end{center}
+\caption{projection over the real axis}
+\label{realprojection}
 \end{figure}
 
-we can then write the retraction as $r_1=m(t)e^{2i\pi\theta(t)}$ where $m(t)$ is the function indicating how the modulus of $r_1(x,t)$ changes and $\theta(t)$ portrays the change of the argument.
-The preimage of this function by $F$, which would give the explicit expression of $\tilde{r}_1$, is $\tilde{r}_1=\sqrt[d]{m(t)}e^{2i\pi(\frac{\theta(t)}{d}+\theta(0))}$.
+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.
+
+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 $\theta(t)=\arctan\frac{(1-t)z_i}{z_r}$.
+
+We can divide $\CC$ en $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}.
+
+\begin{figure}[h]
+\begin{center}
+\begin{tikzpicture}[scale=0.7]
+	\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}
+\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 $k=2,4,...,2d$.
+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}
+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}
+
+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))\\
+	&=&f_k(H(|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:
+\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)}\\
+\end{array}
+\end{equation}
+
+
 
 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 $(x,y,z)\in M_d\cap\{x^d, y^d, z^d\in\RR\}$, 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
@@ -190,9 +261,9 @@ 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 $\tilde{r}_2(x,y,z)=(\sqrt[d]{1-t}x,\sqrt[d]{1-t}y,\sqrt[d]{1-t}z+\sqrt[d]{t})$. 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 permutations of this case 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 $M_1$.
+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:
 \begin{equation*}