elevacion de las retracciones

generaldegree.tex

@@ -8,6 +8,7 @@
 \usetikzlibrary{positioning}
 \usetikzlibrary{arrows.meta}
 \usetikzlibrary{decorations.pathmorphing}
+\usetikzlibrary{shapes.multipart}
 
 \usepackage{hyperref}
 
@@ -79,6 +80,125 @@ 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:
+\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 (-7,-4.5) to (7,-4.5);
+	\node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
+	\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
+	\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
+	\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
+	\node at (-5.5,-6) {$x\geq0$,\\ $y\leq$,\\ $z\leq0$};
+	\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)};
+	\node[right,above] at (3,-4.5) {(0,1,0)};
+	\node[left,above] at (-3,-4.5) {(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:
+\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 $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. 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}
+P_{x,y,z}=\left\{\begin{array}{lr}
+	(x,y,z)&(+,+,+) \\
+	(1,0,0)&(+,-,-) \\
+	(0,1,0)&(-,+,-) \\
+	(0,0,1)&(-,-,+) \\
+	(\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
+	(\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
+	(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.
+
+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}
+	(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}
+	(\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}
+	(\frac{x}{1-z},\frac{0}{1-z},0)=(1,0,0)&(+,+,-) \\
+	(1,0,0)&(+,-,-) \\
+\end{array}\right.
+\end{equation*}
+similarly to the previous case, in this line $x=1-z$.
+
+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 $S_d$.
+
+We have the following diagram:
+\begin{equation*}
+	\begin{array}{ccccc}
+		M_d & \longrightarrow & M_d\cap\{x^d, y^d, z^d\in\RR\}&\longrightarrow& S_d \\
+		\downarrow & & \downarrow & & \downarrow\\
+		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 $\tilde{r}_1(0,t)=0$.
+
+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:
+\begin{figure}[h]
+\begin{center}
+\begin{tikzpicture}[scale=0.7]
+	\draw (-5,0) to (5,0);
+	\draw (0,5) to (0,-5);
+	\draw[fill] (3,2) circle (2pt) node[above]{$x_r+ix_i$};
+	\draw[dashed] (3,2) to (3,0);
+\end{tikzpicture}
+\end{center}
+\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 $\tilde{r}_1=\sqrt[d]{m(t)}e^{2i\pi(\frac{\theta(t)}{d}+\theta(0))}$.
+
+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
+\begin{equation}
+P_{x,y,z}=\left\{\begin{array}{lr}
+	(x^d,y^d,z^d)&(+,+,+) \\
+	(1,0,0)&(+,-,-) \\
+	(0,1,0)&(-,+,-) \\
+	(0,0,1)&(-,-,+) \\
+	(\frac{x^d}{1-z^d},\frac{y^d}{1-z^d},0)&(+,+,-) \\
+	(\frac{x^d}{1-y^d},0,\frac{z^d}{1-y^d})&(+,-,+) \\
+	(0,\frac{y^d}{1-x^d},\frac{z^d}{1-x^d})&(-,+,+) \\
+\end{array}\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 $(-,-,+)$ 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.
+
+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*}
+	\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.
 \end{proof}
 
 \end{document}