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added explicit expression of llifting of first retraction

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Alba Larraya Sancho 2026-05-22 13:28:36 +02:00
parent d16551e906
commit c6d74bbeea
1 changed files with 8 additions and 7 deletions
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generaldegree.tex
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@ -4,6 +4,7 @@
\usepackage{amsthm} \usepackage{amsthm}
\usepackage{amsmath} \usepackage{amsmath}
\usepackage{tikz} \usepackage{tikz}
\usepackage[new]{old-arrows}
\usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart, \usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart,
decorations.markings, calc} decorations.markings, calc}
@ -53,9 +54,9 @@ We will denote:
We then have a commutative diagram: We then have a commutative diagram:
\begin{equation*} \begin{equation*}
\begin{array}{ccc} \begin{array}{ccc}
S_d & \longrightarrow & M_d \\ S_d & \longhookrightarrow & M_d \\
\downarrow & & \downarrow \\ \downarrow & & \downarrow \\
K & \longrightarrow & \PP^2\setminus C_d \\ K & \longhookrightarrow & \PP^2\setminus C_d \\
\end{array} \end{array}
\end{equation*} \end{equation*}
@ -205,7 +206,7 @@ similarly to the previous case, in this line $x=1-z$.
\end{proof} \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 $S_d$. 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} \begin{theorem}
$S_d$ is a strong deformation retract of $M_d$. $S_d$ is a strong deformation retract of $M_d$.
@ -250,7 +251,7 @@ we can then write the retraction as $\bar{r}(x,t)=m(x,t)e^{2i\pi\theta(x,t)}$ wh
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 $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$. 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}. 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.
\begin{figure}[h] \begin{figure}[h]
\begin{center} \begin{center}
@ -317,7 +318,7 @@ Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since:
\bar{r}(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{array}
\end{equation} \end{equation}
We can the write the lifting of $r_1$ as $\tilde{r}_1((x,y,z),t)=(G(x,t), G(y,t), G(z,t))$.
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 $S_d$. 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 $S_d$.
@ -417,9 +418,9 @@ We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertic
\begin{array}{ccccc} \begin{array}{ccccc}
\Delta & \longleftrightarrow & S_1 & & \\ \Delta & \longleftrightarrow & S_1 & & \\
\uparrow& & \uparrow & & \\ \uparrow& & \uparrow & & \\
\Delta^d & \longleftrightarrow & S_d & \longrightarrow & M_d \\ \Delta^d & \longleftrightarrow & S_d & \hookrightarrow & M_d \\
& & \downarrow & & \downarrow \\ & & \downarrow & & \downarrow \\
& & K & \longrightarrow & \PP^2\setminus C_d \\ & & K & \hookrightarrow & \PP^2\setminus C_d \\
\end{array} \end{array}
\end{equation*} \end{equation*}
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. 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.