From 5cd4521f94f2ee094a8bba0f28800853db6733e1 Mon Sep 17 00:00:00 2001 From: Alba Larraya Sancho Date: Wed, 13 May 2026 16:22:03 +0200 Subject: [PATCH] =?UTF-8?q?demostraci=C3=B3n=20de=20la=20equivalencia=20de?= =?UTF-8?q?=20homotopia=20entre=20el=20=20complementario=20de=20la=20curva?= =?UTF-8?q?=20y=20K?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- generaldegree.tex | 93 +++++++++++++++++++++++++++++++++++++++++------ 1 file changed, 82 insertions(+), 11 deletions(-) diff --git a/generaldegree.tex b/generaldegree.tex index ac47d2c..fde911d 100644 --- a/generaldegree.tex +++ b/generaldegree.tex @@ -76,10 +76,10 @@ We are going to define a strong retraction $r_1:M_1\times I \longrightarrow M_1$ % 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 \\ - (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} +\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}] +M_1\times{[0,1]}\rar["r_1"]&M_1\cap\RR^3\\ +(P, t)\rar[mapsto]& t \Re P + (1 - t) \Im P. +\end{tikzcd} \end{equation} Note that for $(x,y,z)\in M_1$, we already have that @@ -96,7 +96,14 @@ and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we g $M_1\cap\RR^3$ 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: + This map is a well-defined deformation retract since $M_1$ has real equations. Note that $M_1\cap\RR^3$ +$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is +\[ +S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\}, +\] +i.e, $S_1$ is the triangular section 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 in Figure~\ref{fig:triangulo}. We will retract each of them separately: + + \begin{figure}[h] \begin{center} \begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}] @@ -123,9 +130,9 @@ and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we g \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,-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} { @@ -144,6 +151,8 @@ and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we g \node[above right] at (B) {(0,1,0)}; \node[above left] at (A) {(1,0,0)}; \end{tikzpicture} +\caption{Triangle $S_1$} +\label{fig:triangulo} \end{center} \end{figure} @@ -297,6 +306,7 @@ Then, we define $G=f_k\circ\bar{r}\circ f_k^{-1}$, where $k$ depends on the poin &=&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} +\label{Gliftedr1} \end{equation} Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since: @@ -311,7 +321,7 @@ Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since: 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 +For a point $(x\xi_1,y\xi_2,z\xi_3)\in M_d\cap\{x^d, y^d, z^d\in\RR\}$, 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)&(+,+,+) \\ @@ -325,9 +335,9 @@ 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 $\tilde{r}_2(x,y,z)=(\sqrt[d]{1-t}x,\sqrt[d]{1-t}y,\sqrt[d]{(1-t)z^d+t})$. For the cases $(+,-,-)$ and $(-,+,-)$ it is equivalent. +In the $(-,-,+)$ case, $r_2((x^d,y^d,z^d), t)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x\xi_1,y\xi_2,z\xi_3)$, for $\xi_1$, $\xi_2$ and $\xi_3$ some $d-$th roots of unity. Since $x,y$ are negative real numbers, the retraction will push them to $0$, while we will push $z\xi_3$ to $\xi_3$. We can write this as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3), t)=(\sqrt[d]{(1-t)x^d}\xi_1,\sqrt[d]{(1-t)y^d}\xi_2,\sqrt[d]{(1-t)z^d+t}\xi_3)$. 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 $M_1$. +Lastly, for the case $(+,+,-)$ and it's permutations, we can write the lifting as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3),t)=(\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1,\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2,\sqrt[d]{(1-t)z^d}\xi_3)$. 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 intersections of the regions. For the intersection of $(+,+,-)$ and $(-,+,-)$, we have that: \begin{equation*} @@ -339,6 +349,67 @@ The first root corresponds to $\tilde{r}_2$ in $(+,+,-)$ and the last one corres 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 $\PP^2\setminus C_d$ and $K$. +\end{theorem} +\begin{proof} + The class representatives we choose in $\PP^2\setminus C_d$ are such that the coordinates to the $d$ power sum $1$. 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 different representatives, they are sent to the same class by $r$. + + Let $p=[x:y:z]\in\PP^2\setminus C_d$ and $\lambda p=[\lambda x:\lambda y: \lambda z]$ be two representatives of some point $p$, since both representatives must be such that $x^d+y^d+z^d=1$, 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,t): G(y,t): G(z,t)] \\ + \tilde{r}_1(\lambda p,t) & = & [G(\lambda x,t): G(\lambda y,t): G(\lambda z,t)] \\ + \end{array} + \end{equation} + If we define $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$ for $x\in\CC$, we get that + \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} + 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. + + For a point $(x,y,z)\in M_d\cap\RR^3$, $\tilde{r}_2$ is defined by the region of $M_1\cap\RR^3$ where $(x^d, y^d, z^d)$ is, and therefore, when passing to the projective, it will depend on where the point $p^d=[x^d: y^d: z^d]$ lies. Since we have that $\lambda$ must be a $d-$th root of unity, we get that both points $p^d$ and $(\lambda p)^d=p$ will lie in the same region. + + In the case that they lie in the regions $(-,-,+)$, $(-,+,-)$ or $(+,-,-)$, the retraction $\tilde{r}_2$ is, in the first case: + \begin{equation} + \tilde{r}_2([x\xi_1:y\xi_2:z\xi_3],t)=\left[\sqrt[d]{(1-t)x^d}\xi_1:\sqrt[d]{(1-t)y^d}\xi_2:\sqrt[d]{(1-t)z^d+t}\xi_3\right] + \end{equation} + Then if we apply it to $[\lambda x\xi_1:\lambda y\xi_2:\lambda z\xi_3]$, for $\lambda $ another root of unity, we get: + \begin{equation} + \tilde{r}_2([\lambda x\xi_1:\lambda y\xi_2:\lambda z\xi_3],t)=\left[\sqrt[d]{(1-t)(\lambda x)^d}\xi_1:\sqrt[d]{(1-t)(\lambda y)^d}\lambda y\xi_2:\sqrt[d]{(1-t)(\lambda z)^d+t}\xi_3\right] + \end{equation} + + The rest of these cases would be equivalent. + + Next, for the cases $(+,+,-)$, $(+,-,+)$ or $(-,+,+)$, the retraction has the form, in the first case: + \begin{equation} + \tilde{r}_2([x\xi_1,y\xi_2,z\xi_3],t)=\left[\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1:\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2:\sqrt[d]{(1-t)z^d}\xi_3\right] + \end{equation} + Applying it to another representative of the same class, we get: + \begin{equation} + \begin{array}{lll} + \tilde{r}_2([\lambda x\xi_1,\lambda y\xi_2,\lambda z\xi_3],t)& = &\\ + & = & \left[\sqrt[d]{(1-t+\frac{t}{1-(\lambda z)^d})(\lambda x)^d}\xi_1:\sqrt[d]{(1-t+\frac{t}{1-(\lambda z)^d})(\lambda y)^d}\xi_2:\sqrt[d]{(1-t)(\lambda z)^d}\xi_3\right]\\ + & = &\left[\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1:\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2:\sqrt[d]{(1-t)z^d}\xi_3\right] + \end{array} + \end{equation} + + We can then see that both representatives will go to the same point by the retraction. + + Lastly, if $[x^d: y^d: z^d]$ is in $(+,+,+)$, the retraction is the identity, and therefore both representatives will stay the same and therefore go to the same class. +\end{proof} + +We have now seen that there is a homotopy equivalence between $\PP^2\setminus C_d$ and $K$. Therefore, to get a CW-complex structure on $\PP^2\setminus C_d$, it suffices to get it for $K$. + 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: