demostración de la equivalencia de homotopia entre el complementario de la curva y K

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: