primeros cambios

complementarioconica.tex

@@ -21,7 +21,8 @@
 \newcommand{\PP}{\mathbb{P}}
 \newcommand{\s}{\mathbb{S}}
 \newcommand{\ZZ}{\mathbb{Z}}
-
+\newcommand{\norma}[1]{\left\lVert#1\right\rVert}
+\newcommand{\uno}{\mathbf{1}}
 
 %opening
 \title{Complement of a projective curve}
@@ -38,21 +39,38 @@
 Given $C$ a proyective curve in $\PP^2$ over $\CC$, we know that the complement of the curve has the homotopy type of a CW-complex of real dimension $2$, our problem is how to find an effective model for said CW-complex.
 
 \section{Complement of a conic projective curve}
-We start with the case of the conic curve of degree $2$. We know that all projective conic curves are topologicaly equivalent to each other, this is, we have automorphisms of the projective plane that take one conic to any other conic.
+We start with the case of the conic curve of degree $2$. We know that all projective conic curves are topologicaly equivalent to each other, i.e., we have automorphisms of the projective plane that take one conic to any other conic.
 
-We take $C=\{x^2+y^2+z^2=0\}$. We have a natural inclusion of $\RR\PP^2$ into $\PP^2\setminus C$, since there are no real solutions of $x^2+y^2+z^2=0$. At the same time, we have $\PP^1\cong \s^2$ is a universal cover of $\RR\PP^2$. We are going to see that the Milnor fiber $X=\{x^2+y^2+z^2=1\}$ is a universal cover of $\PP^2\setminus C$. We have the following diagram:
+We take $C=\{x^2+y^2+z^2=0\}$. We have a natural inclusion of $\RR\PP^2$ into $\PP^2\setminus C$, since there are no real solutions of $x^2+y^2+z^2=0$. At the same time, we have $\PP^1\cong \s^2$ is a universal cover of $\RR\PP^2$. We are going to see that the Milnor fiber $X=\{x^2+y^2+z^2=1\}$ is the universal cover of $\PP^2\setminus C$. We have the following diagram:
 \begin{equation*}
-	\begin{array}{ccc}
-		\PP^1\simeq\s^2 & \longrightarrow & X \\
-		\downarrow & & \downarrow \\
-		\RR\PP^2 & \longrightarrow & \PP^2\setminus C \\
-	\end{array}
+		\begin{tikzcd}
+		\PP^1\simeq\s^2 \rar\dar& X\dar \\
+		\RR\PP^2 \rar& \PP^2\setminus C \\
+	\end{tikzcd}
 \end{equation*}
+First, we see that $\s^2$ is a deformation retract of $X$.  Let us introduce some notation. For $p:=(x,y,z)\in\CC^3$ we denote $\Re p, \Im p\in\RR^3$ the real and imaginary parts of 
+$p$. Let $H: X\times[0,1]\rightarrow X$ given by 
+\[
+H(p,t)=
+\frac{t}{\norma{\Re p}} \Re p + (1 - t) p.
+%(x_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)x_i, y_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)y_i, z_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)z_i)
+\]
+%$ H(x,y,z,t)=(x_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)x_i, y_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)y_i, z_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)z_i) $, where $x=x_r+ix_i$, $y=y_r+iy_i$ and $z=z_r+iz_i$. 
+If $H_t:X\to X$ is given by $H_t(p):=H(p, t)$, then $H_0=\uno_{X}$ and ${H_t}_{\s^2}$ is the inclusion for $t\in[0,1]$.
+%Then, $H(\cdot,0)=id_X$, $H(\cdot,1)\in\s^2$ and for $(x,y,z)\in\s^2$, $H(x,y,z)=id_{\s^2}$.
+Therefore, $\s^2$ is a deformation retract of $X$, and hence $\s^2\simeq X$.
 
-First, we see that $\s^2$ is a deformation retract of $X$. Let $H: X\times[0,1]\rightarrow X$ given by $ H(x,y,z,t)=(x_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)x_i, y_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)y_i, z_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)z_i) $, where $x=x_r+ix_i$, $y=y_r+iy_i$ and $z=z_r+iz_i$. Then, $H(\cdot,0)=id_X$, $H(\cdot,1)\in\s^2$ and for $(x,y,z)\in\s^2$, $H(x,y,z)=id_{\s^2}$. Therefore, $\s^2$ is a deformation retract of $X$, and as such $\s^2\simeq X$.
-
-The covering map $\s^2\rightarrow\RR\PP^2$ given by $(x,y,z)\rightarrow[x:y:z]$ gives a 2 to 1 cover, as does the covering map $X\rightarrow
-\PP^2\setminus C$ given by $(x,y,z)\rightarrow[x:y:z]$. Therefore, we get that $X$ is a universal cover of $\PP^2\setminus C$.
+The horizontal maps of 
+\[
+\begin{tikzcd}
+	\s^2\rar\dar&\dar\RR\PP^2\\ 
+	X \rar& \PP^2\setminus C\\ [-20pt]
+{(x, y, x)}\rar[mapsto]&{[x:y:z]}
+\end{tikzcd}
+\]
+are $2:1$ coves. In particular, $X$ is the universal cover of $\PP^2\setminus C$.
+%The covering map $\s^2\rightarrow\RR\PP^2$ given by $(x,y,z)\rightarrow[x:y:z]$ gives a 2 to 1 cover, as does the covering map $X\rightarrow
+%\PP^2\setminus C$ given by $(x,y,z)\rightarrow[x:y:z]$. Therefore, we get that $X$ is a universal cover of $\PP^2\setminus C$.
 
 We know that for $n\geq 2$, the cover maps induce isomorphisms in the homotopy groups $\pi_n$, and since $\s^2\simeq X$, we get that $\pi_n(\RR\PP^2)\cong\pi_n(\PP^2\setminus C)$. We just need to see that the fundamental groups are also isomorphic.