propuestas

complementarioconica.tex

@@ -61,20 +61,43 @@ If $H_t:X\to X$ is given by $H_t(p):=H(p, t)$, then $H_0=\uno_{X}$ and ${H_t}_{\
 Therefore, $\s^2$ is a deformation retract of $X$, and hence $\s^2\simeq X$.
 
 The horizontal maps of 
-\[
-\begin{tikzcd}
+\begin{equation}
+\begin{tikzcd}\label{eq:2-covers}
 	\s^2\rar\dar&\dar\RR\PP^2\\ 
 	X \rar& \PP^2\setminus C\\ [-20pt]
 {(x, y, x)}\rar[mapsto]&{[x:y:z]}
 \end{tikzcd}
-\]
+\end{equation}
 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.
+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 since the horizontal arrows
+of~\eqref{eq:2-covers} are the universal covers (both $2:1$).
+Hence, the map below induces isomorphisms on all the homotopy groups, and
+then it is a homotopy equivalence.
 
-$\pi_1(\RR\PP^2)=\ZZ$, so we take a non constant generator, such as $[\cos(\pi t): \sin(\pi t): 0]$, and it can be seen that it is also a generator of $\pi_1(\PP^2\setminus C)$. Therefore, both fundamental groups are isomorphic and we can conclude that $\PP^2\setminus C$ and $\RR\PP^2$ have the same homotopy type, so we can give a CW-complex structure on $\PP^2\setminus C$ equal to that of $\RR\PP^2$.
+Actually we can see that $\RR\PP^2$ is  a deformation retract of $\PP^2\setminus C$.
+Let us recall that the monodromy of the cover $X\to\PP^2\setminus C$ is given
+by the antipodal map $\sigma:X\to X$ given by $\sigma(p)=-p$. Note that
+\[
+\sigma(H(t, p)) = -\left(\frac{t}{\norma{\Re p}} \Re p + (1 - t) p\right)=
+\frac{t}{\norma{\Re (-p)}} \Re (-p) + (1 - t) (-p)= H(t,\sigma(p)).
+\]
+Since $H$ is equivariant it descends to a retraction homotopy downstairs.
+
+{\color{red} Propuesta: Hacer el caso \emph{trivial} reemplazando $x^2 + y^2 + z^2$ por
+$x + y + z$. Hacemos la retracción al triángulo 
+\[
+\{(x, y, z)\in\RR^3\mid x + y + z=1,\ x, y, z\geq 0\}.
+\]
+Para la cuártica cambiaría la notación de los símplices. En lugar de numerarlos, usar 
+ternas de signos o algo similar como índices. Eso es fácilmente generalizable a cualquier
+dimensión.}
+
+%$\pi_1(\RR\PP^2)=\ZZ$, so we take a non constant generator, such as $[\cos(\pi t): \sin(\pi t): 0]$, and it can be seen that it is also a generator of $\pi_1(\PP^2\setminus C)$. Therefore, both fundamental groups are isomorphic and we can conclude that $\PP^2\setminus C$ and $\RR\PP^2$ have the same homotopy type, so we can give a CW-complex structure on $\PP^2\setminus C$ equal to that of $\RR\PP^2$.
 
 \section{Complement of the quadric projective curve}
 We study know the case of the quadric curve $C=\{x^4+y^4+z^4=0\}$. In this case, Milnor's fiber is $X=\{x^4 +y^4+z^4=1\}$, and it is a 4 to 1 universal cover of $\PP^2\setminus C$, since $(x,y,z),(ix,iy,iz),(\sqrt{i}x,\sqrt{i}y,\sqrt{i}z), (-ix,-iy,-iz)$ all have the same image $[x:y:z]$.