From bc23c8cf6f1f93a47daf2109cd15e440227715a6 Mon Sep 17 00:00:00 2001 From: Enrique Artal Date: Thu, 5 Mar 2026 21:34:56 +0100 Subject: [PATCH] propuestas --- complementarioconica.tex | 33 ++++++++++++++++++++++++++++----- 1 file changed, 28 insertions(+), 5 deletions(-) diff --git a/complementarioconica.tex b/complementarioconica.tex index 50ee030..7f039c8 100644 --- a/complementarioconica.tex +++ b/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 $[x:y:z]$.