añadir mas cosas

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Alba Larraya Sancho 2025-07-24 12:49:10 +02:00
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@ -49,7 +49,7 @@ We take $C=\{x^2+y^2+z^2=0\}$. We have a natural inclusion of $\RR\PP^2$ into $\
\end{array}
\end{equation*}
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)+(1-t)x_i, y_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+(1-t)y_i, z_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+(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 $\s^2\simeq X$.
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 $\PP^2\setminus C$.
@ -59,7 +59,7 @@ We know that for $n\geq 2$, the cover maps induce isomorphisms in the homotopy g
$\pi_1(\RR\PP^2)=\ZZ$, 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 $\PP^2\setminus 4$, since $(x,y,z),(ix,iy,iz),(\sqrt{i}x,\sqrt{i}y,\sqrt{i}z), (-ix,-iy,-iz)$ all have the same image $[x:y:z]$.
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 $[x:y:z]$.
As in the previous case, we will see that $\PP^2\setminus C$ is homotopy equivalent to some CW-complex. In particular, we will prove that $\PP^2\setminus C$ is homotopy equivalent to $K=\{[x:y:z]\in\PP^2 : x^4,y^4,z^4\in\RR^{\geq 0}\}$. Since for all $ [x:y:z]\in K$, it is not in $C$, we get an inclusion from $K$ to $\PP^2\setminus C$. To find a universal cover of $K$, we start by computing the CW-complex structure of $K$, considering that we can decompose $K$ as the union of the projective subspaces: $\{[x:y:z]:x,y,z\in\RR\}$, $\{[x:y:iz]:x,y,z\in\RR\}$, $\{[x:iy:z]:x,y,z\in\RR\}$, $\{[x:iy:iz]:x,y,z\in\RR\}$. For each of the subspaces, we have the CW-structure of $\RR\PP^2$ with three vertices, six edges and four 2-simplices, and we can identify some edges common to some subspaces. By doing this, we get the following CW-structure for $K$:
\begin{itemize}
@ -163,5 +163,77 @@ As in the previous case, we will see that $\PP^2\setminus C$ is homotopy equival
\end{tabular}
\end{center}
The fundamental group of $K$ is a cyclic group of order 4, with generator $e$. Then we can construct the universal cover as in %Computing the homology of universal covers via effective homology and discrete vector fields.
This way, we get a universal cover whose CW-complex structure is formed by $12$ $0-$simplices, $48$ $1-$simplices and $64$ $2-$simplices.
This way, we get a universal cover whose CW-complex structure is formed by $12$ $0-$simplices, $48$ $1-$simplices and $64$ $2-$simplices. Since
\begin{table}[]
\begin{tabular}{|c|c|c||c|c|c|}
\hline
& $\partial_0$ & $\partial_1$ & & $\partial_0$ & $\partial_1$ \\ \hline
$(1,e_1)$ & $(1,v_1)$ & $(d_2^2,v_2)$ & $(d_2,e_1)$ & $(d_2,v_1)$ & $(d_2^{-1},v_2)$ \\\hline
$(1,e_2)$ & $(1,v_1)$ & $(d_2^{-1},v_2)$ & $(d_2,e_2)$ & $(d_2,v_1)$ & $(1,v_2)$ \\\hline
$(1,e_3)$ & $(1,v_2)$ & $(1,v_1)$ & $(d_2,e_3)$ & $(d_2,v_2)$ & $(d_2,v_1)$ \\\hline
$(1,e_4)$ & $(1,v_2)$ & $(d_2^{-1},v_1)$ & $(d_2,e_4)$ & $(d_2,v_2)$ & $(1,v_1)$ \\\hline
$(1,e_5)$ & $(1,v_2)$ & $(d_2^2,v_0)$ & $(d_2,e_5)$ & $(d_2,v_2)$ & $(d_2^{-1},v_0)$ \\\hline
$(1,e_6)$ & $(1,v_2)$ & $(1,v_0)$ & $(d_2,e_6)$ & $(d_2,v_2)$ & $(d_2,v_0)$ \\v
$(1,e_7)$ & $(1,v_1)$ & $(d_2,v_0)$ & $(d_2,e_7)$ & $(d_2,v_1)$ & $(d_2^2,v_0)$ \\\hline
$(1,e_8)$ & $(1,v_1)$ & $(d_2^{-1},v_0)$ & $(d_2,e_8)$ & $(d_2,v_1)$ & $(1,v_0)$ \\\hline
$(1,e_9)$ & $(1,v_1)$ & $(1,v_0)$ & $(d_2,e_9)$ & $(d_2,v_1)$ & $(d_2,v_0)$ \\\hline
$(1,e_{10})$ & $(1,v_1)$ & $(d_2^2,v_0)$ & $(d_2,e_{10})$ & $(d_2,v_1)$ & $(d_2^{-1},v_0)$ \\\hline
$(1,e_{11})$ & $(1,v_2)$ & $(d_2^{-1},v_0)$ & $(d_2,e_{11})$ & $(d_2,v_2)$ & $(1,v_0)$ \\\hline
$(1,e_{12})$ & $(1,v_2)$ & $(d_2,v_0)$ & $(d_2,e_{12})$ & $(d_2,v_2)$ & $(d_2^2,v_0)$ \\\hline \hline
$(d_2^2,e_1)$ & $(d_2^2,v_1)$ & $(1,v_2)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},v_1)$ & $(d_2,v_2)$ \\\hline
$(d_2^2,e_2)$ & $(d_2^2,v_1)$ & $(d_2,v_2)$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},v_1)$ & $(d_2^2,v_2)$ \\\hline
$(d_2^2,e_3)$ & $(d_2^2,v_2)$ & $(d_2^2,v_1)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},v_2)$ & $(d_2^{-1},v_1)$ \\\hline
$(d_2^2,e_4)$ & $(d_2^2,v_2)$ & $(d_2,v_1)$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},v_2)$ & $(d_2^2,v_1)$ \\\hline
$(d_2^2,e_5)$ & $(d_2^2,v_2)$ & $(1,v_0)$ & $(d_2^{-1},e_5)$ & $(d_2^{-1},v_2)$ & $(d_2,v_0)$ \\\hline
$(d_2^2,e_6)$ & $(d_2^2,v_2)$ & $(d_2^2,v_0)$ & $(d_2^{-1},e_6)$ & $(d_2^{-1},v_2)$ & $(d_2^{-1},v_0)$ \\\hline
$(d_2^2,e_7)$ & $(d_2^2,v_1)$ & $(d_2^{-1},v_0)$ & $(d_2^{-1},e_7)$ & $(d_2^{-1},v_1)$ & $(1,v_0)$ \\\hline
$(d_2^2,e_8)$ & $(d_2^2,v_1)$ & $(d_2,v_0)$ & $(d_2^{-1},e_8)$ & $(d_2^{-1},v_1)$ & $(d_2^2,v_0)$ \\\hline
$(d_2^2,e_9)$ & $(d_2^2,v_1)$ & $(d_2^2,v_0)$ & $(d_2^{-1},e_9)$ & $(d_2^{-1},v_1)$ & $(d_2^{-1},v_0)$ \\\hline
$(d_2^2,e_{10})$ & $(d_2^2,v_1)$ & $(1,v_0)$ & $(d_2^{-1},e_{10})$ & $(d_2^{-1},v_1)$ & $(d_2,v_0)$ \\\hline
$(d_2^2,e_{11})$ & $(d_2^2,v_2)$ & $(d_2,v_0)$ & $(d_2^{-1},e_{11})$ & $(d_2^{-1},v_2)$ & $(d_2^2,v_0)$ \\\hline
$(d_2^2,e_{12})$ & $(d_2^2,v_2)$ & $(d_2^{-1},v_0)$ & $(d_2^{-1},e_{12})$ & $(d_2^{-1},v_2)$ & $(1,v_0)$ \\\hline
\end{tabular}
\end{table}
\begin{table}[]
\begin{tabular}{|c|c|c|c||c|c|c|c|}
\hline
& $\partial_0$ & $\partial_1$ & $\partial_2$ & & $\partial_0$ & $\partial_1$ & $\partial_2$ \\\hline
$(1,T_1)$ & $(1,e_3)$ & $(1,e_5)$ & $(1,e_{10})$ & $(d_2,T_1)$ & $(d_2,e_3)$ & $(d_2,e_5)$ & $(d_2,e_{10})$ \\\hline
$(1,T_2)$ & $(1,e_1)$ & $(1,e_9)$ & $(d_2^2,e_5)$ & $(d_2,T_2)$ & $(d_2,e_1)$ & $(d_2,e_9)$ & $(d_2^{-1},e_5)$ \\\hline
$(1,T_3)$ & $(1,e_3)$ & $(1,e_{6})$ & $(1,e_9)$ & $(d_2,T_3)$ & $(d_2,e_3)$ & $(d_2,e_{6})$ & $(d_2,e_9)$ \\\hline
$(1,T_4)$ & $(1,e_1)$ & $(1,e_{10})$ & $(d_2^2,e_6)$ & $(d_2,T_4)$ & $(d_2,e_1)$ & $(d_2,e_{10})$ & $(d_2^{-1},e_6)$ \\\hline
$(1,T_5)$ & $(1,e_3)$ & $(1,e_{11})$ & $(1,e_8)$ & $(d_2,T_5)$ & $(d_2,e_3)$ & $(d_2,e_{11})$ & $(d_2,e_8)$ \\\hline
$(1,T_6)$ & $(1,e_1)$ & $(1,e_7)$ & $(d_2^2,e_{11})$ & $(d_2,T_6)$ & $(d_2,e_1)$ & $(d_2,e_7)$ & $(d_2^{-1},e_{11})$ \\\hline
$(1,T_7)$ & $(1,e_3)$ & $(1,e_{12})$ & $(1,e_7)$ & $(d_2,T_7)$ & $(d_2,e_3)$ & $(d_2,e_{12})$ & $(d_2,e_7)$ \\\hline
$(1,T_8)$ & $(1,e_1)$ & $(1,e_8)$ & $(d_2^2,e_{12})$ & $(d_2,T_8)$ & $(d_2,e_1)$ & $(d_2,e_8)$ & $(d_2^{-1},e_{12})$ \\\hline
$(1,T_9)$ & $(1,e_4)$ & $(1,e_{12})$ & $(d_2^{-1},e_{10})$ & $(d_2,T_9)$ & $(d_2,e_4)$ & $(d_2,e_{12})$ & $(1,e_{10})$ \\\hline
$(1,T_{10})$ & $(1,e_2)$ & $(1,e_9)$ & $(d_2^{-1},e_{10})$ & $(d_2,T_{10})$ & $(d_2,e_2)$ & $(d_2,e_9)$ & $(1,e_{10})$ \\\hline
$(1,T_{11})$ & $(1,e_4)$ & $(1,e_{11})$ & $(d_2^{-1},e_9)$ & $(d_2,T_{11})$ & $(d_2,e_4)$ & $(d_2,e_{11})$ & $(1,e_9)$ \\\hline
$(1,T_{12})$ & $(1,e_2)$ & $(1,e_{10})$ & $(d_2^{-1},e_{11})$ & $(d_2,T_{12})$ & $(d_2,e_2)$ & $(d_2,e_{10})$ & $(1,e_{11})$ \\\hline
$(1,T_{13})$ & $(1,e_4)$ & $(1,e_5)$ & $(d_2^{-1},e_8)$ & $(d_2,T_{13})$ & $(d_2,e_4)$ & $(d_2,e_5)$ & $(1,e_8)$ \\\hline
$(1,T_{14})$ & $(1,e_2)$ & $(1,e_7)$ & $(d_2^{-1},e_5)$ & $(d_2,T_{14})$ & $(d_2,e_2)$ & $(d_2,e_7)$ & $(1,e_5)$ \\\hline
$(1,T_{15})$ & $(1,e_4)$ & $(1,e_6)$ & $(d_2^{-1},e_7)$ & $(d_2,T_{15})$ & $(d_2,e_4)$ & $(d_2,e_6)$ & $(1,e_7)$ \\\hline
$(1,T_{16})$ & $(1,e_2)$ & $(1,e_8)$ & $(d_2^{-1},e_6)$ & $(d_2,T_{16})$ & $(d_2,e_2)$ & $(d_2,e_8)$ & $(1,e_6)$ \\\hline\hline
$(d_2^2,T_1)$ & $(d_2^2,e_3)$ & $(d_2^2,e_5)$ & $(d_2^2,e_{10})$ & $(d_2^{-1},T_1)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_5)$ & $(d_2^{-1},e_{10})$ \\\hline
$(d_2^2,T_2)$ & $(d_2^2,e_1)$ & $(d_2^2,e_9)$ & $(1,e_5)$ & $(d_2^{-1},T_2)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_9)$ & $(d_2,e_5)$ \\\hline
$(d_2^2,T_3)$ & $(d_2^2,e_3)$ & $(d_2^2,e_{6})$ & $(d_2^2,e_9)$ & $(d_2^{-1},T_3)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_{6})$ & $(d_2^{-1},e_9)$ \\\hline
$(d_2^2,T_4)$ & $(d_2^2,e_1)$ & $(d_2^2,e_{10})$ & $(1,e_6)$ & $(d_2^{-1},T_4)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_{10})$ & $(d_2,e_6)$ \\\hline
$(d_2^2,T_5)$ & $(d_2^2,e_3)$ & $(d_2^2,e_{11})$ & $(d_2^2,e_8)$ & $(d_2^{-1},T_5)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_{11})$ & $(d_2^{-1},e_8)$ \\\hline
$(d_2^2,T_6)$ & $(d_2^2,e_1)$ & $(d_2^2,e_7)$ & $(1,e_{11})$ & $(d_2^{-1},T_6)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_7)$ & $(d_2,e_{11})$ \\\hline
$(d_2^2,T_7)$ & $(d_2^2,e_3)$ & $(d_2^2,e_{12})$ & $(d_2^2,e_7)$ & $(d_2^{-1},T_7)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_{12})$ & $(d_2^{-1},e_7)$ \\\hline
$(d_2^2,T_8)$ & $(d_2^2,e_1)$ & $(d_2^2,e_8)$ & $(1,e_{12})$ & $(d_2^{-1},T_8)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_8)$ & $(d_2,e_{12})$ \\\hline
$(d_2^2,T_9)$ & $(d_2^2,e_4)$ & $(d_2^2,e_{12})$ & $(d_2,e_{10})$ & $(d_2^{-1},T_9)$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_{12})$ & $(d_2^2,e_{10})$ \\\hline
$(d_2^2,T_{10})$ & $(d_2^2,e_2)$ & $(d_2^2,e_9)$ & $(d_2,e_{10})$ & $(d_2^{-1},T_{10})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_9)$ & $(d_2^2,e_{10})$ \\\hline
$(d_2^2,T_{11})$ & $(d_2^2,e_4)$ & $(d_2^2,e_{11})$ & $(d_2,e_9)$ & $(d_2^{-1},T_{11})$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_{11})$ & $(d_2^2,e_9)$ \\\hline
$(d_2^2,T_{12})$ & $(d_2^2,e_2)$ & $(d_2^2,e_{10})$ & $(d_2,e_{11})$ & $(d_2^{-1},T_{12})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_{10})$ & $(d_2^2,e_{11})$ \\\hline
$(d_2^2,T_{13})$ & $(d_2^2,e_4)$ & $(d_2^2,e_5)$ & $(d_2,e_8)$ & $(d_2^{-1},T_{13})$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_5)$ & $(d_2^2,e_8)$ \\\hline
$(d_2^2,T_{14})$ & $(d_2^2,e_2)$ & $(d_2^2,e_7)$ & $(d_2,e_5)$ & $(d_2^{-1},T_{14})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_7)$ & $(d_2^2,e_5)$ \\\hline
$(d_2^2,T_{15})$ & $(d_2^2,e_4)$ & $(d_2^2,e_6)$ & $(d_2,e_7)$ & $(d_2^{-1},T_{15})$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_6)$ & $(d_2^2,e_7)$ \\\hline
$(d_2^2,T_{16})$ & $(d_2^2,e_2)$ & $(d_2^2,e_8)$ & $(d_2,e_6)$ & $(d_2^{-1},T_{16})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_8)$ & $(d_2^2,e_6)$ \\\hline
\end{tabular}
\end{table}
\end{document}