añadido estrutura simplicial K
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Alba Larraya Sancho
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@ -20,6 +20,7 @@
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\newcommand{\BB}{\mathbb{B}}
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\newcommand{\PP}{\mathbb{P}}
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\newcommand{\s}{\mathbb{S}}
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\newcommand{\ZZ}{\mathbb{Z}}
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%opening
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@ -39,15 +40,128 @@ Given $C$ a proyective curve in $\PP^2$ over $\CC$, we know that the complement
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\section{Complement of a conic projective curve}
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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.
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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 now that $\PP^1\times\PP^1\setminus\Delta$ is a universal cover of $\PP^2\setminus C_1$, where $C_1=\{4xz=y^2\}$ is an equivalent curve to $C$ and $\Delta$ is the diagonal of $\PP^1\times\PP^1$. Let $\phi:\CC\times\CC\rightarrow\CC^2$ be the map given by $(t,s)\rightarrow(ts,t+s)$, since $\CC\times\CC\in\PP^1\times\PP^1$ and $\CC^2\in\PP^2$, we can homogenize the coordinates and get a map $\Phi:\PP^1\times\PP^1\rightarrow\PP^2$ be the map given by $([t_1:s_1], [t_2:s_2]) \rightarrow[t_1s_1:t_1s_2+t_2s_1:t_2s_2]$.
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Sea $x=([t:1],[t:1])\in\Delta$, then $\Phi(x)=[t^2:t:1]\in C_1$, so we have a covering map from $\PP^1\times\PP^1\setminus\Delta$ to $\PP^2\setminus C_1$. To see that we have a universal cover, we see that $\PP^1\times\PP^1\setminus\Delta$ is simply connected. For this, we consider the proyection $\PP^1\times\PP^1\setminus\Delta\rightarrow\PP^1$, which has fiber $\PP^1\setminus\{p\}$, with $p$ a point. Then, this fiber has the homotopy type of $\s^2\setminus\{p\}\simeq\DD^2$, which is contractible, and therefore $\PP^1\times\PP^1\setminus\Delta$ has the same homotopy type of $\PP^1$, which is simply connected. We have the following diagram:
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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:
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\begin{equation*}
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\begin{array}{ccc}
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\PP^1\simeq\s^2 & \longrightarrow & \PP^1\times\PP^1\setminus\Delta \\
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\PP^1\simeq\s^2 & \longrightarrow & X \\
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\downarrow & & \downarrow \\
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\RR\PP^2 & \longrightarrow & \PP^2\setminus C \\
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\end{array}
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\end{equation*}
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We have already seen that the spaces in the upper row are homotopy equivalent
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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 $\s^2\simeq X$.
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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 $X\rightarrow
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\PP^2\setminus C$ given by $(x,y,z)\rightarrow[x:y:z]$. Therefore, we get that $X$ is a universal cover of $\PP^2\setminus C$.
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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.
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$\pi_1(\RR\PP^2)=\ZZ$, 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 $\RR\PP^2$.
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\section{Complement of the quadric projective curve}
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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 $[x:y:z]$.
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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 $K$:
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\begin{itemize}
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\item There are three vertices, $v0=[0:0:1]$, $v1=[1:0:0]$ and $v2=[0:1:0]$.
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\item There are twelve 1-simplices, four from $v0$ to $v1$, four from $v0$ to $v2$, two from $v1$ to $v2$ and two from $v2$ to $v1$.
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\item There are sixteen 2-simplices.
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\end{itemize}
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\begin{center}
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\begin{tabular}{|c| c c |}
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\hline
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& $\partial_0$ & $\partial_1$ \\
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\hline\hline
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$e_1$ & $v_1$& $v_2$ \\
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\hline
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$e_2$ & $v_1$& $v_2$ \\
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\hline
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$e_3$ & $v_2$& $v_1$ \\
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\hline
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$e_4$ & $v_2$& $v_1$ \\
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\hline
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$e_5$ & $v_2$& $v_0$ \\
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\hline
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$e_6$ & $v_2$& $v_0$ \\
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\hline
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$e_7$ & $v_1$& $v_0$ \\
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\hline
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$e_8$ & $v_1$& $v_0$ \\
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\hline
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$e_9$ & $v_1$& $v_0$ \\
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\hline
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$e_{10}$ & $v_1$& $v_0$ \\
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\hline
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$e_{11}$ & $v_2$& $v_0$ \\
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\hline
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$e_{12}$ & $v_2$& $v_0$ \\
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\hline
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\end{tabular}
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\end{center}
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\begin{center}
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\begin{tabular}{|c| c c c|}
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\hline
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& $\partial_0$ & $\partial_1$ & $\partial_2$ \\
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\hline\hline
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$T_1$ & $e_3$& $e_5$ & $e_{10}$\\
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\hline
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$T_2$ & $e_1$& $e_9$ & $e_5$\\
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\hline
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$T_3$ & $e_3$& $e_{6}$ & $e_9$\\
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\hline
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$T_4$ & $e_1$& $e_{10}$ & $e_6$ \\
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\hline
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$T_5$ & $e_3$& $e_{11}$ & $e_8$\\
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\hline
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$T_6$ & $e_1$& $e_7$ & $e_{11}$\\
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\hline
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$T_7$ & $e_3$& $e_{12}$ & $e_7$\\
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\hline
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$T_8$ & $e_1$& $e_8$ & $e_{12}$\\
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\hline
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$T_9$ & $e_4$& $e_{12}$ & $e_{10}$\\
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\hline
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$T_{10}$ & $e_2$& $e_9$ & $e_{10}$\\
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\hline
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$T_{11}$ & $e_4$& $e_{11}$ & $e_9$\\
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\hline
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$T_{12}$ & $e_2$& $e_{10}$ & $e_{11}$\\
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\hline
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$T_{13}$ & $e_4$& $e_5$ & $e_8$\\
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\hline
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$T_{14}$ & $e_2$& $e_7$ & $e_5$\\
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\hline
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$T_{15}$ & $e_4$& $e_6$ & $e_7$\\
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\hline
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$T_{16}$ & $e_2$& $e_8$ & $e_6$\\
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\hline
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\end{tabular}
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\end{center}
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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.
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This way, we get a universal cover whose CW-complex structure is formed by $12$ $0-$simplices, $48$ $1-$simplices and $64$ $2-$simplices.
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\end{document}
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