280 lines
17 KiB
TeX
280 lines
17 KiB
TeX
\documentclass[a4paper,10pt]{amsart}
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\usepackage[utf8]{inputenc}
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\usepackage{amsfonts}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{tikz}
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\usepackage{tikz-cd}
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\usetikzlibrary{positioning}
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\usetikzlibrary{arrows.meta}
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\usetikzlibrary{decorations.pathmorphing}
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\usepackage{hyperref}
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\newtheorem{theorem}{Theorem}
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\newtheorem{lemma}{Lemma}
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\newtheorem{definition}{Definition}
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\newcommand{\CC}{\mathbb{C}}
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\newcommand{\RR}{\mathbb{R}}
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\newcommand{\DD}{\mathbb{D}}
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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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\newcommand{\norma}[1]{\left\lVert#1\right\rVert}
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\newcommand{\uno}{\mathbf{1}}
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%opening
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\title{Complement of a projective curve}
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\author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz}
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\begin{document}
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\maketitle
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\begin{abstract}
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\end{abstract}
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\section{Definition of the problem}
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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.
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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, i.e., 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 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:
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\begin{equation*}
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\begin{tikzcd}
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\PP^1\simeq\s^2 \rar\dar& X\dar \\
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\RR\PP^2 \rar& \PP^2\setminus C \\
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\end{tikzcd}
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\end{equation*}
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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
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$p$. Let $H: X\times[0,1]\rightarrow X$ given by
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\[
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H(p,t)=
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\frac{t}{\norma{\Re p}} \Re p + (1 - t) p.
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%(x_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)x_i, y_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)y_i, z_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)z_i)
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\]
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%$ 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 $z=z_r+iz_i$.
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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 $t\in[0,1]$.
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%Then, $H(\cdot,0)=id_X$, $H(\cdot,1)\in\s^2$ and for $(x,y,z)\in\s^2$, $H(x,y,z)=id_{\s^2}$.
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Therefore, $\s^2$ is a deformation retract of $X$, and hence $\s^2\simeq X$.
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The horizontal maps of
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\begin{equation}
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\begin{tikzcd}\label{eq:2-covers}
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\s^2\rar\dar&\dar\RR\PP^2\\
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X \rar& \PP^2\setminus C\\ [-20pt]
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{(x, y, x)}\rar[mapsto]&{[x:y:z]}
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\end{tikzcd}
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\end{equation}
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are $2:1$ coves. In particular, $X$ is the universal cover of $\PP^2\setminus C$.
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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 $\pi_n(\RR\PP^2)\cong\pi_n(\PP^2\setminus C)$.
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%We just need to see that
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The fundamental groups are also isomorphic since the horizontal arrows
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of~\eqref{eq:2-covers} are the universal covers (both $2:1$).
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Hence, the map below induces isomorphisms on all the homotopy groups, and
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then it is a homotopy equivalence.
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Actually we can see that $\RR\PP^2$ is a deformation retract of $\PP^2\setminus C$.
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Let us recall that the monodromy of the cover $X\to\PP^2\setminus C$ is given
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by the antipodal map $\sigma:X\to X$ given by $\sigma(p)=-p$. Note that
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\[
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\sigma(H(t, p)) = -\left(\frac{t}{\norma{\Re p}} \Re p + (1 - t) p\right)=
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\frac{t}{\norma{\Re (-p)}} \Re (-p) + (1 - t) (-p)= H(t,\sigma(p)).
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\]
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Since $H$ is equivariant it descends to a retraction homotopy downstairs.
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{\color{red} Propuesta: Hacer el caso \emph{trivial} reemplazando $x^2 + y^2 + z^2$ por
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$x + y + z$. Hacemos la retracción al triángulo
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\[
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\{(x, y, z)\in\RR^3\mid x + y + z=1,\ x, y, z\geq 0\}.
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\]
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Para la cuártica cambiaría la notación de los símplices. En lugar de numerarlos, usar
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ternas de signos o algo similar como índices. Eso es fácilmente generalizable a cualquier
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dimensión.}
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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 $\PP^2\setminus C$, since $(x,y,z),(ix,iy,iz),(\sqrt{i}x,\sqrt{i}y,\sqrt{i}z), (-ix,-iy,-iz)$ all have the same image $[x:y:z]$.
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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. Since
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\begin{table}[]
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\begin{tabular}{|c|c|c||c|c|c|}
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\hline
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& $\partial_0$ & $\partial_1$ & & $\partial_0$ & $\partial_1$ \\ \hline
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$(1,e_1)$ & $(1,v_1)$ & $(d_2^2,v_2)$ & $(d_2,e_1)$ & $(d_2,v_1)$ & $(d_2^{-1},v_2)$ \\\hline
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$(1,e_2)$ & $(1,v_1)$ & $(d_2^{-1},v_2)$ & $(d_2,e_2)$ & $(d_2,v_1)$ & $(1,v_2)$ \\\hline
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$(1,e_3)$ & $(1,v_2)$ & $(1,v_1)$ & $(d_2,e_3)$ & $(d_2,v_2)$ & $(d_2,v_1)$ \\\hline
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$(1,e_4)$ & $(1,v_2)$ & $(d_2^{-1},v_1)$ & $(d_2,e_4)$ & $(d_2,v_2)$ & $(1,v_1)$ \\\hline
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$(1,e_5)$ & $(1,v_2)$ & $(d_2^2,v_0)$ & $(d_2,e_5)$ & $(d_2,v_2)$ & $(d_2^{-1},v_0)$ \\\hline
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$(1,e_6)$ & $(1,v_2)$ & $(1,v_0)$ & $(d_2,e_6)$ & $(d_2,v_2)$ & $(d_2,v_0)$ \\\hline
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$(1,e_7)$ & $(1,v_1)$ & $(d_2,v_0)$ & $(d_2,e_7)$ & $(d_2,v_1)$ & $(d_2^2,v_0)$ \\\hline
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$(1,e_8)$ & $(1,v_1)$ & $(d_2^{-1},v_0)$ & $(d_2,e_8)$ & $(d_2,v_1)$ & $(1,v_0)$ \\\hline
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$(1,e_9)$ & $(1,v_1)$ & $(1,v_0)$ & $(d_2,e_9)$ & $(d_2,v_1)$ & $(d_2,v_0)$ \\\hline
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$(1,e_{10})$ & $(1,v_1)$ & $(d_2^2,v_0)$ & $(d_2,e_{10})$ & $(d_2,v_1)$ & $(d_2^{-1},v_0)$ \\\hline
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$(1,e_{11})$ & $(1,v_2)$ & $(d_2^{-1},v_0)$ & $(d_2,e_{11})$ & $(d_2,v_2)$ & $(1,v_0)$ \\\hline
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$(1,e_{12})$ & $(1,v_2)$ & $(d_2,v_0)$ & $(d_2,e_{12})$ & $(d_2,v_2)$ & $(d_2^2,v_0)$ \\\hline \hline
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$(d_2^2,e_1)$ & $(d_2^2,v_1)$ & $(1,v_2)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},v_1)$ & $(d_2,v_2)$ \\\hline
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$(d_2^2,e_2)$ & $(d_2^2,v_1)$ & $(d_2,v_2)$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},v_1)$ & $(d_2^2,v_2)$ \\\hline
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$(d_2^2,e_3)$ & $(d_2^2,v_2)$ & $(d_2^2,v_1)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},v_2)$ & $(d_2^{-1},v_1)$ \\\hline
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$(d_2^2,e_4)$ & $(d_2^2,v_2)$ & $(d_2,v_1)$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},v_2)$ & $(d_2^2,v_1)$ \\\hline
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$(d_2^2,e_5)$ & $(d_2^2,v_2)$ & $(1,v_0)$ & $(d_2^{-1},e_5)$ & $(d_2^{-1},v_2)$ & $(d_2,v_0)$ \\\hline
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$(d_2^2,e_6)$ & $(d_2^2,v_2)$ & $(d_2^2,v_0)$ & $(d_2^{-1},e_6)$ & $(d_2^{-1},v_2)$ & $(d_2^{-1},v_0)$ \\\hline
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$(d_2^2,e_7)$ & $(d_2^2,v_1)$ & $(d_2^{-1},v_0)$ & $(d_2^{-1},e_7)$ & $(d_2^{-1},v_1)$ & $(1,v_0)$ \\\hline
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$(d_2^2,e_8)$ & $(d_2^2,v_1)$ & $(d_2,v_0)$ & $(d_2^{-1},e_8)$ & $(d_2^{-1},v_1)$ & $(d_2^2,v_0)$ \\\hline
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$(d_2^2,e_9)$ & $(d_2^2,v_1)$ & $(d_2^2,v_0)$ & $(d_2^{-1},e_9)$ & $(d_2^{-1},v_1)$ & $(d_2^{-1},v_0)$ \\\hline
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$(d_2^2,e_{10})$ & $(d_2^2,v_1)$ & $(1,v_0)$ & $(d_2^{-1},e_{10})$ & $(d_2^{-1},v_1)$ & $(d_2,v_0)$ \\\hline
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$(d_2^2,e_{11})$ & $(d_2^2,v_2)$ & $(d_2,v_0)$ & $(d_2^{-1},e_{11})$ & $(d_2^{-1},v_2)$ & $(d_2^2,v_0)$ \\\hline
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$(d_2^2,e_{12})$ & $(d_2^2,v_2)$ & $(d_2^{-1},v_0)$ & $(d_2^{-1},e_{12})$ & $(d_2^{-1},v_2)$ & $(1,v_0)$ \\\hline
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\end{tabular}
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\end{table}
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\begin{table}[]
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\begin{tabular}{|c|c|c|c||c|c|c|c|}
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\hline
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& $\partial_0$ & $\partial_1$ & $\partial_2$ & & $\partial_0$ & $\partial_1$ & $\partial_2$ \\\hline
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$(1,T_1)$ & $(1,e_3)$ & $(1,e_5)$ & $(1,e_{10})$ & $(d_2,T_1)$ & $(d_2,e_3)$ & $(d_2,e_5)$ & $(d_2,e_{10})$ \\\hline
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$(1,T_2)$ & $(1,e_1)$ & $(1,e_9)$ & $(d_2^2,e_5)$ & $(d_2,T_2)$ & $(d_2,e_1)$ & $(d_2,e_9)$ & $(d_2^{-1},e_5)$ \\\hline
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$(1,T_3)$ & $(1,e_3)$ & $(1,e_{6})$ & $(1,e_9)$ & $(d_2,T_3)$ & $(d_2,e_3)$ & $(d_2,e_{6})$ & $(d_2,e_9)$ \\\hline
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$(1,T_4)$ & $(1,e_1)$ & $(1,e_{10})$ & $(d_2^2,e_6)$ & $(d_2,T_4)$ & $(d_2,e_1)$ & $(d_2,e_{10})$ & $(d_2^{-1},e_6)$ \\\hline
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$(1,T_5)$ & $(1,e_3)$ & $(1,e_{11})$ & $(1,e_8)$ & $(d_2,T_5)$ & $(d_2,e_3)$ & $(d_2,e_{11})$ & $(d_2,e_8)$ \\\hline
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$(1,T_6)$ & $(1,e_1)$ & $(1,e_7)$ & $(d_2^2,e_{11})$ & $(d_2,T_6)$ & $(d_2,e_1)$ & $(d_2,e_7)$ & $(d_2^{-1},e_{11})$ \\\hline
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$(1,T_7)$ & $(1,e_3)$ & $(1,e_{12})$ & $(1,e_7)$ & $(d_2,T_7)$ & $(d_2,e_3)$ & $(d_2,e_{12})$ & $(d_2,e_7)$ \\\hline
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$(1,T_8)$ & $(1,e_1)$ & $(1,e_8)$ & $(d_2^2,e_{12})$ & $(d_2,T_8)$ & $(d_2,e_1)$ & $(d_2,e_8)$ & $(d_2^{-1},e_{12})$ \\\hline
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$(1,T_9)$ & $(1,e_4)$ & $(1,e_{12})$ & $(d_2^{-1},e_{10})$ & $(d_2,T_9)$ & $(d_2,e_4)$ & $(d_2,e_{12})$ & $(1,e_{10})$ \\\hline
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$(1,T_{10})$ & $(1,e_2)$ & $(1,e_9)$ & $(d_2^{-1},e_{10})$ & $(d_2,T_{10})$ & $(d_2,e_2)$ & $(d_2,e_9)$ & $(1,e_{10})$ \\\hline
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$(1,T_{11})$ & $(1,e_4)$ & $(1,e_{11})$ & $(d_2^{-1},e_9)$ & $(d_2,T_{11})$ & $(d_2,e_4)$ & $(d_2,e_{11})$ & $(1,e_9)$ \\\hline
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$(1,T_{12})$ & $(1,e_2)$ & $(1,e_{10})$ & $(d_2^{-1},e_{11})$ & $(d_2,T_{12})$ & $(d_2,e_2)$ & $(d_2,e_{10})$ & $(1,e_{11})$ \\\hline
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$(1,T_{13})$ & $(1,e_4)$ & $(1,e_5)$ & $(d_2^{-1},e_8)$ & $(d_2,T_{13})$ & $(d_2,e_4)$ & $(d_2,e_5)$ & $(1,e_8)$ \\\hline
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$(1,T_{14})$ & $(1,e_2)$ & $(1,e_7)$ & $(d_2^{-1},e_5)$ & $(d_2,T_{14})$ & $(d_2,e_2)$ & $(d_2,e_7)$ & $(1,e_5)$ \\\hline
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$(1,T_{15})$ & $(1,e_4)$ & $(1,e_6)$ & $(d_2^{-1},e_7)$ & $(d_2,T_{15})$ & $(d_2,e_4)$ & $(d_2,e_6)$ & $(1,e_7)$ \\\hline
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$(1,T_{16})$ & $(1,e_2)$ & $(1,e_8)$ & $(d_2^{-1},e_6)$ & $(d_2,T_{16})$ & $(d_2,e_2)$ & $(d_2,e_8)$ & $(1,e_6)$ \\\hline\hline
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$(d_2^2,T_1)$ & $(d_2^2,e_3)$ & $(d_2^2,e_5)$ & $(d_2^2,e_{10})$ & $(d_2^{-1},T_1)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_5)$ & $(d_2^{-1},e_{10})$ \\\hline
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$(d_2^2,T_2)$ & $(d_2^2,e_1)$ & $(d_2^2,e_9)$ & $(1,e_5)$ & $(d_2^{-1},T_2)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_9)$ & $(d_2,e_5)$ \\\hline
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$(d_2^2,T_3)$ & $(d_2^2,e_3)$ & $(d_2^2,e_{6})$ & $(d_2^2,e_9)$ & $(d_2^{-1},T_3)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_{6})$ & $(d_2^{-1},e_9)$ \\\hline
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$(d_2^2,T_4)$ & $(d_2^2,e_1)$ & $(d_2^2,e_{10})$ & $(1,e_6)$ & $(d_2^{-1},T_4)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_{10})$ & $(d_2,e_6)$ \\\hline
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$(d_2^2,T_5)$ & $(d_2^2,e_3)$ & $(d_2^2,e_{11})$ & $(d_2^2,e_8)$ & $(d_2^{-1},T_5)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_{11})$ & $(d_2^{-1},e_8)$ \\\hline
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$(d_2^2,T_6)$ & $(d_2^2,e_1)$ & $(d_2^2,e_7)$ & $(1,e_{11})$ & $(d_2^{-1},T_6)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_7)$ & $(d_2,e_{11})$ \\\hline
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$(d_2^2,T_7)$ & $(d_2^2,e_3)$ & $(d_2^2,e_{12})$ & $(d_2^2,e_7)$ & $(d_2^{-1},T_7)$ & $(d_2^{-1},e_3)$ & $(d_2^{-1},e_{12})$ & $(d_2^{-1},e_7)$ \\\hline
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$(d_2^2,T_8)$ & $(d_2^2,e_1)$ & $(d_2^2,e_8)$ & $(1,e_{12})$ & $(d_2^{-1},T_8)$ & $(d_2^{-1},e_1)$ & $(d_2^{-1},e_8)$ & $(d_2,e_{12})$ \\\hline
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$(d_2^2,T_9)$ & $(d_2^2,e_4)$ & $(d_2^2,e_{12})$ & $(d_2,e_{10})$ & $(d_2^{-1},T_9)$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_{12})$ & $(d_2^2,e_{10})$ \\\hline
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$(d_2^2,T_{10})$ & $(d_2^2,e_2)$ & $(d_2^2,e_9)$ & $(d_2,e_{10})$ & $(d_2^{-1},T_{10})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_9)$ & $(d_2^2,e_{10})$ \\\hline
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$(d_2^2,T_{11})$ & $(d_2^2,e_4)$ & $(d_2^2,e_{11})$ & $(d_2,e_9)$ & $(d_2^{-1},T_{11})$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_{11})$ & $(d_2^2,e_9)$ \\\hline
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$(d_2^2,T_{12})$ & $(d_2^2,e_2)$ & $(d_2^2,e_{10})$ & $(d_2,e_{11})$ & $(d_2^{-1},T_{12})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_{10})$ & $(d_2^2,e_{11})$ \\\hline
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$(d_2^2,T_{13})$ & $(d_2^2,e_4)$ & $(d_2^2,e_5)$ & $(d_2,e_8)$ & $(d_2^{-1},T_{13})$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_5)$ & $(d_2^2,e_8)$ \\\hline
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$(d_2^2,T_{14})$ & $(d_2^2,e_2)$ & $(d_2^2,e_7)$ & $(d_2,e_5)$ & $(d_2^{-1},T_{14})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_7)$ & $(d_2^2,e_5)$ \\\hline
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$(d_2^2,T_{15})$ & $(d_2^2,e_4)$ & $(d_2^2,e_6)$ & $(d_2,e_7)$ & $(d_2^{-1},T_{15})$ & $(d_2^{-1},e_4)$ & $(d_2^{-1},e_6)$ & $(d_2^2,e_7)$ \\\hline
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$(d_2^2,T_{16})$ & $(d_2^2,e_2)$ & $(d_2^2,e_8)$ & $(d_2,e_6)$ & $(d_2^{-1},T_{16})$ & $(d_2^{-1},e_2)$ & $(d_2^{-1},e_8)$ & $(d_2^2,e_6)$ \\\hline
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\end{tabular}
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\end{table}
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\end{document} |