\documentclass[a4paper,10pt]{amsart} \usepackage[utf8]{inputenc} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amsmath} \usepackage{tikz} \usepackage{tikz-cd} \usetikzlibrary{positioning} \usetikzlibrary{arrows.meta} \usetikzlibrary{decorations.pathmorphing} \usepackage{hyperref} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{definition}{Definition} \newcommand{\CC}{\mathbb{C}} \newcommand{\RR}{\mathbb{R}} \newcommand{\DD}{\mathbb{D}} \newcommand{\BB}{\mathbb{B}} \newcommand{\PP}{\mathbb{P}} \newcommand{\s}{\mathbb{S}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\norma}[1]{\left\lVert#1\right\rVert} \newcommand{\uno}{\mathbf{1}} %opening \title{Complement of a projective curve} \author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz} \begin{document} \maketitle \begin{abstract} \end{abstract} \section{Definition of the problem} 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. \section{Complement of a conic projective curve} 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. 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: \begin{equation*} \begin{tikzcd} \PP^1\simeq\s^2 \rar\dar& X\dar \\ \RR\PP^2 \rar& \PP^2\setminus C \\ \end{tikzcd} \end{equation*} 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 $p$. Let $H: X\times[0,1]\rightarrow X$ given by \[ H(p,t)= \frac{t}{\norma{\Re p}} \Re p + (1 - t) p. %(x_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)x_i, y_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)y_i, z_r(\frac{t}{\sqrt{1+x_i^2+y_i^2+z_i^2}}+1-t)+i(1-t)z_i) \] %$ 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 $z=z_r+iz_i$. 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 $t\in[0,1]$. %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 hence $\s^2\simeq X$. The horizontal maps of \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 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. 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 $[x:y:z]$. 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} \item There are three vertices, $v0=[0:0:1]$, $v1=[1:0:0]$ and $v2=[0:1:0]$. \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 $v1$. \item There are sixteen 2-simplices. \end{itemize} \begin{center} \begin{tabular}{|c| c c |} \hline & $\partial_0$ & $\partial_1$ \\ \hline\hline $e_1$ & $v_1$& $v_2$ \\ \hline $e_2$ & $v_1$& $v_2$ \\ \hline $e_3$ & $v_2$& $v_1$ \\ \hline $e_4$ & $v_2$& $v_1$ \\ \hline $e_5$ & $v_2$& $v_0$ \\ \hline $e_6$ & $v_2$& $v_0$ \\ \hline $e_7$ & $v_1$& $v_0$ \\ \hline $e_8$ & $v_1$& $v_0$ \\ \hline $e_9$ & $v_1$& $v_0$ \\ \hline $e_{10}$ & $v_1$& $v_0$ \\ \hline $e_{11}$ & $v_2$& $v_0$ \\ \hline $e_{12}$ & $v_2$& $v_0$ \\ \hline \end{tabular} \end{center} \begin{center} \begin{tabular}{|c| c c c|} \hline & $\partial_0$ & $\partial_1$ & $\partial_2$ \\ \hline\hline $T_1$ & $e_3$& $e_5$ & $e_{10}$\\ \hline $T_2$ & $e_1$& $e_9$ & $e_5$\\ \hline $T_3$ & $e_3$& $e_{6}$ & $e_9$\\ \hline $T_4$ & $e_1$& $e_{10}$ & $e_6$ \\ \hline $T_5$ & $e_3$& $e_{11}$ & $e_8$\\ \hline $T_6$ & $e_1$& $e_7$ & $e_{11}$\\ \hline $T_7$ & $e_3$& $e_{12}$ & $e_7$\\ \hline $T_8$ & $e_1$& $e_8$ & $e_{12}$\\ \hline $T_9$ & $e_4$& $e_{12}$ & $e_{10}$\\ \hline $T_{10}$ & $e_2$& $e_9$ & $e_{10}$\\ \hline $T_{11}$ & $e_4$& $e_{11}$ & $e_9$\\ \hline $T_{12}$ & $e_2$& $e_{10}$ & $e_{11}$\\ \hline $T_{13}$ & $e_4$& $e_5$ & $e_8$\\ \hline $T_{14}$ & $e_2$& $e_7$ & $e_5$\\ \hline $T_{15}$ & $e_4$& $e_6$ & $e_7$\\ \hline $T_{16}$ & $e_2$& $e_8$ & $e_6$\\ \hline \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. 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 $(1,e_6)$ & $(1,v_2)$ & $(1,v_0)$ & $(d_2,e_6)$ & $(d_2,v_2)$ & $(d_2,v_0)$ \\\hline $(1,e_7)$ & $(1,v_1)$ & $(d_2,v_0)$ & $(d_2,e_7)$ & $(d_2,v_1)$ & $(d_2^2,v_0)$ \\\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}