descripcion como modulos
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Alba Larraya Sancho
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{tikz}
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\usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart,
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decorations.markings, calc}
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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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\usetikzlibrary{shapes.multipart}
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\usepackage{hyperref}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\theoremstyle{definition}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{example}[theorem]{Example}
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\newtheorem{theorem}{Theorem}
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\newtheorem{example}{Example}
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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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@ -23,8 +24,6 @@ decorations.markings, calc}
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\newcommand{\s}{\mathbb{S}}
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\newcommand{\ZZ}{\mathbb{Z}}
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\numberwithin{equation}{section}
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\tikzset{->-/.style={decoration={markings, mark=at position #1 with {\arrow{>}}},postaction={decorate}}}
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%opening
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\title{Complement of a projective curve}
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@ -40,134 +39,89 @@ decorations.markings, calc}
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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{Smooth curves}
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Given a smooth curve of $C$ of degree $d$ in $\PP^2$, we know $(\PP^2,C)$ is homemorphic to $(\PP^2,C_d)$ where
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Given a smooth curve of degree $d$ in $\PP^2$, we know it is isomorphic to a curve of the form
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\begin{equation}
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C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\}
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\end{equation}
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Therefore, we will work with $C_d$, from now on.
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We will denote:
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\begin{eqnarray}
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M_d&=&\{(x,y,z)\in\CC^3 |\, x^d+y^d+z^d=1\} \\
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S_d&=&\{(x,y,z)\in M_d |\, x^d, y^d, z^d\in\RR_{\geq0}\} \\
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K &=&\{[x:y:z]\in\PP^2 |\, x^d, y^d, z^d\in\RR_{\geq0}\}
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M_d&=&\{(x,y,z)\in\CC | x^d+y^d+z^d=1\} \\
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S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
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K &=&\{[x:y:z]\in\PP^2 | x^d, y^d, z^d\in\RR_{\geq0}\}
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\end{eqnarray}
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We then have a commutative diagram:
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\begin{equation*}
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\begin{tikzcd}[/tikz/column 2/.append style={anchor=base west}]
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S_d\rar[hookrightarrow, "i"]\dar&M_d\dar[start anchor={[shift={(0,-.2)}]center}, end anchor={[shift={(-.32,.35)}]center}]\\
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K\rar[hookrightarrow, "i_\PP"]&\PP^2\setminus C_d.
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\end{tikzcd}
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\begin{array}{ccc}
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S_d & \longrightarrow & M_d \\
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\downarrow & & \downarrow \\
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K & \longrightarrow & \PP^2\setminus C_d \\
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\end{array}
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\end{equation*}
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The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following result.
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The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma:
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\begin{theorem}
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The space $S_d$ is a strong deformation retract of $M_d$.
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$S_d$ is a strong deformation retract of $M_d$.
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\end{theorem}
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\begin{proof}
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We start by getting a retraction $r:M_d\longrightarrow S_d$ for $d=1$. Note that for $d=1$ the result is trivial since $M_d$ is an affine complex plane (contractible) and $S_d$ is a closed triangle in an affine real plane. The key point is that in this case the retraction is easily constructed and we will lift it
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for $d>1$.
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We start by getting a retraction $r:M_d\longrightarrow S_d$ for $d=1$, and then we'll see that we can lift it.
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We can construct $r$ in two steps:
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\begin{equation*}
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\begin{tikzcd}
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M_1\rar["r_1"]\ar[dr, "r"]&M_1\cap\RR^3\dar["r_2"]\\
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&S_1.
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\end{tikzcd}
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\begin{array}{cccccc}
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r:&M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
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\end{array}
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\end{equation*}
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%Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
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The map $r_1$ is given by:
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Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow S_1$. Then $r_1$ is given by:
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\begin{equation}
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\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
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M_1\rar["r_1"]&M_1\cap\RR^3\\
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P\rar[mapsto]& \Re P.
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\end{tikzcd}
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\begin{array}{ccc}
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r_1: M_1\times I & \longrightarrow & M_1\cap\RR^3 \\
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(x_r+ix_i,y_r+iy_i,z_r+iz_i, t) & \longrightarrow & (x_r+(1-t)ix_i,y_r+(1-t)iy_i,z_r+(1-t)iz_i) \\
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\end{array}
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\end{equation}
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It is a deformation retraction via
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\begin{equation}
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\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
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M_1\times{[0,1]}\rar["H_1"]&M_1\cap\RR^3\\
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(P, t)\rar[mapsto]& t r_1(P) + (1 - t) P.
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\end{tikzcd}
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\end{equation}
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This map is a well-defined deformation retract since $M_1$ has real equations. Note that $M_1\cap\RR^3$
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$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is
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\[
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S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
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\]
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i.e, $S_1$ is the triangular section defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described in Figure~\ref{fig:triangulo}. We will retract each of them as follows.
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The restriction of the retraction to $(+,+,)$ is the identity. For $(+,-,-)$ we sent any point in there
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to $(1,0,0)$ (a similar argument holds when there is only one $+$).
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For $(-,+,+)$, for a given $(x,y,z)$ we consider intersection of the segment $[(x,y,z),(1,0,0)]$ with $x=0$, i.e., $r_2(x,y,z)=(0,\frac{y}{1-x},\frac{z}{1-x})$ (a similar argument holds when there is only one $-$). Since this definition is compatible in the intersection, it defines
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a retraction $r_2:M_1\cap\RR^3\to S_1$ and we construct the deformation using the segments:
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\begin{equation}
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\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
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(M_1\cap\RR^3)\times{[0,1]}\rar["H_1"]&S_1\\
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(P, t)\rar[mapsto]& t r_2(P) + (1 - t) P.
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\end{tikzcd}
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\end{equation}
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$M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)\in\RR_{\geq0}^3|x+y+z=1\}=\{(x,y,z)\in[0,1]^3|x+y+z=1\}$, this is, $S_1$ is a triangular section of the plane defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described below, and we will retract each of them separately:
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\begin{figure}[h]
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\begin{center}
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\begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
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\coordinate (A) at (-3,-4.5);
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\coordinate (B) at (3,-4.5);
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\coordinate (C) at (0,0);
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\fill[green!25!white] (C) to (B) to (A);
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\draw ($2*(C)-(A)$) to ($2*(A)-(C)$);
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\draw ($2*(B)-(A)$) to ($2*(A)-(B)$);
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\draw ($2*(B)-(C)$) to ($2*(C)-(B)$);
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\node at (0,-2.5) {$(+,+,+)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (5.5,0) {$(-,+,+)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (7.5,-6.5) {$(-,+,-)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (0,-8.5) {$(+.+,-)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (-8,-6.5) {$(+.-,-)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
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\node at (-6,0) {$(+.-,+)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
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\node at (0,4) {$(-,-,+)$};% {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
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\foreach \x[evaluate={\y=1-\x}] in {.125, .25, ..., 1}
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{
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\draw[->-=.5,green!75!black] ($1.75*(C) -.75*\x*(A) -.75*\y*(B)$) -- (C);
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\draw[->-=.5,green!75!black] ($1.75*(B) -.75*\x*(A) -.75*\y*(C)$) -- (B);
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\draw[->-=.5,green!75!black] ($1.75*(A) -.75*\x*(C) -.75*\y*(B)$) -- (A);
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}
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\foreach \x[evaluate={\y=1-\x}] in {0.25, .3125, ..., .75}
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{
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\draw[->-=.5,green!75!black] ($1.75*\x*(C) -.75*(A) + 1.75*\y*(B)$) -- ($\x*(C)+ \y*(B)$);
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\draw[->-=.5,green!75!black] ($1.75*\x*(A) -.75*(B) + 1.75*\y*(C)$) -- ($\x*(A)+ \y*(C)$);
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\draw[->-=.5,green!75!black] ($1.75*\x*(A) -.75*(C) + 1.75*\y*(B)$) -- ($\x*(A)+ \y*(B)$);
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}
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%\fill (6, -4.5) circle[radius=.1cm];
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\node[right=2pt] at (C) {(0,0,1)};
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\node[above right] at (B) {(0,1,0)};
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\node[above left] at (A) {(1,0,0)};
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\draw (3,4.5) to (-6,-9);
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\draw (-3,4.5) to (6,-9);
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\draw (-7,-4.5) to (7,-4.5);
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\node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
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\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
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\node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
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\node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
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\node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
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\node[right] at (0,0) {(0,0,1)};
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\node[right,above] at (3,-4.5) {(0,1,0)};
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\node[left,above] at (-3,-4.5) {(1,0,0)};
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\end{tikzpicture}
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\caption{Triangle $K$}
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\label{fig:triangulo}
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\end{center}
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\end{figure}
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%We denote each section as $(\pm,\pm,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ is $(+,+,+)$.
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%We describe the retraction $r_2$ in two steps. First, we define a map $p:M_1\cap\RR^3\to S_1$ as follows
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%
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%The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
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%\begin{equation}
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% \begin{array}{ccc}
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% r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
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% (x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
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% \end{array}
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%\end{equation}
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%Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
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%\begin{equation}
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%P_{x,y,z}=\left\{\begin{array}{lr}
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% (x,y,z)&(+,+,+) \\
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% (1,0,0)&(+,-,-) \\
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% (0,1,0)&(-,+,-) \\
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% (0,0,1)&(-,-,+) \\
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% (\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
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% (\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
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% (0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
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%\end{array}\right.
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%\end{equation}
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We denote each section as $(\pm,\pm,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ would be $(+,+,+)$.
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The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
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\begin{equation}
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\begin{array}{ccc}
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r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
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(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
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\end{array}
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\end{equation}
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Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
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\begin{equation}
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P_{x,y,z}=\left\{\begin{array}{lr}
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(x,y,z)&(+,+,+) \\
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(1,0,0)&(+,-,-) \\
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(0,1,0)&(-,+,-) \\
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(0,0,1)&(-,-,+) \\
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(\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
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(\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
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(0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
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\end{array}\right.
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\end{equation}
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We need to see now that this point is well defined, for this, we'll see that in the borders of the different parts it coincides. We will see this for the part $(+,+,-)$, for the rest of permutations it is equivalent.
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We start with the border between $(+,+,-)$ and $(+,+,+)$, this is, the line with $z=0$. In this case
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@ -332,12 +286,82 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$. We deno
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\end{equation*}
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where $\Delta^d$ is a simplicial complex that we are trying to find and identify with $S_d$, equivalently as what we have for $d=1$.
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Let $\phi: S_d\longleftarrow S_1$ be the map given by $\phi(x,y,z)=(x^d, y^d, z^d)$, then necessarily $\phi^{-1}(\Delta)=\Delta^d$.
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In particular, the $d^3$ vertices of the simplicial complex $\Delta^d$ will be identified to $(\xi_d,0,0)$, $(0,\xi_d,0)$, $(0,0,\xi_d)$ for all $\xi_d$ the $d$-$th$ roots of unity.
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%Let $\phi: S_d\longleftarrow S_1$ be the map given by $\phi(x,y,z)=(x^d, y^d, z^d)$, then necessarily $\phi^{-1}(\Delta)=\Delta^d$.
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%In particular, the $d^3$ vertices of the simplicial complex $\Delta^d$ will be identified to $(\xi_d,0,0)$, $(0,\xi_d,0)$, $(0,0,\xi_d)$ for all $\xi_d$ the $d$-$th$ roots of unity.
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The edges of $\Delta$ can be identified with the lines of $S_1$ parametrized as $(1-t,t,0)$, $(1-t,0,t)$ and $(0,t,1-t)$ with $t\in[0,1]$. Then, the preimages of these lines by $\phi$ are of the form $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$ in the first case and analogous in the rest, where we denote $\sqrt[s]{x}$ the positive real $d$-root for $x$ real, and where $\xi_i$ and $\xi_j$ are two any $d$-$th$ roots of unity. For each of these preimages we can define an edge between the corresponding vertices in $\Delta^d$.
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%The edges of $\Delta$ can be identified with the lines of $S_1$ parametrized as $(1-t,t,0)$, $(1-t,0,t)$ and $(0,t,1-t)$ with $t\in[0,1]$. Then, the preimages of these lines by $\phi$ are of the form $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$ in the first case and analogous in the rest, where we denote $\sqrt[s]{x}$ the positive real $d$-root for $x$ real, and where $\xi_i$ and $\xi_j$ are two any $d$-$th$ roots of unity. For each of these preimages we can define an edge between the corresponding vertices in $\Delta^d$.
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The $2$-faces of the simplicial complex $\Delta^d$ will then be identified to the areas between the lines $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$, $(\sqrt[d]{1-t}\xi_i,0, \sqrt[d]{t}\xi_k)$ and $(0,\sqrt[d]{t}\xi_j, \sqrt[d]{1-t}\xi_k)$.
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%The $2$-faces of the simplicial complex $\Delta^d$ will then be identified to the areas between the lines $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$, $(\sqrt[d]{1-t}\xi_i,0, \sqrt[d]{t}\xi_k)$ and $(0,\sqrt[d]{t}\xi_j, \sqrt[d]{1-t}\xi_k)$.
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$\Delta^d$ will have $3d$ $0-$simplices, $3d^2$ $1-$simplices and $d^3$ $2-$simplices.
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Let $\mu_d=\{\xi| \xi^d=1\}$. We can express $S_d$ as an union
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\begin{equation}
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\begin{array}{rll}
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S_d & = & \{(x,y,z)| x^d+y^d+z^d=1 \text{ and } x^d, y^d, z^d\in\RR_{\geq0}\} \\
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& = & \bigcup_{(\xi_1, \xi_2, \xi_3)\in\mu^3_d} X_{(\xi_1, \xi_2, \xi_3)} \\
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\end{array}
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\end{equation}
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where $X_{(\xi_1, \xi_2, \xi_3)}=\{(x,y,z)| x^d+y^d+z^d=1 \text{ and } \xi_1x, \xi_2y, \xi_3z\geq0\}$.
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Then, we can see that the faces of the simplex are given by $\mathring{X}_{(\xi_1, \xi_2, \xi_3)}$, while the edges are given by $X_{(\xi_1, \xi_2, 1)}\cap\{z=0\}$, $X_{(\xi_1, 1, \xi_2)}\cap\{y=0\}$, $X_{(1, \xi_1, \xi_2)}\cap\{x=0\}$ and the vertices by $X_{(\xi, 1, 1)}\cap\{y=0,z=0\}$, $X_{(1, 1, \xi)}\cap\{x=0,y=0\}$, $X_{(1, \xi, 1)}\cap\{x=0, z=0\}$.
|
||||
We can remark that $X_{(\xi_1, \xi_2, \xi_3)}=(\xi_1, \xi_2, \xi_3)X_{(1,1,1)}$, and therefore that we have a group action of $\mu^3_d$ over $S_d$.
|
||||
|
||||
Let
|
||||
\begin{equation}
|
||||
\begin{array}{rll}
|
||||
L^z_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{z=0\}\\
|
||||
L^y_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{y=0\}\\
|
||||
L^x_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{x=0\}\\
|
||||
V^z_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{x=0, y=0\}\\
|
||||
V^y_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{x=0, z=0\}\\
|
||||
V^x_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{y=0, z=0\}\\
|
||||
\end{array}
|
||||
\end{equation}
|
||||
|
||||
Then, the $2-$simplices, $1-$simplices and $0-$simplices of $\Delta^d$ are given by
|
||||
\begin{equation}
|
||||
\begin{array}{rll}
|
||||
C_2(\Delta^d) & = & \ZZ[\mu^3_d]<X_{(1,1,1)}>\\
|
||||
C_1(\Delta^d) & = & \ZZ[\mu^3_d]/\ZZ[\mu_d]<L^z_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu_d]<L^y_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu_d]<L^x_{(1,1,1)}>\\
|
||||
C_0(\Delta^d) & = & \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^z_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^y_{(1,1,1)}> \bigoplus \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^x_{(1,1,1)}>\\
|
||||
\end{array}
|
||||
\end{equation}
|
||||
|
||||
The border maps are given by:
|
||||
\begin{equation}
|
||||
\begin{array}{lll}
|
||||
\partial X_{(1,1,1)} & = & L^z_{(1,1,1)}-L^y_{(1,1,1)}+L^x_{(1,1,1)}\\
|
||||
\partial L^z_{(1,1,1)} & = & V^y_{(1,1,1)}-V^x_{(1,1,1)}\\
|
||||
\partial L^y_{(1,1,1)} & = & V^z_{(1,1,1)}-V^x_{(1,1,1)}\\
|
||||
\partial L^x_{(1,1,1)} & = & V^z_{(1,1,1)}-V^y_{(1,1,1)}\\
|
||||
\end{array}
|
||||
\end{equation}
|
||||
|
||||
The border maps for the rest of simplices are defined by the group action as:
|
||||
\begin{equation}
|
||||
\begin{array}{lllll}
|
||||
\partial X_{(\xi_1, \xi_2, \xi_3)} & =& (\xi_1, \xi_2, \xi_3)\partial X_{(1,1,1)} & = & (\xi_1, \xi_2, 1)L^z_{(1,1,1)}-(\xi_1, 1, \xi_3)L^y_{(1,1,1)}+(1, \xi_2, \xi_3)L^x_{(1,1,1)}\\
|
||||
\partial L^z_{(\xi_1, \xi_2, 1)} & = & (\xi_1, \xi_2, 1)\partial L^z_{(1,1,1)} & = & (1, \xi_2, 1)V^y_{(1,1,1)}-(\xi_1, 1, 1)V^x_{(1,1,1)}\\
|
||||
\partial L^y_{(\xi_1, 1, \xi_2)} & = & (\xi_1, 1, \xi_2)\partial L^y_{(1,1,1)} & = & (1, 1, \xi_2)V^z_{(1,1,1)}-(\xi_1, 1, 1)V^x_{(1,1,1)}\\
|
||||
\partial L^x_{(1, \xi_1, \xi_2)} & = & (1, \xi_1, \xi_2)\partial L^x_{(1,1,1)} & = & (1, \xi_1, 1)V^z_{(1,1,1)}-(1, 1, \xi_2)V^y_{(1,1,1)}\\
|
||||
\end{array}
|
||||
\end{equation}
|
||||
|
||||
We have then defined $\Delta^d$, the simplicial complex with which we can identify $S_d$. Our last step now is to complete the diagram with a simplicial complex associated to the projective space $K$, which we will call $\Delta^d_\text{proj}$.
|
||||
|
||||
\begin{equation*}
|
||||
\begin{array}{ccccc}
|
||||
\Delta & \longleftrightarrow & S_1 & & \\
|
||||
\uparrow& & \uparrow & & \\
|
||||
\Delta^d & \longleftrightarrow & S_d & \longrightarrow & M_d \\
|
||||
\downarrow & & \downarrow & & \downarrow \\
|
||||
\Delta^d_\text{proj} & \longleftrightarrow & K & \longrightarrow & \PP^2\setminus C_d \\
|
||||
\end{array}
|
||||
\end{equation*}
|
||||
|
||||
Where the map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x,y,z)\longrightarrow[x:y:z]$. Since we are passing to the projective plane,two points $(x\xi_1, y\xi_2, z\xi_3)$ and $(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)$ will be identified, and therefore, so will $X_{(x\xi_1, y\xi_2, z\xi_3)}$ and $X_{(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)}$. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices.
|
||||
|
||||
\begin{example}{}
|
||||
We are going to compute now the simplicial complex for $d=2$, this is, $S_2=\{(x,y,z)\in M_2| x^2, y^2, z^2\in\RR_{\geq 0}\}$.
|
||||
|
||||
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Reference in New Issue
Block a user