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diff --git a/controle-20230412.tex b/controle-20230412.tex index c7dec35..c6100bb 100644 --- a/controle-20230412.tex +++ b/controle-20230412.tex @@ -1,6 +1,7 @@ %% This is a LaTeX document. Hey, Emacs, -*- latex -*- , get it? \documentclass[12pt,a4paper]{article} -\usepackage[francais]{babel} +\usepackage[a4paper,margin=2.5cm]{geometry} +\usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} %\usepackage{ucs} @@ -24,11 +25,11 @@ %\externaldocument{notes-accq205}[notes-accq205.pdf] % \theoremstyle{definition} -\newtheorem{comcnt}{Tout} +\newtheorem{comcnt}{Whatever} \newcommand\thingy{% \refstepcounter{comcnt}\smallskip\noindent\textbf{\thecomcnt.} } -\newcommand\exercice{% -\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercice~\thecomcnt.}\par\nobreak} +\newcommand\exercise{% +\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercise~\thecomcnt.}\par\nobreak} \renewcommand{\qedsymbol}{\smiley} % \newcommand{\id}{\operatorname{id}} @@ -37,6 +38,7 @@ \newcommand{\val}{\operatorname{val}} % \DeclareUnicodeCharacter{00A0}{~} +\DeclareUnicodeCharacter{A76B}{z} % \DeclareMathSymbol{\tiret}{\mathord}{operators}{"7C} \DeclareMathSymbol{\traitdunion}{\mathord}{operators}{"2D} @@ -52,9 +54,9 @@ \newcommand{\spaceout}{\hskip1emplus2emminus.5em} \newif\ifcorrige \corrigetrue -\newenvironment{corrige}% +\newenvironment{answer}% {\ifcorrige\relax\else\setbox0=\vbox\bgroup\fi% -\smallbreak\noindent{\underbar{\textit{Corrigé.}}\quad}} +\smallbreak\noindent{\underbar{\textit{Answer.}}\quad}} {{\hbox{}\nobreak\hfill\checkmark}% \ifcorrige\par\smallbreak\else\egroup\par\fi} % @@ -62,12 +64,12 @@ % \begin{document} \ifcorrige -\title{ACCQ205\\Contrôle de connaissances — Corrigé\\{\normalsize Courbes algébriques}} +\title{ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}} \else -\title{ACCQ205\\Contrôle de connaissances\\{\normalsize Courbes algébriques}} +\title{ACCQ205\\Final exam\\{\normalsize Algebraic curves}} \fi \author{} -\date{12 avril 2023} +\date{2023-04-12} \maketitle \pretolerance=8000 @@ -75,38 +77,31 @@ \vskip1truein\relax -\noindent\textbf{Consignes.} +\noindent\textbf{Instructions.} -Les exercices sont totalement indépendants. Ils pourront être traités -dans un ordre quelconque, mais on demande de faire apparaître de façon -très visible dans les copies où commence chaque exercice. +The different exercises below are completely independent. They can be +answered in any order, but candidates are asked to label very clearly +on their papers where each exercise starts. -La longueur du sujet ne doit pas effrayer : l'énoncé du dernier -exercice est long parce que beaucoup de rappels ont été faits et que -la rédaction des questions cherche à donner tous les éléments -nécessaires pour passer d'une question aux suivantes. - -La difficulté des questions étant variée, il vaut mieux ne pas rester -bloqué trop longtemps. +\medbreak -Si on ne sait pas répondre rigoureusement, une réponse informelle peut -valoir une partie des points. +Answers can be written in English or French. \medbreak -L'usage de tous les documents (notes de cours manuscrites ou -imprimées, feuilles d'exercices, livres) est autorisé. +Use of written documents of any kind (such as handwritten or printed +notes, exercise sheets or books) is authorized. -L'usage des appareils électroniques est interdit. +Use of electronic devices of any kind is prohibited. \medbreak -Durée : 2h +Duration: 2 hours \ifcorrige -Ce corrigé comporte \textcolor{red}{XXX} pages (page de garde incluse). +This answer key has \textcolor{red}{XXX} pages (cover page included). \else -Cet énoncé comporte \textcolor{red}{XXX} pages (page de garde incluse). +This exam has \textcolor{red}{XXX} pages (cover page included). \fi \vfill @@ -123,7 +118,379 @@ Git: \input{vcline.tex} % % +\exercise + +\textit{The goal of this exercise is to study a representation of + lines in $\mathbb{P}^3$.} + +We fix a field $k$. Recall that \emph{points} in $\mathbb{P}^3(k)$ +are given by quadruplets $(x_0{:}x_1{:}x_2{:}x_3)$ of “homogeneous +coordinates” in $k$, not all zero, defined up to a common +multiplicative constant, and that \emph{planes} in $\mathbb{P}^3(k)$ +are of the form $\{(x_0{:}x_1{:}x_2{:}x_3) \in \mathbb{P}^3(k) : u_0 +x_0 + \cdots + u_3 x_3 = 0\}$ (for some $u_0,\ldots,u_3$, not all +zero, defined up to a common multiplicative constant) which we can +denote as $[u_0{:}u_1{:}u_2{:}u_3]$ (a point of the +“dual” $\mathbb{P}^3$). Our goal is to find a representation for +lines. + +It may be convenient, if so desired, to call $\langle w\rangle$ the +point in projective space $\mathbb{P}^{m-1}(k)$ defined by a vector +$w\neq 0$ in $k^m$ (i.e., if $w = (w_0,\ldots,w_m)$ then $\langle w +\rangle = (w_0{:}\cdots{:}w_m)$), that is, the class of $w$ under +collinearity. + +\textbf{(1)} Given $x := (x_0,\ldots,x_3) \in k^4$ and $y := +(y_0,\ldots,y_3) \in k^4$, let us call $x\wedge y := (w_{0,1}, +w_{0,2}, w_{0,3}, w_{1,2}, w_{1,3}, w_{2,3}) \in k^6$ where $w_{i,j} +:= x_i y_j - x_j y_i$. What is $(\lambda x)\wedge(\mu y)$ in relation +to $x\wedge y$? Under what necessary and sufficient condition do we +have $x\wedge y = 0$? What is $x\wedge(\lambda x+\mu y)$ in relation +to $x\wedge y$? + +\begin{answer} +The $w_{i,j}$ are bilinear in $x,y$ (they are $2\times 2$ +determinants) so $(\lambda x)\wedge(\mu y) = \lambda\mu(x\wedge y)$. +Vanishing of $w_{i,j}$ means $(x_i,x_j)$ is proportional to +$(y_i,y_j)$ so vanishing of all the $w_{i,j}$ means precisely that +$x$ or $y$ is zero or that $x$ and $y$ are collinear. Again by +bilinearity, we have $x\wedge(\lambda x+\mu y) = \lambda(x\wedge x) + +\mu(x\wedge y)$ which is just $\mu(x\wedge y)$ since $x\wedge x$ +is $0$. +\end{answer} + +\textbf{(2)} Show that if $V \subseteq k^4$ is a $2$-dimensional +vector subspace, then the set of $x\wedge y$ for $x,y\in V$ is a +$1$-dimensional subspace of $k^6$. + +\begin{answer} +Consider $u,v$ a basis of $V$: then $u\wedge v$ is nonzero, and any +element of $V$ can be written $\lambda u + \mu v$, and then $(\lambda +u + \mu v) \wedge (\lambda' u + \mu' v) = (\lambda \mu' - \lambda' +\mu)(u \wedge v)$ by (1) (or by composition of determinants). In +other words, any $x\wedge y$ with $x,y\in V$ is collinear to $u\wedge +v$, and the latter is nonzero, and since $\lambda \mu' - \lambda' \mu$ +can obviously take every value in $k$, we see that $\{x\wedge y : +x,y\in V\}$ is the line spanned by $u\wedge v$. +\end{answer} + +\textbf{(3)} Deduce from (2) that if $L \subseteq \mathbb{P}^3(k)$ is +a line, then $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$, where $w_{i,j} := x_i y_j - x_j y_i$ as +above, and $(x_0{:}x_1{:}x_2{:}x_3)$ and $(y_0{:}y_1{:}y_2{:}y_3)$ are +two distinct points on $L$, is a well-defined point in +$\mathbb{P}^5(k)$, not depending on the chosen points on $L$ nor on +the homogeneous coordinates representing them. + +\begin{answer} +Calling $\langle w\rangle$ the point in projective space +$\mathbb{P}^{m-1}(k)$ defined by a vector $w\neq 0$ in $k^m$, if $L$ +is a line in $\mathbb{P}^3(k)$ we can see it as $\{\langle v\rangle : +v\in V\}$ for a $2$-dimensional vector subspace $V \subseteq k^4$, and +we have seen that $\langle x\wedge y\rangle \in \mathbb{P}^4(k)$ +exists when $x$ and $y$ are not collinear (so that $x\wedge y \neq +0$), i.e., when $\langle x\rangle \neq \langle y\rangle$, and does not +depend on the $x,y \in V$. +\end{answer} + +The $w_{i,j}$ in question are known as the \textbf{Plücker + coordinates} of $L$. + +\textbf{(4)} Show that any point $(z_0{:}z_1{:}z_2{:}z_3)$ on the +line $L$ as above satisfies +\[ +w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0 = 0 +\tag{$*$} +\] +— and analogously by replacing $0,1,2$ by three distinct coordinates +in $\{0,1,2,3\}$. + +\begin{answer} +Expanding the $3\times 3$ determinant +\[ +\left| +\begin{matrix} +x_0&y_0&z_0\\ +x_1&y_1&z_1\\ +x_2&y_2&z_2\\ +\end{matrix} +\right| +\] +gives $w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0$. If $\langle +z\rangle$ is in the line through $\langle x\rangle$ and $\langle +y\rangle$, meaning the three vectors $x,y,z$ are linearly dependent, +then this determinant is zero. The same holds, of course, for any +other choice of coordinates instead of $0,1,2$. +\end{answer} + +\textbf{(5)} Deduce from (4) that the $w_{i,j}$ satisfy the following +relation: +\[ +w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + w_{0,3} w_{1,2} = 0 +\tag{$\dagger$} +\] + +\begin{answer} +By (4), we have $w_{0,1} x_2 - w_{0,2} x_1 + w_{1,2} x_0 = 0$ and +$w_{0,1} y_2 - w_{0,2} y_1 + w_{1,2} y_0 = 0$. Adding $y_3$ times the +first and $-x_3$ times the second gives the stated +relation ($\dagger$). +\end{answer} + +The projective algebraic variety defined by ($\dagger$) in +$\mathbb{P}^5$ is known as the \textbf{Plücker quadric}. In other +words, we have shown how to associate to any line $L$ in +$\mathbb{P}^3(k)$ a $k$-point on the Plücker quadric. We now consider +the converse. + +\textbf{(6)} Assuming $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$ +satisfies ($\dagger$) (viꝫ. belongs to the Plücker quadric), and +assuming also that $w_{0,3} \neq 0$, show that the two points +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ are meaningful and distinct, and +that the line joining them has the Plücker coordinates +$(w_{0,1}:\cdots:w_{2,3})$ that were given. (\emph{Hint:} +\underline{first} compute $(w_{0,3},w_{1,3},w_{2,3},0) \wedge +(0,w_{0,1},w_{0,2},w_{0,3})$ and use the result, with the Plücker +relation and the fact that $w_{0,3} \neq 0$ to conclude.) + +\begin{answer} +We straightforwardly compute $(w_{0,3},w_{1,3},w_{2,3},0) \wedge +(0,w_{0,1},w_{0,2},w_{0,3}) = (w_{0,3} w_{0,1}, w_{0,3} w_{0,2}, +w_{0,3}^2, \penalty0 w_{1,3} w_{0,2} - w_{2,3} w_{0,1}, w_{1,3} +w_{0,3}, w_{2,3} w_{0,3})$. By Plücker's relation ($\dagger$), +$w_{1,3} w_{0,2} - w_{2,3} w_{0,1} = w_{1,2} w_{0,3}$, so we get +$w_{0,3}$ times $(w_{0,1}, w_{0,2}, w_{0,3}, \penalty0 w_{1,2}, +w_{1,3}, w_{2,3})$. Assuming $w_{0,3} \neq 0$, this is a nonzero +vector, which implies (by question (1)) that +$(w_{0,3},w_{1,3},w_{2,3},0)$ and $(0,w_{0,1},w_{0,2},w_{0,3})$ are +nonzero and non-collinear, so that the two points +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ are meaningful and distinct, and +by the computation we have just done, the Plücker coordinates of the +line joining them is the set of coordinates $(w_{0,1}:\cdots:w_{2,3})$ +that were given. +\end{answer} + +\textbf{(7)} Deduce from (6) that any $(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:} +\penalty0 w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$ +satisfying ($\dagger$) (viꝫ. belonging to the Plücker quadric) is the +set of Plücker coordinates of a (clearly unique) line in +$\mathbb{P}^3(k)$. (\emph{Hint:} what needs to be proved is that the +assumption $w_{0,3} \neq 0$ in (6) is harmless: explain how it can be +arranged by a judicious permutation of coordinates.) + +\begin{answer} +We have seen in (6) that when $w_{0,3} \neq 0$ then +$(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ are the Plücker coordinates of a line +in $\mathbb{P}^3(k)$. But all $w_{i,j}$ cannot be zero (as they are +given in $\mathbb{P}^5$), and we can always permute coordinates in +such a way that any $w_{i,j} \neq 0$, which is sure to exist, becomes +$w_{0,3}$, and the formula $w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + +w_{0,3} w_{1,2} = 0$ is invariant under any permutation of coordinates +(for this is is enough to check a cyclic permutation and a +transposition; keep in mind that $w_{j,i} = -w_{i,j}$ when rewriting +so as $i<j$), so we have confirmed the result in all cases. +\end{answer} + +At this point, we have established a bijection between the set of +lines $L$ in $\mathbb{P}^3(k)$ and the set of $k$-points in the +Plücker quadric defined by ($\dagger$) in $\mathbb{P}^5$; we know how +to compute Plücker coordinates from two distinct points lying on $L$ +(by definition). We now wish to compute Plücker coordinates for a +line defined as the the intersection of two plines. + +\textbf{(8)} Rephrase (4) to deduce that, if $L$ is a line with +Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$, then the planes +$[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ and $[0 : w_{2,3} : {-w_{1,3}} + : w_{1,2}]$ contain $L$. Now consider these as points in the +dual $\mathbb{P}^3$: show that the Plücker coordinates of the line +$L^*$ joining the two points in question are: $[w_{2,3} : {-w_{1,3}} : + w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$, provided $w_{1,2} \neq +0$. + +\begin{answer} +The relation ($*$) of (4), namely $w_{1,2} z_0 - w_{0,2} z_1 + w_{0,1} +z_2 = 0$, means precisely that $(z_0{:}z_1{:}z_2{:}z_3)$ is on the +plane $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$. Shifting coordinates +cyclically, it is also on the plane $[0 : w_{2,3} : {-w_{1,3}} : + w_{1,2}]$. Computing the Plücker coordinates as defined in +questions (1) to (3) for the line through these two (dual) points +gives $[w_{1,2} w_{2,3} : {-w_{1,2} w_{1,3}} : w_{1,2}^2 : {w_{0,2} + w_{1,3} - w_{0,1} w_{2,3}} : {-w_{0,2} w_{1,2}} : w_{0,1} + w_{1,2}]$ (provided not all are zero). By Plücker's +relation ($\dagger$), $w_{0,2} w_{1,3} - w_{0,1} w_{2,3} = w_{0,3} +w_{1,2}$, and now we can divide all coordinates by $w_{1,2}$ if it is +nonzero, giving $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$ (we already know that these six coordinates +cannot vanish simultaneously, so $w_{1,2} \neq 0$ is the only +condition we need). +\end{answer} + +\textbf{(9)} Deduce from (8) that if $L$ is a line in +$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1} : w_{0,2} : +w_{0,3} : \penalty0 w_{1,2} : w_{1,3} : w_{2,3})$, and if $L^*$ +denotes the “dual” line in the dual $\mathbb{P}^3$, that is, the line +consisting of all points corresponding to planes containing $L$, then +$L^*$ has (dual) Plücker coordinates $[w_{2,3} : {-w_{1,3}} : w_{1,2} + : w_{0,3} : {-w_{0,2}} : w_{0,1}]$. (\emph{Hint:} the only thing +that needs to be proved is that the assumption $w_{1,2} \neq 0$ in (8) +is harmless: explain how it can be arranged by a judicious permutation +of coordinates.) + +\begin{answer} +We have seen in (8) that when $w_{1,2} \neq 0$ then the line $L^*$ +dual to $L$ is given by $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$. But all $w_{i,j}$ cannot be zero +(question (3)), and we can always permute coordinates in such a way +that any $w_{i,j} \neq 0$, which is sure to exist, becomes $w_{1,2}$, +and the formula $(w_{0,1} : w_{0,2} : w_{0,3} : \penalty0 w_{1,2} : +w_{1,3} : w_{2,3}) \mapsto [w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$ is covariant under any permutation of +coordinates (for this is is enough to check a cyclic permutation and a +transposition), so we have confirmed the result in all cases. +\end{answer} + +\textbf{(10)} If $[u_0{:}u_1{:}u_2{:}u_3]$ and +$[v_0{:}v_1{:}v_2{:}v_3]$ are distinct planes in $\mathbb{P}^3(k)$, +how can we compute the Plücker coordinates of their line of +intersection? (\emph{Hint:} first describe the Plücker coordinates of +the line $L^*$ joining the corresponding points in the “dual” +$\mathbb{P}^3$, and then apply the result of (9), together with +projective duality.) + +\begin{answer} +Le line $L^*$ joining the points $[u_0{:}u_1{:}u_2{:}u_3]$ and +$[v_0{:}v_1{:}v_2{:}v_3]$ of the dual $\mathbb{P}^3$ is given by the +Plücker coordinates $w_{i,j} = u_i v_j - u_j v_i$. We then obtain the +Plücker coordinates of $L$ as $(w_{2,3} : {-w_{1,3}} : w_{1,2} : +w_{0,3} : {-w_{0,2}} : w_{0,1})$ (the formula given in (9) for +computing the Plücker coordinates of $L^*$ from those of $L$: it is +involutive and projective duality ensures that it also computes the +Plücker coordinates of $L$ from those of $L^*$). +\end{answer} + +\textbf{(11)} Explain how, by expanding a $4\times 4$ determinant +expressing the fact that four points $(x_0{:}x_1{:}x_2{:}x_3)$, +$(y_0{:}y_1{:}y_2{:}y_3)$, $(z_0{:}z_1{:}z_2{:}z_3)$ and +$(p_0{:}p_1{:}p_2{:}p_3)$ in $\mathbb{P}^3$ are coplanar, we can +obtain a formula for the plane through a line $L$ defined by its +Plücker coordinates and a point $(z_0{:}z_1{:}z_2{:}z_3)$ not situated +on $L$. (It is not required to go through the full computations.) + +\begin{answer} +Consider the $4\times 4$ determinant +\[ +\left| +\begin{matrix} +x_0&y_0&z_0&p_0\\ +x_1&y_1&z_1&p_1\\ +x_2&y_2&z_2&p_2\\ +x_3&y_3&z_3&p_3\\ +\end{matrix} +\right| +\] +whose vanishing expresses the fact that $x,y,z,p$ are in a common +hyperplane in $k^4$, i.e., that the points $\langle x\rangle$, +$\langle y\rangle$, $\langle z\rangle$ and $\langle p\rangle$ are +coplanar. Expanding it with respect to the final column $p$ gives a +linear condition for $p$ to be in the plane $P$ spanned by $\langle +x\rangle$, $\langle y\rangle$, $\langle z\rangle$, whose coefficients +are $3\times 3$ determinants, which are the coordinates of the +plane $P$ in the dual $\mathbb{P}^3$. Expanding those $3\times 3$ +determinants with respect to the final column $z$ writes them in terms +of the Plücker coordinates of the line $L$ through $\langle x\rangle$, +and $\langle y\rangle$, and the point $\langle z\rangle$. + +To be precise (although this was not asked), we get: +\[ +\begin{aligned} +\relax [\; +& - w_{1,2} z_3 + w_{1,3} z_2 - w_{2,3} z_1 \\ +:\; +& w_{2,3} z_0 + w_{0,2} z_3 - w_{0,3} z_2 \\ +:\; +& w_{0,3} z_1 - w_{1,3} z_0 - w_{0,1} z_3 \\ +:\; +& w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0 +\;] +\end{aligned} +\] +(the last coordinate being precisely given by ($*$)). +\end{answer} + +For your additional information: if $L$ and $L'$ are two lines in +$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$ +and $(w'_{0,1}:\cdots:w'_{2,3})$ respectively, then $L$ and $L'$ meet +in a common point (or equivalently, belong to a common plane) +iff\footnote{This relation (the “polarization” of the quadratic + relation ($\dagger$)) can be interpreted by saying that the line + in $\mathbb{P}^5$ joining the two points $(w_{0,1}:\cdots:w_{2,3})$ + and $(w'_{0,1}:\cdots:w'_{2,3})$ on the Plücker quadric is entirely + contained in said quadric.} +\[ +w_{0,1} w'_{2,3} - w_{0,2} w'_{1,3} + w_{0,3} w'_{1,2} ++ w_{2,3} w'_{0,1} - w_{1,3} w'_{0,2} + w_{1,2} w'_{0,3} = 0 +\tag{$\ddagger$} +\] +(it is not required to prove this). + +\textbf{(12)} Briefly summarize all of the above, emphasizing how we +have obtained formulæ allowing algorithmic computation of all possible +geometric constructions between points, lines and planes +in $\mathbb{P}^3$. + +\begin{answer} +For projective subspaces in $\mathbb{P}^3$ we can represent: +\begin{itemize} +\item \textbf{points} by their homogeneous coordinates + $(x_0{:}x_1{:}x_2{:}x_3)$ (which are arbitrary not all zero, defined + up to a common multiplicative constant), +\item \textbf{lines} by their Plücker coordinates + $(w_{0,1} : w_{0,2} : w_{0,3} : \penalty0 + w_{1,2} : w_{1,3} : w_{2,3})$ (which are not all zero, and subject + to the sole condition ($*$) of belonging to the Plücker quadric, + defined up to a common multiplicative constant), or equivalently by + their dual Plücker coordinates which are the same up to a + permutation and some changes of sign, $[w_{2,3} : {-w_{1,3}} : + w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$, +\item \textbf{planes} by their dual coordinates + $[u_0{:}u_1{:}u_2{:}u_3]$ (which are arbitrary not all zero, defined + up to a common multiplicative constant). +\end{itemize} +We can then compute: +\begin{itemize} +\item whether a point lies on a line by checking the relation ($*$) + and all its permutations of coordinates (cyclic permutations + suffice), +\item whether a point lies on a plane by the relation $u_0 x_0 + + \cdots + u_3 x_3 = 0$, +\item whether a line lies in a plane by checking whether the dual + point of the plane lies on the dual line (this gives $w_{2,3} u_2 + + w_{1,3} u_1 + w_{0,3} u_0 = 0$ and cyclic permutations), +\item the line joining two distinct points by computing the Plücker + coordinates as $2\times 2$ determinants as defined in question (1), +\item the plane though a line and a point not lying on it by the + formula found as explained in question (11), +\item the intersection line of two distinct planes by computing dual + Plücker coordinates as explained in question (10), +\item the point of intersection of a line and a plane by taking the + dual of the formula for the plane through a line and a point, +\item a sample point on a line as $(w_{0,3} : w_{1,3} : w_{2,3} : 0)$ + or some coordinate permutation thereof (as shown in question (6)), +\item a sample plane through a line dually to the previous point, + namely $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ (as shown in + question (8)), +\item whether two lines meet, or equivalently, belong to a common + plane, by using the criterion stated before (12), in which case + their point of intersection can be computed by intersecting one of + the lines with a sample plane through the other (as explained in the + previous items), and their common plane can be computed dually. +\end{itemize} +\end{answer} |