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Hey, Emacs, -*- latex -*- , get it? +\documentclass[12pt,a4paper]{article} +\usepackage[a4paper,margin=2.5cm]{geometry} +\usepackage[english]{babel} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +%\usepackage{ucs} +\usepackage{times} +% A tribute to the worthy AMS: +\usepackage{amsmath} +\usepackage{amsfonts} +\usepackage{amssymb} +\usepackage{amsthm} +% +\usepackage{mathrsfs} +\usepackage{wasysym} +\usepackage{url} +% +\usepackage{graphics} +\usepackage[usenames,dvipsnames]{xcolor} +\usepackage{tikz} +\usetikzlibrary{matrix,calc} +\usepackage{hyperref} +% +%\externaldocument{notes-accq205}[notes-accq205.pdf] +% +\theoremstyle{definition} +\newtheorem{comcnt}{Whatever} +\newcommand\thingy{% +\refstepcounter{comcnt}\smallskip\noindent\textbf{\thecomcnt.} } +\newcommand\exercise{% +\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercise~\thecomcnt.}\par\nobreak} +\renewcommand{\qedsymbol}{\smiley} +% +\newcommand{\id}{\operatorname{id}} +\newcommand{\alg}{\operatorname{alg}} +\newcommand{\ord}{\operatorname{ord}} +\newcommand{\val}{\operatorname{val}} +% +\DeclareUnicodeCharacter{00A0}{~} +\DeclareUnicodeCharacter{A76B}{z} +% +\DeclareMathSymbol{\tiret}{\mathord}{operators}{"7C} +\DeclareMathSymbol{\traitdunion}{\mathord}{operators}{"2D} +% +\DeclareFontFamily{U}{manual}{} +\DeclareFontShape{U}{manual}{m}{n}{ <-> manfnt }{} +\newcommand{\manfntsymbol}[1]{% + {\fontencoding{U}\fontfamily{manual}\selectfont\symbol{#1}}} +\newcommand{\dbend}{\manfntsymbol{127}}% Z-shaped +\newcommand{\danger}{\noindent\hangindent\parindent\hangafter=-2% + \hbox to0pt{\hskip-\hangindent\dbend\hfill}} +% +\newcommand{\spaceout}{\hskip1emplus2emminus.5em} +\newif\ifcorrige +\corrigetrue +\newenvironment{answer}% +{\ifcorrige\relax\else\setbox0=\vbox\bgroup\fi% +\smallbreak\noindent{\underbar{\textit{Answer.}}\quad}} +{{\hbox{}\nobreak\hfill\checkmark}% +\ifcorrige\par\smallbreak\else\egroup\par\fi} +% +% +% +\begin{document} +\ifcorrige +\title{ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}} +\else +\title{ACCQ205\\Final exam\\{\normalsize Algebraic curves}} +\fi +\author{} +\date{2023-04-12} +\maketitle + +\pretolerance=8000 +\tolerance=50000 + +\vskip1truein\relax + +\noindent\textbf{Instructions.} + +This exam consists of a single lengthy problem. Although the +questions depend on each other, they have been worded in such a way +that the necessary information for subsequent questions is given in +the text. Thus, failure to answer one question should not make it +impossible to proceed to later questions. + +\medbreak + +Answers can be written in English or French. + +\medbreak + +Use of written documents of any kind (such as handwritten or printed +notes, exercise sheets or books) is authorized. + +Use of electronic devices of any kind is prohibited. + +\medbreak + +Duration: 2 hours + +\ifcorrige +This answer key has 7 pages (cover page included). +\else +This exam has 4 pages (cover page included). +\fi + +\vfill +{\noindent\tiny +\immediate\write18{sh ./vc > vcline.tex} +Git: \input{vcline.tex} +\immediate\write18{echo ' (stale)' >> vcline.tex} +\par} + +\pagebreak + + +% +% +% + +\textit{The goal of this problem is to study a representation of lines + in $\mathbb{P}^3$.} + +\smallskip + +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. + +{\footnotesize 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. \par} + +\bigskip + +\textbf{(1)} Given $x := (x_0,\ldots,x_3) \in k^4$ and $y := +(y_0,\ldots,y_3) \in k^4$, let us define $x\wedge y := (w_{0,1}, +w_{0,2}, w_{0,3}, w_{1,2}, w_{1,3}, w_{2,3}) \in k^6$ 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 where $(x_0{:}x_1{:}x_2{:}x_3)$ 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} + +\bigskip + +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 to $-x_3$ times the second gives the stated +relation ($\dagger$). +\end{answer} + +\bigskip + +The projective algebraic variety defined by ($\dagger$) in +$\mathbb{P}^5$ is known as the \textbf{Plücker quadric}. In other +words, we have shown above how to associate to any line $L$ in +$\mathbb{P}^3(k)$ a $k$-point $(w_{0,1}:\cdots:w_{2,3})$ 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 +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ in $\mathbb{P}^3(k)$ 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 then 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} + +\bigskip + +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 that is described as the the intersection of two planes. + +\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 $[0 : w_{2,3} : {-w_{1,3}} + : w_{1,2}]$ both contain $L$. Now consider these as points in the +dual $\mathbb{P}^3$ and 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 $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$. +\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 $L^*$), in other words $(u_2 +v_3 - u_3 v_2 : {-u_1 v_3 + u_3 v_1} : u_1 v_2 - u_2 v_1 : u_0 v_3 - +u_3 v_0 : {-u_0 v_2 + u_2 v_0} : u_0 v_1 - u_1 v_0)$ +\end{answer} + +\bigskip + +\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: just +explain how it would work.) + +\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). + +\bigskip + +\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 thereof), +\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} + +\bigskip + +\centerline{\hbox to3truecm{\hrulefill}} + +\medskip + +\textbf{(13)} Independently of all of the above, compute the number of +lines in $\mathbb{P}^3(\mathbb{F}_q)$ (for example by counting the +number of pairs of distinct points in $\mathbb{P}^3(\mathbb{F}_q)$ and +the number of pairs of distinct points in a given line +$\mathbb{P}^1(\mathbb{F}_q)$), where $q$ is a prime power and +$\mathbb{F}_q$ denotes the finite field with $q$ elements. + +\begin{answer} +There are $\frac{q^4-1}{q-1} = q^3 + q^2 + q + 1$ points in +$\mathbb{P}^3(\mathbb{F}_q)$. There are thus $(q^3 + q^2 + q + 1)(q^3 ++ q^2 + q) = q (q^3 + q^2 + q + 1) (q^2 + q + 1)$ pairs of distinct +points in $\mathbb{P}^3(\mathbb{F}_q)$, each defining a line. There +are $\frac{q^2-1}{q-1} = q + 1$ points in +$\mathbb{P}^2(\mathbb{F}_q)$. There are thus $q(q + 1)$ pairs of +distinct points defining any given line. Thus, there are $(q (q^3 + +q^2 + q + 1) (q^2 + q + 1)) / (q (q+1)) = (q^2 + q + 1)(q^2 + 1)$ +lines in $\mathbb{P}^3(\mathbb{F}_q)$ (that is, $q^4 + q^3 + 2q^2 + q ++ 1$). +\end{answer} + +\textbf{(14)} Deduce from (13) and (1)–(7) the number of +$\mathbb{F}_q$-points on the hypersurface of degree $2$ (“quadric”) +$\{X_0 X_3 + X_1 X_4 + X_2 X_5 = 0\}$ in $\mathbb{P}^5(\mathbb{F}_q)$ +with coordinates $(X_0:\cdots:X_5)$. Assuming $q \equiv 1 \pmod{4}$, +deduce the number of $\mathbb{F}_q$-points on the hypersurface of +degree $2$ (“quadric”) $\{Z_0^2 + \cdots + Z_5^2 = 0\}$ in +$\mathbb{P}^5(\mathbb{F}_q)$ with coordinates $(Z_0:\cdots:Z_5)$ +(\emph{hint:} $q \equiv 1 \pmod{4}$ means $-1$ is a square in +$\mathbb{F}_q$, so we can factor $Z^2 + Z^{\prime2}$). + +\begin{answer} +We have seen in (1)–(7) that $\mathbb{F}_q$-points on the Plücker +quadric are in bijection with lines in $\mathbb{P}^3(\mathbb{F}_q)$, +and in (13) that there are $(q^2 + q + 1)(q^2 + 1) = q^4 + q^3 + 2q^2 ++ q + 1$ of them. Thus, there are that many points on the Plücker +quadric. The equation of the latter can be written in the form $X_0 +X_3 + X_1 X_4 + X_2 X_5 = 0$ by a simple linear coordinate change, +i.e., projective transformation ($X_0 = w_{0,1}$, $X_1 = w_{0,2}$, +$X_2 = w_{0,3}$, $X_3 = w_{2,3}$, $X_4 = -w_{1,3}$ and $X_5 = +w_{1,2}$) which certainly does not change the number of points. + +As for $Z_0^2 + \cdots + Z_5^2 = 0$, if we call $\sqrt{-1}$ some fixed +square root of $-1$ (which exists when $q \equiv 1 \pmod{4}$ as this +means that the Legendre symbol $(\frac{-1}{q})$ is $1$), we can write +$Z_0^2 + Z_1^2 = (Z_0+\sqrt{-1}\,Z_1) (Z_0-\sqrt{-1}\,Z_1)$ and +similarly for $Z_2^2 + Z_3^2$ and $Z_4^2 + Z_5^2$, and since the +linear transformation $X_0 = Z_0+\sqrt{-1}\,Z_1$, $X_1 = +Z_2+\sqrt{-1}\,Z_3$, $X_2 = Z_4+\sqrt{-1}\,Z_5$, $X_3 = +Z_0-\sqrt{-1}\,Z_1$, $X_4 = Z_2-\sqrt{-1}\,Z_3$, $X_5 = +Z_4-\sqrt{-1}\,Z_5$ is invertible, there are still the same number of +points. +\end{answer} + +\bigskip + +\centerline{\hbox to3truecm{\hrulefill}} + +\medskip + +(This question is more difficult; it is independent of (13)\&(14).) + +\textbf{(15)} Let $h \in k[t_0,t_1,t_2,t_3]$ be a homogeneous +polynomial, so that it defines a Zariski closed set (hypersurface) $X +:= \{h(x_0,x_1,x_2,x_3) = 0\}$ in $\mathbb{P}^3$. Show that the of +lines contained in $X$ defines a Zariski closed subset $Y$ of the +Plücker quadric in $\mathbb{P}^5$. (To be completely clear, this +means\footnote{Here $k^{\alg}$ denotes the algebraic closure of $k$, + but feel free to assume that $k$ is algebraically closed ($k = + k^{\alg}$) in this question.}: there is a Zariski closed set $Y$ in +$\mathbb{P}^5$, defined over $k$ and contained in the Plücker quadric +$Q$ (defined by $\dagger$), such that, for $w \in Q(k^{\alg})$, we +have $w \in Y(k^{\alg})$ if and only if $L_w \subseteq X(k^{\alg})$, +where $L_w$ denotes the line in $\mathbb{P}^3(k^{\alg})$ having +Plücker coordinates $w$.) + +The important part of this question is: how can we compute equations +for $Y$ given the equation $h=0$ of $X$? + +\begin{answer} +We have seen in question (6) that, so long as $w_{0,3} \neq 0$, the +line $L_w$ defined by $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $Q$ is the line through +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$. In other words, $\lambda,\mu$ it +is the line consisting of points $(\lambda w_{0,3} : \lambda w_{1,3} + +\mu w_{0,1} : \lambda w_{2,3} + \mu w_{0,2} : \mu w_{0,3})$. This +line is included in $X$ iff $h(\lambda w_{0,3}, \lambda w_{1,3} + \mu +w_{0,1}, \lambda w_{2,3} + \mu w_{0,2}, \mu w_{0,3}) = 0$ for all +$\lambda,\mu$ in $k^{\alg}$, which just means that, seen as a +polynomial in $\lambda,\mu$ now seen as two \emph{indeterminates}, +this is identically zero. But this is a homogeneous polynomial in +$\lambda,\mu$ whose coefficients are homogeneous polynomials in the +$w_{i,j}$, so equating all these coefficients to $0$ gives homogeneous +equations in the $w_{i,j}$ (coordinates in $\mathbb{P}^5$) for $L_w$ +to be included in $X$. This only holds so long as $w_{0,3} \neq 0$ +(when it is zero, some of the resulting equation will be trivial); +however, at least one $w_{i,j}$ must be zero in any case, so writing +the corresponding equations for all permutations of coordinates, +together with the equation ($\dagger$) of the Plücker quadric itself +(to ensure that the $w_{i,j}$ do correspond to a line in +$\mathbb{P}^3$) gives us the equations of the desired $Y$. + +To illustrate that this is actually algorithmic, the following Sage +code computes the equations for the set $Y$ of lines inside the +“diagonal cubic surface” $X := \{x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0\}$: +{\fontsize{8}{10}\relax +\begin{verbatim} +sage: R.<x0,x1,x2,x3,w01,w02,w03,w12,w13,w23,u,v> = PolynomialRing(QQ,12) +sage: xvars = [x0,x1,x2,x3] +sage: wvars = [[0,w01,w02,w03],[-w01,0,w12,w13],[-w02,-w12,0,w23],[-w03,-w13,-w23,0]] +sage: # Plücker equation: +sage: plucker = w01*w23 - w02*w13 + w03*w12 +sage: # Equation of the surface X: +sage: h = x0^3 + x1^3 + x2^3 + x3^3 +sage: deg = h.degree() +sage: # All possible permutations of (0,1,2,3): +sage: perm4 = [(j0,j1,j2,j3) for j0 in range(4) for j1 in range(4) for j2 in range(4) +....: for j3 in range(4) if len(set([j0,j1,j2,j3]))==4] +sage: # Generate the ideal I of the set Y of lines in X, as above: +sage: I = R.ideal([plucker] + [h.subs(dict([(xvars[j0],wvars[j0][j3]*u), (xvars[j1],w +....: vars[j1][j3]*u+wvars[j0][j1]*v), (xvars[j2],wvars[j2][j3]*u+wvars[j0][j2]*v), ( +....: xvars[j3],wvars[j0][j3]*v)])).coefficient({u:deg-k,v:k}) for k in range(deg) fo +....: r (j0,j1,j2,j3) in perm4]) +sage: # Compute its radical: +sage: I0 = I.radical() +sage: # This really means Y is 0-dimensional in projective space: +sage: I0.dimension() +7 +sage: # This computes its number of geometric points (i.e., geometric lines on X): +sage: hp = I0.hilbert_polynomial() ; hp.leading_coefficient()*factorial(hp.degree()) +27 +\end{verbatim} +\par}\noindent (notation is as above except that $\lambda,\mu$ have +been called \texttt{u},\texttt{v}); the above code proves that thare +are $27$ geometric lines in the surface $\{x_0^3 + x_1^3 + x_2^3 + +x_3^3 = 0\} \subseteq \mathbb{P}^3$ (over $\mathbb{Q}$). +\end{answer} + + + + +% +% +% +\end{document} |