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diff --git a/controle-20230412.tex b/controle-20230412.tex index c6100bb..9332301 100644 --- a/controle-20230412.tex +++ b/controle-20230412.tex @@ -79,9 +79,11 @@ \noindent\textbf{Instructions.} -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. +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 @@ -99,9 +101,9 @@ Use of electronic devices of any kind is prohibited. Duration: 2 hours \ifcorrige -This answer key has \textcolor{red}{XXX} pages (cover page included). +This answer key has 6 pages (cover page included). \else -This exam has \textcolor{red}{XXX} pages (cover page included). +This exam has 3 pages (cover page included). \fi \vfill @@ -118,10 +120,10 @@ Git: \input{vcline.tex} % % -\exercise +\textit{The goal of this problem is to study a representation of lines + in $\mathbb{P}^3$.} -\textit{The goal of this exercise 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 @@ -140,8 +142,10 @@ $w\neq 0$ 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. +\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 call $x\wedge y := (w_{0,1}, +(y_0,\ldots,y_3) \in k^4$, 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 @@ -193,6 +197,8 @@ exists when $x$ and $y$ are not collinear (so that $x\wedge y \neq depend on the $x,y \in V$. \end{answer} +\bigskip + The $w_{i,j}$ in question are known as the \textbf{Plücker coordinates} of $L$. @@ -233,15 +239,17 @@ w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + w_{0,3} w_{1,2} = 0 \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 +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 how to associate to any line $L$ in -$\mathbb{P}^3(k)$ a $k$-point on the Plücker quadric. We now consider -the converse. +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 $\mathbb{P}^5(k)$ @@ -252,7 +260,7 @@ $(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ 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 +(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} @@ -295,18 +303,20 @@ 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 defined as the the intersection of two plines. +line defined 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 $[0 : w_{2,3} : {-w_{1,3}} : w_{1,2}]$ contain $L$. Now consider these as points in the -dual $\mathbb{P}^3$: show that the Plücker coordinates of the line +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 $w_{1,2} \neq 0$. @@ -320,13 +330,11 @@ cyclically, it is also on the plane $[0 : w_{2,3} : {-w_{1,3}} : 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 + 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). + {-w_{0,2}} : w_{0,1}]$. \end{answer} \textbf{(9)} Deduce from (8) that if $L$ is a line in @@ -369,16 +377,21 @@ 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 $L^*$). +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.) +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 @@ -437,6 +450,8 @@ w_{0,1} w'_{2,3} - w_{0,2} w'_{1,3} + w_{0,3} w'_{1,2} \] (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 @@ -470,7 +485,7 @@ We can then compute: \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), + 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 @@ -492,6 +507,66 @@ We can then compute: \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} + + % |