géométrie projective: Exemple de calculs

Résumé:
\bullet On se donne
une forme linéaire non nulle T \in E^* et on considère l’hyperplan vectoriel E_{\infty} défini par T. On pose P(E_{\infty}) =D_{\infty} (c’est un hyperplan projectif) et X = P(E )\setminus D_{\infty }

L’application qui à \overrightarrow{u} \in E_{\infty} et \bar{x}\in X associe \overrightarrow{u}.\bar{x} = \overline{x + T(x)\overrightarrow{u}} est bien définie. C’est une opération de E_{\infty} sur X , simplement transitive, de sorte qu’elle fait de X un espace affine sous E_{\infty}.
Si \bar{a}, \bar{b} sont deux points de X, le vecteur \overrightarrow{ab} est le vecteur de E_{\infty} défini par \overrightarrow{ab}=\dfrac{b}{T(b)}-\dfrac{a}{T(a)} , il est indépendant du choix des représentants des points.

Situation 1 (T=x+y+z)
Soit a(0,1,1) et b(2,0,1), \overrightarrow{ab}=(\dfrac{2}{3},-\dfrac{1}{2}, -\dfrac{1}{6})

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Soient \bar a , \bar b deux points de P(E), la formule (a \wedge  b)(c)=[a,b,c] définit une forme linéaire sur E^* qui est une équation de (\bar a \bar b)

Pour f,g dans E^*, le vecteur f\wedge g de E défini par la formule h(f\wedge g) = [f,g,h] aura pour image dans P(E) le point d’intersection des deux droites définies par f et g.

Une formule fondamentale :(a\wedge b)\wedge(c\wedge d) = [a,c,d]b-[b,c,d]a.

\bullet On appelle droite affine la trace surX d’une droite projective distincte de D_{\infty}. Une telle droite D est l’image d’un plan vectoriel V de E, distinct de E_{\infty}. La droite vectorielle associée est l’intersection
\overrightarrow{D} = E_{\infty}\cap V et son image dans D_{\infty} est l’unique point à l’infini de D (appelé \underline{\text{direction}} de D).

Situation 1 (T=x+y+z)
Soit a(0,1,1) et b(2,0,1)
a\wedge b donne l’équation de (ab) soit x+2y-2z=0, et pour déterminer la direction ,comme :

(1,2,-2)\wedge(1,1,1)=(4,-3,-1), la direction de (ab) sera le point de D_{\infty}   (-4,3,1),

Deux droites D et D' sont dites parallèles si les droites projectives associées ont même point à l’infini.
Soit T l’équation de D_{\infty} et F, G les droites d’équations f et g. Les droites F, G sont parallèles si et seulement si on a  T(f \wedge g) = 0 .

Si f est une forme définissant une droite F et si a, b sont deux points, les droites F et (ab) sont parallèles si et seulement si on a T(b)f(a) = T(a)f(b).

\bullet Dans le plan affine, on peut définir (indépendamment de toute structure métrique) la \underline{ \text{mesure}\; \text{algébrique}} d’un vecteur de direction donnée.

On suppose qu’on s’est donné une équation f \in E^* de V . Alors, le vecteur \overrightarrow{i} = f \wedge T est un vecteur non nul de\overrightarrow{D } (qu’on appelle un \underline{ \text{vecteur}\; \text{directeur}} de D), et si a, b sont deux points de D, on appelle mesure algébrique du vecteur \overrightarrow{ab} relativement à \overrightarrow{i} le nombre \lambda=\overline{ab} défini par la formule : \overrightarrow{ab}=\lambda \overrightarrow{i}

Exemples : situation 1 (x+y+z=0)

Calcul du birapport: (voir article du même nom)

Soient \bar a (1,1.5,1),\bar b(3,2.5,1) ,\bar c(4,3,1) ,\bar d (8,5,1) , quatre points de la droite D image d’un plan vectoriel V d’équation x-2y+2z=0 , (f).

Méthode 1 : (la plus courante)

[a,b,c,d]=\dfrac{\overline{ac}}{\overline{ad}}\div\dfrac{\overline{bc}}{\overline{bd}}

Méthode 2 : Si V est muni d’une base e_1 , e_2, et si les vecteurs image de ces points ont pour coordonnées respectives \begin{pmatrix}  a_1 \\  a_2  \end{pmatrix} ,\begin{pmatrix}  b_1 \\  b_2  \end{pmatrix} ,\begin{pmatrix}  c_1 \\  c_2  \end{pmatrix} et \begin{pmatrix}  d_1 \\  d_2  \end{pmatrix} dans cette base, on a aussi :

[a,b,c,d]=\dfrac{\begin{bmatrix}c_1 & a_1 \\c_2 & a_2\end{bmatrix}.\begin{bmatrix}d_1 & b_1 \\d_2 & b_2\end{bmatrix} }{\begin{bmatrix}c_1 & b_1 \\c_2 & b_2\end{bmatrix}.\begin{bmatrix}d_1 & a_1 \\d_2 & a_2\end{bmatrix}}

Méthode 3 : (Très pratique )

Si f et g sont les équations de deux droites passant par c et d, alors:

[a,b,c,d]=\dfrac{f(a)g(b)}{f(b)g(a)}

Vérification :

(f \wedge T)=(-2\times(1)-1\times 2,-(1\times 1 - 1\times 2), 1 \times1-1\times(-2)) et \overrightarrow{i}(-4,1,3) est un vecteur directeur de D.

\overrightarrow{ac}=\dfrac{c}{T(c)}-\dfrac{a}{T(a)}=\dfrac{1}{8}(4,3,1)-\dfrac{1}{3.5}(1,1.5,1)

=(\dfrac{12}{56},\dfrac{-3}{56}, \dfrac{-9}{56})=-\dfrac{3}{56}\overrightarrow{i}

donc \overline{ac}=-\dfrac{3}{56}

\overrightarrow{bd}=\dfrac{d}{T(d)}-\dfrac{b}{T(b)}=\dfrac{1}{14}(8,5,1)-\dfrac{1}{6.5}(3,2.5,1)

=(\dfrac{20}{182},\dfrac{-5}{182}, \dfrac{-15}{182})=-\dfrac{5}{182}\overrightarrow{i}

donc \overline{bd}=-\dfrac{5}{182}

On aura aussi , \overline{ad}=-\dfrac{7}{98} et \overline{bc}=-\dfrac{1}{104} et pour finir:

[a,b,c,d]=\dfrac{-\dfrac{3}{56}}{-\dfrac{7}{98}}\div\dfrac{-\dfrac{1}{104}}{-\dfrac{5}{182}}=\dfrac{15}{7}

Pour la méthode 2 on choisit e_1(2,1,0) , e_2(-2,0,1) comme base de V

Considérons maintenant les vecteurs, a(1,1.5,1) image de \bar a, a a pour coordonnées dans (e_1 , e_2), \begin{pmatrix}  1.5 \\  1  \end{pmatrix} , b aura pour coordonnées \begin{pmatrix}  2.5 \\  1  \end{pmatrix} , c \begin{pmatrix}  3 \\  1  \end{pmatrix} et d \begin{pmatrix}  5 \\  1  \end{pmatrix}

donc [a,b,c,d]=\dfrac{\begin{bmatrix}3 & 1.5 \\1& 1\end{bmatrix}.\begin{bmatrix}5 & 2.5 \\1 & 1\end{bmatrix} }{\begin{bmatrix}3 & 2.5 \\1& 1\end{bmatrix}.\begin{bmatrix}5 & 1.5 \\1 & 1\end{bmatrix}}=\dfrac{1.5 \times 2.5}{3.5 \times 0.5}=\dfrac{15}{7} !

Pour illustrer la méthode 3,  on choisit (au hasard ?)  f: x+2y-10z=0 et g:x-y-3z=0

on a bien f(c)=g(d)=0 et

[a,b,c,d]=\dfrac{f(a)g(b)}{f(b)g(a)}

=\dfrac{(-6)\times(-2.5)}{(-2) \times(-3.5)}=\dfrac{15}{7}.

Un autre exemple: parmi les applications projectives par excellence on trouve les incidences et les perspectives:

Soit D une droite de P(E) et m un point de P(E), avec m \notin D.

On appelle incidence l’ application i : m^ * \rightarrow D qui à une droite d passant par m , associe l’unique point d’ intersection de d et de D, montrons à travers un exemple que cette application est une homographie.

Soit m(3,2,1) et D:x+y-8z=0, d a pour équation \alpha x+\beta y +\gamma z=0 avec 3\alpha + 2\beta +\gamma =0 soit d: \alpha x+\beta y + (-3\alpha-2\beta)z=0

Prenons e_1^* (1,0,-3) et e_2^*(0,1,-2) comme base de m^ *, et e_1 (3,5,1) et e_2(6,2,1) comme base de V dont D est l’image, le point d’ intersection de d et de D est donné par le calcul (D\wedge d)

=(1\times(-3\alpha-2\beta)-\beta\times (-8),-(1\times (-3\alpha-2\beta) - \alpha\times (-8)), 1 \times\beta-1\times\alpha)

=(-3\alpha+6\beta, -5\alpha+2\beta, \beta-\alpha)=-\alpha e_1 +\beta e_2

…puis les coordonnées exactes du point d’intersection seront =(\dfrac{-3\alpha+6\beta}{ \beta-\alpha},\dfrac{ -5\alpha+2\beta}{\beta-\alpha}, 1)

Exemple, avec d:2x-y-4z=0, (\alpha=2,\beta=-1), les coordonnées du point cherché sont :

(\dfrac{-6-6}{ -3},\dfrac{ -10-2}{-3}, 1)=(4,4,1) !

i est bien une homographie car i(\bar x)=\bar {u(x)}u est l’application linéaire de m^* dans V définie par
u(\alpha e_1 ^*+\beta e_2 ^*)=-\alpha e_1 +\beta e_2 .

On appelle perspective de centre m de D sur D_0.l’application p_m : D\rightarrow D_0 qui à x \in D associe l’unique point d’ intersection de (mx) et de D_0, montrons à travers un exemple que cette application est une homographie.

D=x-y+z=0 , D_0:x-2y+8z=0 et m(2,1,1) .

Prenons u (1,1,0) et v(1,0,-1) comme base de V dont D est l’image , et u' (8,0,-1) et v'(8,4,1) comme base de V' dont D_0 est l’image

Soit x un point de D et \alpha u+ \beta v son image dans V, (mx) aura pour équation (m\wedge x)

=(1\times(-\beta)-1\times \alpha,-(2\times (-\beta) - 1\times (\alpha +\beta)), 2\times\alpha-1\times (\alpha +\beta))

=(-\alpha-\beta, \alpha+3\beta, \alpha-\beta)

Puis on calcule (mx \wedge D_0) et on obtient =(10\alpha+22\beta, 9\alpha+7\beta,\alpha-\beta)=-\dfrac{10\alpha+22\beta}{8} u' +\dfrac{9\alpha+7\beta}{8} v'

…puis les coordonnées exactes du point d’intersection seront =(\dfrac{10\alpha+22\beta}{ \alpha-\beta},\dfrac{ 9\alpha+7\beta}{\alpha-\beta}, 1)

p_m est bien une homographie car p_m(\bar x)=\bar {u(x)}u est l’application linéaire de V dans V' définie par
la matrice \begin{pmatrix} -\frac{10}{8} & -\frac{22}{8} \\ \frac{9}{4} & \frac{7}{4} \end{pmatrix}

Remarque , on aurait pu remarquer qu’une perspective n’est rien d’autre que la composée de deux incidences…

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