% for print: \documentclass[DIV16,halfparskip]{scrartcl} % for ebook: %\documentclass[DIV35,halfparskip,papersize,paper=123mm:93mm,8pt]{scrartcl} \usepackage[utf8]{inputenc} \usepackage{booktabs} \usepackage{eucal} \usepackage{amssymb} \usepackage[pdftex,bookmarks,colorlinks,urlcolor=blue]{hyperref} \newcommand{\nullcm}{\mathcal C} \newcommand{\nullconic}{$\nullcm$ } \newcommand{\C}{\!\!:\!\!} \begin{document} \title{Universal Hyperbolic Geometry} \subtitle{Summary of N. J. Wildberger's online lecture series\\ \url{http://www.youtube.com/watch?v=EvP8VtyhzXs}} \author{Martin Pitt (\texttt{martin@piware.de})} \date{} \maketitle \section{Notation} \begin{description} \item [Conics:] intersections of a (double-sided) cone with a plane: point, circle, elipse, parabola, hyperbola (consists of two separate parts) \item [Null conic:] Single fixed given circle or other conic; symbol \nullconic \item [Points:] points have lower-case variables ($a$, $p$); points on \nullconic have lower-case greek variables ($\alpha$, $\eta$); the product of two points $ab$ is the line that goes through $a$ and $b$\\ algebraic coordinate notation: $a = [x]$ (one-dimensional), $a = [x,y]$ (two-dimensional) \item [Lines:] lines have upper-case variables ($A$, $B$); tangents on \nullconic have upper-case greek variables ($\Gamma$); the product of two lines $AB$ is their intersection, i.~e. a point\\ algebraic coordinate notation: $A := (a\C b\C c)$ representing $ax+by=c$ \end{description} \section{Polarity} \url{http://www.youtube.com/watch?v=AjVM5Q-pvjw} \begin{description} \item [Pole:] The pole $a$ of a line $A$ is the inversion point of $A$'s closest point to \nullconic. $\bf a = A^\perp$ \item [Polar:] The line $A$ is the polar of the pole point $a$. $\bf A = a^\perp$ \end{description} \textbf{Construction of polar of $a$:} \begin{itemize} \item Find null points $\alpha, \beta, \gamma, \delta$ on \nullconic so that $(\alpha\beta)(\gamma\delta) = a$ (i.~e. select two secants of \nullconic which pass through $a$) \item Define $b := (\alpha\delta)(\beta\gamma)$ and $c := (\alpha\gamma)(\beta\delta)$, i.~e. the two intersections of the other diagonals of the quadrangle $\alpha\beta\gamma\delta$ \item Then $A = a^\perp = bc$. Also, $b^\perp = ac$ and $c^\perp = ab$ (three-fold symmetry of polars and poles produced by the three intersections of diagonals) \end{itemize} \textbf{Construction of pole of $A$:} Choose any two points $b, c\in A$. Then $a = (b^\perp c^\perp)$. \begin{description} \item [Polar Independence Theorem:] The polar $A = a^\perp$ does not depend on the choice of lines through $a$, i.~e. the choice of $\alpha, \beta, \gamma, \delta$. \item [Polar Duality Theorem:] For any two points $a$ and $b$: $a\in b^\perp \Leftrightarrow b\in a^\perp$ \end{description} \pagebreak[2] \textbf{Construction of polar of null point $\gamma$:} \nopagebreak[2] \begin{itemize} \item Chose any two secants $A$, $B$ that pass through $\gamma$ \item Construct their poles $A^\perp$, $B^\perp$ \item Polar $\Gamma = \gamma^\perp$ is the tangent $(A^\perp B^\perp)$ \end{itemize} \section{Harmonic conjugates} \url{http://www.youtube.com/watch?v=t7oXlrcPBb4} Four collinear points $a, b, c, d$ are a \textbf{harmonic range} if $a$ and $c$ are \textbf{harmonic conjugates} to $b$, $d$, i.~e. if they divide $\overline{bc}$ internally and externally by the same ratio: \[ \frac{\vec{ab}}{\vec{ad}} = -\frac{\vec{cb}}{\vec{cd}} \] In that case, $b$ and $d$ are then harmonic conjugates to $a$ and $c$: \[ \frac{\vec{ba}}{\vec{bc}} = -\frac{\vec{da}}{\vec{dc}} \] Note: $\vec{ab}$ measures displacements on a linear scale, not distances. (affine geometry only, no units) \begin{description} \item[Harmonic Ranges Theorem:] The image of a harmonic range under a projection from a point onto another line is another harmonic range.\\ $\to$ harmonic ranges are not dependent on the choice of a scale (affine geometry), but are really part of projective geometry \end{description} If $a, b, c, d$ are a harmonic range, and $p$ a point not on the line $abcd$, then the four lines $ap, bp, cp, dp$ are a \textbf{harmonic pencil}. By the previous theorem, the intersections of any line through these four lines are harmonic ranges. \begin{description} \item [Harmonic Pole/polar Theorem:] For a point $a$ and any secant through $a$ that meets \nullconic at two points $\beta$, $\gamma$ there is a point $c = (\beta\gamma)a^\perp$. The points $a, \beta, c, \delta$ are a harmonic range. \item [Harmonic Bisectors Theorem:] If $C$, $D$ are the two bisectors of two non-parallel lines $A$, $B$, then $A, C, B, D$ is a harmonic pencil. \item [Harmonic Vectors Theorem:] If $\vec{a}, \vec{b}$ are linearly independent vectors, then the lines spanned by $\vec{a}, \vec{b}$ are harmonic conjugates to the lines spanned by $\vec{a}+\vec{b}, \vec{a}-\vec{b}$ \item [Harmonic Quadrangle Theorem:] If $\overline{pqrs}$ is a quadrangle, find the two intersections $a = (ps)(qr)$ and $c = (pq)(rs)$ and then $b = (pr)(ac)$, $d = (sq)(ac)$. Then $a, b, c, d$ are a harmonic range. \end{description} \section{Cross ratios} \url{http://www.youtube.com/watch?v=JJbh0iJ1Agc} \textbf{Cross ratio} of four collinear points $a, b, c, d$: \[ R(a,b:c,d) := \frac{\vec{ac}}{\vec{ad}} ~ / ~ \frac{\vec{bc}}{\vec{bd}} \quad = \quad \frac{a-c}{a-d} ~ / ~ \frac{b-c}{b-d}\] $a, b, c, d$ are a harmonic range if $R(a,b:c,d) = -1$. \begin{description} \item[Cross-ratio Transformation Theorem:] If $R(a,b:c,d) = \lambda$ then \[ R(b,a:c,d) = R(a,b:d,c) = \frac{1}{\lambda}\] and \[ R(a,c:b,d) = R(d,b:c,a) = 1-\lambda \] Four points determine 24 possible cross ratios, but only 6 will generally be different (permutations of $\lambda$, $\frac{1}{\lambda}$, $1-\lambda$, $\frac{1}{1-\lambda}$, $\frac{\lambda-1}{\lambda}$, $\frac{\lambda}{\lambda-1}$). \item[Cross-ratio Theorem:] The cross ratio is invariant under projection from a point $p$ to another line $L$: With $a':=(pa)L$, $b':=(pb)L$, $c':=(pc)L$, $d':=(pd)L$: \[ R(a,b:c,d) = R(a',b':c',d') \] Therefore the cross-ratio can be transferred to lines: \[ R(pa, pb, pc, pd) := R(a,b,c,d) \] \item [Chasles Theorem:] If $\alpha, \beta, \gamma, \delta$ are fixed points on a null conic \nullconic and $\eta$ a fifth point on \nullconic, then $R(\alpha\eta,\beta\eta:\gamma\eta,\delta\eta)$ is independent of the choice of $\eta$. \end{description} \section{Introduction to hyperbolic geometry} \url{http://www.youtube.com/watch?v=UXQas-B5ObQ} \subsection{Definitions} \begin{description} \item [Hyperbolic geometry]: Geometry on a projective plane and a single fixed given circle \nullconic; only tool is a straightedge \item [Duality:] Terminology of pole and polar get replaced by simply \textbf{duality}: the dual of a point $a$ is the line $a^\perp$, and vice versa; points and lines are completely dual concepts \item [Line perpendicularity:] $A\perp B \Leftrightarrow B^\perp \in A \Leftrightarrow A^\perp \in B$ ($A$ is p. to $B$ if $A$ passes through the dual of $B$) \item [Point perpendicularity:] $a\perp b \Leftrightarrow a\in b^\perp \Leftrightarrow b\in a^\perp$ ($a$ is p. to $b$ if $a$ lies on the dual of $b$) \item [Quadrance between points:] $q(a_1,a_2) := R(a_1,b_2:a_2,b_1)$ with $b_1:=(a_1a_2)a_1^\perp$ and $b_2:=(a_1a_2)a_2^\perp$ \item [Spread between lines:] $S(A_1,A_2) := R(A_1,B_2:A_2,B_1)$ with $B_1:=(A_1A_2)A_1^\perp$ and $B_2:=(A_1A_2)A_2^\perp$ \end{description} \subsection{Basic theorems} \begin{description} \item [Quadrance--Spread duality:] \( q(a_1,a_2) = S(a_1^\perp,a_2^\perp) \) \item [Pythagoras:] If $a_1a_3 \perp a_2a_3$, and $q_1 = q(a_2,a_3)$, $q_2 = q(a_1,a_3)$, $q_3 = q(a_1,a_2)$: \( q_3 = q_1 + q_2 - q_1q_2 \) If $A_1A_3 \perp A_2A_3$, and $S_1 = S(A_2,A_3)$, $S_2 = S(A_1,A_3)$, $S_3 = S(A_1,A_2)$: \( S_3 = S_1 + S_2 - S_1S_2 \) \item [Triple Spread/Quadrance:]\hfill\\ If $a_1, a_2, a_3$ are collinear: \( (q_1+q_2+q_3)^2 = 2(q_1^2+q_2^2+q_3^2) + 4q_1q_2q_3 \) If $A_1, A_2, A_3$ are concurrent: \( (S_1+S_2+S_3)^2 = 2(S_1^2+S_2^2+S_3^2) + 4S_1S_2S_3 \) \item [Spread Law:] For a triangle: $\frac{S_1}{q_1} = \frac{S_2}{q_2} = \frac{S_3}{q_3}$ \item [Cross law:] $(q_1q_2S_3 - (q_1 + q_2 + q_3) + 2)^2 = 4(1-q_1)(1-q_2)(1-q_3)$ \item [Relation to Beltrami-Klein model:]\hfill\\ For points $a_1, a_2$ inside \nullconic: \( q(a_1,a_2) = -\sinh^2 d(a_1,a_2) \) For lines $A_1, A_2$ inside \nullconic: \( S(A_1,A_2) = \sin^2 \angle(A_1,A_2) \) \end{description} \section{Calculations with Cartesian coordinates} \url{http://www.youtube.com/watch?v=YDGUnGGkaTs}, \url{http://www.youtube.com/watch?v=XomxP2pxYnw} \begin{description} \item [Point/line duality:] \(a=[x_0,y_0] \quad\Leftrightarrow\quad a^\perp=(x_0:y_0:1) \) \item [Point on null circle:] Parameterized by $t\in\mathbb{Q}$: \( e(t) := \left[\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right] \)\\ This reaches any point except $[-1,0]$, which corresponds to $e(\infty)$ \item [Line through points:] \( [x_1,y_1] [x_2,y_2] = (y_1-y_2 : x_2-x_1 : x_2y_1-x_1y_2) \) \item [Line through null points:] \( e(t_1)e(t_2) = (1-t_1t_2 : t_1+t_2 : 1+t_1t_2) \) \item [Line relation to \nullconic:] The line $(a:b:c)$ \begin{itemize} \item is tangent to \nullconic $\Leftrightarrow a^2+b^2=c^2$ \item meets \nullconic at two points $\Leftrightarrow a^2+b^2-c^2$ is a square \item does not meet \nullconic $\Leftrightarrow a^2+b^2-c^2$ is a non-square (negative or no rational root) \end{itemize} \item [Quadrance:] If $a_1=[x_1,y_1]$ and $a_2=[x_2,y_2]$, then \( q(a_1,a_2) = 1 - \frac{(x_1x_2 + y_1y_2 - 1)^2}{(x_1^2+y_1^2-1)(x_2^2+y_2^2-1)} \) \item [Spread:] If $L_1=(l_1:m_1:n_1)$ and $L_2=(l_2:m_2:n_2)$, then \( S(L_1,L_2) = 1 - \frac{(l_1l_2 + m_1m_2 - n_1n_2)^2} {(l_1^2+m_1^2-n_1^2)(l_2^2+m_2^2-n_2^2)} \) \item [Point perpendicularity:] \( [x_1,y_1]\perp[x_2,y_2] ~\Leftrightarrow~ x_1x_2 + y_1y_2 - 1 = 0 \) \item [Line perpendicularity:] \( (l_1:m_1:n_1)\perp(l_2:m_2:n_2) ~\Leftrightarrow~ l_1l_2 + m_1m_2 - n_1n_2 = 0 \) \end{description} \section{Projective homogeneous coordinates} \url{http://www.youtube.com/watch?v=gzalbNLcwGI}, \url{http://www.youtube.com/watch?v=KhxVy75NetE} In affine geometry, lines from a 3D object through the projective plane are parallel, and projections maintain linear spacing. In projective geometry, they meet in two points (``observer'' or ''horizon''), and projections distort linear spacing. In \textbf{one-dimensional geometry}, the projective vector space is an one-dimensional subspace of two-dimensional $\mathbb{V}^2$, i. e. a line through $[0,0]$). This is specified by a proportion $[x:y]$ (\textbf{projective point}); canonically, either $[x:1]$, or $[x:0]$ for the special case of the subspace being the x axis (representing ``point at $\infty$'') In \textbf{two-dimensional geometry}, the projective plane $\mathbb{P}^2$ is described with a three-dimensional vector space $\mathbb{V}^3$, \textbf{projective points} $a=[x:y:z]$ (lines through the origin) and \textbf{projective lines} \\$A=(l:m:n) \Leftrightarrow lx+my-nz = 0$ (planes through the origin). In particular, we introduce a \textbf{viewing plane} $y_{z=1}$. Then any projective point $[x:y:z], z\neq 0$ meets the viewing plane at $[\frac{x}{z}:\frac{y}{z}:1]$. Any projective point $[x:y:0]$ corresponds to the two-dimensional equivalent $[x:y]$, representing the ``point at $\infty$ in the direction $[x:y]$''. So the projective plane $\mathbb{P}^2$ corresponds to the affine plane $\mathbb{A}^2$ plus the projective line $\mathbb{P}^1$ for the points at infinity. Any projective line $(l:m:n), l,m \neq 0$ meets the viewing plane at the line $lx+my=n$ (which is $(l:m:n)$ in the viewing plane). Coordinates in the viewing plane are called $X:=\frac{x}{z}$ and $Y:=\frac{y}{z}$. The unit circle $X^2+Y^2=1$ in the viewing plane corresponds to the cone $(\frac{x}{z})^2 + (\frac{y}{z})^2 = 1 \Rightarrow x^2+y^2-z^2=0$ in $\mathbb{A}^3$. \section{Calculations with homogenous coordinates} \url{http://www.youtube.com/watch?v=tk58sBLWzHk}, \url{http://www.youtube.com/watch?v=N2T0bg_DJLQ}, \url{http://www.youtube.com/watch?v=PSFr6_EhchI} \begin{description} \item[(Hyperbolic) Point:] a proportion $a:=[x\C y\C z]$ with $x,y,z\in\mathbb{Q}$ and not all zero \item[(Hyperbolic) Line:] a proportion $L:=(l\C m\C n)$ with $l,m,n\in\mathbb{Q}$ and not all zero \item[Duality:] \(a=[x\C y\C z] \Leftrightarrow a^\perp=(x\C y\C z) \qquad L=(l\C m\C n) \Leftrightarrow L^\perp=[l\C m\C n] \) \item[Line/Point Incidence:] [$x$:$y$:$z$] lies on ($l$:$m$:$n$) $\Leftrightarrow$ ($l$:$m$:$n$) goes through [$x$:$y$:$z$] $\Leftrightarrow$ $lx+my-nz=0$ \item[Line through two points:] \([x_1\C y_1\C z_1][x_2\C y_2\C z_2] = (y_1z_2-y_2z_1 : z_1x_2-z_2x_1 : x_2y_1-x_1y_2) \) \item[Point on two lines:] \( (l_1\C m_1\C n_1) (l_2\C m_2\C n_2) = [m_1n_2-m_2n_1 : n_1l_2-n_2l_1 : l_2m_1-l_1m_2] \) \item [Quadrance:] If $a_1=[x_1\C y_1\C z_1]$ and $a_2=[x_2\C y_2\C z_2]$, then \( q(a_1,a_2) = 1 - \frac{(x_1x_2 + y_1y_2 - z_1z_2)^2}{(x_1^2+y_1^2-z_1^2)(x_2^2+y_2^2-z_2^2)} \) $a_1 \perp a_2 \;\Leftrightarrow\; q(a_1,a_2) = 1$ \item [Spread:] If $L_1=(l_1:m_1:n_1)$ and $L_2=(l_2:m_2:n_2)$, then \( S(L_1,L_2) = 1 - \frac{(l_1l_2 + m_1m_2 - n_1n_2)^2} {(l_1^2+m_1^2-n_1^2)(l_2^2+m_2^2-n_2^2)} \) $L_1 \perp L_2 \;\Leftrightarrow\; S(L_1,L_2) = 1$ \end{description} \section{Triangle geometry} \url{http://www.youtube.com/watch?v=PbA4Js3qKOQ}, \url{http://www.youtube.com/watch?v=Drs8hUPzRP0} \begin{description} \item [Side:] A side $\overline{a_1a_2}$ is a set of two points $\{a_1,a_2\}$. \item [Vertex:] A vertex $\overline{L_1L_2}$ is a set of two lines $\{L_1,L_2\}$. \item [Couple:] A couple $\overline{aL}$ is a set consisting of a point and a line: $\{a, L\}$. A couple is \textbf{dual} if $a = L^\perp$. \item [Triangle:] A triangle $\overline{a_1a_2a_3}$ is a set of three non-collinear points $\{a_1, a_2, a_3\}$. A triangle is \textbf{dual} if one of its points is dual to its opposite side. \item [Dual trilateral:] $\overline{a_1a_2a_3}^\perp := \overline{a_1^\perp a_2^\perp a_3^\perp}$ (similar for dual triangle) \item [Trilateral:] A trilateral $\overline{L_1L_2L_3}$ is a set of three non-concurrent lines $\{L_1, L_2, L_3\}$. \item [Altitude Line Theorem:] For any non-dual couple $\overline{aL}$ there is an unique line $N$ passing through $a$ and perpendicular to $L$, called \textbf{altitude}. $N=aL^\perp$ \item [Altitude Point Theorem:] For any non-dual couple $\overline{aL}$ there is an unique point $n$ which lies on $L$ and is perpendicular to $a$, called \textbf{altitude point}. $n=a^\perp L = N^\perp$ \item[Triangle Altitudes:] In a triangle $\overline{abc}$, altitudes are determined in the usual way: e. g. the altitude of $a$ goes through $a$ and is perpendicular to $bc$. Thus the altitude of a is $a(bc)^\perp$. \item [Orthocenter:] Point $h$ where the three altitudes meet; always exists \item [Ortholine:] Line $H$ on which the three altitude points are collinear; $H=h^\perp$ \item [Desargues Theorem:] In the projective plane, if two triangles $\overline{a_1a_2a_3}$ and $\overline{b_1b_2b_3}$ are perspective from a point $p$ ($a_1b_1$, $a_2b_2$, $a_3b_3$ are concurrent) then they are perspective from a line $L$ (where $(a_1a_2)(b_1b_2)$, $(a_2a_3)(b_2b_3)$, $(a_3a_1)(b_3b_1)$ are collinear). \textbf{Desargues polarity:} $L = \hat{p}$ \item [Definitions relative to fixed triangle:] \textbf{Orthic axis} $S=\hat{h}$, \textbf{Orthostar} $s=S^\perp$, \textbf{Ortho-axis} $A=hs$, \textbf{Ortho-axis point} $a=A^\perp$ (lies on ortholine) \item [Orthic triangle:] Triangle $\overline{b_1b_2b_3}$ of the base points of altitudes of triangle $\overline{a_1a_2a_3}$; $b_1=(a_1h)(a_2a_3)$ \item [Base center:] A triangle and its dual orthic triangle are perspective from some point, called the \textbf{base center}, i. e. the intersection of all lines through a triangle point with its corresponding dual orthic triangle point. It lies on the ortho-axis $A=hs$. \item [Base triple orthocenter theorem:] Suppose that the triangle $\overline{a_1a_2a_3}$ has the orthic triangle $\overline{b_1b_2b_3}$. Suppose that $h_1, h_2, h_3$ are the respective orthocenters of $\overline{a_1b_2b_3}$, $\overline{a_2b_1b_3}$, and $\overline{a_3b_1b_2}$. Then the orthocenter of $\overline{h_1h_2h_3}$ is the base center $b$ of $\overline{a_1a_2a_3}$. Also, $b$ is the center of perspectivity between $\overline{a_1a_2a_3}$ and $\overline{h_1h_2h_3}$. \item [Quadrea:] $A(\overline{a_1a_2a_3}) = q_1q_2S_3 = q_2q_3S_1 = q_1q_3S_2 = $\\ $-\frac{\left|\begin{array}{ccc}x_1 & y_1 & z_1\\x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3\end{array}\right|^2} {(x_1^2 + y_1^2 - z_1^2)(x_2^2 + y_2^2 - z_2^2)(x_3^2 + y_3^2 - z_3^2)}$ (most important triangle invariant) \item [Equilateral Triangle Theorem:] If a triangle has three equal non-zero quadrances $q$, then it also has three equal spreads $S$, and $(1-Sq)^2=4(1-S)(1-q)$. \item [Thales Theorem:] Immediately following from the spread law: in a right triangle with $S_3=1$:\\ $S_1=\frac{q_1}{q_3}$ and $S_2=\frac{q_2}{q_3}$. Corollary: If $h_3$ is the quadrance of the altitude of $a_3$ to $a_1a_2$, then the quadrea $A=h_3q_3$. (Similar for the other two altitudes) \item [Napier's rules:] In a right triangle ($S_3=1)$, if any two of $S_1, S_2, q_1, q_2, q_3$ are known, the other three follow from Pythagoras' and Thales' theorems. \end{description} \section{Null points and null lines} \url{http://www.youtube.com/watch?v=IhEXH5etvog} \begin{description} \item [Null point:] A point $a=[x\C y\C z]$ is null $\Leftrightarrow$ $a$ is incident with $a^\perp$ $\Leftrightarrow$ $x^2+y^2-z^2=0$ \item [Null line:] A line $L=(l\C m\C n)$ is null $\Leftrightarrow$ $L$ is incident with $L^\perp$ $\Leftrightarrow$ $l^2+m^2-n^2=0$ \item [Null point parameterization:] \( \alpha = e(t\C u) := [u^2-t^2 : 2ut : u^2+t^2] \) \item [Null line parameterization:] \( \Phi = E(t\C u) := (u^2-t^2 : 2ut : u^2+t^2) \) \item [Join of null points:] \( e(t_1\C u_1)e(t_2\C u_2) = (u_1u_2-t_1t_2 : t_1u_2+t_2u_1 : u_1u_2 + t_1t_2) \) \item [Meet of null lines:] \( E(t_1\C u_1)E(t_2\C u_2) = [u_1u_2-t_1t_2 : t_1u_2+t_2u_1 : u_1u_2 + t_1t_2] \) \end{description} \section{Reflections} \url{http://www.youtube.com/watch?v=faPCRHyzPGM}, \url{http://www.youtube.com/watch?v=elDCJmDQBfc} Unlike the Euclidean plane, the projective plane is not orientable. A reflection $\sigma_a$ in a point $a$ is the same as a reflection $\sigma_L$ in a line if $a=L^\perp$. Lines are reflected to lines ($M=\sigma_aL$), null points to null points, null lines to null lines. \begin{description} \item[Construction:] To determine the reflection $c=b\sigma_a$ of a point $b$ in a point $a$: \begin{itemize} \item Choose a line through $b$ which crosses \nullconic in two points $\beta_1$ and $\beta_2$. ($c$ does not depend on this choice) \item Determine the two other intersections $\gamma_1$ and $\gamma_2$ with \nullconic of the two lines $a\beta_1$ and $a\beta_2$. \item Now $c=(ab)(\gamma_1\gamma_2)$. \end{itemize} \item[Reflection matrix:] A point $a=[x\C y\C z]$ defines a reflection matrix \( m_a = \left[\begin{array}{cc}y & x+z\\ x-z & -y\end{array}\right] \) For any $m_a$: $m_a^2 = \mathbf{1}$, so that $m_a^{-1} = m_a$ For null points $\alpha=e(t:u)$: \( m_\alpha = \left[\begin{array}{cc}tu & u^2 \\ -t^2 & -tu\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right] \left[\begin{array}{c}t \\ u\end{array}\right] \left[\begin{array}{cc}t & u\end{array}\right] \) For any $m_\alpha$: $m_\alpha^2 = \mathbf{0}$, $\mathrm{det}\;m_\alpha = 0$ \item [Reflection of null points:] With projective linear algebra, a reflection at the point $a=[x\C y\C z]$ sends the null point $\alpha_1=e(t_1\C u_1)$ to another null point $\alpha_2=e(t_2\C u_2)$. Then $[t_2 u_2] = [t_1 u_1]\: m_a$. Also, $m_{\alpha_2} = m_\alpha m_{\alpha_1} m_\alpha$ (reflection matrix conjugation theorem). \item [Reflection of an arbitrary point:] $c=b\sigma_a \;\Leftrightarrow\; m_c = m_a m_b m_a$ \item [Null reflection theorem:] If $\alpha$ is a null point, then $b\sigma_\alpha = \alpha$ for any point $b$. $m_\alpha m_b m_\alpha = m_\alpha$. \item [Matrix perpendicularity theorem:] For any points $a, b$: \(a\perp b \;\Leftrightarrow\; \mathrm{tr}(m_am_b) = 0 \) \item [Reflection preserves perpendicularity:] For any two points $b, c$, and a non-null point $a$:\\ \(b\perp c \;\Leftrightarrow\; b\sigma_a \perp c\sigma_a\) \item [Reflection preserves lines:] If $a$ is a non-null point, then $b, c, d$ are collinear $\Leftrightarrow$ $b\sigma_a, c\sigma_a, d\sigma_a$ are collinear. \end{description} \section{Midpoints and bisectors} \url{http://www.youtube.com/watch?v=gYqp_m7at2} \begin{description} \item [Midpoint:] The non-null point $a$ is a midpoint of the side $\overline{bc} \;\Leftrightarrow\; b\sigma_a = c \;\Leftrightarrow\; c\sigma_a = b$\\ In general there are two different points with that property for $\overline{bc}$, if both points are interior or exterior. \item [Geometrical construction of midpoints:] For points $b, c$: \begin{itemize} \item Construct $(bc)^\perp$ \item Construct the lines $b(bc)^\perp$ and $c(bc)^\perp$, yielding four null points $\alpha$, $\beta$, $\gamma$, $\delta$ \item The other two diagonal points of $\overline{\alpha\beta\gamma\delta}$ are the two midpoints of $\overline{bc}$. \end{itemize} Sometimes that constructions does not work because $b(bc)^\perp$ and $c(bc)^\perp$ don't meet \nullconic. In that case: \begin{itemize} \item Construct $b^\perp$ and $c^\perp$, which meet \nullconic in four null points $\alpha$, $\beta$, $\gamma$, $\delta$ \item The other two diagonal points of $\overline{\alpha\beta\gamma\delta}$ are the two midpoints of $\overline{bc}$. \end{itemize} \item [Bisector:] $A$ is a bisector of the vertex $\overline{BC}$ $\Leftrightarrow$ $A^\perp$ is a midpoint of side $\overline{B^\perp C^\perp}$ \end{description} \section{The J function, SL(2) and the Jacobi identity} \url{http://www.youtube.com/watch?v=f68eYuDCsjw} \begin{description} \item[SL(2)]: Lie algebra/group of 2x2 matrices $A=\small\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ with $tr(A) = 0$ and a \textbf{bracket} operation\linebreak $[A,B]:= AB-BA$ (closed operation as $tr(AB-BA) = tr(AB)-tr(BA) = 0$) \item[Properties of bracket operation:] $[]$ is not associative, and anticommutative ($[A,B] = -[B,A]$) \item[Jacobi identity:] $[[A,B],C] + [[B,C],A] + [[C,A],B] = 0$ \item[Projective matrix algebra:] If $a,b$ are projective matrices, scaling with non-zero scalars does not change the matrix. $m+n$ is \emph{not} well defined. The product $ab$ and bracket $[a,b] = ab-ba$ are well defined. \item[Bracket theorem:] If $a$ and $b$ are distinct points:$[m_a,m_b] = m_c$ with $c=(ab)^\perp$\\ The bracket operation gives a multiplication of points and computes their join: $[a,b] \simeq (ab)^\perp$. It is commutative, as the negative of a projective matrix is identical to the matrix. \item [Meaning of Jacobi identity:] For three points $a, b, c$ the term $[[a,b],c]$ is the altitude point (dual of the altitude) of $c$ to $ab$. As all three altitudes/altitude points sum up to 0, they are linearly dependent and thus concurrent/collinear. $\Rightarrow$ simplified proof that ortholine/orthocenter always exist. \end{description} \section{Miscellaneous} \begin{description} \item [Pappus' Theorem:] If $a_1, a_2, a_3$ are collinear and $b_1, b_2,b_3$ are collinear:\\ $c_1 := (a_2b_3)(a_3b_2)$, $c_2:=(a_1b_3)(a_3b_1)$, $c_3 := (a_2b_3)(a_3b_2)$ are collinear. \item [Zero Quadrance Theorem:] If $a_1, a_2$ are distinct points, then $q(a_1,a_2) = 0$ $\Leftrightarrow$ $a_1a_2$ is a null line. \item [Zero Spread Theorem:] If $L_1, L_2$ are distinct lines, then $S(L_1,L_2) = 0$ $\Leftrightarrow$ $L_1L_2$ is a null point. \item [Right Parallax Theorem:] If a right triangle $\overline{a_1a_2a_3}$ has spreads $S_1=0$ (i. e. $a_1$ is a null point), $S_2 := S \neq 0$, and $S_3=1$, then it will have only one defined quadrance $q=q(a_2,a_3)=\frac{S-1}{S}$. \item [Isosceles Parallax Theorem:] If $\overline{a_1a_2a_3}$ is a non-null isosceles triangle with $S_1=0$ (i. e. $a_1$ is a null point) and $S_2 = S_3 := S$, then $q=q(a_2,a_3)=\frac{4(S-1)}{S^2}$. \end{description} \end{document}