## Models for non-Euclidean Gemetry

\textit{$\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Introduction} As mentioned several times, Hilbert established a set of axioms for both Euclidean and non-Euclidean geometry. These axiom sets differ only in the Parallel Postulate as formulated by Playfair. By showing that both Euclidean and non-Euclidean geometries have models in Cartesian (analytic) geometry, he proved that the Parallel Postulate is \textit{independent} from absolute geometry. This finally settled the 2200 year old controversy going back to Euclid. In the previous section, we showed how a model for a Euclidean axiom system, namely Birkhoff's, can be constructed in Cartesian geometry. We will accept the much more difficult theorem stated above. It is more important for you to understand how to use the models to discover and prove theorems in non-Euclidean geometry, than to develop non-Euclidean geometry \textit{synthetically} as Gauss, Bolyai and Lobachevsky did. But even that is not possible to do rigorously in an introductory course. Instead we will pick out \textit{certain aspects} of this study, because it illustrates the relevant methods and ideas. In particular, many of you are taking MA402 because you will, when you are high school teachers, have to teach an approach to non-Euclidean geometry that happens to be favored by the author of whatever textbook your school district favors. In order to prepare you to do this, we will consider several approaches that could be taken, but only their beginnings. Hvidsten's text follows \texbf{all} approaches just beyond where your high school text is likely to stop. Unless you are very short of cash, I suggest you keep Hvisdsten's book for future reference. In this course we will use the approach of Hvidsten's Chapter 8, but only the first four sections (17 pages !). This choice is dictated by the following considerations: \begin{itemize} \item It uses the algebra of \textit{complex numbers} to treat the geometry of the Euclidean plane in which the models for non-Euclidean geometry are placed. \item Most future high school teachers graduate from college without having taken a course on complex numbers. By at least learning the geometry of complex numbers, which is much easier than the differential and integral calculus of complex numbers, this loss is mitigated. \item Hvidsten treats both Euclidean and non-Euclidean plane geometry from the \textit{transformational} viewpoint. This is the way all modern higher geometry has been treated for the past century, high school textbooks notwithstanding. \item While this is the \textit{steepest} approach to the plateau we need to achieve, it is also the \textit{shortest} path there. This makes the best use of the pedagogical resources of a university math course. \end{itemize} \subsection{The Role of GEX} Although the Geometry Explorer software is a superb tool for experimentation, it cannot provide proofs for any theorems. The operation of GEX is based on the analytic geometry that does provide rigorous proofs. For example (see the figure above), start with two circles with centers and a radial points, hencforth denoted by $crcl(A,|AF|), crcl(C,|CD|)$ respectively. Locate their intersections and drop the perpendicular from one to the line connecting the centers of the circle. Note that the perpendicular passes through the other intersection. Complete $\triangle FAC $ and measure $\angle F$. Wiggle one circle until this angle is as close to a right angle as possible. Now focus your attention on this \textit{Thales Figure}. The three right triangles in Thales' Figure are similar. The proof of this fundamental theorem is developed in an Exercise below. Conclude that \begin{eqnarray*} \mbox{short/long =} |AH|/|FH| & = & |FH|/|HC| \\ |AH||HC| & = & |FH|^2 \\ \mbox{short/hypot =} |AH|/|AF| & = & |AF|/|AC| \\ |AH||AC| & = & |AF|^2 \mbox{= radius circle}_a^2 \\ |AH| &= & \frac{r_a^2}{|AC|} \\ c &= & \frac{r_a^2}{h} \\ \end{eqnarray*} The two points $H$ and $C$ are said to be \textit{inverses} relative to to $crcl{A;B}$. Notice that if $crcl{A,B}$ is the unit circle, then the distances from the origin, $ hc = 1$ are inverses in the sense of \textit{reciprocals.} \subsection{Construction of the Inverse of an Inside Point} Given $crcl(A,D)$ and any point $H$ inside the circle. To construct its inverser, $C$, reconstruct Thales' Figure from different starting points. Erec the perpendicular to $ray(A;H)$ at $H$, intersecting the circle at $F$. Now construct a right angle to $AF$ at $F$. The inverse point $F$ is where the other leg of the \textit{gnomon} crosses the $ray(A;H)$. \subsection{Construct the Inverse of an Outside Point.} To do this construction precisely, with ruler and compass or with GEX, is not entirely trivial. But for an adequate approximate solution with a \textit{gnomon} is very easy. Given $crcl(O,R)$ (change of notation is intentional.) Let $Y$ be a point outside the circle. Position your gnomon with one edge going through $O$, the other edge going through $Y$, and the right angled corner on the circle. This is the Thales right triangle, with hypotenuse on the line $\ell_{OY}$. Drop the altitude (again by placing your gnomon intelligently) to $X$ on the hypotenuse. That is the inverse point. To do this precisely we apply the classical construction the Greeks called \textit{constructing the mean and extreme ratio.} \subsection{Construction of the Mean and Extreme} Given two lengths (represented by segments) $x,y$, find their geometric mean $ h = \sqrt{xy} $. We saw one solution to this above. But here is a clean, simple one, using just Thales' Figure. Bisect a segment of lenght $ x+y $, erect a semicircle abover this segment as diameter, and erect the perpendicular to it at the point separating the diameter into lengths $x$ and $y$. Where the altitude crosses the semicircle is the right angle of the inscribed right angle. Hence it is the geometry mean. Given two lengths (represented by segments) $h,y$, find the length $x$, solving the equation $ h = \sqrt{xy}$. (1) On a segment of length $y$, erect a seqment of length $h$ perpendicular to it. In the figure this point is $C$. (2) Erect the perpendicular to $BC$ at $C$ to (3) where it crosses the linge $\ell_{AB}$. This completes Thales Figure once again, and we have found $x$. These two constructions are useful to find inverses of points in a given circle, whether the starting point is inside or outside. Be sure you complete this construction in your journal. \section{The Poincare Disk Model} The document, titled "The Poincar\'e Models of the Hyperbolic Plane" , provided here in .pdf format, uses the concept of \textit{inversion in circles}, which is a generalization of \textit{reflection in lines}. Hvidsten introduces this idea in Chapter 2 and it is the basis of his exposition of non-Euclidean geometry in Chapter 7. In Chapter 8, Hvidsten covers much of the same material using a different appraoch, one using complex numbers. Although we shall take the \textit{Complex Number approach}, it is useful for you to have some familiarity with the \texit{Cartesian approach} using inversions in circles. The reason for taking the complex number approach is that it is no more difficult to learn a bit of the geometry of complex numbers than to work one's way through inversive geometry in a purely analytic way. It is one of the more spectacular successes in mathematics, that complex numbers, originally invented to solve polynomial equations (like the familiar quadratic formula) should be so well suited to describe plane geometry as well. \subsection{What you should study now.} Chapter 8 is quite difficult and requires time. So we shall not pursue the .pdf notes on the Poincare models beyond sections 1 and 2, but only up to section 2.1, for the time being. \section{The Klein Disk Model} The document, titled "The Beltrami-Klein Model of Hyperbolic Geometry" , provided here in .pdf format, takes a different approach than the previous, one proper to \texit{synthetic geometry}. Analytic (or Cartesian) geometry was such a huge success in mathematics after the 17th century, that the beauty of the classical way of doing geometry in the style of Euclid was neglected. Geometers coined the term \textit{synthetic} as opposed to \textit{analytic} geometry for the classical approach. And synthetic geometry involves construction with ruler and compass. So here we present non-Euclidean geometry in the style that was more familiar to the actual discoverers of non-Euclidean geometry, Gauss, Lobachevsky and Bolyai. \subsection{Euclid's First Postulate} The very first construction in Euclid is to connect two Points with a Line. (Note the capitals refer to the primitives in an axiomatic system, to be interpereted in the model.) In the Klein model, this is trivial, because, by definition the Line connecting two Points is just the secant of the disk through the two points, excluding the end points. In the Poincar\'e disk, the Line $\ell_AB$ is the arc of a circle through the points which is also perpendicular to the circle. The critical theorem to use here is the following, which we shall prove later. \begin{theorem} Given a circle $K$, another circle $C$ is perpendicular to $K$ if and only if it is its own inverse in $K$. \end{theorem} So now, given Points $P,Q$ in the Poincare model, we invert one of them, to get $P'$ outside the Poincare circle. Recall the high school geometry construction of a circle through three given points $P,Q,P'$. Its center is where the perpendicular bisector of secant $PQ$ meets the perpendicular bisector of $P'Q$. \subsection{Euclid's Third Postulate} Solving the problem of dropping a Perpendicular from a Point to a Line depends first and foremost on the interpretation of Perpendicular in the model. For the Klein model, this is fairly odd. In the picture, let $crcl(A,|AB|)$ be the Klein disk. The secant $BC$ is a Line. Construct tangents to the circle at the endpoints of the secant, intersecting at $J$ in the figure. This is called the \texit{pole of the secant}. We define another secant, say $MN$ to be a Line Perpendicular to the original Line, as long as its extensions outside goes through the pole $J$. \textbf{Exercise} Show by experimentation that if Line $MN$ is Perpendicular to $BC$, as just defined, then Line $BC$ is also Perpendicular to Line $MN$. This reflexivity of ``being perpendicular" is not at all obvious. \subsection{What you should study now.} From the .pdf notes titled the \textit{Beltrami-Klein Projective Model}, you need only study the first two sections for now.