Non-euclidean geometry

In a sense, projective geometry is non-euclidean geometry, since it lacks rulers and protractors -- there is no measurement of distance or angles in projective geometry. But in a more exact sense, projective geometry includes euclidean geometry as one of several metric geometries within itself. One of the outstanding achievements of 19th century geometry was the discovery of these new geometries and how they all are connected to one another in a thoroughly organic way.

The discovery of other metric geometries besides euclidean geometry was provoked by the attempt to prove Euclid's Fifth Postulate: Through a given point in a given plane, there is exactly one line parallel to a given line. Finally, several mathematicians simultaneously realized that it is possible to replace this postulate with alternative versions, and that the resulting axiom systems yield consistent geometries. There are basically two such alternatives: if there are no parallels, this leads to elliptic geometry; if there are infinitely many such parallels, this leads to hyperbolic geometry. There are many different ways to describe these geometries. Here we describe one way based on projective geometry.

NOTE: The following material concentrates on the non-euclidean cases of elliptic and hyperbolic geometry. It is also possible to derive euclidean geometry in the same context, but it is more complicated and it is not done here. You can however, continue to rely on what you know about euclidean geometry to calculate distances and angles -- even when coordinates are given using projective, homogeneous coordinates.

Quadratic forms

Using homogeneous coordinates of $\mathbb{P}^n$ , it is possible to define a quadratic form $Q: \mathbb{P}^n \rightarrow \mathbb{R}$ just as a quadratic form is defined on $\mathbb{R}^n$ . Such a Q can be realized as a symmetric matrix of size (n+1), and then $<x,y>_Q := Q(x, y) = x^t~Q~y$ . A quadratic form is a generalized inner product; when $Q=Id$ , the result is the standard inner product.

One can ask, which projective transformations leave the quadratic form invariant? In terms of matrices, let $M \in PGL(n+1)$ be the matrix of a projective transformation. Then M preserves Q if and only if $M^t~Q~M~=\lambda Q, \lambda \neq 0$ . (Proof: exercise).

Pole and Polar with respect to a quadric

Given a quadratic form Q one can define a polarity (see introduction ) which maps points to hyper-planes and vice-versa. Let $x_0$ be a fixed point. Then define the polar set of $x_0$ by the linear condition $<x,x_0>_Q = 0$ . Denote this map from a point to its polar set by $P(x)$

Claim: The polar set of a point is a hyperplane. Proof: The condition defining the set is linear in the coordinates of the point $x_0$ . In fact, $P(x_0) = Qx_0$ gives the plane coordinates of $P(x_0)$ .

Define the pole of a hyperplane in exactly the same way but use the adjoint quadratic form $Q'~:={Q^{-1}}^t$ (inverse transpose). In all the cases we deal with here, $Q'=Q$ . One can similarly show that the pole of a hyperplane is a point. Call this map also $P$ .

Claim: $P^2 = Id$ . Proof: Obvious.

Points satisfying $<x,x>_Q = 0$ are incident with their polar, that is, lie on their polar; and vice-versa, and are called self-polar. The polar plane of a self-polar point x is the hyperplane tangent to the quadric surface $<x,x>_Q=0$ at the point x. One can show that as the point moves away from the surface in one direction, the polar moves away in the other direction.

Example

If $M=(1,0,1)$ then the polar line in elliptic geometry is given by the equation $<M,x>_Q=0$ , which leads to $1x+0y+1z=0$ . Expressed in line coordinates, $m=(1,0,1)$ . In the hyperbolic case, we get $-1x+0y+1z=0$ . In line coordinates, $m=(-1,0,1)$ . Conclusion: In elliptic geometry, the polar line of a point is obtained by interpreting the coordinates of the point as line coordinates; the same is true in hyperbolic geometry, except all but the last coordinate are negated.

Exercise Investigate the movement of the polar line of M as M moves according to $M(t)=(t,0,1)$ in both spherical and hyperbolic geometry. Consider the movement in the standard picture of $\mathbb{P}^2$ , as $\mathbb{R}^2$ extended with a line of ideal points.

Elliptic geometry

If one chooses $Q=Id$ , one obtains the familiar inner product, but this time applied to homogeneous coordinates of $\mathbb{P}^n$ . The matrices that preserve this quadratic form are just the familiar elements of $O(n+1)$ , the rotation/reflection group of dimension (n+1). But, instead of acting on $\mathbb{R}^{n+1}$ , here they act on points of $\mathbb{P}^n$ . Topologically, this is one hemisphere of a sphere, with opposite boundary points identified.

Defining a metric in elliptic geometry

One can define the norm of a point as ${||x||}_Q := \sqrt{<x,x>_Q}$ . The set of points of norm 1 in $\mathbb{P}^n$ form a hemisphere with opposite boundary points identified.

Without going into details, one can then define a distance function on points of $\mathbb{P}^n$ : $d_Q(x,y) := {\cos}^{-1}(\frac{<x,y>_Q}{{||x||}_Q~{||y||}_Q})$ . This is the familiar formula for the measure of the angle between the two points $x,y \in S^n$ . However, in this case, $x,y \in \mathbb{P}^n$ . (See above).

Technically, elliptic geometry is $\mathbb{P}^n$ equipped with the invariant quadric $Q=Id$ . Spherical geometry, on the other hand, is obtained from elliptic by taking two copies of $\mathbb{P}^n$ with this quadric, and gluing them together along their ideal lines to get a sphere.

Example

Exercise: Find the distance between the points $P=(1,0,0)$ and $Q=(.5,.5,.5)$ in elliptic geometry.

Solution: Note the coordinates given can be arbitrary homogeneous coordinates. The formula gives: $d(P,Q)={\cos}^{-1}(\frac{<m,n>_Q}{{||m||}_Q~{||n||}_Q})=\cos^{-1}(\frac{.5}{\sqrt{.75}})=54.74$ degrees $= .955$ radians

Hyperbolic geometry

If instead of $Q=Id$ one chooses the quadratic form with signature (n,1) which has n 1's and one -1 along the diagonal, one obtains hyperbolic geometry. For example, for $n=2$ one has the diagonal matrix $(1,1,-1)$ .

One obtains the same results if one uses the quadratic form Q' with signature (1,n) with n -1's and one 1 on the diagonal. We will try to indicate in the following investigations how one verifies this equivalence for particular formulas. (The original version of these notes used the quadric Q' here.)

Defining a metric in hyperbolic geometry

For reason which will hopefully become clear in what follows, the model for hyperbolic geometry is provided by the points where $<x,x>_Q~<~ 0$ .

One can again define the norm of a point as $||x|| := \sqrt{<x,x>_Q}$ . The set of points of norm i (!) in $\mathbb{R}^{n+1}$ form a hyperboloid of one sheet. For $n=2$ this has equation $x^2+y ^2-z^2=1$ .

WIthout going into details, the resulting distance formula is: $d_Q(x,y) := {cosh}^{-1}(\frac{<x,y>_Q}{{||x||}_Q~{||y||}_Q})$ , where here the hyperbolic cosine appears instead of the cosine.

Peculiarities of Geometry of Hyperbolic Plane in Projective Model
- In contrast to elliptic geometry, there are real points in projective space for which $<x,x>_Q=0$ . Example: In n=2, these points form the surface $x^2+y^2-z^2=0$ , a cone. Working with dehomogenized coordinates (the plane $z=1$ ), this reduces to the circle $x^2+y^2=1$ . In this so-called projective model, the hyperbolic plane $H^2$ itself is defined to be the interior of this circle . The points on the circle have undefined distance from any point of $H^2$ , as one can verify. This situation did not arise in the elliptic case since no point of $\mathbb{P}^n$ satisfies $<x,x>_Q=0$ .
- Any hyperbolic isometry must preserve this quadric surface.
- The points in the unit disk have imaginary norm! For example, the point $(0,0,1)$ has norm $\sqrt(-1) = i$ . One might think it would be better to use the alternative quadric Q' introduced above, where this situation does not arise. But the following exercise shows this is unnecessary. So, one can work with imaginary norm without coming into problems when calculating distance and angle.
- Any two points in the unit disk have negative inner product.
- Exercise Show that $d_Q(x,y) = d_{Q'}(x,y)$ where Q has signature (n,1) and Q'=-Q.

Example

Exercise: Find the distance between the points $x=(0,0,1)$ and $y=(.5,0,1)$ in hyperbolic plane geometry.

Solution: The formula gives: $d(x,y)={cosh}^{-1}(\frac{<x,y>_Q}{{||x||}_Q~{||y||}_Q})=cosh^{-1}(\frac{1}{\sqrt{.75}})=.549...$ . Note that this value is slightly larger than .5, the Euclidean distance between the two points.

Angle measurement in elliptic and hyperbolic geometry

The formulas for angle measurement in the two non-euclidean geometries can be simply expressed. First, assume we are in n=2 and we want to measure the angle between two lines m and n. Assume further that the intersection of the two llines lies within the unit disk, that is, lies within the model for hyperbolic plane. Then the angle formula in both geometries is the same: $\angle (m,n) :={\cos}^{-1}(\frac{<m,n>_Q}{{||m||}_Q~{||n||}_Q})$ . (Of course, the quadric form Q is different in the two cases, hence the angles that one computes differ in the two geometries).

Example

Exercise: Find the angle between the lines $m=(1,0,0)$ and $n=(1,1,-.5)$ in elliptic, in euclidean, and in hyperbolic plane geometry.

Solution: Elliptic geometry:Tthe formula yields $\angle(m,n)={\cos}^{-1}(\frac{<m,n>_Q}{{||m||}_Q~{||n||}_Q})=cos^{-1}(\frac{1}{\sqrt{2.25}})=48.18..$ degrees.

Euclidean geometry: the angle is clearly 45 degrees.

Hyperbolic geometry: The formula gives: $angle(m,n)={\cos}^{-1}(\frac{<m,n>_Q}{{||m||}_Q~{||n||}_Q})=cosh^{-1}(\frac{1}{\sqrt{1.75}})=40.8....$ degrees.

Non-euclidean Reflections

A non-euclidean reflection in a plane is an orientation-reversing isometry which fixes the plane point-wise and preserves the quadratic form associated to the geometry. The formula for a non-euclidean reflection is similar to the one we already saw for Euclidean reflections. Let $m'$ be the polar point of the plane m. Then the desired reflection is the harmonic homology with center $m'$ and axis $m$ . That is, ${x}' = {x} - 2 m' \frac{<x,m>}{<m',m>}$ . Important: the inner product in this formula is the ordinary inner product, not the one induced by the quadratic form Q. If M is the matrix representing this harmonic homology, then $m_{ij} = \delta_{ij}-\frac{2{m'}_i m_j}{<m',m>}$ .

Example

Exercise:Calculate the hyperbolic reflection in the line $x=.5$ , and confirm that it preserves the quadratic form. Confirm that the center is fixed, and the axis is point-wise fixed.

Solution: The plane coordinates for the plane are $m=(1,0,-.5)$ and the polar point is $m'=(-1,0,-.5)$ . The scale factor appearing in the denominator is $<m,m'>=-\frac{3}{4}$ (ordinary scalar product!). Using the formula above, we get a matrix form $M= \begin{pmatrix} 1-2(-1)(-\frac{4}{3}) & 0 & -2(-1)(-.5)(-\frac{4}{3}) \\ 0 & 1-0 & 0 \\ -2(-.5)(1)(-\frac{4}{3}) & 0 & 1-2(-.5)(-.5)(-\frac{4}{3} ) \end{pmatrix} = \begin{pmatrix} -\frac{5}{3} & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ -\frac{4}{3} & 0 & \frac{5}{3} \end{pmatrix}$

$M^t~Q~M=\begin{pmatrix} -\frac{5}{3} & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ -\frac{4}{3} & 0 & \frac{5}{3} \end{pmatrix} \begin{pmatrix} -1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} -\frac{5}{3} & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ -\frac{4}{3} & 0 & \frac{5}{3} \end{pmatrix}= \begin{pmatrix} \frac{5}{3} & 0 & \frac{4}{3} \\ 0 & -1 & 0 \\ -\frac{4}{3} & 0 & \frac{5}{3} \end{pmatrix} \begin{pmatrix} -\frac{5}{3} & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ -\frac{4}{3} & 0 & \frac{5}{3} \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$M(m') = -m'$ , which is projectively equivalent to $m'$ . To show that the axis is fixed, apply the transpose of M to m and one gets $M^t(m)~=~-m$ which is projectively equivalent to $m$ .

Non-euclidean trigonometry

In Euclidean plane geometry, triangles play an important role, probably because any geometric area can be well-approximated by triangles, the simplest polygon. There is a fully-developed theory of triangles, known as trigonometry, with many famous trigonometric formulae. In particular, the three angles and three edge lengths of a triangle are related by a variety of formula. In general, if three of these quantities are known, the other three can be computed (with famous exceptions such as similar triangles,etc).

In general, the three vertices of a triangle are represented by $A,B,C$ and the lengths of the opposite edges (whether in euclidean or non-euclidean geometry) are represented by $a,b,c$ .

Non-euclidean trigonometry also exists, to be exact, both spherical and hyperbolic versions are available. We begin with spherical trigonometry.

Spherical Trigonometry

There are two main formulae in spherical trigonometry.

Spherical law of cosines:
$\cos(c) = \cos(a) \cos(b) + \sin(a) \sin(b) \cos(C)$
Dual spherical law of cosines:
$\cos(C) = -\cos(A) \cos(B) + \sin(A) \sin(B) \cos(c)$

Of course, these formulas also work if (A,B,C) are permuted (of course (a,b,c) has to be permuted in the same way).

Example

Given $A=90^\circ, B=60^\circ, C=45^\circ$ solve for $a,b,c$

Solution: Use dual spherical law of cosines. We get

$\cos(a) = \frac{\cos(A)+\cos(B)\cos(C)}{\sin{B}\sin{C}} = \frac{0+\frac{1}{2}\frac{\sqrt(2)}{2}}{\frac{\sqrt(3)}{2} \frac{\sqrt(2)}{2}} = \frac{1}{\sqrt(3)}$ , or $a=54.74^\circ$ .
$\cos(b) = \frac{\cos(B)+\cos(A)\cos(C)}{\sin{A}\sin{C}} = \frac{\frac{1}{2}+0\frac{\sqrt(2)}{2}}{1 \frac{\sqrt(2)}{2}} = \frac{1}{\sqrt(2)}$ , or $a=45^\circ$ .
$\cos(c) = \frac{\cos(C)+\cos(A)\cos(B)}{\sin{A}\sin{B}} = \frac{\frac{\sqrt(2)}{2}+0\frac{\sqrt(3)}{2}}{1 \frac{\sqrt(3)}{2}} = \sqrt(\frac{2}{3})$ , or $a=35.26^\circ$ .

HyperbolicTrigonometry

There are also two main formulae in hyperbolic trigonometry.

Hyperbolic law of cosines:
$\cosh(c) = \cosh(a) \cosh(b) - \sinh(a) \sinh(b) \cos(C)$
Dual hyperbolic law of cosines:
$\cos(C) = -\cos(A) \cos(B) + \sin(A) \sin(B) \cosh(c)$

In the same way as above, these formulas also work if (A,B,C) are permuted (of course (a,b,c) has to be permuted in the same way).

Notice that the side lengths always appear with hyperbolic trigonometric functions; while the triangle angles always appear with the ordinary trig functions.

Example

Given $A=90^\circ, B=45^\circ, C=30^\circ$ solve for $a$

Solution: Use dual hyperbolic law of cosines. We get

$\cosh(a) = \frac{\cos(A)+\cos(B)\cos(C)}{\sin{B}\sin{C}} = \frac{0+\frac{\sqrt(2)}{2}\frac{\sqrt(3)}{2}}{\frac{\sqrt(2)}{2}\frac{1}{2} } = \sqrt(3)$ , or $a=1.146$ .

NonEuclideanGeometry