MATH325: Lecture 17


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Curvature

As you've probably realized by now, there are two possible ways that our Axiom 5 can fail. Axiom 5 says that given a point and a straight line, there exists an unique line through that point that is parallel to the given line.

On the sphere, this axiom is not true because there is no line parallel to the original line.

In the hyperbolic plane, this axiom is not true because there can be more than one line through that point that does not intersect the given line.

The textbook has a wonderful, detailed discussion of the geometry of the hyperbolic plane and the different ways two lines can fail to intersect. However, it seems reasonable to first discuss simpler ways in which the hyperbolic plane is similar to and differs from the sphere and the plane.

One immediately obvious difference is that the surface of a sphere closes in on itself while a plane extends infinitely in every direction. A plane has more surface area than a sphere. Similarly, a hyperbolic plane can be thought of as having more surface area than a plane.

Planes are infinite; what might it mean to have more surface area than a plane?

Simply put, there's more stuff around each point in a hyperbolic plane than there is around a point in a plane. More precisely, in the plane a circle with radius 1cm has circumference 2πcm. On a sphere, a circle with radius 1cm will have a circumference of less than 2πcm, because the sphere is curving back in on itself. The size of the sphere determines how sharply the sphere curves and so how much less than 2πcm the circumference will be.

In a hyperbolic plane, the circumference of a circle with radius 1cm will be more than 2πcm. Some surfaces that have this property include horses' saddles, Pringles potato chips, and the graph of z=x2-y2.

The Bertrand-Diquet-Puiseux theorem says that the Gaussian Curvature K of a surface at a point P can be defined as:

K = limr->0 (2πr - C(r))*(3/(πr3)),

where C(r) is the circumference of a circle of radius r centered at P. (For more information and a more readable version of this formula, see Wikipedia.)

On a plane, 2πr - C(r) is zero for every radius r. On a sphere, 2πr - C(r) is positive and so is the curvature K. On a hyperbolic plane 2πr - C(r) is negative and so is the curvature. A hyperbolic plane is a surface with constant (the same everywhere) negative curvature.

Poincaré Disk Model

Drawing figures on the surface of a superball was hard; drawing anything complicated on Pringles would be nearly impossible. The first hurdle we must overcome when studying the hyperbolic plane is finding a satisfactory way to sketch the figures we're working with.

There are three popular methods for representing figures on a hyperplane by drawing them on a plane. Each method has drawbacks, just as each flat map of the earth has flaws. (The area of Greenland is a little over 2 million square miles kilometers that of South America is nearly 20 million square kilometers. However, they appear to be the same size in this image.)

The first model introduced by our text book seems to be the most popular one; it is the Poincaré disk model (discovered by Riemann). In this model, the entire hyperplane is shrunk to fit inside a circle. You may wish to think of this as like looking at the hyperbolic plane through a fisheye lens -- figures at the center of the disk appear almost normal, while offcenter figures are shrunken and distorted.

This article illustrates an application of the Poincaré disk model to viewing large trees of data -- translating the origin to different nodes of the tree focuses attention on those nodes, while the remainder of the tree is "marginalized".

The tiling of the hyperplane by congruent squares on page 183 is a good illustration of this distortion. Each four sided figure inside the disk is a hyperbolic square. The central figure does indeed look squarelike, but the five other hyperbolic squares around each of its vertices do not look like squares as we know them.

Just as the areas of South America and Greenland were distorted in the picture of the earth's surface, the areas of the figures in the Poincaré disk are distorted. The advantages of the Poincaré disk model are that it's attractive (see pages 190-191) and that it does not distort the angles between hyperbolic lines (even though it does distort distances and lengths.)

As you can see from the picture on page 183, hyperbolic lines are represented as arcs of circles in the Poincaré disk. Specifically, lines in the hyperbolic plane correspond to arcs of circles that intersect the boundary of the disk at right angles. Each diameter of the Poincaré disk corresponds to a hyperbolic line, as do each of the circular arcs shown on page 183.

The two other popular models of the hyperplane are the Klein-Beltrami model, in which lines are modeled as chords of a circle and angles are distorted, and the Poincaré upper half-plane model (discovered by Beltrami) which is, in essence, obtained by stretching the boundary of the Poincaré disk to cover the x-axis; the resulting model of the hyperbolic plane covers the upper half-plane. The tiling of the hyperplane by regular pentagons shown in this Wikipedia article seems to indicate that angles are distorted in the Poincaré upper half-plane model.

Axioms

Our first four axioms for plane geometry remain true for the hyperplane -- unlike in the sphere, each pair of points in the hyperbolic plane determines a unique hyperbolic line. Axioms 6 through 8 also remain true, so in particular our SSS, ASA and SAS theorems for triangle congruence should continue to be true in the hyperbolic plane.

Axiom 5, of course, is not true in the hyperbolic plane. Given any line l and any point P not on l, there exist two distinct lines l1 and l2 which do not intersect l.

Angle-Angle-Angle

In spherical geometry, we claimed that two triangles whose angles were all congruent were congruent triangles (angle-angle-angle). Having proven that the area of a triangle was dependent on its angles, this seemed like a reasonable conclusion.

For hyperbolic triangles, a similar formula governs their area:

Area(triangle ABC) = π - (m∠A + m∠B + m∠C).

We will not prove this, but by assuming that the sum of the angles of a hyperbolic triangle is less than 180° (or π radians) we will prove the angle-angle-angle formula for hyperbolic geometry.

Theorem 6.3.1: In the hyperbolic plane, for any triangle ABC, we have:

m∠A + m∠B + m∠C < 180°.

The book does not prove this theorem and neither will we. Drawing triangles on Pringles is unlikely to be helpful. The java applet on this web page might provide some insight, or might not.

Corollary 6.3.2: The sum of the angles in a quadrilateral is less than 360°.

Discussion of proof of Corollary: There are two ways to subdivide a quadrilateral into two triangles. Either way, the sum of the angles of the two triangles equals the sum of the angles of the quadrilateral. Since the triangles' angles sum to less than 180° each, the quadrilateral's angles must sum to less than 360°.

Theorem 6.3.3: Suppose triangle ABC is similar to triangle A'B'C' (i.e. corresponding angles are congruent). Then triangle ABC is congruent to triangle A'B'C'.

Proof: Since m∠BAC = m∠B'A'C', axioms 6, 7 and 8 allow us to isometrically move A' to A, the ray A'B' to ray AB, and ray A'C' to ray AC. This is illustrated in Figure 6, where the images of B' and C' under the isometry are labeled as B'' and C''.

We will prove this theorem by contradiction -- assume that triangles ABC and A'B'C' are not congruent. Then either B'' ≠ B or C'' ≠ C. By relabeling the vertices if necessary we can assume B'' ≠ B. We'll also assume that B'' lies between B and A on ray AB; if this is an incorrect assumption we could reverse the roles of B and B'', and the rest of the proof could remain the same.

Segment BC cannot cross segment B''C'' because such a crossing would create two triangles with angle sum greater than 180°, which can't exist in the hyperbolic plane. Therefore BCC''B'' forms a quadrilateral.

Inside this quadrilateral, the angle at B is the supplement of the angle at B'' and the angle at C is the supplement of the angle at C''. This means that the sum of the interior angles of the quadrilateral sum to 360°, which is a contradiction to corollary 6.3.2.

We have discovered a contradiction to our assumption that the two triangles are not congruent (B'' ≠ B). Therefore triangle ABC must be congruent to triangle A'B'C'.//