Chapter 2: non-Euclidean geometry

13 July 2022

Exercise 2.1

Easy. Show that if Euclid’s form of the parallel postulate holds, then Playfair’s conclusion of the uniqueness of parallels must follow.

[In other words, show Playfair’s axiom from Euclid’s postulates.] This is intuitively obvious, but Euclid never makes anything easy, of course. A reference can be found at David Joyce’s site. Presumably RP was looking for the intuitive explanation, since reconstructing Euclid from first principles is not “easy” by any means.

Solution

Let $$\overline{AB}$$ be a line. Let $$C$$ be a point not on $$\overline{AB}.$$ Construct a line perpendicular to $$\overline{AB}$$ passing through $$C$$ (by Euclid I.12.) Call the intersection point $$D.$$ Assume WLOG that $$A \not = D.$$ Then the angle $$\angle ADC$$ is right (by definition 10,) as is its complement $$\angle ADE.$$ By Euclid I.12, there is a unique perpendicular $$\overline{CF}$$ to $$\overline{CD}$$ passing through $$C,$$ which necessarily forms right angles $$\angle FCD, \angle GCD.$$ Any other line through $$C$$ will form one acute angle; but this angle together with the right angle on the same side is less than two right angles. By Euclid’s parallel postulate, such a line must intersect $$\overline{AB}.$$

That $$\overline{CF} \parallel \overline{AB}$$ is proved in Euclid I.28.

Exercise 2.2

Medium. …an expression for the hyperbolic distance between two points $$A$$ and $$B$$ inside the [Poincaré disc] […] is $\log \frac{QA \cdot PB}{QB \cdot PA},$ where $$P$$ and $$Q$$ are the point where the Euclidean circle (i.e., hyperbolic straight line) through $$A$$ and $$B$$ orthogonal to the bounding circle meets the bounding circle and where $$QA$$, etc., refer to Euclidean distances.

If you want to include the $$C$$ of Lambert’s area formula,i.e., $$\pi - (\alpha + \beta + \gamma) = C\Delta,$$ where $$\alpha, \beta, \gamma$$ are the angles of a hyperbolic triangle and $$\Delta$$ its area. just multiply the above distance expression by $$C^{-1/2}.$$

Can you see a simple reason why?

Solution

Consider a hyperbolic triangle. $$C$$ is a constant that varies with the geometry, and represents the ratio of the triangle’s (scale-invariant) angular deficit to its area; think of it as being measured in units of $$\mathrm{distance}^{-2}.$$ (Remember, radians are unitless.)

The distance expression given is a (logarithm of) a ratio of distances, and is thus unitless. Dimensional analysis dictates that the proper “scaling factor” should be in units of distance; hence, $$C^{-1/2}.$$

Exercise 2.3

Easy. …prove that, according to [the distance formula of Exercise 2.2], if $$A, B,$$ and $$C$$ are three successive points on a hyperbolic straight line, then the hyperbolic distances $$AB$$, etc. satisfy $$AB + BC = AC.$$

Solution

Let $$P, Q$$ be the endpoints of the common hyperbolic line, as described in Exercise 2.2. Then

\begin{align*} AB + BC &= \log \frac{QA \cdot PB}{QB \cdot PA} + \log \frac{QB \cdot PC}{QC \cdot PB} \newline &= \log \left(\frac{QA \cdot PB}{QB \cdot PA} \cdot \frac{QB \cdot PC}{QC \cdot PB}\right) \newline &= \log \frac{QA \cdot PC}{QC \cdot PA} = AC. \end{align*}

Exercise 2.4

Medium. [The projective representation of hyperbolic geometry] can be obtained from the conformal one by means of an expansion radially out from the centre by an amount given by $\frac{2R^2}{R^2 + r_c^2},$ where $$R$$ is the radius of the bounding circle and $$r_c$$ is the Euclidean distance out from the centre of the bounding circle of a point in the conformal representation. Show this.

Solution

Consider a conformal hyperbolic geometry of radius $$R,$$ represented as an origin-centered circle in $$\mathbb R^2.$$ Let $$p_c$$ be a point inside the circle with polar coordinates $$(r_c, \theta).$$

Using Beltrami’s spherical model, consider the projection from the south pole $$s = (0, 0, -R)$$ through $$p_c$$ to a point $$p_s$$ on the northern hemisphere. Evidently $$p_s = tp_c + (1 - t)s$$ for some $$t \in \mathbb R^+$$. Solving, we find that $t = \frac{2R^2}{R^2 + r_c^2}.$

Projecting $$p_s$$ straight down onto the equatorial disc then gives $$p_p = (tr_c, \theta),$$ as desired.

Exercise 2.5

Easy. Two important properties of stereographic projection […] are that it is conformal, so that it preserves angles, and that it sends circles on the sphere to circles (or, exceptionally, to straight lines) on the plane.

Assuming these two stated properties […] show that Beltrami’s hemispheric representation is conformal, with hyperbolic “straight lines” as vertical semicircles.

Solution

The hemispheric representation is reached from the conformal representation by stereographic projection. Hence it must be conformal as well.

The claim about semicircles follows from the injectivity of stereographic projection; it must send circular arcs (or straight lines) in the plane to circles on the sphere. Since projecting these circular arcs downwards produces (Euclidean) straight lines, these circles must be “vertical.”

Exercise 2.6

Hard. Can you see how to prove [the two properties of stereographic projection mentioned in Exercise 2.5]?

Solution

Conformality follows from the calculation done in Exercise 2.4. In particular, in cylindrical coordinates, stereographic projection is the mapping $(r_c, \theta, 0) \mapsto (tr_c, \theta, (t - 1)R).$

Showing that circles on the sphere stereographically project to circles in the plane is considerably trickier.I am at a loss as to the geometric argument which Penrose alludes to.

Exercise 2.7

Hard. See if you can [construct a tessellation of the hyperbolic plane] with hyperbolic regular pentagons and squares.

Solution

RP is most likely referring to the quasiregular tiling $$\{5 | 4\}$$ and/or $$\{4 | 5\},$$ viewable on David Joyce’s site.At some point I intend to write a simple JavaScript viewer for these tilings. They’re so beautiful.

Exercise 2.8

Hard. Try to prove [the formula for the area of a spherical triangle] $\Delta = R^2(\alpha + \beta + \gamma - \pi),$ basically using only symmetry arguments and the fact that the total area of the sphere is $$4\pi R^2.$$

Solution

Note that the area of a wedge bounded by two “great semicircles” separated by an angle $$\theta$$ is $\frac{\theta}{2\pi} \cdot 4\pi R^2 = 2\theta R^2.$

The total area of the three wedges formed by the three angles of the triangle is then $$2R^2(\alpha + \beta + \gamma).$$ Doubling this area gives the total area of the sphere plus $$4$$ times the triangle’s area. (Why?) In other words, $4\Delta + 4\pi R^2 = 4R^2 (\alpha + \beta + \gamma).$ Solving for $$\Delta$$ produces the desired formula.