## Table of Contents

## 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.See this discussion on math.SE. 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.

https://www.geogebra.org/3d/jx5y5eec https://web.archive.org/web/20210928090500/http://www.cs.bsu.edu/homepages/hfischer/math345/stereo.pdf

## 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.