17. Hyperbolic geometry¶

Chapter 15 showed that a geometry with many parallels is as consistent as Euclid's, and Chapter 16 worked out the sphere, where there are none. This chapter builds the hyperbolic plane itself, so that you can compute in it: measure distances and angles, find areas, draw its tilings, and solve its triangles. You will meet it in three pictures (the Poincaré disc, the upper half-plane and the hyperboloid), prove that a triangle's area is \(\pi\) minus its angle sum, find Lobachevsky's angle of parallelism, and see why the same geometry describes the velocities of special relativity.

In this chapter

You will learn:

the Poincaré disc and upper half-plane models: their lines, angles and distance
the distance formula \(\cosh d = 1 + \frac{|z - w|^2}{2\,\mathrm{Im}\, z\, \mathrm{Im}\, w}\), and that the isometries are real Möbius maps (with proofs)
that a triangle's area is \(\pi - (\alpha + \beta + \gamma)\) (full proof by integration)
limiting parallels, ultraparallels and the angle of parallelism \(\cos\Pi(d) = \tanh d\)
regular tilings \(\{p, q\}\) and when they exist; Escher's Circle Limit prints
the hyperboloid model, and how it describes velocities in special relativity
hyperbolic trigonometry: the laws of cosines and sines, and Pythagoras with \(\cosh\)

You need: Chapter 13 (inversion, Möbius maps, cross-ratio), Chapter 15 (neutral geometry, defect, models) and Chapter 16 (spherical trigonometry, for comparison). The hyperbolic functions \(\cosh x = \frac{e^x + e^{-x}}{2}\), \(\sinh x = \frac{e^x - e^{-x}}{2}\), \(\tanh x = \frac{\sinh x}{\cosh x}\), with \(\cosh^2 x - \sinh^2 x = 1\), and integrals of one variable, as in MAT137 Unit 9.

Levels: 17.1 and 17.2 are advanced; 17.3 and 17.4 are advanced to expert; 17.5 is advanced; 17.6 and 17.7 are expert.

17.1 The Poincaré disc and the upper half-plane¶

Level: advanced. How can you draw a plane in which a point off a line has infinitely many parallels? The Klein model of Chapter 15 does it with straight chords, but distorts angles. Henri Poincaré, in 1882, found two models in which angles are drawn true. He came to them through complex analysis: his own famous account says the idea struck him as he stepped onto a bus, that the transformations he had been using for certain functions were the same as those of non-Euclidean geometry.

The upper half-plane \(\mathbb{H} = \{z = x + iy : y > 0\}\).

Points are complex numbers with positive imaginary part. The real axis is not part of the plane; it is the boundary at infinity, together with one more ideal point \(\infty\).
Lines (geodesics) are vertical half-lines \(\{x = a, \ y > 0\}\) and semicircles centred on the real axis.
Angles are the ordinary Euclidean angles between curves.
Lengths are measured with the rule \(ds = \dfrac{|dz|}{y}\): a tiny step of Euclidean length \(|dz|\) at height \(y\) has hyperbolic length \(|dz|/y\). Near the real axis, everything is very long.

The Poincaré disc \(\mathbb{D} = \{z : |z| < 1\}\).

Points are inside the unit circle; the circle itself is the boundary at infinity.
Lines are diameters and arcs of circles that meet the unit circle at right angles.
Angles are Euclidean angles.
Lengths use \(ds = \dfrac{2|dz|}{1 - |z|^2}\).

The two are the same geometry. The Möbius map of Chapter 13

\[ C(z) = \frac{z - i}{z + i} \]

sends \(\mathbb{H}\) onto \(\mathbb{D}\): it sends \(i \mapsto 0\), and for real \(x\), \(|x - i| = |x + i|\), so the real axis goes to the unit circle. It keeps angles and sends generalised circles to generalised circles; it sends lines and circles perpendicular to the real axis to ones perpendicular to the unit circle; and a computation shows it carries the length rule of \(\mathbb{H}\) to that of \(\mathbb{D}\). So any fact proved in one model holds in the other. We mostly compute in \(\mathbb{H}\), where formulas are simplest, and draw in \(\mathbb{D}\), where the pictures are most symmetric.

Why "lines". We define lines as shortest paths for the length rule. That vertical half-lines are shortest is proved in 17.2; semicircles then follow because isometries carry vertical lines onto semicircles. Through any two points of \(\mathbb{H}\) passes exactly one such curve: if they have the same real part, a vertical line; otherwise the perpendicular bisector of the Euclidean segment between them meets the real axis at one point \(c\), and the semicircle about \(c\) through both points is the line.

Example. The line through \(i\) and \(1 + i\): the perpendicular bisector of the segment from \(i\) to \(1 + i\) is \(x = \tfrac12\), so the centre is \(c = \tfrac12\), and the radius is \(|i - \tfrac12| = \sqrt{1/4 + 1} = \tfrac{\sqrt5}{2}\). It meets the boundary at \(\tfrac12 \pm \tfrac{\sqrt5}{2}\), that is at \(-0.618\) and \(1.618\).

17.2 Distance, isometries and circles¶

Level: advanced. Vertical distance. The length of the vertical segment from \(ia\) to \(ib\) (\(0 < a < b\)) is

\[ \int_a^b \frac{dy}{y} = \ln\frac{b}{a} . \]

So \(i\), \(ei\), \(e^2 i\), \(e^3 i\), ... are equally spaced, each a distance \(1\) from the next, and the real axis (\(a \to 0\)) is infinitely far away.

It is the shortest path. Any path from \(ia\) to \(ib\), given as \(z(t) = x(t) + iy(t)\), has length

\[ \int \frac{|z'(t)|}{y(t)} \, dt \ \ge \ \int \frac{|y'(t)|}{y(t)} \, dt \ \ge \ \int_a^b \frac{dy}{y} = \ln\frac{b}{a}, \]

since \(|z'| \ge |y'|\), with equality only if \(x\) is constant and \(y\) never goes back down. So the vertical segment is the unique shortest path. \(\square\)

Isometries. Which maps of \(\mathbb{H}\) keep the length rule?

Theorem 17.1. Every Möbius map \(f(z) = \dfrac{az + b}{cz + d}\) with \(a, b, c, d\) real and \(ad - bc = 1\) maps \(\mathbb{H}\) onto itself and preserves hyperbolic length.

Proof. Compute the imaginary part. Multiply top and bottom by \(\overline{cz + d} = c\bar z + d\):

\[ f(z) = \frac{(az + b)(c\bar z + d)}{|cz + d|^2} = \frac{ac|z|^2 + adz + bc\bar z + bd}{|cz + d|^2}, \]

so \(\mathrm{Im} f(z) = \dfrac{(ad - bc)\,\mathrm{Im}\, z}{|cz + d|^2} = \dfrac{\mathrm{Im}\, z}{|cz + d|^2} > 0\). And \(f'(z) = \dfrac{a(cz + d) - c(az + b)}{(cz + d)^2} = \dfrac{1}{(cz + d)^2}\). A tiny step \(dz\) goes to \(f'(z)\,dz\), so its hyperbolic length becomes

\[ \frac{|f'(z)|\,|dz|}{\mathrm{Im} f(z)} = \frac{|dz|/|cz + d|^2}{\mathrm{Im}\, z/|cz + d|^2} = \frac{|dz|}{\mathrm{Im}\, z} . \ \square \]

These maps form the group \(\mathrm{PSL}(2, \mathbb{R})\) (real matrices of determinant \(1\), with \(M\) and \(-M\) giving the same map). Adding the reflection \(z \mapsto -\bar z\) gives all isometries of \(\mathbb{H}\) (we state this without proof). Examples: translations \(z \mapsto z + b\), scalings \(z \mapsto kz\) (\(k > 0\)), and the "inversion" \(z \mapsto -1/z\).

The isometries are plentiful: you can move any point to any other (\(x + iy \mapsto (z - x)/y\) sends it to \(i\)), and the maps \(z \mapsto \dfrac{z\cos\theta + \sin\theta}{-z\sin\theta + \cos\theta}\) fix \(i\) and turn the directions at \(i\) through the angle \(2\theta\). So the hyperbolic plane looks the same at every point and in every direction, just like the Euclidean plane and the sphere.

Theorem 17.2 (distance formula). For \(z, w \in \mathbb{H}\),

\[ \cosh d(z, w) = 1 + \frac{|z - w|^2}{2\,\mathrm{Im}\, z \,\mathrm{Im}\, w} . \]

Proof. Both sides are unchanged by the isometries of Theorem 17.1. The left side because they are isometries. The right side because \(\mathrm{Im} f(z) = \mathrm{Im}\, z/|cz + d|^2\), as computed, and the identity \(f(z) - f(w) = \dfrac{z - w}{(cz + d)(cw + d)}\) (check by putting the difference over a common denominator and using \(ad - bc = 1\)) gives \(|f(z) - f(w)|^2 = |z - w|^2/(|cz + d|^2|cw + d|^2)\); the factors cancel. Now move \(z\) to \(i\) and then turn about \(i\) until \(w\) lies on the imaginary axis above it, \(w = ib\) with \(b \ge 1\). There \(d = \ln b\) and the right side is \(1 + \frac{(b - 1)^2}{2b} = \frac{b^2 + 1}{2b} = \frac{b + b^{-1}}{2} = \cosh(\ln b)\). \(\square\)

Examples. \(d(i, 2i) = \ln 2 \approx 0.693\). \(d(i, 1 + i)\): \(\cosh d = 1 + \frac{1}{2} = 1.5\), so \(d = \operatorname{arcosh} 1.5 \approx 0.9624\). The Euclidean distance is \(1\) in both cases, but the hyperbolic ones differ, because the length rule depends on the height. (Check with the formula for the first: \(1 + \frac{1}{2 \cdot 2} = 1.25 = \cosh(\ln 2) = \frac{2 + 1/2}{2}\).)

In the disc. The distance from \(0\) to a point at Euclidean distance \(r\) is

\[ d(0, r) = \int_0^r \frac{2\,dt}{1 - t^2} = \ln\frac{1 + r}{1 - r} = 2\operatorname{artanh} r . \]

So \(r = \tanh(d/2)\). The point \(r = \tfrac12\) is at distance \(\ln 3 \approx 1.10\); \(r = 0.9\) at \(\ln 19 \approx 2.94\); \(r = 0.99\) at \(\ln 199 \approx 5.29\).

Circles. The hyperbolic circle with centre \(i\) and radius \(\rho\) is the set of \(z = x + iy\) with \(\cosh\rho = 1 + \frac{x^2 + (y - 1)^2}{2y}\). Multiply out: \(x^2 + y^2 - 2y\cosh\rho + 1 = 0\), that is

\[ x^2 + (y - \cosh\rho)^2 = \sinh^2\rho . \]

It is a Euclidean circle, with Euclidean centre \(i\cosh\rho\) (higher than \(i\)) and radius \(\sinh\rho\). Hyperbolic circles look like ordinary circles in both models, but their hyperbolic centre is pulled towards the boundary.

17.3 Angle sum and area¶

Level: advanced to expert. Area in \(\mathbb{H}\) uses the same scaling as length, squared: a tiny rectangle of Euclidean area \(dx\,dy\) at height \(y\) has hyperbolic area

\[ dA = \frac{dx\,dy}{y^2} . \]

(The isometries of Theorem 17.1 keep it: their Jacobian is \(|f'(z)|^2 = 1/|cz + d|^4\), and \(y^2\) also changes by \(1/|cz + d|^4\).)

Theorem 17.3. A triangle with angles \(\alpha, \beta, \gamma\) in the hyperbolic plane has area

\[ \text{Area} = \pi - (\alpha + \beta + \gamma) . \]

In particular the angle sum is less than \(\pi\), and the area is exactly the defect of Chapter 15. The proof comes in two steps.

Step 1: one vertex at infinity. A triangle may have ideal vertices on the boundary at infinity, where its two sides meet at angle \(0\). Take a triangle with one ideal vertex and finite angles \(\alpha\) and \(\beta\). Move its ideal vertex to \(\infty\) (an isometry). Its two sides through \(\infty\) are vertical lines, and its third side is a semicircle; translate and scale (isometries again) so that the semicircle is the unit circle. Then the finite vertices are \(A = e^{i(\pi - \alpha)}\) and \(B = e^{i\beta}\): the angle at \(A\) between the vertical and the circle is \(\alpha\) because the radius to \(A\) makes the angle \(\alpha\) with the real axis, and the tangent is perpendicular to the radius.

The triangle is the region above the arc and between the verticals \(x = -\cos\alpha\) and \(x = \cos\beta\). Its area is

\[ \int_{-\cos\alpha}^{\cos\beta} \int_{\sqrt{1 - x^2}}^{\infty} \frac{dy}{y^2} \, dx = \int_{-\cos\alpha}^{\cos\beta} \frac{dx}{\sqrt{1 - x^2}} . \]

Substitute \(x = \cos\theta\), \(dx = -\sin\theta\,d\theta\), \(\sqrt{1 - x^2} = \sin\theta\): the integral becomes \(\int_{\beta}^{\pi - \alpha} d\theta = \pi - \alpha - \beta\). So a triangle with one ideal vertex has area \(\pi - \alpha - \beta\). With \(\alpha = \beta = 0\) (all three vertices ideal) the area is \(\pi\): every ideal triangle has area \(\pi\), however "large" it looks.

Step 2: any triangle. Let \(ABC\) have angles \(\alpha, \beta, \gamma\). Extend the side \(AC\) beyond \(C\) to its ideal endpoint \(\Omega\). Then the triangle \(AB\Omega\) (one ideal vertex) is made of \(ABC\) and the triangle \(CB\Omega\). Let \(\varepsilon = \angle CB\Omega\). The angles are:

in \(AB\Omega\): \(\alpha\) at \(A\), \(\beta + \varepsilon\) at \(B\), \(0\) at \(\Omega\);
in \(CB\Omega\): \(\pi - \gamma\) at \(C\) (the supplement of \(\gamma\), on the straight line \(AC\Omega\)), \(\varepsilon\) at \(B\), \(0\) at \(\Omega\).

By Step 1,

\[ \text{Area}(ABC) = \big(\pi - \alpha - \beta - \varepsilon\big) - \big(\pi - (\pi - \gamma) - \varepsilon\big) = \pi - \alpha - \beta - \gamma . \ \square \]

Consequences.

The angle sum of every triangle is less than \(\pi\), and the area of every triangle is less than \(\pi\).
AAA congruence. Two triangles with the same angles have the same area; more is true, they are congruent (this follows from the hyperbolic law of cosines for angles in 17.7, which determines the sides from the angles). There are no similar triangles of different sizes.
Polygons. Cutting into triangles, a convex polygon with \(n\) sides and angles \(\theta_1, \ldots, \theta_n\) has area \((n - 2)\pi - (\theta_1 + \cdots + \theta_n)\).
Scale. Our models have curvature \(-1\). Scaling all lengths by \(R\) gives the plane of curvature \(-1/R^2\), in which the area of a triangle is \(R^2\) times its defect: Lambert's discovery of 1766, now a theorem.

Example. A triangle with angles \(30^\circ\), \(45^\circ\) and \(60^\circ\) has area \(\pi - \frac{3\pi}{4} = \frac{\pi}{4} \approx 0.785\).

17.4 Many parallels¶

Level: advanced to expert. Take a line \(\ell\) and a point \(P\) not on it. In the disc, put \(P\) at the centre \(0\) (an isometry can always do this). Lines through \(0\) are diameters. A diameter meets \(\ell\) only if it passes through the arc of \(\ell\); the two diameters that end exactly at the two ideal endpoints of \(\ell\) meet \(\ell\) only "at infinity", which is not in the plane.

So the lines through \(P\) fall into three kinds:

those that meet \(\ell\) (diameters between the two special ones, on the side of \(\ell\));
two limiting parallels (also called asymptotic parallels), the diameters through the endpoints of \(\ell\); they get closer and closer to \(\ell\) without meeting it;
infinitely many ultraparallels, all the rest; each of these and \(\ell\) have a common perpendicular, and they move apart in both directions.

The angle of parallelism. Let \(Q\) be the foot of the perpendicular from \(P\) to \(\ell\), and \(d = PQ\). The angle \(\Pi(d)\) between \(PQ\) and a limiting parallel depends only on \(d\).

Theorem 17.4 (Lobachevsky). \(\cos\Pi(d) = \tanh d\), or equivalently \(\tan\dfrac{\Pi(d)}{2} = e^{-d}\).

Proof. With \(P = 0\) and \(Q\) on the positive real axis, \(Q = r\) with \(r = \tanh(d/2)\) (17.2). The line \(\ell\) is perpendicular to the real axis at \(r\), so it is an arc of a circle with centre \(c\) on the real axis, \(c > 1\), radius \(s\), meeting the unit circle at right angles. Orthogonality means the radii at a meeting point are perpendicular, so \(c^2 = 1 + s^2\). And the arc passes through \(r\): \(c - s = r\). From \(c^2 - s^2 = 1\) we get \(c + s = 1/r\), so \(c = \frac12(r + 1/r)\).

The ideal endpoints of \(\ell\) are where the two circles meet: \(x^2 + y^2 = 1\) and \((x - c)^2 + y^2 = s^2\) give \(1 - 2cx + c^2 = s^2\), so \(x = \frac{1 + c^2 - s^2}{2c} = \frac{1}{c}\). The limiting parallel is the diameter to this point, at angle \(\Pi\) from the real axis, so

\[ \cos\Pi = \frac{1}{c} = \frac{2r}{1 + r^2} = \frac{2\tanh(d/2)}{1 + \tanh^2(d/2)} = \tanh d, \]

by the double angle formula for \(\tanh\). The second form: \(\tan^2(\Pi/2) = \frac{1 - \cos\Pi}{1 + \cos\Pi} = \frac{1 - \tanh d}{1 + \tanh d} = e^{-2d}\). \(\square\)

What it says. As \(d \to 0\), \(\Pi(d) \to 90^\circ\): very close to \(\ell\), the parallels are nearly unique, as in Euclid. As \(d \to \infty\), \(\Pi(d) \to 0\): far away, almost every line through \(P\) misses \(\ell\). For example \(\Pi(1) = \arccos(\tanh 1) = \arccos 0.7616 \approx 40.4^\circ\). The function \(\Pi\) also explains the "natural unit of length" of Chapter 15: the distance at which \(\Pi(d) = 45^\circ\) (namely \(d = \operatorname{artanh}(1/\sqrt2) \approx 0.881\)) is defined by an angle, so it is a length the geometry singles out.

17.5 Hyperbolic tilings¶

Level: advanced. A regular tiling \(\{p, q\}\) covers the plane with regular \(p\)-gons, \(q\) meeting at each corner. In the Euclidean plane there are three, \(\{3, 6\}\), \(\{4, 4\}\) and \(\{6, 3\}\) (Chapter 3). On the sphere the regular tilings are the Platonic solids blown up onto their circumscribed sphere (Chapter 11).

Theorem 17.5. The tiling \(\{p, q\}\) (with \(p, q \ge 3\)) exists in the hyperbolic plane exactly when

\[ \frac1p + \frac1q < \frac12 . \]

Proof. At each corner, \(q\) equal angles fill a full turn, so each angle of the \(p\)-gon is \(2\pi/q\). By 17.3 its area is \((p - 2)\pi - p \cdot \frac{2\pi}{q}\), which must be positive: \((p - 2)q > 2p\), which rearranges (divide by \(2pq\)) to \(\frac1p + \frac1q < \frac12\). Conversely, if the inequality holds, a regular \(p\)-gon with angle \(2\pi/q\) exists: put the centre at \(0\) in the disc and the vertices at radius \(t\); as \(t\) grows from \(0\) towards \(1\), the angle decreases continuously from the Euclidean value \((p - 2)\pi/p\) (for tiny polygons) towards \(0\) (vertices near infinity), so it passes through \(2\pi/q\), which lies in between exactly when the inequality holds. Reflecting this polygon in its sides, again and again, fills the plane without gaps or overlaps (the proof that it never overlaps is Poincaré's polygon theorem, which we do not prove). \(\square\)

So there are infinitely many hyperbolic regular tilings: \(\{7, 3\}\), \(\{3, 7\}\), \(\{4, 5\}\), \(\{5, 4\}\), \(\{8, 3\}\), and so on. The three kinds of geometry correspond to \(\frac1p + \frac1q\) being greater than, equal to, or less than \(\frac12\).

The figure shows \(\{7, 3\}\), coloured by rings. Each heptagon has angles \(120^\circ\) and area \(5\pi - 7 \cdot \frac{2\pi}{3} = \frac{\pi}{3}\). Its circumradius \(R\) satisfies \(\cosh R = \cot\frac{\pi}{7}\cot\frac{\pi}{3} \approx 1.199\) (a formula from hyperbolic trigonometry, 17.7), so in the disc its corners are at Euclidean radius \(\tanh(R/2) \approx 0.30\). The figure was drawn by exactly the process in the proof: start with the central heptagon and reflect, by inversion in the circles of its sides (Chapter 13).

Rings grow exponentially. Count the tiles in successive rings: they grow roughly by a fixed factor each time. In the Euclidean plane, the number of tiles within distance \(r\) grows like \(r^2\); in the hyperbolic plane it grows like \(e^r\). The circumference of a hyperbolic circle of radius \(r\) is \(2\pi\sinh r\) and its area is \(2\pi(\cosh r - 1)\): for \(r = 5\), about \(466\) and \(460\), against \(31.4\) and \(78.5\) in the Euclidean plane.

Escher's circle limits. The Dutch artist M. C. Escher saw a figure of a hyperbolic tiling in a 1957 paper by the geometer H. S. M. Coxeter, wrote to him, and made four woodcuts, Circle Limit I to IV (1958 to 1960), filling the Poincaré disc with fish, and with angels and devils. In Circle Limit III the white curves along which the fish swim are not hyperbolic lines: Coxeter showed in 1979 that they are equidistant curves, arcs that meet the boundary at about \(80^\circ\), not at the right angle a line would need. Escher had placed them with a craftsman's precision.

Where hyperbolic surfaces come from. A hyperbolic polygon with its sides glued in pairs gives a closed surface. A regular octagon with angles \(45^\circ\) (\(\{8, 8\}\) tiling) glued in the pattern of a two-holed doughnut gives a hyperbolic two-holed torus; every closed surface with two or more holes carries a hyperbolic geometry (Chapter 20). Natural objects also show hyperbolic-like crinkling: lettuce leaves, coral and sea slugs grow more surface than a flat sheet allows, and Daina Taimina's crocheted models of the hyperbolic plane (from 1997) make the geometry something you can hold.

17.6 The hyperboloid model and relativity¶

Level: expert. The sphere sits in space as the set of points at "distance \(1\)" from the origin. Is there a surface in space that is the hyperbolic plane in the same way? Not with the ordinary distance (Hilbert proved in 1901 that the complete hyperbolic plane cannot sit in Euclidean space with its distances kept), but yes with a different notion of length.

Minkowski space. On \(\mathbb{R}^3\) with coordinates \((x_1, x_2, x_3)\), use the Lorentz form

\[ \langle \mathbf{x}, \mathbf{y} \rangle = x_1 y_1 + x_2 y_2 - x_3 y_3 . \]

The hyperboloid model is the upper sheet

\[ \mathcal{L} = \{ \mathbf{x} : \langle \mathbf{x}, \mathbf{x} \rangle = -1, \ x_3 > 0 \}, \]

that is \(x_3 = \sqrt{1 + x_1^2 + x_2^2}\). Its distance is given by

\[ \cosh d(\mathbf{x}, \mathbf{y}) = -\langle \mathbf{x}, \mathbf{y} \rangle, \]

just as on the unit sphere \(\cos d = \mathbf{x} \cdot \mathbf{y}\). Its lines are its intersections with planes through the origin, just as great circles are on the sphere. Its isometries are the linear maps keeping the Lorentz form and the upper sheet, the group \(\mathrm{O}^+(2, 1)\), just as rotations are for the sphere.

Example. The points \((0, 0, 1)\) and \((\sinh d, 0, \cosh d)\) lie on \(\mathcal{L}\) (since \(\sinh^2 d - \cosh^2 d = -1\)), and \(-\langle \cdot, \cdot \rangle = \cosh d\): they are at distance \(d\). The curve \(u \mapsto (\sinh u, 0, \cosh u)\) is a line traced at unit speed.

Back to the discs. Project \(\mathcal{L}\) from the point \((0, 0, -1)\) onto the plane \(x_3 = 0\): the point \(\mathbf{x}\) goes to \(\frac{x_1 + ix_2}{1 + x_3}\), inside the unit disc. This is the Poincaré disc, and it is the exact analogue of stereographic projection (Chapter 13). Check with the example: the point at distance \(d\) goes to \(\frac{\sinh d}{1 + \cosh d} = \tanh\frac d2\), which is what 17.2 says. Project instead from the origin onto the plane \(x_3 = 1\): \(\mathbf{x}\) goes to \(\frac{x_1 + ix_2}{x_3}\), and lines (planes through the origin) become straight chords. This is the Klein disc of Chapter 15, with the point at distance \(d\) going to \(\tanh d\).

Relativity. In special relativity (units with the speed of light \(c = 1\)), the spacetime of one time and two space dimensions is Minkowski space with \(x_3\) as time. An observer moving with velocity \(\mathbf{v}\) (with \(|\mathbf{v}| < 1\)) has four-velocity \(\gamma(\mathbf{v}, 1)\), where \(\gamma = 1/\sqrt{1 - |\mathbf{v}|^2}\). It satisfies \(\langle \gamma(\mathbf{v}, 1), \gamma(\mathbf{v}, 1)\rangle = \gamma^2(|\mathbf{v}|^2 - 1) = -1\): the possible velocities form exactly the hyperboloid \(\mathcal{L}\). Lorentz transformations, the changes of observer, are its isometries.

The hyperbolic distance between two four-velocities has a physical meaning: \(\cosh d = -\langle u_1, u_2\rangle\) is the Lorentz factor \(\gamma\) of their relative velocity, and \(d\) itself is the rapidity, with relative speed \(\tanh d\). For motion along one line, rapidities add, because distances along a line add. So if a rocket moves at \(0.6\) (60% of light speed) relative to a station, and fires a probe forward at \(0.6\) relative to itself, the probe's speed relative to the station is not \(1.2\) but

\[ \tanh\left(2\operatorname{artanh} 0.6\right) = \frac{0.6 + 0.6}{1 + 0.6 \times 0.6} = \frac{1.2}{1.36} \approx 0.882 . \]

This is the relativistic velocity addition formula, \(\frac{u + v}{1 + uv}\), which is the addition formula for \(\tanh\). For velocities in different directions, velocity "addition" follows hyperbolic triangles, and its peculiar features (it is not commutative, and successive boosts in different directions produce a rotation, the Thomas precession) are features of hyperbolic geometry. The link was noticed early in the history of relativity, by Arnold Sommerfeld and Vladimir Varićak around 1909 to 1910.

17.7 Hyperbolic trigonometry¶

Level: expert. Now solve triangles. Take a triangle with sides \(a, b, c\) opposite angles \(A, B, C\), in the plane of curvature \(-1\).

Theorem 17.6 (hyperbolic law of cosines).

\[ \cosh c = \cosh a \cosh b - \sinh a \sinh b \cos C . \]

Proof. Copy the spherical proof of Chapter 16 on the hyperboloid. Put \(C\) at \((0, 0, 1)\). The point \(A\) is at distance \(b\); turn about the \(x_3\)-axis (an isometry) so that \(A = (\sinh b, 0, \cosh b)\). The point \(B\) is at distance \(a\) from \(C\), in the direction making angle \(C\) with the direction to \(A\): \(B = (\sinh a\cos C, \sinh a\sin C, \cosh a)\). (At \((0, 0, 1)\) the hyperboloid is horizontal and the model's angles there are Euclidean, which is why \(C\) appears as an ordinary angle in the \(x_1x_2\)-plane.) Then

\[ \cosh c = -\langle A, B\rangle = -\sinh b\sinh a\cos C + \cosh b\cosh a . \ \square \]

The formula is the spherical one with \(\cos \to \cosh\), \(\sin \to \sinh\) and a sign change, an instance of a general rule: the hyperbolic plane is a "sphere of imaginary radius", as Lambert guessed, since \(\cos(ix) = \cosh x\) and \(\sin(ix) = i\sinh x\).

Hyperbolic Pythagoras. If \(C = 90^\circ\):

\[ \cosh c = \cosh a \cosh b . \]

Example. Legs \(a = b = 1\): \(\cosh c = \cosh^2 1 \approx 2.3811\), so \(c \approx 1.5134\), longer than the Euclidean \(\sqrt2 \approx 1.4142\). (On the sphere the hypotenuse came out shorter.) For small triangles, \(\cosh x \approx 1 + x^2/2\), and the formula becomes \(1 + c^2/2 \approx 1 + a^2/2 + b^2/2\): ordinary Pythagoras again.

Right triangles, more. In a right triangle with right angle at \(C\), one also has

\[ \cos A = \frac{\tanh b}{\tanh c}, \qquad \sin A = \frac{\sinh a}{\sinh c} . \]

(These follow from the law of cosines, the law of sines below, and Pythagoras.) In the example, \(\cos A = \tanh 1/\tanh 1.5134 \approx 0.8391\), so \(A = B \approx 32.95^\circ\). The angle sum is \(90 + 2 \times 32.95 = 155.89^\circ\), and the area is the defect, \(24.11^\circ \approx 0.421\). Let \(c \to \infty\) with \(b\) fixed: \(\tanh c \to 1\) and \(\cos A \to \tanh b\), which is Lobachevsky's angle of parallelism once more.

Theorem 17.7 (hyperbolic law of sines).

\[ \frac{\sin A}{\sinh a} = \frac{\sin B}{\sinh b} = \frac{\sin C}{\sinh c} . \]

Proof. As in Chapter 16: solve the law of cosines for \(\cos C\), compute \(\sin^2 C = 1 - \cos^2 C\), and check that \(\sin^2 C/\sinh^2 c\) equals an expression symmetric in \(a, b, c\), namely \(\frac{2\cosh a\cosh b\cosh c + 1 - \cosh^2 a - \cosh^2 b - \cosh^2 c}{\sinh^2 a\sinh^2 b\sinh^2 c}\). \(\square\)

The law of cosines for angles.

\[ \cos C = -\cos A\cos B + \sin A \sin B\cosh c . \]

Given the three angles, this determines each side, which proves AAA congruence (17.3). In particular, the circumradius of the regular heptagon in 17.5 comes from the right triangle formed by the centre, a vertex and a midpoint of a side: its angles are \(\pi/7\) at the centre and \(\pi/3\) at the vertex, and the formula \(\cosh(\text{hypotenuse}) = \cot A \cot B\) for right triangles (from this law with \(C = \pi/2\)) gives \(\cosh R = \cot\frac\pi7\cot\frac\pi3\).

Why anyone cares now. Beyond its role in the history of mathematics, hyperbolic geometry is everywhere in modern mathematics: in complex analysis (every Riemann surface with enough topology is a quotient of \(\mathbb{H}\)), in number theory (modular forms live on \(\mathbb{H}\), and the group \(\mathrm{SL}(2, \mathbb{Z})\) acts on it), in three-dimensional topology (Thurston's geometrization, Chapter 21), and in computer science, where trees, which grow exponentially like hyperbolic circles, embed in the hyperbolic plane with little distortion; machine-learning researchers now use such "hyperbolic embeddings" for hierarchical data.

Check yourself¶

1. Find the hyperbolic distance from \(i\) to \(4i\) in the upper half-plane, and check with the distance formula.

Along the vertical line, \(d = \ln 4 \approx 1.386\). Formula: \(\cosh d = 1 + \frac{9}{2 \cdot 1 \cdot 4} = \frac{17}{8}\), and \(\cosh(\ln 4) = \frac{4 + 1/4}{2} = \frac{17}{8}\). They agree.

2. Find the hyperbolic distance from \(2 + i\) to \(-2 + i\). Is the horizontal segment between them the shortest path?

\(\cosh d = 1 + \frac{16}{2} = 9\), so \(d = \operatorname{arcosh} 9 \approx 2.887\). The horizontal segment has hyperbolic length \(\int_{-2}^{2} dx/1 = 4\), longer. The shortest path is the semicircle centred at \(0\) through both points, of radius \(\sqrt5\), which rises to height \(\sqrt5\), where lengths are cheaper.

3. Show that \(f(z) = -1/z\) maps \(\mathbb{H}\) to itself and is an isometry. Where does it send \(i\) and \(2i\)?

It is \(\frac{0z - 1}{1z + 0}\), with real coefficients and \(ad - bc = 0 - (-1)(1) = 1\), so Theorem 17.1 applies. It sends \(i \mapsto -1/i = i\) and \(2i \mapsto -1/(2i) = i/2\). Check: \(d(i, 2i) = \ln 2\) and \(d(i, i/2) = \ln 2\).

4. A hyperbolic triangle has angles \(30^\circ\), \(45^\circ\) and \(60^\circ\). What is its area? Can a hyperbolic triangle have angles \(60^\circ\), \(60^\circ\), \(60^\circ\)?

Area \(= \pi - \frac{3\pi}{4} = \frac{\pi}{4} \approx 0.785\). No: angles \(60^\circ, 60^\circ, 60^\circ\) add up to \(180^\circ\), which would give area \(0\). Every hyperbolic triangle has angle sum strictly less than \(180^\circ\).

5. What is the area of a regular hyperbolic pentagon with all angles \(90^\circ\)? Which tiling does it make?

Area \(= (5 - 2)\pi - 5 \cdot \frac{\pi}{2} = \frac{\pi}{2}\). Four right angles fill a full turn, so four meet at each corner: the tiling \(\{5, 4\}\). Check: \(\frac15 + \frac14 = \frac{9}{20} < \frac12\).

6. Find the angle of parallelism \(\Pi(d)\) for \(d = 0.5\) and \(d = 2\), and the distance \(d\) at which \(\Pi(d) = 60^\circ\).

\(\cos\Pi = \tanh d\). For \(d = 0.5\): \(\tanh 0.5 \approx 0.4621\), \(\Pi \approx 62.5^\circ\). For \(d = 2\): \(\tanh 2 \approx 0.9640\), \(\Pi \approx 15.4^\circ\). For \(\Pi = 60^\circ\): \(\tanh d = \frac12\), \(d = \operatorname{artanh}\frac12 = \frac12\ln 3 \approx 0.549\) (check: \(\tan 30^\circ = 1/\sqrt3 = e^{-d}\) gives \(d = \ln\sqrt3\), the same).

7. A hyperbolic right triangle has legs \(2\) and \(3\). Find its hypotenuse, and compare with the Euclidean value.

\(\cosh c = \cosh 2 \cosh 3 \approx 3.7622 \times 10.0677 \approx 37.877\), so \(c \approx \operatorname{arcosh} 37.877 \approx 4.327\). Euclidean: \(\sqrt{13} \approx 3.606\). For large triangles, \(c \approx a + b - \ln 2\) (here \(5 - 0.693 = 4.307\)): the hypotenuse is nearly as long as the two legs together, because the legs spread apart exponentially fast.

8. Which of the tilings \(\{3, 3\}\), \(\{4, 4\}\), \(\{3, 7\}\), \(\{5, 3\}\), \(\{6, 4\}\), \(\{4, 6\}\) are spherical, Euclidean or hyperbolic?

Compare \(\frac1p + \frac1q\) with \(\frac12\). \(\{3, 3\}\): \(\frac23 > \frac12\), spherical (the tetrahedron). \(\{4, 4\}\): equal, Euclidean (squares). \(\{3, 7\}\): \(\frac{10}{21} < \frac12\), hyperbolic. \(\{5, 3\}\): \(\frac{8}{15} > \frac12\), spherical (the dodecahedron). \(\{6, 4\}\) and \(\{4, 6\}\): \(\frac{5}{12} < \frac12\), hyperbolic.

9. In the Poincaré disc, how far from the centre is the point at Euclidean distance \(0.5\)? At \(0.999\)? Which Euclidean radius is at hyperbolic distance \(10\)?

\(d = \ln\frac{1 + r}{1 - r}\). For \(0.5\): \(\ln 3 \approx 1.099\). For \(0.999\): \(\ln 1999 \approx 7.60\). For \(d = 10\): \(r = \tanh 5 \approx 0.99991\). In a drawing of the disc 10 cm across, every point more than 10 units from the centre lies within 0.005 mm of the edge.

10. A spaceship moves at \(0.9\) of light speed relative to Mars and launches a probe forward at \(0.9\) relative to itself. How fast is the probe relative to Mars? What are the rapidities?

\(\frac{0.9 + 0.9}{1 + 0.81} = \frac{1.8}{1.81} \approx 0.99448\). The rapidity of \(0.9\) is \(\operatorname{artanh} 0.9 \approx 1.472\); the rapidities add to \(2.944\), and \(\tanh 2.944 \approx 0.99448\). (Real spacecraft are far slower: at Mars-transfer speeds of tens of km/s, \(v\) is about \(10^{-4}\), and the correction is negligible.)

11. Show that a hyperbolic circle of radius \(\rho\) has circumference \(2\pi\sinh\rho\), using the disc model.

Put the centre at \(0\). The circle is the Euclidean circle of radius \(r = \tanh(\rho/2)\). Its hyperbolic length is \(\frac{2}{1 - r^2} \cdot 2\pi r = 2\pi \cdot \frac{2\tanh(\rho/2)}{1 - \tanh^2(\rho/2)} = 2\pi\sinh\rho\), by the double angle formula \(\sinh\rho = \frac{2\tanh(\rho/2)}{1 - \tanh^2(\rho/2)}\).

12. Prove that every ideal triangle (all three vertices on the boundary) can be moved by an isometry to the one with vertices \(-1\), \(1\), \(\infty\) in \(\mathbb{H}\). Conclude that all ideal triangles are congruent.

A real Möbius map sends any three distinct points of \(\mathbb{R} \cup \{\infty\}\) to any other three, by Theorem 13.9 of Chapter 13 with real numbers throughout (the map then has real coefficients). If its determinant is negative, follow it by \(z \mapsto -\bar z\) to correct the orientation, which still fixes the set \(\{-1, 1, \infty\}\). An isometry carries lines to lines, so the triangle goes onto the standard ideal triangle. Hence all ideal triangles are congruent, and all have area \(\pi\).

What you learned¶

The upper half-plane (\(ds = |dz|/y\)) and the Poincaré disc (\(ds = 2|dz|/(1 - |z|^2)\)) show hyperbolic geometry with true angles; lines are vertical lines and semicircles, or diameters and arcs orthogonal to the boundary.
Real Möbius maps of determinant \(1\) are isometries, and \(\cosh d = 1 + |z - w|^2/(2\,\mathrm{Im}\, z\,\mathrm{Im}\, w)\).
A triangle has area \(\pi - (\alpha + \beta + \gamma)\), so angle sums are less than \(\pi\) and ideal triangles have area \(\pi\).
Through a point off a line pass two limiting parallels and infinitely many ultraparallels; \(\cos\Pi(d) = \tanh d\).
\(\{p, q\}\) tiles the hyperbolic plane exactly when \(\frac1p + \frac1q < \frac12\); circles and tilings grow exponentially.
The hyperboloid \(\langle\mathbf{x}, \mathbf{x}\rangle = -1\) in Minkowski space is the hyperbolic plane; it is the space of velocities of special relativity, and rapidities add like distances.
Hyperbolic trigonometry is spherical trigonometry with \(\cosh\) and \(\sinh\): \(\cosh c = \cosh a\cosh b - \sinh a\sinh b\cos C\).

Next: Chapter 18 leaves distance and angle behind altogether, and asks what is left of geometry when only lines and their meetings count.