Non-Euclidean geometry: spherical and hyperbolic

MotivationH

The postulate that would not be proved

Euclid's Elements (c. 300 BCE) builds plane geometry from five postulates. The first four are short and obvious: a line can be drawn between any two points, extended indefinitely, a circle drawn with any centre and radius, and all right angles are equal. The fifth is long and different. In the equivalent form popularised by John Playfair (1795):

Parallel postulate. Through a point not on a given line there is exactly one line that does not meet it.

For two thousand years mathematicians tried to prove the fifth postulate from the other four — Ptolemy and Proclus in Alexandria, Ibn al-Haytham in Cairo, Omar Khayyam in Isfahan, Naṣīr al-Dīn al-Ṭūsī at Maragha, Giovanni Saccheri in Pavia, Johann Heinrich Lambert in Berlin. All of them failed, and many of their "proofs" quietly assumed something equivalent to what they were proving.

Meanwhile, navigators had been living in a non-Euclidean world all along. Fly from the North Pole straight down to the equator, turn left through $90^\circ$ , fly a quarter of the way around the Earth, turn left through $90^\circ$ again, and fly straight home: you have traced a triangle with three right angles. On a sphere the angles of a triangle add up to more than $180^\circ$ — and the excess measures its area.

In the 1820s, Carl Friedrich Gauss, János Bolyai and Nikolai Lobachevsky independently took the bold step: suppose the postulate is false, with many parallels through a point, and see what follows. The result was not a contradiction but a new geometry, the hyperbolic plane, in which triangles have angle sums less than $180^\circ$ . By the 1860s–1880s, Eugenio Beltrami, Felix Klein and Henri Poincaré had built concrete models of it — and with them proved that the parallel postulate can never be derived from the others. This lesson uses Euclidean geometry and its axioms, trigonometry, and vectors with the dot product; the cross product is recalled below.

IntuitionH

Three geometries, one picture

Drag the vertices of a triangle in each of the three geometries and watch its angle sum. In the plane it is always $180^\circ$ . On the sphere it is larger, by an amount proportional to the area. In the hyperbolic plane (drawn in the Poincaré disc, where straight lines appear as circular arcs meeting the boundary at right angles) it is smaller, again by an amount proportional to the area.

Triangles on curved surfaces

Start with the triangle with three right angles: it covers exactly one eighth of the sphere.

Drag a corner to move it; drag anywhere else to turn the sphere. The sides are arcs of great circles — the shortest paths on a sphere.

∠A90.0°
∠B90.0°
∠C90.0°

∠A + ∠B + ∠C = 270.0°

Angle excess (sum − 180°): 90.0° = 1.571 rad
Area, measured numerically: 1.571 R²
Share of the whole sphere (area ÷ 4πR²): 12.50 %

∠A + ∠B + ∠C − π = Area / R²On a sphere of radius R, the excess in radians equals the area divided by R². The area is computed separately, by numerical integration, so the match is a check rather than an assumption.

∠A + ∠B + ∠C − π = K · Area

Flat plane	K = 0	∠A+∠B+∠C = 180°
Sphere	K = 1/R²	∠A+∠B+∠C > 180°
Hyperbolic plane	K = −1	∠A+∠B+∠C < 180°

This is a first glimpse of the Gauss–Bonnet theorem: for a triangle whose sides are geodesics (shortest paths), the angle sum minus π equals the total curvature inside it. Here the curvature K is the same everywhere; on other surfaces you add up (integrate) K over the area.

The table under the widget records a single law for all three: angle sum $-\, \pi = K \cdot \text{area}$ , where $K$ is the curvature: $+1/R^2$ on a sphere of radius $R$ , $0$ in the plane and $-1$ in the standard hyperbolic plane. Small triangles have small area, so in all three geometries tiny triangles look Euclidean — which is why the Earth looks flat to someone standing on it.

DefinitionsH

Definitions

Definition— great circle, spherical distance

On the sphere $S_R = \{x \in \R^3 : |x| = R\}$ , a great circle is the intersection of $S_R$ with a plane through the centre $O$ . The distance between two points is the length of the shorter great-circle arc joining them: $R$ times the angle $\angle AOB$ , so for unit vectors $\cos d(A, B) = A \cdot B$ . Great circles are the "straight lines" (geodesics) of the sphere: the shortest paths between points. Two points are antipodal if they are opposite ends of a diameter.

Definition— cross product

For $u, v \in \R^3$ , the cross product is $u \times v = (u_2 v_3 - u_3 v_2,\ u_3 v_1 - u_1 v_3,\ u_1 v_2 - u_2 v_1)$ . It is perpendicular to both $u$ and $v$ , and its length is $|u|\,|v| \sin\theta$ , where $\theta$ is the angle between them. So for a great circle through $A$ and $C$ , the vector $C \times A$ is a normal to its plane.

Definition— spherical triangle, angle, lune

A spherical triangle $ABC$ has as sides the great-circle arcs $BC$ , $CA$ , $AB$ (each shorter than a semicircle). Its angle at $C$ is the angle between the tangent lines of the two sides at $C$ — equivalently, the angle between the planes $OCA$ and $OCB$ , or between their normals $C \times A$ and $C \times B$ . A lune of angle $\alpha$ is the region between two half great circles from a point to its antipode, meeting at angle $\alpha$ .

Definition— upper half-plane model

The hyperbolic plane in the upper half-plane model is $\mathbb H = \{z = x + iy \in \C : y > 0\}$ with lengths measured by $ds = |dz|/y$ : a curve $z(t)$ , $a \le t \le b$ , has hyperbolic length $\int_a^b |z'(t)| / \operatorname{Im} z(t)\, dt$ , and the distance $d(z, w)$ is the infimum of the lengths of curves from $z$ to $w$ . Areas are measured by $dx\,dy/y^2$ . Angles are the usual Euclidean angles (the model is conformal).

Definition— hyperbolic lines, parallels

The lines (geodesics) of $\mathbb H$ are the vertical half-lines $\{\operatorname{Re} z = c\}$ and the semicircles centred on the real axis. The real axis together with $\infty$ is the boundary at infinity: it is not part of $\mathbb H$ . Two lines that do not meet are parallel; they are limiting parallels if they meet at a point of the boundary and ultraparallel otherwise.

Definition— Poincaré disc model

The Poincaré disc is the open unit disc $\mathbb D = \{|w| < 1\}$ , with lines the diameters and the circular arcs meeting the unit circle at right angles. It is the image of $\mathbb H$ under the Cayley transform $w = (z - i)/(z + i)$ , which carries the half-plane model onto the disc model, lines to lines, preserving angles; the boundary circle is the boundary at infinity.

Definition— Möbius transformation, excess, defect

A Möbius transformation of $\mathbb H$ is a map $z \mapsto \dfrac{az + b}{cz + d}$ with $a, b, c, d \in \R$ and $ad - bc = 1$ . The angle excess of a spherical triangle with angles $\alpha, \beta, \gamma$ is $\alpha + \beta + \gamma - \pi$ ; the defect of a hyperbolic triangle is $\pi - (\alpha + \beta + \gamma)$ .

Worked examplesH

Examples

Example— Three right angles

The North Pole $N$ and the points $(0^\circ, 0^\circ)$ and $(0^\circ, 90^\circ\mathrm{E})$ on the equator form a spherical triangle with three right angles: angle sum $\frac{3\pi}{2}$ , excess $\frac{\pi}{2}$ . It is one of the eight congruent triangles into which the three coordinate planes cut the sphere, so its area is $\frac18 \cdot 4\pi R^2 = \frac{\pi}{2} R^2 = R^2 \cdot \text{excess}$ — Girard's theorem in its simplest case.

Example— Great circles versus parallels of latitude

Madrid and New York both lie at about $40^\circ$ N. The circle of latitude through them is not a great circle (its plane does not pass through the Earth's centre), so it is not the shortest route: the great-circle route bends north over the Atlantic, and long-distance flights follow such curving routes. The equator is the only circle of latitude that is a line of spherical geometry. Meridians are great circles, and any two of them meet at the poles: on the sphere there are no parallel lines at all.

Example— Distances in the half-plane

Along the imaginary axis, the curve $z(t) = it$ from $i$ to $iy$ ( $y > 1$ ) has length $\int_1^y dt/t = \ln y$ . As $y \to \infty$ or $y \to 0$ , the distance from $i$ grows without bound: the boundary is infinitely far away, even though it looks close in the picture. Horizontal distances shrink with height: the points $i$ and $1 + i$ are at hyperbolic distance $\operatorname{arcosh}\frac32 \approx 0.96$ , but $100i$ and $1 + 100i$ are only about $0.01$ apart.

The Poincaré disc. Through the point P there are infinitely many hyperbolic lines (solid arcs) that never meet the horizontal line L. The two dashed arcs are the limiting parallels: they meet L only at the boundary circle, which is infinitely far away.

ProofH

Spherical geometry

Take $R = 1$ (for radius $R$ , lengths scale by $R$ and areas by $R^2$ ). Let $A, B, C$ be unit vectors, with sides $a = d(B, C)$ , $b = d(C, A)$ , $c = d(A, B)$ and angles $\alpha, \beta, \gamma$ at $A, B, C$ .

Theorem— Theorem (spherical law of cosines)

$\cos c = \cos a \cos b + \sin a \sin b \cos \gamma$ .

Proof

The normals to the planes $OCA$ and $OCB$ are $C \times A$ and $C \times B$ , with lengths $|C \times A| = \sin b$ and $|C \times B| = \sin a$ , and $\gamma$ is the angle between them, so $(C \times A) \cdot (C \times B) = \sin b \sin a \cos\gamma$ . On the other hand, Lagrange's identity $(u \times v)\cdot(w \times x) = (u \cdot w)(v \cdot x) - (u \cdot x)(v \cdot w)$ with $C \cdot C = 1$ gives

(C \times A) \cdot (C \times B) = A \cdot B - (A \cdot C)(B \cdot C) = \cos c - \cos b \cos a .

Equating the two expressions gives the law.

∎

For small triangles, $\cos x \approx 1 - x^2/2$ and $\sin x \approx x$ ; substituting and keeping terms of second order turns the spherical law into the Euclidean law of cosines $c^2 = a^2 + b^2 - 2ab\cos\gamma$ .

Theorem— Lemma (area of a lune)

A lune of angle $\alpha$ on the unit sphere has area $2\alpha$ .

Proof

Rotations about the axis through the lune's two vertices preserve area and carry lunes to lunes of the same angle, so the area of a lune depends only on its angle, say $L(\alpha)$ ; it is additive ( $L(\alpha + \beta) = L(\alpha) + L(\beta)$ , placing lunes side by side) and increasing. Since $n$ lunes of angle $2\pi/n$ tile the sphere, $L(2\pi/n) = 4\pi/n$ , and by additivity $L(2\pi m/n) = 4\pi m/n$ . An increasing function that agrees with $\alpha \mapsto 2\alpha$ on all rational multiples of $2\pi$ equals it everywhere.

∎

Theorem— Theorem (Girard, 1629)

A spherical triangle on a sphere of radius $R$ has area $R^2(\alpha + \beta + \gamma - \pi)$ .

Proof

Take $R = 1$ and write $A', B', C'$ for the antipodes of $A, B, C$ . The three great circles $BC$ , $CA$ , $AB$ cut the sphere into eight triangles: $\Delta = ABC$ , its antipodal image $\Delta' = A'B'C'$ (of the same area, since $x \mapsto -x$ is an isometry), and six others. The two great circles through $A$ bound a lune of angle $\alpha$ containing $\Delta$ ; the great circle $BC$ cuts it into $\Delta$ and the triangle $A'BC$ . So

L_A = \Delta \cup A'BC, \quad L_B = \Delta \cup AB'C, \quad L_C = \Delta \cup ABC',

and their antipodal lunes are $L_A' = \Delta' \cup AB'C'$ , $L_B' = \Delta' \cup A'BC'$ , $L_C' = \Delta' \cup A'B'C$ . Each of the six other triangles appears in exactly one of these six lunes, while $\Delta$ appears in three and $\Delta'$ in three. Adding areas (boundaries have zero area), using the lemma:

2(2\alpha + 2\beta + 2\gamma) = 4\pi + 2\operatorname{area}(\Delta) + 2\operatorname{area}(\Delta') = 4\pi + 4 \operatorname{area}(\Delta),

so $\operatorname{area}(\Delta) = \alpha + \beta + \gamma - \pi$ .

∎

Consequences. Every spherical triangle has angle sum greater than $\pi$ . Two triangles with the same angles have the same area — and in fact are congruent: on the sphere there are no similar triangles of different sizes, and so no scale models. There are no rectangles (a quadrilateral with four right angles would have angle sum $2\pi$ , so zero area). And any two great circles meet — their planes through $O$ share a line, which meets the sphere in two antipodal points — so there are no parallels.

Spherical geometry is not, however, Euclidean geometry with the fifth postulate replaced: two antipodal points lie on infinitely many great circles (every meridian joins the poles), so the first postulate's uniqueness fails too, and lines are closed curves of finite length. Identifying each point with its antipode gives elliptic geometry on the projective plane, in which two points determine exactly one line and any two lines meet in exactly one point.

ProofH

Hyperbolic geometry

Theorem— Proposition (vertical segments are shortest)

For $0 < a < b$ , the hyperbolic distance from $ia$ to $ib$ is $\ln(b/a)$ , and the vertical segment is the only shortest curve.

Proof

For any curve $z(t) = x(t) + iy(t)$ from $ia$ to $ib$ ,

\int \frac{\sqrt{x'^2 + y'^2}}{y}\,dt \;\ge\; \int \frac{|y'|}{y}\,dt \;\ge\; \int \frac{y'}{y}\,dt = \ln b - \ln a,

with equality only if $x' \equiv 0$ and $y' \ge 0$ : the curve runs straight up the imaginary axis. The vertical segment has length exactly $\ln(b/a)$ .

∎

Theorem— Theorem (Möbius transformations are isometries)

If $g(z) = \frac{az + b}{cz + d}$ with $a, b, c, d$ real and $ad - bc = 1$ , then $g$ maps $\mathbb H$ onto itself and preserves hyperbolic lengths, distances, angles and areas.

Proof

Multiplying numerator and denominator by $\overline{cz + d}$ ,

\operatorname{Im} g(z) = \frac{\operatorname{Im}\big((az + b)(c\bar z + d)\big)}{|cz + d|^2} = \frac{(ad - bc)\operatorname{Im} z}{|cz + d|^2} = \frac{\operatorname{Im} z}{|cz + d|^2},

so $g$ maps $\mathbb H$ to $\mathbb H$ ; its inverse $z \mapsto \frac{dz - b}{-cz + a}$ is of the same form, so it is a bijection. Its derivative is $g'(z) = \frac{1}{(cz + d)^2}$ , so

\frac{|g'(z)|\,|dz|}{\operatorname{Im} g(z)} = \frac{|dz|/|cz + d|^2}{\operatorname{Im} z/|cz + d|^2} = \frac{|dz|}{\operatorname{Im} z}:

the length element is preserved, hence the length of every curve, hence distances. A holomorphic map with non-zero derivative preserves angles, and the area element $dx\,dy/y^2$ transforms by $|g'(z)|^2 / (\operatorname{Im} g / \operatorname{Im} z)^2 = 1$ .

∎

Möbius transformations map circles and lines of $\C$ to circles and lines and preserve angles, so they map hyperbolic lines to hyperbolic lines. They are plentiful: $z \mapsto z + t$ ( $t \in \R$ ), $z \mapsto \lambda z$ ( $\lambda > 0$ ) and $z \mapsto -1/z$ generate them all, and they can move any point to any point and any line to any line — in particular, any line to the imaginary axis. Together with the reflection $z \mapsto -\bar z$ , they form the whole isometry group of $\mathbb H$ . Combining this with the proposition: every hyperbolic line is the image of the imaginary axis, so the lines of $\mathbb H$ are exactly the vertical half-lines and semicircles centred on $\R$ , and any two points are joined by exactly one line.

Theorem— Theorem (distance formula)

For $z, w \in \mathbb H$ : $\displaystyle \cosh d(z, w) = 1 + \frac{|z - w|^2}{2 \operatorname{Im} z \operatorname{Im} w}$ .

Proof

Both sides are unchanged by Möbius transformations: the left because they are isometries, the right because, by the formulas above, $|g(z) - g(w)|^2 = \frac{|z - w|^2}{|cz + d|^2 |cw + d|^2}$ and $\operatorname{Im} g(z) \operatorname{Im} g(w) = \frac{\operatorname{Im} z \operatorname{Im} w}{|cz + d|^2 |cw + d|^2}$ . So we may move $z$ and $w$ onto the imaginary axis: $z = ia$ , $w = ib$ with $a < b$ . Then the right side is $1 + \frac{(b - a)^2}{2ab} = \frac{a^2 + b^2}{2ab} = \frac12\left(\frac ba + \frac ab\right) = \cosh \ln\frac ba$ , which is $\cosh d(ia, ib)$ by the proposition.

∎

Theorem— Theorem (infinitely many parallels)

Through a point $p$ not on a hyperbolic line $L$ there pass infinitely many lines that do not meet $L$ .

Proof

Move $L$ to the imaginary axis by an isometry; reflecting in it by $z \mapsto -\bar z$ if necessary and then scaling by $z \mapsto z/\operatorname{Re} p$ , we may assume $p = 1 + it$ with $t > 0$ . For $c \in \R$ , the semicircle centred at $c$ through $p$ has radius $r$ with $r^2 = (1 - c)^2 + t^2$ . It meets $L$ at a point $iy$ ( $y > 0$ ) exactly when $c^2 + y^2 = (1 - c)^2 + t^2$ , i.e. $y^2 = 1 + t^2 - 2c$ . So for every $c \ge \frac{1 + t^2}{2}$ there is no such $y$ : each of these infinitely many semicircles is a line through $p$ that misses $L$ . (The vertical line $\operatorname{Re} z = 1$ is another.)

∎

Exactly two of the lines through $p$ are limiting parallels (here $c = \frac{1+t^2}{2}$ , meeting $L$ at the boundary point $0$ , and the vertical line, meeting it at $\infty$ ); all lines between them are ultraparallel.

Theorem— Theorem (area of a hyperbolic triangle)

A hyperbolic triangle with angles $\alpha, \beta, \gamma$ has area $\pi - (\alpha + \beta + \gamma)$ . In particular every triangle has angle sum less than $\pi$ and area less than $\pi$ .

Proof

Step 1: one vertex at infinity. Consider a "triangle" with one vertex at $\infty$ (where its two vertical sides meet, at angle $0$ ) and two vertices on the unit circle at $e^{i(\pi - \alpha)}$ and $e^{i\beta}$ , where its angles are $\alpha$ and $\beta$ (the angle between a vertical line and the unit circle at $e^{i\theta}$ is $\theta$ or $\pi - \theta$ ). 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}} = \arcsin(\cos\beta) + \arcsin(\cos\alpha) = \pi - \alpha - \beta .

Any triangle with one vertex on the boundary can be moved to this position: send that vertex to $\infty$ , then use $z \mapsto \lambda z + t$ to make the opposite side the unit circle.

Step 2: the general triangle. Extend the side $AB$ beyond $B$ until it reaches the boundary at an ideal point $D$ . Let $\delta$ be the angle $\angle BCD$ . The triangle $ACD$ has angles $\alpha$ , $\gamma + \delta$ , $0$ , and the triangle $BCD$ has angles $\pi - \beta$ , $\delta$ , $0$ ; and $ABC = ACD \setminus BCD$ . By step 1,

\operatorname{area}(ABC) = \big(\pi - \alpha - \gamma - \delta\big) - \big(\pi - (\pi - \beta) - \delta\big) = \pi - \alpha - \beta - \gamma .

∎

Letting all three vertices run off to the boundary gives an ideal triangle, with all angles $0$ and area exactly $\pi$ — the largest possible. As on the sphere, two triangles with the same angles have the same area, and in fact are congruent: hyperbolic geometry has no similar triangles of different sizes. One also has the hyperbolic law of cosines $\cosh c = \cosh a \cosh b - \sinh a \sinh b \cos\gamma$ (proved, for example, in Anderson, ch. 5), which for small triangles again reduces to the Euclidean one.

ExplanationH

Why the parallel postulate is independent

To show that a statement $P$ cannot be proved from axioms $\Sigma$ , it is enough to find a model: a structure in which all of $\Sigma$ holds but $P$ fails. Any proof of $P$ from $\Sigma$ would be valid in every model, so it would prove something false in this one.

Make "Euclid's other axioms" precise as Hilbert's axioms (1899) for incidence, betweenness, congruence and continuity, without the parallel axiom — so-called neutral geometry. Interpret "point" as a point of $\mathbb H$ , "line" as a hyperbolic line, "between" and "congruent" via hyperbolic distance. One checks that every axiom of neutral geometry holds: two points lie on exactly one line (shown above); lines are infinitely long and separate the plane; there are enough isometries (the Möbius transformations and reflections) to move any segment or angle to any position, which gives the congruence axioms; and the continuity axioms hold because $\mathbb H$ is built from real numbers. (Full verifications are in Greenberg, ch. 7, and Hartshorne, ch. 7.) But the parallel postulate fails, by the theorem on infinitely many parallels.

Hence the parallel postulate is independent of the other axioms. More precisely, the argument is one of relative consistency: the model is constructed inside the real numbers — inside Euclidean geometry — so if Euclidean geometry is consistent, so is hyperbolic geometry. Neither geometry is "the true one" as a matter of logic; which one describes a given situation is a question about that situation. (See model theory for the general method, and first-order logic for what "proof" means here.)

Spherical geometry is not a model of neutral geometry: some of the other axioms fail there too (lines are finite and two antipodal points lie on many lines). Indeed, neutral geometry proves that the angle sum of a triangle is at most $\pi$ (the Saccheri–Legendre theorem), so only the hyperbolic alternative is compatible with Euclid's first four postulates and the continuity axioms.

Formal mathematicsF

Spheres and half-planes in Lean

For spherical geometry we use vectors in $\R^3$ (in Mathlib, Fin 3 → ℝ with the dot product ⬝ᵥ and the cross product ⨯₃). Lagrange's identity is checked coordinate by coordinate; the spherical law of cosines then follows exactly as in the human proof, with the angle $\gamma$ at $C$ specified through the normals $C \times A$ and $C \times B$ :

lagrange_identityLean 4Proved here from basic factsFormally verified

/-- **Lagrange's identity** for the cross product in `ℝ³`:
`(u × v) · (w × x) = (u · w)(v · x) - (u · x)(v · w)`, checked coordinate by coordinate.
(Mathlib has it, for any commutative ring, as `cross_dot_cross`.) -/
theorem lagrange_identity (u v w x : Fin 3 → ℝ) :
    (u ⨯₃ v) ⬝ᵥ (w ⨯₃ x) = (u ⬝ᵥ w) * (v ⬝ᵥ x) - (u ⬝ᵥ x) * (v ⬝ᵥ w) := by
  simp only [cross_apply, dotProduct, Fin.sum_univ_three]
  simp
  ring