Part IX · Horizons Chapter 58 of 60

Non-Euclidean geometries

A travel diary through three worlds. In the plane the angles of a triangle add up to exactly 180°, on a sphere to more, and in Lobachevsky’s world to less, and there infinitely many parallels pass through a point.

Years 1–2 60 minutes

Builds on: 57 · Topology

You will learn

why the fifth postulate can’t be derived from the others, and how a geometry in which it fails is built
find the area of a triangle on a sphere and in the Lobachevsky plane from its angles
explain why the Earth has no map without distortion, and why GPS clocks need relativity

Line IX · Horizons

You are here: 58 · Non-Euclidean geometries

Interchanges 16 · Euclid's Elements

The last chapter left us holding an octagon. If its sides are glued by the word $aba^{-1}b^{-1}cdc^{-1}d^{-1}$, you get a pretzel with two holes, and all eight corners meet at one point. For there to be neither a gap nor a fold around it, each angle must be $45^\circ$. But in the ordinary plane any octagon has angles adding up to $1080^\circ$, and it can’t have eight angles of $45^\circ$. Another plane is needed. The search for it began two thousand years before anyone needed a pretzel, with Euclid’s fifth postulate.

This chapter is a travel diary through three worlds. The first is the ordinary plane, where the angles of a triangle add up to exactly $180^\circ$. The second is the sphere, where they add up to more. The third is the Lobachevsky plane, where they add up to less, and that’s where our octagon will be found.

Entry one. At home: the postulate they wanted to prove

In the chapter on the Elements the fifth postulate looked like a theorem Euclid had forgotten to prove. We also found there that it’s equivalent to Playfair’s axiom: at most one parallel passes through a point off a line. The first twenty-eight propositions of the Elements manage without it, and among them are the exterior angle theorem and its corollary: any two angles of a triangle add up to less than $180^\circ$. For two thousand years mathematicians tried to derive the postulate from the other four.

Absolute geometry is everything that can be proved without the fifth postulate, from the other axioms (in rigorous form, from Hilbert’s axioms without the parallel axiom). The name was coined by János Bolyai: the theorems of absolute geometry hold whatever the fate of the fifth postulate.

The most persistent assault was made by the Italian Jesuit Giovanni Saccheri. In 1733 his book Euclid Freed of Every Flaw came out. Saccheri built the quadrilateral that now bears his name and went through the possibilities for its upper angles.

A Saccheri quadrilateral is a quadrilateral $ABCD$ whose angles $A$ and $B$ at the base are right angles and whose sides $AD$ and $BC$ are equal. Its upper angles $C$ and $D$ are equal to each other: the quadrilateral is symmetric about the perpendicular bisector of its base.

If the upper angles are right angles, the fifth postulate holds. Saccheri honestly refuted the hypothesis of the obtuse angle. But the hypothesis of the acute angle wouldn’t give way: he derived consequence after consequence from it, each stranger than the last, but found no contradiction, and in the end declared it “repugnant to the nature of the straight line”. There was no contradiction in it. Saccheri, without knowing it, was the first to explore a non-Euclidean geometry, and he renounced it. One obstacle, though, he set up securely, and he proved it without the fifth postulate.

In absolute geometry the angles of any triangle add up to at most $180^\circ$.

The angle of the triangle at the vertex $A$.

The part of it cut off by the median $AE$ on the side of $B$; after the construction it will be the angle at the new vertex $F$.

The other part, on the side of $C$: the angle of the new triangle at the vertex $A$.

Example: for $\alpha = 60^\circ$ one of the parts is at most $30^\circ$. After ten repetitions of the construction, a triangle with the same angle sum will have an angle of at most $60^\circ / 2^{10} \approx 0.06^\circ$.

The trick is to rebuild the triangle without changing its angle sum, but making one of its angles half as large. The construction is the same as in the proof of the exterior angle theorem in chapter 16.

Let the angles of the triangle $ABC$ add up to $\alpha + \beta + \gamma$.

Mark the midpoint $E$ of the side $BC$ and extend the median $AE$ beyond $E$ by the same length: $EF = AE$.

The triangles $ABE$ and $FCE$ are congruent by two sides and the included angle: $BE = CE$, $AE = FE$, and the angles at $E$ are vertically opposite. So $\angle CFE = \angle BAE = \alpha'$ and $\angle FCE = \angle ABE = \beta$. The fifth postulate isn’t needed: the congruence of triangles and of vertical angles is proved without it.

Add up the angles of the triangle $AFC$. At $A$ the angle $\alpha''$ remains, at $F$ the angle $\alpha'$, at $C$ the angle $\gamma + \beta$. The sum $\alpha'' + \alpha' + \gamma + \beta = \alpha + \beta + \gamma$ is the same. And the two angles $\alpha'$ and $\alpha''$ together make up $\alpha$, so one of them is at most $\frac{\alpha}{2}$.

Repeat the construction $n$ times, each time from the vertex with the smallest angle. We get a triangle with the same angle sum and an angle of at most $\frac{\alpha}{2^n}$. If the sum were $180^\circ + \varepsilon$ with $\varepsilon > 0$, take $n$ such that $\frac{\alpha}{2^n} < \varepsilon$ (this can be done by the Archimedean axiom). Then the other two angles together exceed $180^\circ$, which is forbidden by the corollary of the exterior angle theorem (chapter 16). So the sum is at most $180^\circ$.

Two options remain: the angle sum is exactly $180^\circ$ or less. Saccheri and later Adrien-Marie Legendre showed that if even one triangle has sum exactly $180^\circ$, then all do, and then the fifth postulate holds. So a non-Euclidean geometry, if it exists, is a world where every triangle has angle sum less than $180^\circ$. Johann Heinrich Lambert, in a work of 1766 (printed only in 1786), noticed that under this hypothesis the shortfall from $180^\circ$ behaves like an area, and suggested that such a geometry might live on an “imaginary sphere”. He was closer to the truth than anyone before Lobachevsky.

Entry two. The sphere: no parallels

Before setting off to where angles fall short, let’s visit where there are too many. We live on a ball, and navigators have long known its geometry.

A great circle is the section of a sphere by a plane through its centre: the equator, any meridian. The shortest path between two points of a sphere runs along an arc of a great circle. Lines that play the role of straight lines on a surface, being shortest paths between their nearby points, are called geodesics. On a sphere they’re the great circles.

A plane from Astana to Vancouver flies almost over the North Pole: the northernmost point of the great circle is at $84^\circ$ N. This route is about $8760$ km long, while keeping a constant compass bearing would take almost $11{,}800$ km. Even Madrid and New York, lying at almost the same latitude, are joined by a great circle that arcs north, up to $46^\circ$, and saves about $170$ km.

Drag the vertices over the globe; the scene can be rotated. “Three right angles” is a triangle with vertices at the pole and on the equator. Choose a flight: the solid line is the great circle, the dashed one the constant bearing.

Take great circles as the straight lines. Any two of them meet, and in two diametrically opposite points, so there are no parallels at all. The angles of a triangle add up to more than $180^\circ$: the triangle with vertices at the North Pole and at two points of the equator a quarter-turn apart has three right angles. It takes up an eighth of the Earth’s surface, about $63.8$ million square kilometres.

The sphere breaks more than the fifth postulate. Infinitely many “lines” pass through two opposite points, lines are finite and closed, and even the exterior angle theorem fails. So the sphere isn’t a model of absolute geometry, and the Saccheri–Legendre theorem doesn’t apply to it. A geometry in which lines are finite and any two meet was described by Bernhard Riemann in a famous lecture of 1854. And on a sphere the excess of the angle sum over $180^\circ$ measures area exactly.

The spherical excess of a triangle is the difference $\alpha + \beta + \gamma - \pi$ between its angle sum and that of a plane triangle. A lune of angle $\alpha$ is the part of the sphere between two halves of great circles that leave a point at angle $\alpha$ and meet again at the opposite point.

The area of a triangle on a sphere of radius $R$ with angles $\alpha$, $\beta$, $\gamma$ (in radians) is $S = (\alpha + \beta + \gamma - \pi)R^2$.

The angle at the vertex $A$: the angle between the tangents to the sides, in radians.

The angle at the vertex $B$.

The angle at the vertex $C$.

The angle sum of a plane triangle; the bracket is the excess.

The square of the radius: on a large sphere the same angle sum gives a larger area.

Example: for the triangle with three right angles the excess is $\frac{3\pi}{2} - \pi = \frac{\pi}{2}$, and $S = \frac{\pi}{2}R^2$, exactly an eighth of the sphere’s area $4\pi R^2$. For the Earth, $R = 6371$ km, that’s $\frac{\pi}{2} \cdot 6371^2 \approx 63.8$ million km². A small triangle is almost flat: a triangle on Earth with sides $69$, $85$ and $107$ km has an excess of about $15''$, fifteen seconds of arc.

The trick is to cover the sphere with three pairs of lunes and count how many times each piece is covered.

The area of a lune is proportional to its angle. Lunes with equal angles can be brought together by a rotation, a lune of angle $\alpha + \beta$ is cut into lunes of angles $\alpha$ and $\beta$, and the area grows with the angle, so it depends linearly on the angle. An angle of $2\pi$ gives the whole sphere, of area $4\pi R^2$ (chapter 20), and a lune of angle $\alpha$ has area $\frac{\alpha}{2\pi} \cdot 4\pi R^2 = 2\alpha R^2$.

Extend the sides $AB$ and $AC$ to great circles. They cut out a lune of angle $\alpha$ containing our triangle, and an identical opposite lune at the point opposite $A$. Together, $4\alpha R^2$.

Likewise the sides at the vertex $B$ give a pair of lunes of area $4\beta R^2$.

And the sides at $C$ a pair of area $4\gamma R^2$.

The six lunes cover the whole sphere. The triangle $ABC$ is covered three times, once by the lune at each vertex; the opposite triangle of the same area also three times; the rest of the sphere exactly once.

The total area of the lunes is the area of the sphere plus two extra covers of both triangles: $4(\alpha + \beta + \gamma)R^2 = 4\pi R^2 + 4S$. Hence $S = (\alpha + \beta + \gamma - \pi)R^2$.

The angles of a triangle on a sphere of radius $3$ are $90^\circ$, $90^\circ$ and $60^\circ$. Find its area (you may answer in terms of $\pi$, for example 2pi).

The excess is $90^\circ + 90^\circ + 60^\circ - 180^\circ = 60^\circ = \frac{\pi}{3}$. The area is $\frac{\pi}{3} \cdot 3^2 = 3\pi \approx 9.42$. Such a triangle is a piece of an “orange segment” between two meridians $60^\circ$ apart, from the pole to the equator.

Entry three. Kazan, 1826

In February 1826 Nikolai Lobachevsky, a professor at Kazan University, presented a paper on the foundations of geometry to the physics and mathematics section. The manuscript hasn’t survived. In 1829–1830 his work “On the Foundations of Geometry” appeared in the journal Kazan Messenger. Lobachevsky did what Saccheri and Lambert hadn’t dared to: he took the negation of the fifth postulate as an axiom and set about building a new geometry without waiting for a contradiction. The contradiction never came, and the geometry turned out harmonious, with its own trigonometry. His colleagues in Kazan didn’t understand it, and the reviews in the capital were mocking.

Independently of him, the Hungarian János Bolyai reached the same conclusions; his “Appendix” came out in 1832 as a supplement to his father’s textbook. The father sent the book to his friend Gauss, who replied that to praise it would be to praise himself: he had come to the same ideas many years before. Gauss really hadn’t published them; in an 1829 letter to Bessel he admitted he feared “the clamour of the Boeotians”, the noise of those who don’t understand.

Through a point off a line pass at least two lines that don’t meet the given line.

The Lobachevsky plane, or hyperbolic plane, is a plane in which all the axioms of absolute geometry hold, with Lobachevsky’s axiom in place of the fifth postulate.

How can you see such a plane if it doesn’t fit into ours? Draw a map of it, as one draws a map of the Earth. The most convenient map was devised by Henri Poincaré in 1882.

The Poincaré disk is a map of the Lobachevsky plane inside a disc. The points are the points inside the disc; the edge doesn’t belong to the plane. The lines are the diameters of the disc and arcs of circles perpendicular to its edge. Angles between lines on the map are the same as on the plane itself, while distances are distorted: the closer to the edge, the more they’re compressed.

Reflections in this model are inversions in the circles on which its lines lie; they were promised by the chapter on symmetry. Inversion preserves angles and takes circles perpendicular to the edge to circles of the same kind, so lines go to lines. The distance from the centre to a point of the map lying at distance $r$ from the centre is given by the model’s formula.

Grows from $1$ to $2$ as the point of the map goes from the centre to the edge.

Tends to zero at the edge: the denominator is what makes the edge infinitely far away.

Example: a point halfway to the edge, $r = 0.5$, is at distance $\ln 3 \approx 1.10$; at $r = 0.9$ it’s already $\ln 19 \approx 2.94$, at $r = 0.99$, $\ln 199 \approx 5.29$. You can never reach the edge. Here the curvature of the plane is $-1$; more on curvature in entry six.

Entry four. A world in a disc

Put a line $a$ and a point $P$ off it on the map. Each line through $P$ either meets $a$ or doesn’t, and the boundary between these two families is two special lines that meet $a$ only on the edge of the disc, at points at infinity.

Turn the line through $P$ by moving the point $Q$: the orange line meets $a$, the blue one doesn’t. The two highlighted lines are the limiting parallels, and the thin lines are part of the infinite family of lines through $P$ that don’t meet $a$.

The angle of parallelism $\Pi(d)$ is the angle between the perpendicular dropped from the point $P$ to the line $a$ and the limiting parallel through $P$, the line that meets $a$ only at a point at infinity. It depends only on the length $d$ of the perpendicular.

In the Lobachevsky plane of curvature $-1$ the angle of parallelism at distance $d$ satisfies $\tan\frac{\Pi(d)}{2} = e^{-d}$.

The angle between the perpendicular to the line and the limiting parallel.

The distance from the point to the line.

Example: at $d = 1$ we get $\tan\frac{\Pi}{2} = e^{-1} \approx 0.368$ and $\Pi \approx 40.4^\circ$; at $d = 2$ it’s already $15.4^\circ$. Near the line $\Pi(d)$ is almost $90^\circ$, as in Euclid: on a small scale the Lobachevsky plane can hardly be told from the ordinary one.

Let’s arrange the picture conveniently. By a motion, take the line $a$ to the horizontal diameter and the foot of the perpendicular from $P$ to the centre of the model. Then $P$ lies on the vertical radius: $P = (0, t)$.

The perpendicular from $P$ to $a$ is the vertical radius $PO$, and by the model’s formula its length is $d = \ln\frac{1 + t}{1 - t}$.

The limiting parallel through $P$ goes to the end $U = (1, 0)$ of the line $a$. It’s an arc of a circle perpendicular to the edge at the point $U$, so the radius $ZU$ of this circle is tangent to the edge, and its centre $Z$ lies on the vertical line $x = 1$: $Z = (1, k)$.

The point $P$ lies on the same circle: $ZP = ZU = k$, that is, $1 + (k - t)^2 = k^2$. Hence $k = \frac{1 + t^2}{2t}$.

The angle $\Pi$ is between the downward direction $PO$ and the tangent to the arc at $P$. The tangent is perpendicular to the radius $ZP$, which points along $(-1, t - k) = \frac{1}{2t}(-2t, t^2 - 1)$, so it points along $(1 - t^2, -2t)$. So $\cos\Pi = \frac{2t}{1 + t^2}$ and $\sin\Pi = \frac{1 - t^2}{1 + t^2}$.

By the half-angle formula $\tan\frac{\Pi}{2} = \frac{\sin\Pi}{1 + \cos\Pi} = \frac{1 - t^2}{(1 + t)^2} = \frac{1 - t}{1 + t} = e^{-d}$.

How many lines that don’t meet a line $a$ pass through a point $P$ off it in the Lobachevsky plane?

Lobachevsky’s axiom requires at least two, and it follows that there are infinitely many: every line through $P$ making an angle between $\Pi(d)$ and $180^\circ - \Pi(d)$ with the perpendicular to $a$ doesn’t meet $a$. The extreme ones are the two limiting parallels.

At what distance from a line is the angle of parallelism $60^\circ$ (curvature $-1$)? The answer can be written with a logarithm, for example ln(2), or as a number to three decimal places.

By Lobachevsky’s formula $e^{-d} = \tan 30^\circ = \frac{1}{\sqrt3}$, so $d = \ln\sqrt3 = \frac{\ln 3}{2} \approx 0.549$. The closer the point is to the line, the closer the angle of parallelism is to a right angle.

In the same disc it’s easy to see why triangles here fall short of angles.

The angles of any triangle in the Lobachevsky plane add up to less than $180^\circ$.

The trick is to put one vertex at the centre of the model: then two sides become straight segments, and the triangle can be compared with an ordinary one.

Motions of the Lobachevsky plane take any point to any point, so let the vertex $A$ be the centre of the model. Motions change neither angles nor lengths, so the angle sum stays the same.

The sides $AB$ and $AC$ lie on diameters, that is, they’re drawn as straight segments. The angle $\alpha$ between them on the map is the same as in reality.

The side $BC$ is an arc of a circle with centre $Z$ perpendicular to the edge: $|Z|^2 = 1 + \rho^2$, where $\rho$ is the circle’s radius. For points $X$ of the circle $|X - Z|^2 = \rho^2$, so $X \cdot Z = \frac{1 + |X|^2}{2}$. For the midpoint $M$ of the chord $BC$ we get $M \cdot Z = \frac{2 + |B|^2 + |C|^2}{4} > \frac{|B|^2 + 2B \cdot C + |C|^2}{4} = |M|^2$, because $B \cdot C < 1$. So $Z$ lies beyond the chord $BC$, on the side away from $A$, and the arc $BC$ bulges towards $A$.

So the arc runs inside the ordinary triangle $ABC$, and its tangents at the points $B$ and $C$ turn inwards from the chord. The angles $\beta$ and $\gamma$ are smaller than the angles of the ordinary triangle with the same vertices (dashed).

The angle at $A$ coincides, and the other two are smaller than in the ordinary triangle, whose angles add up to $180^\circ$. So $\alpha + \beta + \gamma < 180^\circ$.

And here’s the answer to the promise in the chapter on the Elements: Lobachevsky’s geometry has no squares.

In the Lobachevsky plane the angles of a quadrilateral add up to less than $360^\circ$. In particular, there are no quadrilaterals with four right angles: no rectangles and no squares.

Cut the quadrilateral along a diagonal.

The diagonal $AC$ cuts the quadrilateral $ABCD$ into the triangles $ABC$ and $ACD$.

The angles at $B$ and $D$ each go entirely to one of the triangles, while the diagonal splits the angles at $A$ and $C$ into two parts, one in each triangle. So the angle sum of the quadrilateral equals the angle sums of the two triangles together.

Each of these sums is less than $180^\circ$, so together they’re less than $360^\circ$. Four right angles would give exactly $360^\circ$. That’s why the upper angles of a Saccheri quadrilateral are acute: the “hypothesis of the acute angle” turned out to be the truth of this world.

Drag the vertices of the triangle: the angle sum is always less than $180^\circ$, and the shortfall is the area. Switch to the tilings and drag the picture: you’ll be walking across the Lobachevsky plane.

Entry five. Area is the shortfall

Lambert guessed right: the bigger the triangle, the more it falls short of $180^\circ$, and the shortfall adds up like area. If a triangle is cut in two by a segment from a vertex, the two angles at the foot of the segment add up to $180^\circ$, and the shortfalls of the parts add up to the shortfall of the whole.

The defect of a triangle in the Lobachevsky plane is the shortfall $\pi - (\alpha + \beta + \gamma)$. A triangle can also have vertices on the edge of the model, at infinity; the angle at such a vertex is zero. A triangle all three of whose vertices are at infinity is called ideal.

In the Lobachevsky plane of curvature $-\frac{1}{R^2}$ the area of a triangle with angles $\alpha$, $\beta$, $\gamma$ is $S = (\pi - \alpha - \beta - \gamma)R^2$.

The angle at the vertex $A$.

The angle at the vertex $B$.

The angle at the vertex $C$.

The angle sum of a plane triangle; the bracket is the defect.

The scale linked to the curvature $-\frac{1}{R^2}$; in the Poincaré disk above, $R = 1$.

Example: for a triangle with angles $30^\circ$, $45^\circ$ and $60^\circ$ at $R = 1$ the area is $\pi - \frac{3\pi}{4} = \frac{\pi}{4} \approx 0.79$. The area of any triangle is less than $\pi R^2$, the area of an ideal one, however far apart the vertices move. The formula is the same as on the sphere, with the opposite sign.

Gauss’s proof. We’ll need three properties of the model: motions preserve areas; by a motion any point can be taken to any point, and any three points at infinity to any three; and the area of an ideal triangle is $\pi$ (its computation is in the box below).

Write $\Delta(\theta)$ for the area of a triangle with one ordinary vertex, where the angle is $\pi - \theta$, and two vertices at infinity. All such triangles with the same $\theta$ are congruent: put the ordinary vertex at the centre, the sides become diameters at angle $\pi - \theta$, and the third side is determined uniquely.

Take an ideal triangle and a point $P$ inside it, and join $P$ to the three vertices at infinity. This gives three triangles of type $\Delta$ with angles $\varphi_1 + \varphi_2 + \varphi_3 = 2\pi$ at $P$, that is, $\theta_1 + \theta_2 + \theta_3 = 3\pi - 2\pi = \pi$. So $\Delta(\theta_1) + \Delta(\theta_2) + \Delta(\theta_3) = \pi$ for any positive $\theta_i$ adding up to $\pi$. For $u(\theta) = \Delta(\theta) - \theta$ this means $u(\theta_1) + u(\theta_2) + u(\theta_3) = 0$. Comparing the triples $(x, y, \pi - x - y)$ and $\bigl(\pi - x - y, \frac{x + y}{2}, \frac{x + y}{2}\bigr)$, we get $u(x) + u(y) = 2u\bigl(\frac{x + y}{2}\bigr)$: a continuous $u$ is linear, $u(\theta) = k\theta + m$. As $\theta \to 0$ the triangle flattens into a line, and $u \to 0$, so $m = 0$; then the first equation gives $k\pi = 0$. We get $\Delta(\theta) = \theta$.

Now the triangle $ABC$. Extend its sides in turn: $AB$ beyond $B$, $BC$ beyond $C$, $CA$ beyond $A$, all the way to the edge. The three points at infinity form an ideal triangle, which is cut into $ABC$ and three triangles of type $\Delta$: at the vertex $A$ the angle between the extension of $CA$ and the side $AB$ is $\pi - \alpha$, and likewise at $B$ and $C$.

Add up the areas: $\pi = S + \Delta(\alpha) + \Delta(\beta) + \Delta(\gamma) = S + \alpha + \beta + \gamma$, that is, $S = \pi - \alpha - \beta - \gamma$. For curvature $-\frac{1}{R^2}$ all areas are multiplied by $R^2$.

Why the area of an ideal triangle is π

It’s easier to compute in another map, the upper half-plane $y > 0$, where the lines are vertical rays and semicircles centred on the $x$-axis, and the area of an element $dx\,dy$ is $\frac{dx\,dy}{y^2}$ (the Poincaré disk is carried to this map by an angle-preserving fractional linear map). All ideal triangles are congruent, so take the triangle with vertices $-1$, $1$ and infinity: above the semicircle $x^2 + y^2 = 1$ between the rays $x = \pm 1$. Its area is

$$\int_{-1}^{1}\int_{\sqrt{1 - x^2}}^{\infty}\frac{dy}{y^2}\,dx = \int_{-1}^{1}\frac{dx}{\sqrt{1 - x^2}} = \arcsin 1 - \arcsin(-1) = \pi.$$

In the $\{5, 4\}$ tiling from the widget above, four regular pentagons meet at each vertex, so their angles are right angles. Find the area of such a pentagon (curvature $-1$; you may answer in terms of $\pi$).

The diagonals from one vertex cut the pentagon into three triangles, and the defects add up: the area is $(5 - 2)\pi - 5 \cdot \frac{\pi}{2} = 3\pi - \frac{5\pi}{2} = \frac{\pi}{2} \approx 1.571$. In the ordinary plane a regular pentagon with right angles is impossible: its angles there are $108^\circ$.

Now back to the octagon. A regular octagon with angles of $45^\circ$ does exist in the Lobachevsky plane: such octagons, eight at each vertex, tile the plane, and you can see them in the widget above. Its area, by the same formula, is the octagon’s defect, $(8 - 2)\pi - 8 \cdot \frac{\pi}{4} = 4\pi$. Gluing its sides, we get a pretzel on which Lobachevsky’s geometry holds. And the area $4\pi$ equals $-2\pi\chi$ for the pretzel with $\chi = -2$: a special case of the Gauss–Bonnet theorem, which links the curvature of a surface with its Euler characteristic.

The main question remained: what if Lobachevsky’s geometry hides a contradiction after all, just very far away? In 1868 Eugenio Beltrami built the first model, inside Euclidean geometry, like the Poincaré disk. All of Lobachevsky’s axioms hold in the model, and if a contradiction followed from them, it would translate into a contradiction in ordinary Euclidean geometry. So the fifth postulate doesn’t follow from the others: two thousand years of attempts to prove it were doomed. Beltrami’s models were developed by Felix Klein (1871) and Poincaré (1882).

Entry six. What an ant can feel

Can an ant living on a surface tell a sphere from a plane and from the Lobachevsky plane without going outside? Yes: let it trace a circle, measuring off the same distance $r$ from a point in all directions and counting the length of the resulting line. In the plane it gets $2\pi r$. On a sphere of radius $R$ the points at distance $r$ from the pole lie on a parallel of radius $R\sin\frac{r}{R}$, and the length is smaller: $2\pi R\sin\frac{r}{R}$. In the Lobachevsky plane it’s larger: $2\pi R\sinh\frac{r}{R}$. For a circle of radius $1000$ km on the Earth the shortfall is about $26$ km out of $6283$.

The Gaussian curvature of a surface at a point is a number $K$ that shows how much shorter or longer the circles around that point are than flat ones.

The length of a circle of radius $r$ in the plane.

The length of a circle of radius $r$ on the surface: all the points at distance $r$ from the given one, measured along the surface.

The shortfall grows like the cube of the radius; we divide by it to get a number.

Example: on a sphere $2\pi r - 2\pi R\sin\frac{r}{R} \approx 2\pi R \cdot \frac{(r/R)^3}{6} = \frac{\pi r^3}{3R^2}$ (chapter 30: $\sin x = x - \frac{x^3}{6} + \dots$), so $K = \frac{1}{R^2}$. The plane has $K = 0$, the Lobachevsky plane $K = -\frac{1}{R^2}$. The formula was found by Joseph Bertrand, Charles Diguet and Victor Puiseux in 1848.

Gauss’s main discovery was the “remarkable theorem” (Theorema Egregium) of 1827. The curvature of a surface lying in space can also be computed from outside: take the two principal directions of bending at a point and multiply the curvatures of the sections along them. Gauss proved that this product coincides with the curvature the ant measures from inside. So it doesn’t change if the surface is bent without stretching. That’s why you hold a slice of pizza bent across: flat dough has curvature zero, and if the slice is curved in one direction, it can’t be curved in the other, so the tip doesn’t droop. For the same reason a sheet of paper can be rolled into a tube, but it can’t be wrapped round a ball without creases.

No piece of a sphere can be mapped onto the plane so that all distances measured along the sphere are preserved.

Compare the lengths of circles: on a sphere they’re shorter than in the plane, while a map without distortion of distances must preserve lengths as well.

Take a point $N$ on a sphere of radius $R$ and all the points at distance $r$ from it, measured along the sphere. They form a circle of radius $R\sin\frac{r}{R}$; in the drawing it’s seen from the side, squashed. Its length is $2\pi R\sin\frac{r}{R}$.

Since $\sin x < x$ for $x > 0$, this length is less than $2\pi r$.

Suppose a map preserves distances. Then the circle goes to points at distance exactly $r$ from the image of $N$, that is, onto a flat circle of radius $r$, with different points going to different ones. A continuous one-to-one image of a circle in a circle is the whole circle: otherwise it would be an arc, and a circle isn’t homeomorphic to an arc (chapter 57). The length of a curve is the limit of the lengths of inscribed polygonal lines, and the map preserves the lengths of the links. So the length of the image is $2\pi R\sin\frac{r}{R}$.

But the length of a flat circle of radius $r$ is $2\pi r$, and we got a smaller number. A contradiction.

No map of a piece of a sphere preserves both angles and ratios of areas.

If a map preserves angles, then at each point small figures are only rotated and stretched by the same factor $k$ in all directions (we used this property of conformal maps in chapter 34). The areas of small figures are multiplied by $k^2$. If moreover the map preserves ratios of areas, then $k^2$ is the same at all points, and after rescaling we can take $k = 1$. Then small segments keep their length, and with them the lengths of all curves and all distances along the surface. The previous theorem forbids this.

Cartographers choose what to sacrifice. In 1859 the French cartographer Nicolas Auguste Tissot proposed drawing on a map what identical small circles on the globe turn into. On the Mercator map (1569), which preserves angles, the circles stay circles but swell towards the poles: that’s why Greenland looks comparable to Africa, although it’s about $14$ times smaller. On an equal-area map the circles keep their area but get squashed.

Identical circles of radius $7^\circ$ on the globe and their images in four projections. Drag the highlighted circle: below you can see how many times areas are stretched and by how much angles are distorted.

Entry seven. Clocks in orbit

In 1915 Albert Einstein completed the general theory of relativity. In it gravity is the curvature of spacetime, and the geometry that Gauss and Riemann had built for surfaces became the language of physics. This isn’t an abstract matter: you carry its consequence in your pocket.

GPS satellites fly at an altitude of about $20{,}200$ km. There gravity is weaker than at the Earth’s surface, and the clocks on a satellite run faster: in a day they gain about $45$ microseconds. Because of the satellite’s speed, about $3.9$ km/s, special relativity makes them lose about $7$ microseconds. The net result is about $38$ microseconds a day. Light travels about $11$ km in that time, and without a correction the positioning error would build up by about $10$ km a day. That’s why the reference frequency of the satellite clocks is set low in advance: $10.22999999543$ MHz instead of $10.23$ MHz, and in orbit they keep in step with clocks on the ground.

The fifth postulate isn’t a theorem but a choice. Remove it, and three independent geometries remain: the plane, the sphere and the Lobachevsky plane. They’re distinguished by curvature, which can be measured from inside: by the angle sum of a triangle or by the length of a circle.

Where next

The Lobachevsky plane has a property the sphere lacks: nearby lines diverge. Send two lines out from one point at a small angle: in the plane the distance between them grows in proportion to the distance travelled, but in Lobachevsky’s geometry it grows exponentially. In 1898 Jacques Hadamard studied billiard balls rolling on surfaces of negative curvature and realised that a tiny error in the direction of the shot quickly makes the ball’s path unpredictable. The rules of motion are simple and exact, yet the distant future can’t be predicted. It was the first chaotic system in mathematics. Why the same happens with the weather, with a population of rabbits and with a double pendulum is the subject of the next chapter.