Lesson 00

How to use this course

For two thousand years geometers tried to prove Euclid’s fifth postulate, the axiom about parallel lines, from the other axioms. Nikolai Lobachevsky was the first to publish the opposite conclusion: a full geometry in which the postulate is false, developed as carefully as Euclid’s own. He did it in a provincial Russian university, in Russian, in a journal few mathematicians read, and he spent the rest of his life defending and extending it while also running that university. János Bolyai reached the same geometry independently and published it in 1832. This course follows Lobachevsky’s life and works through the mathematics in roughly the order he developed it.

Prerequisites for Part II

Euclidean plane geometry at the level of Euclid’s Elements, Book I.
Trigonometry, including spherical trigonometry for Lesson 09, and hyperbolic functions \(\sinh,\cosh,\tanh\).
One-variable calculus; improper integrals for Lesson 09.
Helpful but optional: the upper half-plane model \(ds^2=k^2(dx^2+dy^2)/y^2\), used in several solutions to check results.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. Lobachevsky worked synthetically, with figures and limits. To keep the formulas short we use modern notation, with a length constant \(k\) (the space has curvature \(-1/k^2\)). Lobachevsky usually chose his unit of length so that \(k=1\).

Reference

Timeline

1792	Born 1 December (20 November, Old Style) in Nizhny Novgorod, one of three sons of Ivan Maksimovich Lobachevsky, a clerk in a land-surveying office, and Praskovia Alexandrovna.
c. 1800	After the father’s death his mother moves the family to Kazan.
1802–07	Kazan Gymnasium, on a government scholarship.
1807	Enters the new Kazan University. Studies under Martin Bartels, a friend of Gauss.
1811	Master’s degree in physics and mathematics.
1814	Lecturer (adjunct). Extraordinary professor 1816, full professor 1822.
1819–26	Curator M. L. Magnitsky’s repressive administration of the university.
1820–25	Dean of the physico-mathematical faculty.
1823	Completes the textbook Geometriya; it is printed in its original form only in 1909.
1825–35	Head librarian of the university.
1826	23 February (11 February, Old Style): asks the faculty to hear his new geometry, the Exposition succincte des principes de la géométrie. The text is lost.
1827	Elected rector of Kazan University.
1829–30	O nachalakh geometrii (On the principles of geometry) in the Kazanskii vestnik: the first published non-Euclidean geometry.
1830	Leads the university through a cholera epidemic.
1832	Marries Varvara Alexeyevna Moiseyeva. Bolyai’s Appendix appears in Hungary.
1834	The journal Syn otechestva publishes a lampoon of his geometry. Papers on algebra and on trigonometric series.
1835–38	Imaginary Geometry (1835), Application of Imaginary Geometry to Certain Integrals (1836), New Principles of Geometry (1835–38).
1837	French version of Imaginary Geometry in Crelle’s journal. Raised to the hereditary nobility.
1840	Geometrische Untersuchungen zur Theorie der Parallellinien, Berlin.
1842	Elected a corresponding member of the Göttingen Society of Sciences on Gauss’s recommendation. Great fire of Kazan.
1846	Removed as rector; assistant trustee of the Kazan educational district until 1855.
1855	Nearly blind, dictates Pangeometry in Russian and French.
1856	Dies 24 February (12 February, Old Style) in Kazan, aged 63.
1868	Eugenio Beltrami gives a concrete model of Lobachevsky’s plane.
1897	First Lobachevsky Prize of the Kazan Physico-Mathematical Society, to Sophus Lie.

Part I · Formation

Lesson 01

Nizhny Novgorod to Kazan, 1792–1811

Nikolai Ivanovich Lobachevsky was born on 1 December 1792 in Nizhny Novgorod, on the Volga. His father, Ivan Maksimovich, was a clerk in an office connected with land surveying. He died when Nikolai was seven, leaving his widow, Praskovia Alexandrovna, with three small sons and little money. Around 1800 she moved the family down the Volga to Kazan, the largest city of the empire’s eastern provinces.

In 1802 the three boys entered the Kazan Gymnasium on government scholarships. In 1804 Tsar Alexander I founded Kazan University as one of several new universities, and it drew its first students from the gymnasium. Lobachevsky entered in 1807, aged fourteen, intending at first to study medicine. He soon turned to mathematics and physics.

Bartels

The new university recruited professors from Germany. The most important for Lobachevsky was Martin Bartels, who had been an assistant teacher at Gauss’s primary school in Brunswick and had helped the boy Gauss into the Gymnasium. At Kazan, among other courses, he taught a history of mathematics based on Jean-Étienne Montucla’s, which Lobachevsky attended. He became the young man’s protector at a university where the students were closely supervised.

It has often been suggested that Gauss’s private ideas on parallels reached Lobachevsky through Bartels. Historians of Kazan, notably B. L. Laptev, examined the evidence and found that Bartels knew nothing of Gauss’s non-Euclidean results. The two discoveries were independent. In 1811 Lobachevsky received his master’s degree in physics and mathematics and stayed on at the university.

Lesson 02

A young professor, 1811–1826

Lobachevsky rose quickly: adjunct (lecturer) in 1814, extraordinary professor in 1816, full professor in 1822. He taught mathematics, physics and astronomy, and was dean of the physico-mathematical faculty from 1820 to 1825. From 1825 he was also head librarian, and for a time ran the observatory.

The Magnitsky years

In 1819 Mikhail Magnitsky was appointed curator of the Kazan educational district. He saw the university as a source of irreligion and imposed religious supervision of teaching, dismissals and censorship. Several professors left, Bartels among them. Lobachevsky stayed and clashed with Magnitsky repeatedly, while taking on more and more of the university’s administration. Magnitsky was dismissed in 1826.

The parallel postulate

Lobachevsky had been working on the theory of parallels since at least 1815. Like many before him, he first tried to prove Euclid’s fifth postulate. In 1823 he completed a geometry textbook, Geometriya, in which he remarked that no rigorous proof of the postulate had yet been found. The book was not printed; its original text was published only in 1909. Somewhere between 1823 and 1826 he changed direction. Instead of hunting for a contradiction in the negation of the postulate, he set out to build the geometry that negation describes.

His predecessors had come close. Giovanni Girolamo Saccheri (1733) and Johann Heinrich Lambert (1766) had derived many strange consequences from the negation, and Adrien-Marie Legendre had proved important partial results (Lesson 04). All three expected or claimed a contradiction. Lobachevsky’s step was to expect none.

Lesson 03

The lecture of 1826 and the paper of 1829

On 23 February 1826 (11 February in the Russian calendar), at a session of the physico-mathematical department, Lobachevsky asked to read a paper titled Exposition succincte des principes de la géométrie avec une démonstration rigoureuse du théorème des parallèles. It was passed to colleagues for a report that never appeared. The manuscript has not survived. Kazan now treats the date as the birthday of non-Euclidean geometry.

Its contents went into O nachalakh geometrii (On the principles of geometry), printed in instalments in the Kazanskii vestnik, the university’s journal, in 1829 and 1830. This is the first published account of a non-Euclidean geometry. It contains the definition of parallels, the angle of parallelism, the trigonometric formulas and applications to areas, volumes and integrals.

Reception

The paper was sent to the St Petersburg Academy of Sciences, the authority in Russian mathematics. Mikhail Ostrogradsky, its leading analyst, judged it unworthy of the Academy’s attention. In 1834 the magazine Syn otechestva printed an anonymous piece ridiculing it. Lobachevsky answered by writing more: three long works in Russian between 1835 and 1838, a French version in Crelle’s Journal für die reine und angewandte Mathematik in 1837, and a short book in German in 1840. That book reached Gauss.

DiscussionOstrogradsky was one of the best mathematicians in Russia. Suggest reasons why a strong analyst in 1830 might dismiss a geometry that contradicted Euclid, and what evidence would have been needed to convince him.

Part II · The mathematics

Lesson 04

Absolute geometry and the parallel problem

For the general reader. Euclid built geometry on five postulates. The fifth, about parallel lines, looked less obvious than the rest, and for centuries mathematicians tried to derive it from the other four. Before anyone could see that it was independent, they had to learn how much geometry survives without it.

Goals

State the parallel postulate and some of its equivalents.
Prove the Saccheri–Legendre theorem: in absolute geometry a triangle’s angles sum to at most \(\pi\).
Understand why “at most” is the best one can do without the fifth postulate.

Absolute geometry

Absolute (or neutral) geometry is what follows from the axioms of incidence, order, congruence and continuity, in the form later made rigorous by David Hilbert, without any parallel axiom. Euclid’s first 28 propositions belong to it. For example, Proposition I.16 says an exterior angle of a triangle exceeds each opposite interior angle, so any two angles of a triangle sum to less than \(\pi\). Proposition I.27 shows that parallels exist: two lines making equal alternate angles with a transversal never meet.

Playfair’s form of the parallel postulate

Given a line \(l\) and a point \(P\) not on \(l\), there is at most one line through \(P\) that does not meet \(l\).

In absolute geometry this is equivalent to Euclid’s fifth postulate, to the statement that some triangle has angle sum \(\pi\), to the existence of a rectangle, and to the existence of two similar triangles that are not congruent (John Wallis’s substitute). Lobachevsky replaced it by its negation: through \(P\) there are at least two lines not meeting \(l\).

Theorem (Saccheri 1733, Legendre 1794)

In absolute geometry, the sum of the angles of every triangle is at most \(\pi\). If one triangle has angle sum exactly \(\pi\), then every triangle does.

Proof of the first statement

Let \(ABC\) have angle sum \(S\) and let \(\alpha\) be the angle at \(A\). Let \(M\) be the midpoint of \(BC\) and extend \(AM\) to \(E\) with \(ME=AM\). Triangles \(AMB\) and \(EMC\) are congruent (side, angle, side), so triangle \(AEC\) has the same angle sum \(S\), and its angles at \(A\) and \(E\) add up to \(\alpha\). One of them is at most \(\alpha/2\). Repeating \(n\) times gives a triangle with angle sum \(S\) and one angle at most \(\alpha/2^n\). If \(S=\pi+\varepsilon\) with \(\varepsilon>0\), choose \(n\) with \(\alpha/2^n<\varepsilon\). The other two angles then sum to more than \(\pi\), contradicting I.16. Archimedes’ axiom is used to choose \(n\).

The defect of a triangle is \(\delta=\pi-(A+B+C)\ge0\). Defect is additive: if a triangle is cut into two by a segment from a vertex, the defects of the pieces add up to the defect of the whole. Legendre’s second statement follows from additivity by building larger and larger triangles from one with zero defect. So there are only two possibilities: every triangle has zero defect (Euclid), or every triangle has positive defect. Lobachevsky developed the second.

Exercise 4.1Prove that defect is additive: if \(D\) lies on side \(BC\) of triangle \(ABC\), then \(\delta(ABC)=\delta(ABD)+\delta(ADC)\).

Show solution

Write \(\angle ADB+\angle ADC=\pi\). Then \(\delta(ABD)+\delta(ADC)=2\pi-(\angle BAD+\angle B+\angle ADB)-(\angle DAC+\angle C+\angle ADC)=2\pi-\pi-(\angle A+\angle B+\angle C)=\delta(ABC)\), since \(\angle BAD+\angle DAC=\angle A\).

Exercise 4.2Assuming Playfair’s postulate, prove that every triangle has angle sum \(\pi\).

Show solution

Through \(A\) draw the line \(m\) making with \(AB\) an angle equal to \(\angle B\) on the side away from \(C\) (alternate angles). By I.27, \(m\) does not meet \(BC\). By Playfair it is the only such line, so the line through \(A\) making alternate angle \(\angle C\) with \(AC\) is also \(m\). The three angles at \(A\) along \(m\) are \(\angle B,\angle A,\angle C\) and fill a straight angle, so the sum is \(\pi\).

Exercise 4.3Show that the doubling construction keeps the angle sum: in the proof above, check that triangle \(AEC\) has the same angle sum as \(ABC\).

Show solution

From the congruence, \(\angle MEC=\angle MAB\) and \(\angle MCE=\angle MBA\). The angles of \(AEC\) are \(\angle EAC=\angle MAC\), \(\angle AEC=\angle MAB\) and \(\angle ACE=\angle ACB+\angle MBA\). Their sum is \(\angle MAC+\angle MAB+\angle C+\angle B=\angle A+\angle B+\angle C\).

Lesson 05

Parallels and the angle of parallelism

For the general reader. Stand at a point near a straight road and turn slowly from the direction that points straight at the road. At first your line of sight hits the road. At some critical angle it stops hitting it. Lobachevsky’s insight was that this critical angle depends on how far you are from the road, and that this dependence encodes the whole geometry.

Goals

State Lobachevsky’s definition of parallel lines.
Understand the function \(\Pi(x)\) and its formula.
Derive equivalent forms and limiting behaviour.

The definition

Let \(P\) be a point at distance \(x\) from a line \(l\), with foot of the perpendicular \(Q\). Rays from \(P\) either meet \(l\) or do not. By continuity there is a boundary ray on each side of \(PQ\). Lobachevsky called the lines containing these boundary rays the parallels to \(l\) through \(P\), one for each direction along \(l\). All lines through \(P\) lying between the two parallels (on the side away from \(l\)) also fail to meet \(l\); today they are called ultraparallel. In Euclid’s geometry the two parallels coincide.

Each parallel makes the same angle with \(PQ\), which depends only on \(x\). Lobachevsky called it the angle of parallelism and wrote it \(\Pi(x)\). Parallelism in a given direction is symmetric and transitive, and parallel lines approach each other without limit in that direction.

Theorem (Lobachevsky)

In the geometry where the parallel postulate fails, there is a constant \(k>0\) such that for every \(x>0\)

\[ \tan\frac{\Pi(x)}{2}=e^{-x/k}. \]

Thus \(\Pi\) decreases strictly from \(\pi/2\) (as \(x\to0\)) to \(0\) (as \(x\to\infty\)). Every acute angle is the angle of parallelism of exactly one distance.

Lobachevsky obtained the formula in his 1829 paper and gave a cleaner derivation in the Geometrische Untersuchungen of 1840, using the horosphere (Lesson 06). Equivalent forms, used constantly below, are

\[ \sin\Pi(x)=\frac{1}{\cosh(x/k)},\qquad \cos\Pi(x)=\tanh(x/k),\qquad \cot\Pi(x)=\sinh(x/k). \]

The last statement of the theorem is striking. In this geometry an angle determines a length. There is an absolute unit of length, just as there is an absolute unit of angle in Euclid’s geometry. Gauss had noticed this consequence privately years earlier.

Exercise 5.1Derive the three equivalent forms from \(\tan(\Pi/2)=e^{-x/k}\).

Show solution

Put \(t=\tan(\Pi/2)=e^{-u}\) with \(u=x/k\). Then \(\sin\Pi=\frac{2t}{1+t^2}=\frac{2}{e^{u}+e^{-u}}=\frac1{\cosh u}\) and \(\cos\Pi=\frac{1-t^2}{1+t^2}=\frac{e^u-e^{-u}}{e^u+e^{-u}}=\tanh u\). Dividing gives \(\cot\Pi=\sinh u\).

Exercise 5.2For which distance is \(\Pi(x)=\pi/4\)? For which is \(\Pi(x)=\pi/3\)?

Show solution

\(\tan(\pi/8)=\sqrt2-1\), so \(e^{-x/k}=\sqrt2-1\) and \(x=k\ln(1+\sqrt2)=k\operatorname{arsinh}1\approx0.881\,k\). For \(\pi/3\): \(\cot(\pi/3)=1/\sqrt3=\sinh(x/k)\), so \(x=k\operatorname{arsinh}(1/\sqrt3)=k\ln\sqrt3\approx0.549\,k\).

Exercise 5.3Show that \(\frac{\pi}{2}-\Pi(x)=\frac{x}{k}+O(x^3)\) as \(x\to0\). What does this mean physically if \(k\) is very large?

Show solution

\(\tan(\tfrac\pi2-\Pi)=\cot\Pi=\sinh(x/k)\), so \(\tfrac\pi2-\Pi=\arctan\sinh(x/k)=x/k+O(x^3)\). For distances much smaller than \(k\) the two parallels are almost the same line, so local measurements cannot easily distinguish the geometry from Euclid’s.

Exercise 5.4In the upper half-plane with \(ds^2=(dx^2+dy^2)/y^2\) (so \(k=1\)), take \(l\) to be the unit semicircle and \(P=(0,e^{x})\), at distance \(x\) from \(l\) along the imaginary axis. Verify Lobachevsky’s formula.

Show solution

The parallels through \(P\) are the geodesics from \(P\) to the endpoints \(\pm1\). The one ending at \(1\) is a circle centred at \((c,0)\) with \((1-c)^2=c^2+e^{2x}\), so \(c=(1-e^{2x})/2<0\). At \(P\) its tangent is perpendicular to the radius \((-c,e^{x})\), so the angle \(\theta\) it makes with the vertical satisfies \(\tan\theta=e^{x}/(-c)=2e^{x}/(e^{2x}-1)=1/\sinh x\). Hence \(\cot\theta=\sinh x\), which is \(\cot\Pi(x)\).

Lesson 06

Horocycles and the horosphere

For the general reader. A circle of enormous radius looks almost straight. In Euclid’s geometry, letting the radius go to infinity gives a straight line. In Lobachevsky’s geometry it gives a new curve, the horocycle. Its three-dimensional version, the horosphere, carries an ordinary Euclidean geometry, hidden inside the non-Euclidean space. Lobachevsky used it as a bridge between the two.

Goals

Define horocycles and horospheres.
State the theorem that the horosphere is Euclidean.
Use concentric horocycles to see where \(e^{-x/k}\) comes from.

Definitions

Fix a pencil of parallel lines, all parallel in the same direction. A horocycle (Lobachevsky’s “limit circle”) is a curve that cuts every line of the pencil at right angles. It is the limit of circles through a fixed point whose centres run off to infinity along one of the lines. Rotating a horocycle about one of its axes gives a horosphere (“limit sphere”). Bolyai called the same objects L-lines and F-surfaces.

Theorem (Lobachevsky; Bolyai 1832)

On a horosphere, with the horocycles lying on it playing the role of straight lines, the geometry is Euclidean: horocyclic triangles have angle sum \(\pi\), and the Euclidean trigonometric formulas hold for them.

The idea of the proof: horocycles on a horosphere behave like lines, since two points of the horosphere lie on exactly one of them. Each horocycle on the horosphere lies in a plane containing an axis, and two such horocycles fail to meet exactly when their planes are parallel. Because all the axes are parallel to one another, through a point of the horosphere there is exactly one horocycle missing a given one, so Playfair’s postulate holds there. Lobachevsky then combined Euclidean trigonometry on the horosphere with spherical trigonometry (which is the same in absolute geometry as in Euclid’s) to derive the formulas of his plane. Eugenio Beltrami rediscovered the Euclidean nature of the horosphere in 1868.

Lemma (concentric horocycles)

Let two horocycles belong to the same pencil, at distance \(x\) apart along the axes. Then arcs cut off between the same two axes have lengths \(s\) and \(s'\) with \(s'=s\,e^{-x/k}\), the shorter arc lying further in the direction of parallelism.

That the ratio is an exponential follows from a functional equation: ratios multiply when distances add, and the ratio is monotone in \(x\). The constant \(k\) is defined by this lemma. Lobachevsky combined the lemma with right triangles on the horosphere to reach \(\tan\frac{\Pi(x)}{2}=e^{-x/k}\).

Exercise 6.1In the upper half-plane with \(ds^2=k^2(dx^2+dy^2)/y^2\), the horizontal lines \(y=c\) are horocycles centred at \(\infty\). Verify the lemma for \(y=1\) and \(y=e^{t}\), with the vertical lines \(x=0\) and \(x=1\) as axes.

Show solution

Along \(y=c\), \(ds=k\,dx/c\), so the arcs have lengths \(k\) and \(k e^{-t}\). Along a vertical line \(ds=k\,dy/y\), so the distance between the horocycles is \(k t\). With \(x=kt\) the ratio is \(e^{-t}=e^{-x/k}\).

Exercise 6.2Show that a horocyclic arc of length \(s\) and the straight segment (chord) of length \(d\) joining its ends satisfy \(s=2k\sinh\frac{d}{2k}\).

Show solution

Take \(k=1\) and the arc from \((0,1)\) to \((s,1)\) on \(y=1\). The half-plane distance formula gives \(\cosh d=1+\frac{s^2}{2}\), so \(s^2=2(\cosh d-1)=4\sinh^2\frac d2\). Rescale lengths by \(k\) for general \(k\). The arc is always longer than the chord, as it should be.

Exercise 6.3In the upper half-space \(ds^2=k^2(dx^2+dy^2+dz^2)/z^2\), show that the plane \(z=1\) is a horosphere whose intrinsic geometry is Euclidean.

Show solution

The vertical lines are geodesics all parallel towards \(\infty\), and \(z=1\) meets them at right angles, so it is a horosphere. Restricted to \(z=1\) the metric is \(k^2(dx^2+dy^2)\), a flat Euclidean metric.

Lesson 07

Hyperbolic trigonometry

For the general reader. Trigonometry is the art of computing a triangle from some of its parts. Lobachevsky found the complete set of formulas for his geometry. They look like the familiar ones with ordinary sines and cosines of the sides replaced by hyperbolic functions, and they turn into the familiar ones for small triangles.

Goals

State the right-triangle formulas, the law of sines and the law of cosines.
Translate between Lobachevsky’s \(\Pi\)-notation and hyperbolic functions.
Check the Euclidean limit.

Lobachevsky wrote his formulas using \(\Pi\) of the sides. For example his law of sines reads \(\sin A\tan\Pi(a)=\sin B\tan\Pi(b)\). Since \(\tan\Pi(a)=1/\sinh(a/k)\), this is the modern form below.

Theorem (hyperbolic trigonometry)

For a right triangle with legs \(a,b\), hypotenuse \(c\) and acute angles \(A,B\) opposite \(a,b\):

\[ \cosh\tfrac ck=\cosh\tfrac ak\cosh\tfrac bk,\qquad \sin A=\frac{\sinh(a/k)}{\sinh(c/k)},\qquad \cos A=\frac{\tanh(b/k)}{\tanh(c/k)},\qquad \tan A=\frac{\tanh(a/k)}{\sinh(b/k)}. \]

For any triangle with sides \(a,b,c\) and opposite angles \(A,B,C\):

\[ \begin{aligned} \frac{\sin A}{\sinh(a/k)}&=\frac{\sin B}{\sinh(b/k)}=\frac{\sin C}{\sinh(c/k)},\\ \cosh\tfrac ck&=\cosh\tfrac ak\cosh\tfrac bk-\sinh\tfrac ak\sinh\tfrac bk\cos C,\\ \cos C&=-\cos A\cos B+\sin A\sin B\cosh\tfrac ck. \end{aligned} \]

The third formula has no Euclidean analogue: the angles of a triangle determine its sides. It expresses again the absolute unit of length. As \(k\to\infty\), \(\cosh(a/k)\approx1+a^2/2k^2\) and \(\sinh(a/k)\approx a/k\), and the first two general formulas become the Euclidean laws of sines and cosines.

Proof sketch

Lobachevsky placed a right triangle \(ABC\) in space, erected parallels to a line through its vertices, and cut them with the horosphere through \(A\). The result was a Euclidean right triangle on the horosphere and a spherical triangle at \(A\). The lemma of Lesson 06 related the lengths on the horosphere to the sides of \(ABC\). Equating the two descriptions gave the right-triangle formulas, and the general ones follow by dropping an altitude.

Exercise 7.1Expand \(\cosh(c/k)=\cosh(a/k)\cosh(b/k)\) to second order to recover Pythagoras. Then show that for every right triangle \(c^2>a^2+b^2\).

Show solution

To second order \(1+\frac{c^2}{2k^2}=1+\frac{a^2+b^2}{2k^2}\), which is Pythagoras. For the inequality put \(k=1\) and \(g(t)=\ln\cosh\sqrt t\). Then \(g'(t)=\tanh u/(2u)\) with \(u=\sqrt t\), and \(\tanh u/u\) is strictly decreasing, so \(g\) is strictly concave with \(g(0)=0\). Hence \(g(a^2)+g(b^2)>g(a^2+b^2)\), that is \(\cosh c=\cosh a\cosh b>\cosh\sqrt{a^2+b^2}\), and \(c^2>a^2+b^2\). Check: \(a=b=1\) gives \(\cosh c=\cosh^21\approx2.381\) and \(c\approx1.528>\sqrt2\).

Exercise 7.2An equilateral triangle has side \(a\) and angle \(\alpha\). Show \(\cos\alpha=\dfrac{\cosh(a/k)}{1+\cosh(a/k)}\) and find the limits of \(\alpha\) as \(a\to0\) and \(a\to\infty\).

Show solution

With \(k=1\) and \(C=\alpha\), the law of cosines gives \(\cosh a=\cosh^2a-\sinh^2a\cos\alpha\), so \(\cos\alpha=\frac{\cosh a(\cosh a-1)}{(\cosh a-1)(\cosh a+1)}=\frac{\cosh a}{\cosh a+1}\). As \(a\to0\), \(\cos\alpha\to\frac12\), so \(\alpha\to\pi/3\). As \(a\to\infty\), \(\cos\alpha\to1\), so \(\alpha\to0\).

Exercise 7.3Show that the dual law of cosines with \(C=\pi/2\) gives \(\cosh(c/k)=\cot A\cot B\), and deduce that a right triangle with \(A=B=\pi/4\) is impossible.

Show solution

With \(\cos C=0\) the formula for \(\cos C\) gives \(0=-\cos A\cos B+\sin A\sin B\cosh(c/k)\), so \(\cosh(c/k)=\cot A\cot B\). If \(A=B=\pi/4\) then \(\cosh(c/k)=1\) and \(c=0\). Right triangles need \(A+B<\pi/2\).

Lesson 08

Area, defect and the geometry of the sky

For the general reader. If space follows Lobachevsky’s geometry, big triangles have noticeably smaller angle sums than small ones. Lobachevsky was the first to treat the choice between geometries as a question for measurement. He used the largest triangles available, those formed by the Earth’s orbit and distant stars.

Goals

State the area–defect theorem and the formulas for circles.
Follow Lobachevsky’s astronomical argument.
Turn a parallax measurement into a lower bound on \(k\).

Theorem (area)

The area of a triangle with angles \(A,B,C\) is \(k^2(\pi-A-B-C)\). In particular no triangle has area greater than \(\pi k^2\). A circle of radius \(r\) has circumference \(2\pi k\sinh(r/k)\) and area \(4\pi k^2\sinh^2\!\big(\tfrac{r}{2k}\big)\).

Proportionality of area to defect follows from additivity of both (Exercise 4.1) together with a continuity argument. Lambert had already conjectured it from the hypothesis of the acute angle. The constant of proportionality is fixed by comparing with the circle formulas, which Lobachevsky derived by integration.

Is space Euclidean?

In the 1829 paper Lobachevsky took a triangle formed by two positions of the Earth on its orbit and a star, and asked how large its defect could be. He used published parallaxes, including those claimed for Sirius, Rigel and 28 Eridani, and found that the defect was smaller than any observational error. He concluded that for practical purposes Euclidean geometry is accurate, and that a difference, if any, could appear only on astronomical scales. The logic is sound even though the parallaxes he used were later found to be unreliable. Gauss is said to have measured a large terrestrial triangle in his survey of Hanover with the same question in mind.

Proposition (a parallax bound)

Let the Sun \(A\) and the Earth \(B\) be distance \(a\) apart and let a star \(S\) lie on the perpendicular to \(AB\) at \(A\). An observer who computes a parallax \(p=\tfrac\pi2-\angle ABS\) in the Euclidean way always finds \(\tan p>\sinh(a/k)\). Hence a star of measured parallax \(p\) gives \(k>a/\operatorname{arsinh}(\tan p)\approx a/p\).

Proof. The ray \(BS\) meets the perpendicular at \(A\), so \(\angle ABS<\Pi(a)\). Then \(p>\tfrac\pi2-\Pi(a)\) and \(\tan p>\cot\Pi(a)=\sinh(a/k)\). In Euclid’s geometry \(p\) is the angle at the star and tends to 0 for distant stars. In Lobachevsky’s it never drops below \(\tfrac\pi2-\Pi(a)\): every star, however far, shows a minimum parallax.

Exercise 8.1A star is measured to have parallax \(p=0.001''\) with baseline \(a=1\) astronomical unit. What lower bound on \(k\) follows, in astronomical units and parsecs?

Show solution

\(p=0.001\times4.848\times10^{-6}\) rad \(\approx4.85\times10^{-9}\). Then \(k>a/p\approx2.06\times10^{8}\) AU. Since one parsec is \(206\,265\) AU, \(k>1000\) pc. (Modern astrometry measures parallaxes much smaller still, and cosmology bounds spatial curvature far more tightly.)

Exercise 8.2Show that the area of a circle eventually exceeds that of any triangle, and compare circumference with area for large \(r\).

Show solution

\(4\pi k^2\sinh^2(r/2k)\to\infty\), while triangles have area below \(\pi k^2\); a circle exceeds this once \(\sinh(r/2k)>\tfrac12\). For large \(r\), circumference \(\approx\pi k e^{r/k}\) and area \(\approx\pi k^2e^{r/k}\), so area \(\approx k\times\) circumference. Almost all of the disc lies within a distance of order \(k\) of its boundary.

Exercise 8.3An ideal triangle has all three vertices at infinity. Show that all ideal triangles have the same area, and compute it.

Show solution

Parallel lines meet “at infinity” at angle 0, so all three angles are 0 and the area is \(k^2(\pi-0)=\pi k^2\), the supremum of triangle areas.

Lesson 09

Imaginary geometry and definite integrals

For the general reader. Lobachevsky called his system “imaginary geometry” because its formulas are those of geometry on a sphere whose radius is an imaginary number. He hoped this algebraic link would convince doubters. He also used his geometry as a calculating tool: computing a length, area or volume in two different ways gave him the values of many definite integrals.

Goals

Pass from spherical to hyperbolic trigonometry by an imaginary radius.
Prove Lobachevsky’s integral formula.
Meet the Lobachevsky function and hyperbolic volumes.

The imaginary sphere

On a sphere of radius \(R\), the spherical law of cosines is \(\cos\frac cR=\cos\frac aR\cos\frac bR+\sin\frac aR\sin\frac bR\cos C\). Put \(R=ik\). Since \(\cos(-iu)=\cosh u\) and \(\sin(-iu)=-i\sinh u\), the formula becomes \(\cosh\frac ck=\cosh\frac ak\cosh\frac bk-\sinh\frac ak\sinh\frac bk\cos C\), the hyperbolic law. Every formula of spherical trigonometry turns into one of Lobachevsky’s in this way. He presented this in Imaginary Geometry (1835) and its French version (1837) as evidence that his system was consistent if spherical geometry was. The argument was suggestive rather than a proof; a proper consistency proof came only with Beltrami’s model in 1868.

An integral formula

Theorem (Lobachevsky’s integral formula)

Let \(f\) be continuous on \(\mathbb R\), with \(f(x+\pi)=f(x)\) and \(f(\pi-x)=f(x)\). Then

\[ \int_0^\infty\frac{\sin^2x}{x^2}\,f(x)\,dx=\int_0^{\pi/2}f(x)\,dx. \]

Proof sketch. The hypotheses make \(f\) even, so the left side is half the integral over \(\mathbb R\). Cut \(\mathbb R\) into intervals \([n\pi,(n+1)\pi]\), substitute \(x=t+n\pi\), and use periodicity of \(\sin^2\) and \(f\):

\[ \int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}f(x)\,dx=\int_0^\pi\sin^2t\,f(t)\sum_{n\in\mathbb Z}\frac{1}{(t+n\pi)^2}\,dt=\int_0^\pi f(t)\,dt, \]

by the partial-fraction expansion \(\sum_n(t+n\pi)^{-2}=1/\sin^2t\). Finally \(\int_0^\pi f=2\int_0^{\pi/2}f\) by the symmetry \(f(\pi-x)=f(x)\).

Volumes

Lobachevsky computed volumes of hyperbolic polyhedra and expressed them through the function \(L(x)=-\int_0^x\ln\cos u\,du\). In modern work one uses the closely related Lobachevsky function

\[ Л(\theta)=-\int_0^\theta\ln|2\sin u|\,du. \]

By a formula of John Milnor, an ideal tetrahedron in hyperbolic space (curvature \(-1\)), whose dihedral angles at opposite edges are \(\alpha,\beta,\gamma\) with \(\alpha+\beta+\gamma=\pi\), has volume \(Л(\alpha)+Л(\beta)+Л(\gamma)\). Such volumes are now central in the geometry of three-manifolds and in knot theory, through the work of William Thurston.

Exercise 9.1Use the integral formula to evaluate \(\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}\,dx\) and \(\displaystyle\int_0^\infty\frac{\sin^2x\,|\cos x|}{x^2}\,dx\).

Show solution

With \(f=1\): \(\int_0^{\pi/2}1\,dx=\pi/2\). With \(f=|\cos x|\), which has period \(\pi\) and satisfies \(|\cos(\pi-x)|=|\cos x|\): \(\int_0^{\pi/2}\cos x\,dx=1\).

Exercise 9.2Show that \(Л(\pi/2)=0\) and that \(Л\) is odd and has period \(\pi\).

Show solution

\(Л(\pi/2)=-\frac\pi2\ln2-\int_0^{\pi/2}\ln\sin u\,du=-\frac\pi2\ln2+\frac\pi2\ln2=0\), using the classical value \(\int_0^{\pi/2}\ln\sin u\,du=-\frac\pi2\ln2\). Oddness: the integrand is even. Periodicity: since \(|\sin u|\) has period \(\pi\), \(Л(\theta+\pi)-Л(\theta)=-\int_0^\pi\ln|2\sin u|\,du\), and by the symmetry \(\sin(\pi-u)=\sin u\) this equals \(2Л(\pi/2)=0\).

Exercise 9.3The regular ideal tetrahedron has all dihedral angles \(\pi/3\). Express its volume using \(Л\), and estimate it numerically.

Show solution

\(V=3Л(\pi/3)\approx1.01494\). It is the largest volume of any tetrahedron in hyperbolic 3-space, and twice it is the volume of the complement of the figure-eight knot.

Lesson 10

Analysis and algebra

For the general reader. Lobachevsky was a working analyst and algebraist as well as a geometer. Two of his ideas outside geometry deserve to be better known: a modern definition of a function, and a fast method for finding the roots of polynomial equations.

Goals

See Lobachevsky’s general concept of function and his distinction between continuity and differentiability.
Learn the root-squaring method for polynomial roots.

Functions

In a paper on the convergence of trigonometric series, in the first volume of the Kazan Uchenye zapiski (1834), Lobachevsky defined a function as a correspondence assigning a number to each number in a range, given in any way at all, not necessarily by a formula. Peter Gustav Lejeune Dirichlet gave the same definition independently in 1837, and it is usually credited to him. Lobachevsky also distinguished carefully between continuity and differentiability, at a time when many authors assumed that a continuous function must have a derivative almost everywhere.

Root squaring

In his Algebra (1834) Lobachevsky proposed separating the roots of a polynomial by repeatedly squaring them. Germinal Dandelin had published the same idea in 1826, and Karl Gräffe developed it independently in 1837. It is called the Dandelin–Gräffe method in the West and Lobachevsky’s method in Russia.

Root-squaring step

If \(p(x)=\prod_{i=1}^n(x-r_i)\), then \((-1)^np(x)p(-x)=q(x^2)\), where \(q(y)=\prod_{i=1}^n(y-r_i^2)\). If \(|r_1|>|r_2|>\cdots>|r_n|\) and \(q_m(y)=y^n+c_1y^{n-1}+\cdots+c_n\) is the polynomial after \(m\) steps, then \(|r_1|^{2^m}\approx|c_1|\) and \(|r_i|^{2^m}\approx|c_i/c_{i-1}|\), with relative error tending to 0 as \(m\to\infty\).

The reason: after \(m\) steps the roots \(r_i^{2^m}\) are widely separated, so each elementary symmetric function is dominated by its largest term, \(c_i\approx\pm r_1^{2^m}\cdots r_i^{2^m}\).

Exercise 10.1Apply two root-squaring steps to \(p(x)=x^2-3x+2\) and estimate the larger root from the result.

Show solution

\(p(x)p(-x)=(x^2+2)^2-9x^2=x^4-5x^2+4\), so \(q_1(y)=y^2-5y+4\), roots \(1,4\). Again: \((y^2+4)^2-25y^2=y^4-17y^2+16\), so \(q_2(y)=y^2-17y+16\), roots \(1,16\). Then \(|r_1|^4\approx17\), \(|r_1|\approx17^{1/4}\approx2.03\); the true root is 2, and further steps improve the estimate.

Exercise 10.2Give an example of a function on \([-1,1]\) that is continuous everywhere but not differentiable at infinitely many points.

Show solution

Take \(f(x)=x\,|\sin(\pi/x)|\) for \(x\ne0\) and \(f(0)=0\). It is continuous (at 0 because \(|f(x)|\le|x|\)), and not differentiable at each \(x=1/n\), where \(\sin(\pi/x)\) changes sign and the absolute value creates a corner.

Part III · Life and legacy

Lesson 11

Rector of Kazan, 1827–1846

In 1827, a year after Magnitsky’s fall, the professors elected Lobachevsky rector. He was 34. The new curator of the district, Mikhail Musin-Pushkin, worked closely with him, and Lobachevsky held the office for nineteen years. He oversaw the construction of a new library, an observatory, laboratories and medical buildings, and he founded the university’s scientific journal, the Uchenye zapiski, in 1834. His 1828 speech On the most important subjects of education set out a humane ideal: education should develop all a person’s faculties, not just memory and obedience.

He was a hands-on administrator. He led the university through the cholera epidemic of 1830 and through the great fire of 1842, which destroyed much of Kazan, and directed the rebuilding afterwards. In 1832 he married Varvara Alexeyevna Moiseyeva. They had many children, of whom seven survived to adulthood. In 1837 he was raised to the hereditary nobility for his services to the state.

Gauss

The German book of 1840 reached Gauss, who had held similar views privately since the 1810s and published none of them. In a letter to the astronomer Heinrich Christian Schumacher of 28 November 1846 he wrote that Lobachevsky had followed a different route from his own, handling the subject masterfully and in a genuinely geometric spirit. Gauss, who was learning Russian in his sixties, read Lobachevsky’s Russian writings, and in 1842 he proposed him as a corresponding member of the Göttingen Society of Sciences. The election was the nearest thing to recognition of his geometry that Lobachevsky received in his lifetime. Gauss never commented on non-Euclidean geometry in print. His opinion became known only when his letters to Schumacher were published in the 1860s.

DiscussionLobachevsky spent nineteen years running a university while developing a geometry almost no one accepted. Did the administrative work help or hinder the mathematics? Consider security, reputation, time, and access to a printing press.

Lesson 12

Blindness, Pangeometry, recognition

In 1846 the ministry ended Lobachevsky’s rectorship, officially on grounds of health and length of service. He became assistant trustee of the Kazan educational district, a post that kept him in administration but out of the university. The loss hurt him deeply. His eldest son, his favourite, died soon after, his finances were strained by an estate he had bought at Slobodka, and a progressive illness slowly took his sight.

He kept working. In 1855, by then blind, he dictated a final summary of his geometry, Pangeometry, in Russian and in French. It covers the trigonometry, the computation of lengths, areas and volumes, and the application to definite integrals. It appeared in the Uchenye zapiski in 1855 and in a French edition in 1856. Lobachevsky died in Kazan on 24 February 1856, aged 63, and was buried in the Arskoe cemetery.

Recognition

The tide turned a decade later. Gauss’s letters to Schumacher, published from 1860, showed that the greatest mathematician of the age had taken the new geometry seriously. Jules Hoüel translated the 1840 book into French in 1866. In 1868 Eugenio Beltrami showed that Lobachevsky’s plane geometry is realized on surfaces of constant negative curvature and in what is now the Beltrami–Klein model, proving it as consistent as Euclid’s. Felix Klein, who attended Karl Weierstrass’s 1870 seminar in Berlin on the subject, fitted it into projective geometry in 1871 and named it hyperbolic geometry. Henri Poincaré found it in the theory of automorphic functions in 1882. George Bruce Halsted’s English translation of the Geometrische Untersuchungen followed in 1891.

Legacy

Mathematics: hyperbolic geometry is now one of the central structures of mathematics, in complex analysis, group theory, the topology of three-manifolds and the study of Riemann surfaces. The Lobachevsky function computes hyperbolic volumes.
Philosophy of science: Lobachevsky was the first to publish the view that the geometry of physical space is an empirical question. Bernhard Riemann’s 1854 lecture generalized the idea, and Einstein’s general relativity made it physics.
Honours: the Kazan Physico-Mathematical Society’s Lobachevsky Prize, first awarded in 1897 to Sophus Lie; Lobachevsky State University of Nizhny Novgorod; the lunar crater Lobachevskiy and the asteroid 1858 Lobachevskij. William Kingdon Clifford called him the Copernicus of geometry.

Final projectWrite a two-page account, for a strong undergraduate, of how Lobachevsky would have computed \(\Pi(x)\) from the horosphere, and explain why the existence of an absolute unit of length is a consequence of the failure of the parallel postulate.

Reference

Connections

People linked to Nikolai Lobachevsky across this collection. A name without a link has no page here yet. See Nikolai Lobachevsky in the network of all 337 lives →

Teachers and advisors

Martin Bartels

Students

Nikolai Brashman

Influenced by

Influenced

Colleagues and collaborators

Carl Friedrich Gauss 1840-1855
Peter Gustav Lejeune Dirichlet

Rivals

János Bolyai
Mikhail Ostrogradsky

Also mentioned in this life

Fields

geometry 32 lives
non-euclidean geometry 3 lives
analysis 20 lives
algebra 29 lives

On CoursesHub

Non-Euclidean Geometry · History of Mathematics & Physics

Reference

Sources and further reading

Biography

J. J. O’Connor and E. F. Robertson, Nikolai Ivanovich Lobachevsky, MacTutor History of Mathematics.
B. A. Rosenfeld, Lobachevsky, Nikolai Ivanovich, Dictionary of Scientific Biography.
Nikolai Lobachevsky, Wikipedia.
H. K. Strick, Nikolai Ivanovich Lobachevsky, MacTutor (translated from Spektrum der Wissenschaft).
V. F. Kagan, N. Lobachevsky and his Contribution to Science, Foreign Languages Publishing House, Moscow, 1957.
B. L. Laptev, N. I. Lobachevsky (in Russian), Kazan, 1976.

Mathematics

N. I. Lobachevsky, Geometrische Untersuchungen zur Theorie der Parallellinien, Berlin, 1840. English translation by G. B. Halsted, Geometrical Researches on the Theory of Parallels, Austin, 1891.
N. I. Lobachevsky, Pangeometry, edited and translated by A. Papadopoulos, European Mathematical Society, 2010.
R. Bonola, Non-Euclidean Geometry: A Critical and Historical Study of its Development, 1912; Dover reprint (includes Halsted’s translations of Lobachevsky and Bolyai).
J. Gray, Ideas of Space: Euclidean, Non-Euclidean and Relativistic, 2nd ed., Oxford, 1989.
J. J. Gray, Worlds Out of Nothing: A Course in the History of Geometry in the 19th Century, 2nd ed., Springer, 2010.
A. Papadopoulos and G. Théret, On Lobachevsky’s trigonometric formulae, arXiv:1311.6034.
J. Milnor, Hyperbolic geometry: the first 150 years, Bull. Amer. Math. Soc. 6 (1982) 9–24.
M. J. Greenberg, Euclidean and Non-Euclidean Geometries, 4th ed., W. H. Freeman, 2008.
E. Beltrami, Saggio di interpretazione della geometria non-euclidea, Giornale di Matematiche 6 (1868).