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János Bolyai

Bolyai János

A course in two strands: the life of the Hungarian army engineer who, at twenty, built a geometry without Euclid’s parallel postulate, and the 24-page Appendix of 1832 in which he set out an “absolutely true science of space”.

Born 15 December 1802, Kolozsvár Died 27 January 1860, Marosvásárhely Appendix published 1832 Level advanced undergraduate → graduate
A pencil of parallel lines running to one point at infinity, and the curves Bolyai called L-lines (horocycles) cutting them all at right angles. Between two lines, arcs of successive L-lines shrink by the factor e−x/k.
Lesson 00

How to use this course

János Bolyai published one work in his lifetime: a 24-page appendix, in Latin, to his father’s mathematics textbook. In it he did two things at once. He developed the geometry in which Euclid’s parallel postulate fails, independently of Nikolai Lobachevsky, who had published the same geometry in Russian a few years earlier. He also proved theorems that hold whether the postulate is true or false, a part of geometry he called the absolute science of space. This second idea is Bolyai’s own, and much of Part II is devoted to it.

Prerequisites for Part II
  • Euclidean plane and solid geometry, and spherical trigonometry.
  • Hyperbolic functions and one-variable calculus.
  • For Lesson 09, basic algebra of fields. For Lesson 10, some familiarity with Riemannian metrics helps.
  • The companion course on Lobachevsky covers the angle of parallelism and hyperbolic trigonometry in more detail. The two can be read in either order.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. We follow Bolyai’s own notation where it helps: \(\Sigma\) is the system in which Euclid’s parallel axiom holds, \(S\) the system in which it fails, and \(k\) the constant of \(S\) (Bolyai wrote \(i\)), so that the space has curvature \(-1/k^2\).

Reference

Timeline

1796–99His father, Farkas Bolyai, studies at Göttingen and becomes a close friend of Carl Friedrich Gauss.
1802Born 15 December in Kolozsvár, Transylvania (now Cluj-Napoca, Romania), son of Farkas Bolyai and Zsuzsanna Benkő.
1804Farkas becomes professor of mathematics, physics and chemistry at the Calvinist College in Marosvásárhely (now Târgu Mureș).
1816Farkas asks Gauss to take János into his household as a pupil; Gauss does not take up the request.
1817Completes his studies at the Marosvásárhely college.
1818–22Imperial and Royal Engineering Academy, Vienna.
1820His father warns him to leave the theory of parallels alone.
1823Joins the army engineering corps; posted to Temesvár. 3 November: writes to his father that he has created a new world out of nothing.
1825Shows his father a manuscript on absolute space.
1826Posted to Arad. Gives a German draft to his former teacher Wolter von Eckwehr; it is never returned.
1831Farkas sends Gauss a printed copy of the Appendix (June); the letter goes unanswered.
1832The Appendix scientiam spatii absolute veram exhibens appears with volume 1 of Farkas’s Tentamen. 6 March: Gauss replies to Farkas.
1833Retired from the army on a pension, after postings to Lemberg and Olmütz. Returns to Marosvásárhely.
1834Settles on the family estate at Domáld with Rozália Kibédi Orbán.
1837He and his father enter the Jablonowski Society prize contest in Leipzig on imaginary quantities; neither wins.
1848Reads Lobachevsky’s Geometrische Untersuchungen (1840) and writes extensive critical notes.
1849Marries Rozália Kibédi Orbán; they separate a few years later.
1856Death of Farkas Bolyai.
1860Dies 27 January in Marosvásárhely, of pneumonia, aged 57.
1867–68Richard Baltzer draws attention to the Appendix; Jules Hoüel translates it into French. Beltrami’s model (1868).
1891English translation by George Bruce Halsted.
1905First Bolyai Prize of the Hungarian Academy of Sciences, to Henri Poincaré.
Part I · Formation
Lesson 01

Father and son, 1802–1818

János Bolyai was born on 15 December 1802 in Kolozsvár, the main city of Transylvania, then part of the Habsburg lands of the Hungarian crown. He was born in the house of his mother’s family. His mother, Zsuzsanna Benkő, came from a Kolozsvár family; she died in 1821.

Farkas Bolyai

His father dominates the first half of the story. Farkas Bolyai, from a minor noble family of Transylvania, had studied at Göttingen from 1796 to 1799, where he became a close friend of the young Gauss. The two walked and talked mathematics, and among their topics was the theory of parallels. In 1804 Farkas became professor of mathematics, physics and chemistry at the Calvinist College in Marosvásárhely, where he stayed for the rest of his career. He wrote poetry and plays as well as mathematics, and spent years trying to prove Euclid’s parallel postulate. He sent one attempted proof to Gauss in 1804, and Gauss pointed out its flaw.

Farkas taught János himself. The boy learned mathematics fast. He also began the violin at seven and became good enough to play difficult concert pieces. He spent only a few years in regular school before completing the course of the Marosvásárhely college in 1817. In 1816 Farkas wrote to Gauss asking whether János could live in his house in Göttingen and study with him. Gauss did not take up the request.

DiscussionFarkas Bolyai had spent much of his life on the parallel postulate without success, yet his son succeeded. What did Farkas give János, in method and in the questions he asked, and what may have held Farkas back?

Lesson 02

Vienna, 1818–1823

In 1818 Bolyai entered the Imperial and Royal Engineering Academy in Vienna, which trained officers for the army’s engineering corps. The course was meant to take seven years; he completed it in four, graduating in 1822, and spent a further year in advanced study. He was reputed to be the best swordsman and dancer in the army, and he neither smoked nor drank.

By 1820 he was working on the parallel postulate. His father, alarmed, wrote begging him to give it up. He compared the subject to a bottomless night that had swallowed his own peace of mind, and warned that it could take away János’s health and happiness. János did not give it up. Like his father he first sought a proof. Unlike his father, he began to study what would follow if the postulate were false, and he found not a contradiction but a coherent geometry.

Between 1820 and 1824 he worked out most of a complete system: the parallel as a limit of intersecting lines (Lesson 04), the curves and surfaces that carry the structure of the new geometry, above all the surface on which Euclid’s geometry holds (Lesson 05), and the exponential law that ties lengths to angles.

Lesson 03

A new world out of nothing, 1823–1832

In September 1823 Bolyai joined the engineering corps as a sublieutenant and was posted to the fortress of Temesvár (now Timișoara). On 3 November 1823 he wrote to his father in Hungarian that he had discovered things so wonderful that he was astonished, and that he had created a new, different world out of nothing. He promised to write it up as soon as he had put it in order.

In February 1825 he showed his father a manuscript. Farkas was not enthusiastic. In 1826 Bolyai was posted to Arad and gave a German version to his former teacher, Captain Wolter von Eckwehr, who never returned it. In 1830 he was posted to Lemberg (now Lviv) and in 1832 to Olmütz (now Olomouc).

The Appendix

In 1831 Farkas, whose two-volume textbook Tentamen juventutem studiosam in elementa matheseos purae… introducendi was going through the press, urged his son to write the work up so that it could appear with it. The son wrote it in Latin, under the title Appendix scientiam spatii absolute veram exhibens: an appendix showing the absolutely true science of space, independent of the truth or falsity of Euclid’s Axiom XI. The title promised also a geometric quadrature of the circle in case the axiom is false. The text is 24 pages long and extremely compressed.

Separate copies were printed before the book. In June 1831 Farkas sent one to Gauss, without reply. He wrote again in January 1832. The Tentamen, with the Appendix bound into its first volume, appeared in 1832. Bolyai did not know that Lobachevsky had already published the same geometry in Kazan in 1829–30.

Part II · The mathematics
Lesson 04

Parallels in absolute geometry

For the general reader. Bolyai’s first move was to define parallel lines without deciding in advance how many there are. A parallel is the borderline case between lines that meet a given line and lines that do not. With that definition, much can be proved about parallels that is true in Euclid’s geometry and in the new one alike.

Goals
  • State Bolyai’s definition of asymptotic parallelism.
  • Prove that it is independent of the point chosen and symmetric.
  • Understand the two systems \(\Sigma\) and \(S\).

The definition

Let \(AM\) be a ray and \(B\) a point not on its line. Rotate a ray \(BN\) about \(B\), starting from the direction of \(BA\). At first it meets \(AM\). Bolyai calls \(BN\) parallel to \(AM\), written \(BN\parallel AM\), if it does not meet \(AM\) but every ray from \(B\) inside the angle \(ABN\) does. Continuity of the line (Dedekind’s axiom, in modern terms) guarantees that exactly one such ray exists on the side of \(M\).

Theorem (Appendix, opening sections)

In absolute geometry, parallelism of rays in Bolyai’s sense has these properties:

  1. If \(BN\parallel AM\) and \(C\) is any point of the line \(BN\), then the ray from \(C\) in the direction of \(N\) is parallel to \(AM\).
  2. (Symmetry) If \(BN\parallel AM\), then \(AM\parallel BN\).
  3. (Transitivity) If two rays are parallel to a third in the same direction, they are parallel to each other.

These are the properties needed to speak of a “point at infinity” common to all rays parallel to a given one in a given direction. Bolyai then split geometry in two. If the angle between \(BA\) and the parallel \(BN\) is always a right angle, Euclid’s Axiom XI holds and we are in system \(\Sigma\). If it is acute for some position, it is acute for all, and we are in system \(S\). He proved as much as he could for both systems simultaneously and called the result absolute.

Why symmetry needs proof

The definition of \(BN\parallel AM\) looks at rays from \(B\); the statement \(AM\parallel BN\) is about rays from \(A\). To prove it, one takes a ray \(AC\) inside the angle \(BAM\) and must show it meets \(BN\). Bolyai’s argument compares the two configurations by a congruence that moves \(A\) towards \(B\) and then uses Pasch’s axiom, which says a line entering a triangle through one side must leave through another. Only the axioms of order, congruence and continuity are used, never the parallel postulate. Lobachevsky proved the same facts in his own way.

Exercise 4.1In system \(\Sigma\) (Euclid), show that Bolyai’s parallel through \(B\) is the usual parallel, and that the parallel on the side of \(M\) and the parallel on the other side form one line.

Show solution
In Euclid a ray from \(B\) meets the line \(AM\) exactly when it is not parallel to it in the usual sense. The limiting position is therefore the unique Euclidean parallel through \(B\). The two limiting rays both lie on this line, pointing in opposite directions.

Exercise 4.2In the upper half-plane model (geodesics are vertical half-lines and semicircles centred on the real axis), find all rays from \(B=(0,1)\) parallel, in Bolyai’s sense, to the vertical ray \(x=2\) pointing upwards.

Show solution
Rays parallel to it share its endpoint at \(\infty\). The only geodesic from \(B\) to \(\infty\) is the vertical ray \(x=0\), \(y>1\). So there is exactly one, as the definition requires. Its downward-pointing partner is not parallel to the upward ray; parallelism has a direction.

Exercise 4.3In the same model, show that transitivity holds for the rays ending at \(0\) on the real axis.

Show solution
Rays parallel in the direction of the point at infinity \(0\) are exactly the geodesic rays ending at \(0\): semicircles through \(0\) and the vertical ray \(x=0\). If two rays end at \(0\) and a third also ends at \(0\), the first two share their endpoint, so they are parallel.
Lesson 05

The L-line and the F-surface

For the general reader. Bolyai’s most original discovery was a surface, inside the new non-Euclidean space, on which ordinary Euclidean geometry holds exactly. With it he could carry the whole of Euclid’s trigonometry into the new world and compute there.

Goals
  • Define L-lines and F-surfaces.
  • State the theorem that the F-surface is Euclidean.
  • Derive the exponential law and the formula \(\cot\frac u2=e^{y/k}\).

Definitions

Take a ray \(AM\) and consider all points \(B\) such that the parallel from \(B\) to \(AM\) makes with \(BA\) the same angle that \(AM\) makes with \(AB\). In the plane they form a curve he called an L-line; rotated about \(AM\) they form a surface he called an F-surface. Today these are the horocycle and the horosphere. The rays parallel to \(AM\) through points of the curve are its axes; the L-line meets every axis at a right angle. In system \(\Sigma\) an L-line is a straight line and an F-surface a plane.

Theorem (Appendix, §21)

On an F-surface, with L-lines playing the role of straight lines, Euclid’s Axiom XI holds and all of Euclidean plane geometry and trigonometry is valid. This is true in both systems \(\Sigma\) and \(S\).

The key step is that two L-lines on an F-surface lie in planes through axes, and that these planes meet or fail to meet according as the L-lines do. Because all axes are parallel to one another, Playfair’s axiom on the surface follows from the parallelism theorems of Lesson 04. Bolyai drew the striking conclusion that Euclid’s geometry is not lost in system \(S\): it lives on every F-surface.

Theorem (Appendix, §§24–29)

In system \(S\) there is a constant \(k>0\) with the following properties. If two concentric L-lines are a distance \(y\) apart along their axes, arcs between the same two axes have ratio \(e^{y/k}\). If \(u\) is the angle that the parallel makes with the perpendicular from a point at distance \(y\) from a line, then

\[ \cot\frac u2=e^{y/k}. \]

Bolyai wrote \(Y=e^{y/i}\) and stated the second formula as \(Y=\cot\frac u2\). It is the same as Lobachevsky’s \(\tan\frac{\Pi(y)}2=e^{-y/k}\). The exponential comes from the fact that the ratio \(f(y)\) of arcs satisfies \(f(y_1+y_2)=f(y_1)f(y_2)\) and is monotone.

Exercise 5.1Prove that a monotone function \(f:(0,\infty)\to(0,\infty)\) with \(f(y_1+y_2)=f(y_1)f(y_2)\) has the form \(f(y)=e^{y/k}\) for some constant \(k\ne0\) (or is identically 1).

Show solution
Let \(g=\ln f\). Then \(g\) is additive and monotone. Additivity gives \(g(q)=qg(1)\) for rational \(q>0\). Monotonicity squeezes \(g(y)\) between \(g(q)\) and \(g(q')\) for rationals \(q<y<q'\), so \(g(y)=yg(1)\). Put \(1/k=g(1)\); if \(g(1)=0\) then \(f\equiv1\), which is the Euclidean case.

Exercise 5.2Show that \(\cot\frac u2=e^{y/k}\) is equivalent to \(\cot u=\sinh\frac yk\).

Show solution
Put \(t=\cot\frac u2=e^{y/k}\). Then \(\cot u=\frac{\cot^2(u/2)-1}{2\cot(u/2)}=\frac{t-t^{-1}}{2}=\sinh\frac yk\).

Exercise 5.3In system \(\Sigma\) the ratio of arcs of concentric L-lines is always 1. Explain why this corresponds to “\(k=\infty\)”, and what happens to the formula \(\cot\frac u2=e^{y/k}\) in that limit.

Show solution
In \(\Sigma\) L-lines are parallel straight lines, and the segments cut off on them by two parallel axes are equal, so the ratio is 1 for all \(y\). As \(k\to\infty\), \(e^{y/k}\to1\) and \(\cot\frac u2\to1\), so \(u\to\pi/2\): the parallel is perpendicular to the perpendicular, as in Euclid.
Lesson 06

The absolute law of sines

For the general reader. Bolyai wanted theorems that are true no matter which geometry holds. His most elegant one is a law of sines. Instead of comparing the sides of a triangle with the sines of its angles, it compares the circumferences of circles whose radii are the sides. That formulation is true in Euclid’s geometry, in the new geometry, and needs no decision between them.

Goals
  • State Bolyai’s absolute law of sines.
  • Specialize it to the Euclidean and hyperbolic cases.
  • Compute the circumference of a circle in system \(S\).

Write \(\bigcirc r\) for the circumference of a circle of radius \(r\).

Theorem (Appendix, §25)

In absolute geometry, for every rectilinear triangle with sides \(a,b,c\) and opposite angles \(A,B,C\),

\[ \bigcirc a:\bigcirc b:\bigcirc c=\sin A:\sin B:\sin C. \]

Proof idea. Take a right triangle \(ABC\), right-angled at \(C\), and erect the perpendicular to its plane at \(A\), together with parallel axes through \(B\) and \(C\). The F-surface through \(A\) with these axes cuts out a Euclidean right triangle (Lesson 05), and the planes through the axes cut out a spherical triangle at \(A\). Comparing them shows that \(\sin A\) equals the ratio of the circles on the sphere and F-surface that correspond to \(a\) and \(c\). Bolyai proved the corresponding law for spherical triangles (§26) without the parallel postulate as well. The general case follows by dropping an altitude.

Theorem (Appendix, §30)

In system \(S\), \(\bigcirc r=\pi k\big(e^{r/k}-e^{-r/k}\big)=2\pi k\sinh\frac rk\). In system \(\Sigma\), \(\bigcirc r=2\pi r\).

Combining the two theorems, the law of sines in \(S\) becomes \(\dfrac{\sin A}{\sinh(a/k)}=\dfrac{\sin B}{\sinh(b/k)}=\dfrac{\sin C}{\sinh(c/k)}\), Lobachevsky’s formula, while in \(\Sigma\) it becomes the familiar Euclidean law. Spherical trigonometry is thus absolute: it does not depend on the parallel postulate. Lobachevsky used the same fact.

Exercise 6.1Use the absolute law of sines to prove that in an isosceles triangle (in either system) the base angles are equal if and only if the opposite sides are equal.

Show solution
\(\bigcirc\) is strictly increasing in \(r\) in both systems. If \(a=b\) then \(\sin A=\sin B\), and since \(A+B<\pi\) (Euclid I.17, absolute), \(A=B\). Conversely, \(A=B\) gives \(\bigcirc a=\bigcirc b\), so \(a=b\).

Exercise 6.2Show that a circle of radius \(r\) in system \(S\) has circumference larger than \(2\pi r\), and estimate the relative excess for \(r=k/10\).

Show solution
\(2\pi k\sinh(r/k)=2\pi r\big(1+\frac{r^2}{6k^2}+\cdots\big)>2\pi r\). For \(r=k/10\) the excess is about \(1/600\approx0.17\%\).

Exercise 6.3Check the circumference formula in the Poincaré disc \(ds^2=\dfrac{4k^2(dx^2+dy^2)}{(1-x^2-y^2)^2}\), for the circle centred at the origin of Euclidean radius \(\rho\).

Show solution
The hyperbolic radius is \(r=\int_0^\rho\frac{2k\,dt}{1-t^2}=2k\operatorname{artanh}\rho\), so \(\rho=\tanh\frac r{2k}\). The circumference is \(2\pi\rho\cdot\frac{2k}{1-\rho^2}\). With \(\rho=\tanh s\), \(\frac{2\rho}{1-\rho^2}=\sinh 2s\), so the circumference is \(2\pi k\sinh\frac rk\).
Lesson 07

Constructing the parallel

For the general reader. In Euclid’s geometry it is easy to draw the parallel to a line through a point. In Bolyai’s geometry the parallel is a limiting line, and it is not obvious that it can be drawn with ruler and compass at all. Bolyai found a construction that works in both geometries at once.

Goals
  • Learn Bolyai’s construction of the parallel.
  • Prove it with hyperbolic trigonometry.
  • Use the angle of parallelism to pass between angles and lengths.
Construction (Bolyai, Appendix)

Given a line \(l\) and a point \(P\) not on it: drop the perpendicular \(PQ\) to \(l\); draw the line \(m\) through \(P\) perpendicular to \(PQ\); choose any point \(R\) on \(l\) other than \(Q\), and drop the perpendicular \(RS\) from \(R\) to \(m\). The circle with centre \(P\) and radius \(QR\) meets the segment \(RS\) at a point \(T\). Then the ray \(PT\) is the parallel to \(l\) through \(P\) in the direction of \(R\).

In system \(\Sigma\), \(PQRS\) is a rectangle, so \(PS=QR\). Then \(T=S\), and \(PT\) is the line \(m\), the Euclidean parallel. In system \(S\), \(PQRS\) is a Lambert quadrilateral: three right angles, at \(Q\), \(P\) and \(S\), and an acute angle at \(R\). Then \(PS<QR<PR\), so \(T\) lies strictly between \(S\) and \(R\).

Proof in system \(S\)

Take \(k=1\), \(PQ=d\), \(QR=x\). We use two right-triangle formulas (Lobachevsky course, Lesson 07): \(\cosh(\text{hyp})=\cosh(\text{leg}_1)\cosh(\text{leg}_2)\), and \(\cos(\text{angle})=\tanh(\text{adjacent leg})/\tanh(\text{hyp})\), \(\sin(\text{angle})=\sinh(\text{opposite leg})/\sinh(\text{hyp})\).

  1. Triangle \(PQR\), right-angled at \(Q\): \(\cosh PR=\cosh d\cosh x\) and \(\sin\angle QPR=\sinh x/\sinh PR\).
  2. Since \(\angle QPS\) is right, \(\angle SPR=\tfrac\pi2-\angle QPR\). Triangle \(PSR\), right-angled at \(S\): \(\cos\angle SPR=\tanh PS/\tanh PR\). Hence \[ \tanh PS=\tanh PR\,\sin\angle QPR=\frac{\sinh x}{\cosh PR}=\frac{\tanh x}{\cosh d}. \]
  3. Triangle \(PST\), right-angled at \(S\), with hypotenuse \(PT=x\): \(\cos\angle SPT=\tanh PS/\tanh x=1/\cosh d\).
  4. So \(\sin\angle QPT=\cos\angle SPT=1/\cosh d=\sin\Pi(d)\). Both angles are acute, so \(\angle QPT=\Pi(d)\).

The proof shows more: the construction does not depend on the choice of \(R\). In Euclid’s limit \(d/k\to0\) the angle tends to a right angle, matching \(T=S\).

Exercise 7.1Using step 2, show that \(PS<QR\) in system \(S\), so the circle in the construction really does meet the segment \(RS\).

Show solution
\(\tanh PS=\tanh x/\cosh d<\tanh x\) because \(\cosh d>1\), and \(\tanh\) is increasing, so \(PS<x=QR\). Also \(x<PR\) since \(\cosh PR=\cosh d\cosh x>\cosh x\). As a point moves from \(S\) to \(R\), its distance from \(P\) increases continuously from \(PS\) to \(PR\), so it equals \(x\) at some \(T\).

Exercise 7.2Take \(k=1\), \(d=1\), \(x=2\). Compute \(PS\), \(\angle SPT\) and \(\angle QPT\), and check against \(\tan\frac{\Pi(1)}2=e^{-1}\).

Show solution
\(\tanh PS=\tanh2/\cosh1\approx0.9640/1.5431\approx0.6247\), so \(PS\approx0.733\). \(\cos\angle SPT=1/\cosh1\approx0.6481\), so \(\angle SPT\approx0.8657\) and \(\angle QPT\approx\pi/2-0.8657\approx0.7051\). And \(2\arctan e^{-1}\approx2\times0.3525=0.7051\). They agree.

Exercise 7.3Conversely, given an acute angle \(\alpha\), find the distance \(d\) with \(\Pi(d)=\alpha\), and say how the construction above could be run backwards to produce it.

Show solution
\(\cos\Pi(d)=\tanh d\), so \(d=\operatorname{artanh}(\cos\alpha)\). Geometrically: draw the angle \(\alpha\) at \(P\) between rays \(PA\) and \(PB\), and the perpendicular to \(PA\) that is parallel to \(PB\); the foot of that perpendicular is at distance \(d\) from \(P\).
Lesson 08

Area and the quadrature of the circle

For the general reader. To square the circle is to construct, with ruler and compass, a square with the same area as a given circle. Ferdinand von Lindemann proved in 1882 that this is impossible in Euclid’s geometry. Bolyai showed that in the new geometry a circle and a square of equal area can both be constructed, a result he advertised in the title of the Appendix.

Goals
  • State the area formulas for triangles, polygons and circles in system \(S\).
  • Find a square and a circle of equal area.
  • Understand why this does not contradict Lindemann.
Theorem (area in system \(S\))

A triangle with angles \(A,B,C\) has area \(k^2(\pi-A-B-C)\). More generally a simple \(n\)-gon with angles \(\alpha_1,\dots,\alpha_n\) has area \(k^2\big((n-2)\pi-\sum\alpha_i\big)\). A circle of radius \(r\) has area \(4\pi k^2\sinh^2\frac r{2k}=2\pi k^2\big(\cosh\frac rk-1\big)\).

The polygon formula follows from the triangle formula by triangulation. The circle formula follows from \(\bigcirc r=2\pi k\sinh(r/k)\) (Lesson 06) by integrating \(\int_0^r\bigcirc\rho\,d\rho\). Bolyai derived the area of a triangle from its defect at the end of the Appendix and then turned to the quadrature.

Squaring the circle in \(S\)

A regular quadrilateral (a “square”: four equal sides and four equal angles) with angle \(\alpha<\pi/2\) has area \(k^2(2\pi-4\alpha)\). As \(\alpha\) runs from \(\pi/2\) down to \(0\), this runs from \(0\) to \(2\pi k^2\). Circles of every area exist. So for each square there is a circle of the same area, and conversely each circle of area below \(2\pi k^2\) has a square of equal area. The question is whether both can be drawn with ruler and compass. Bolyai showed that for suitable pairs they can, using his constructions of parallels and the angle of parallelism. A modern analysis by Davis shows that in general one cannot construct the square from a given circle, or the circle from a given square: the two figures must be built separately, as Bolyai did.

This does not contradict Lindemann. In Euclid’s geometry, squaring a circle of radius 1 requires the length \(\sqrt\pi\), which is transcendental. In system \(S\) the area of a square is a multiple of \(k^2\) by an angle, and the circle’s area involves \(\sinh^2\), so the question becomes whether certain angles and lengths are constructible, which for suitable choices they are.

Exercise 8.1Show that the square with angles \(\pi/4\) and the circle with \(\sinh\frac r{2k}=\frac12\) have the same area, and find that area and the radius \(r\).

Show solution
Square: \(k^2(2\pi-4\cdot\frac\pi4)=\pi k^2\). Circle: \(4\pi k^2\cdot\frac14=\pi k^2\). The radius is \(r=2k\operatorname{arsinh}\frac12=2k\ln\frac{1+\sqrt5}{2}\approx0.962\,k\), twice the logarithm of the golden ratio.

Exercise 8.2Show that no triangle has area \(\geq\pi k^2\), and that every circle of radius \(r>0.962\,k\) has area larger than every triangle.

Show solution
The defect \(\pi-A-B-C\) is less than \(\pi\) because the angles are positive. By Exercise 8.1 the circle of radius \(r_0=2k\operatorname{arsinh}\frac12\approx0.962\,k\) has area exactly \(\pi k^2\), and area increases with radius.

Exercise 8.3Show that a regular hexagon with all angles \(2\pi/3\) cannot exist in system \(S\), and find the angle of a regular hexagon of area \(\pi k^2\).

Show solution
A hexagon with angles \(2\pi/3\) would have area \(k^2(4\pi-6\cdot\frac{2\pi}3)=0\), impossible. For area \(\pi k^2\): \(4\pi-6\alpha=\pi\), so \(\alpha=\pi/2\). Regular right-angled hexagons exist in system \(S\); they do not in Euclid’s plane.
Lesson 09

Complex numbers as pairs

For the general reader. In the 1830s many mathematicians were uneasy about “imaginary” numbers such as \(\sqrt{-1}\). Bolyai proposed to remove the mystery by defining a complex number as an ordered pair of ordinary numbers with suitable rules for adding and multiplying. William Rowan Hamilton published the same idea in Dublin in 1837.

Goals
  • Define complex numbers as pairs of reals.
  • Verify the field axioms.
  • See the context of the Jablonowski prize question.

In 1837 the Jablonowski Society in Leipzig held a prize contest on the geometric construction of imaginary quantities, and both Bolyais sent entries. János argued in his Responsio that the question was misconceived: imaginary quantities need no geometric construction, only a rigorous arithmetic definition. Neither entry won. The judges found the Bolyais’ solutions too involved.

Definition and theorem (Bolyai; Hamilton 1837)

On \(\mathbb R^2\) define \((a,b)+(c,d)=(a+c,\,b+d)\) and \((a,b)\cdot(c,d)=(ac-bd,\,ad+bc)\). These operations make \(\mathbb R^2\) a field. The map \(a\mapsto(a,0)\) embeds \(\mathbb R\) as a subfield, and \(i=(0,1)\) satisfies \(i^2=-1\). Every element is uniquely \(a+bi\).

The point of the construction is logical, not computational. If the real numbers exist, the complex numbers exist, and there is nothing imaginary about them. The same strategy later served to construct the reals from the rationals and the integers from the natural numbers. Bolyai pursued a broader programme of founding all of mathematics on axioms, begun in 1834 and never finished.

Exercise 9.1Find the multiplicative inverse of \((a,b)\neq(0,0)\) and verify that it works.

Show solution
\((a,b)^{-1}=\Big(\dfrac{a}{a^2+b^2},\,\dfrac{-b}{a^2+b^2}\Big)\). Then \((a,b)\cdot(a,-b)/(a^2+b^2)=\big((a^2+b^2),\,(-ab+ab)\big)/(a^2+b^2)=(1,0)\).

Exercise 9.2Prove that the multiplication is associative.

Show solution
\(\big((a,b)(c,d)\big)(e,f)=(ace-bde-adf-bcf,\ acf-bdf+ade+bce)\). Computing \((a,b)\big((c,d)(e,f)\big)=(a,b)(ce-df,\,cf+de)\) gives \((ace-adf-bcf-bde,\ acf+ade+bce-bdf)\), the same.

Exercise 9.3Show that there is no ordering of the field \(\mathbb R^2\) compatible with addition and multiplication. Why does this not matter for Bolyai’s purpose?

Show solution
In an ordered field squares are \(\ge0\), so \(1=1^2>0\) and \(-1=i^2\ge0\), contradiction. Bolyai needed only a consistent arithmetic, not an order: complex numbers measure rotations and amplitudes, not sizes on a line.
Lesson 10

Is the new geometry consistent?

For the general reader. Neither Bolyai nor Lobachevsky proved that their geometry contains no contradiction. They had built a large, coherent theory, but a contradiction might lie further on. Bolyai worried about this in later life. The question was settled in 1868, eight years after his death, when Eugenio Beltrami found a model.

Goals
  • Describe the Beltrami–Klein and Poincaré models.
  • Understand relative consistency.
  • Compute in a model to confirm Bolyai’s formulas.
Theorem (Beltrami 1868; Klein 1871; Poincaré 1882)

Let \(D\) be the open unit disc. Take as “points” the points of \(D\), as “lines” the chords of the circle (open at the ends), and define the distance between points \(X,Y\) whose chord has endpoints \(A,B\) by

\[ d(X,Y)=\frac k2\left|\ln\frac{|XB|\,|YA|}{|XA|\,|YB|}\right|. \]

With the appropriate notion of angle, all of Hilbert’s axioms of plane geometry hold except the parallel axiom, which fails. Hence, if Euclidean geometry is consistent, so is Bolyai–Lobachevsky geometry.

Every theorem of system \(S\) becomes a theorem about chords in a Euclidean disc. A contradiction in \(S\) would therefore yield a contradiction in Euclidean geometry. In the Poincaré disc, lines are arcs of circles orthogonal to the boundary, angles are Euclidean angles, and the metric is \(ds^2=4k^2(dx^2+dy^2)/(1-x^2-y^2)^2\). The hero image of this page is drawn in it. Beltrami also observed that small pieces of the plane are realized on the pseudosphere, a surface in Euclidean space with constant curvature \(-1/k^2\).

Exercise 10.1In the Beltrami–Klein disc with \(k=1\), compute the distance from the centre \(O\) to the point \(X=(t,0)\), \(0<t<1\). Show that it tends to infinity as \(t\to1\).

Show solution
The chord is the diameter with \(A=(-1,0)\), \(B=(1,0)\). Then \(|OB|=|OA|=1\), \(|XB|=1-t\), \(|XA|=1+t\), and \(d(O,X)=\frac12\ln\frac{1+t}{1-t}=\operatorname{artanh}t\to\infty\).

Exercise 10.2In the Beltrami–Klein disc, draw a chord \(l\) and a point \(P\) not on it. Describe the two parallels to \(l\) through \(P\) and show that there are infinitely many lines through \(P\) not meeting \(l\).

Show solution
Let \(A,B\) be the endpoints of \(l\) on the circle. The chords \(PA\) and \(PB\), extended to the circle, do not meet \(l\) inside the disc (they meet it only at the excluded boundary points), and are the limiting cases: Bolyai’s parallels. Every chord through \(P\) that lies between \(PA\) and \(PB\), in the pair of vertical angles not containing \(l\), also misses \(l\), and there are infinitely many such chords.

Exercise 10.3The pseudosphere is the surface of revolution of the tractrix, with metric \(ds^2=du^2+e^{-2u}dv^2\) for \(u>0\). Show that the substitution \(y=e^{u}\) turns it into a piece of the upper half-plane metric with \(k=1\).

Show solution
\(dy=e^u du\), so \(du^2=dy^2/y^2\), and \(e^{-2u}dv^2=dv^2/y^2\). Hence \(ds^2=(dv^2+dy^2)/y^2\), the half-plane metric, restricted to \(y>1\) and with \(v\) periodic (since \(v\) is the angle of rotation).
Part III · Life and legacy
Lesson 11

Gauss’s letter and after, 1832–1848

Gauss received the Appendix early in 1832. On 14 February he wrote to his former student Christian Ludwig Gerling that he had received a little paper from Hungary on non-Euclidean geometry in which he found all his own ideas and results, and that he considered the young geometer a genius of the first order. On 6 March 1832 he wrote to Farkas. He said that he could not praise the work, because to praise it would be to praise himself: its whole content, the path János had taken and the results he had reached, coincided almost entirely with his own meditations of the past thirty to thirty-five years. He added that he had not intended to publish these ideas in his lifetime, and that he was glad the son of his old friend had spared him the trouble of writing them down. He suggested a problem for János: find the volume of a tetrahedron in the new geometry.

Farkas thought this high praise. János took it as a disaster. It deprived him of the priority he believed was his, and he felt that Gauss, if he had found these results, should have published them. He published nothing more. In 1833 the army retired him on a pension as a semi-invalid, after postings in Lemberg and Olmütz. He returned to Marosvásárhely and lived with his father, but the two quarrelled. From 1834 he lived on the small family estate at Domáld with Rozália Kibédi Orbán. They had children and married in 1849, separating a few years later.

In 1837 came the disappointment of the Jablonowski contest (Lesson 09). In 1848 Farkas obtained Lobachevsky’s Geometrische Untersuchungen of 1840 and gave it to his son. János read it closely and wrote extensive notes. He praised parts of it, including the derivation of the trigonometric formulas, which he called masterly. He also suspected for a time that “Lobachevsky” might be a name used by Gauss himself. The book prompted him to resume work on his own geometry and on the question of its consistency.

DiscussionGauss’s reply to Farkas was honest by his own lights. Was it fair to János? Compare it with how Gauss treated Lobachevsky (see the Lobachevsky course, Lesson 11), and consider what Gauss could have done differently in 1832.

Lesson 12

Last years and legacy

In 1846 Bolyai moved back to Marosvásárhely. He worked intermittently on absolute geometry, spherical trigonometry and Gauss’s tetrahedron problem, with notes dated as late as 1856. He also devoted much effort to a theory of all knowledge and to a doctrine of happiness, the Üdvtan. His father died in 1856. János died in Marosvásárhely on 27 January 1860, of pneumonia, aged 57, and was buried there.

He left more than 20,000 pages of manuscripts, now in the Teleki–Bolyai Library in Târgu Mureș. No authentic portrait of him survives; the face shown in some encyclopedias and on a Hungarian stamp is not his.

Recognition

The Appendix remained almost unknown until Richard Baltzer mentioned it in the second edition of his Elemente der Mathematik in 1867. Baltzer’s correspondent Jules Hoüel, who had translated Lobachevsky in 1866, published a French translation of the Appendix in 1868. The same year Beltrami gave his model (Lesson 10), and Felix Klein’s papers of 1871 placed the geometry of Lobachevsky and Bolyai inside projective geometry. George Bruce Halsted’s English translation appeared in 1891. The first Bolyai Prize of the Hungarian Academy of Sciences went to Henri Poincaré in 1905, and the second to David Hilbert in 1910.

Legacy

  • Mathematics: hyperbolic geometry, now basic in complex analysis, topology and group theory. Bolyai’s idea of an absolute geometry, the theorems common to Euclid’s and the hyperbolic system, is now standard as neutral geometry.
  • Method: the Appendix is one of the first works to explore systematically what follows from a set of axioms when one of them is replaced by its negation.
  • Honours: Babeș-Bolyai University in Cluj-Napoca (named in 1959 after the merger of the Bolyai University with the Babeș University), the Bolyai Institute of Mathematics at Szeged, the lunar crater Bolyai, and the asteroid 1441 Bolyai.

Final projectChoose one theorem of the Appendix that holds in absolute geometry and one that holds only in system \(S\). For each, write a page explaining the statement, a proof sketch, and what it says in the Euclidean limit \(k\to\infty\).

Reference

Connections

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

Teachers and advisors

  • Farkas Bolyai 1802-1817

Influenced by

Influenced

Colleagues and collaborators

Rivals

Friends

Family

  • Farkas Bolyai

Also mentioned in this life

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Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, János Bolyai, MacTutor History of Mathematics.
  2. Bolyai, János (Johann), Dictionary of Scientific Biography, via Encyclopedia.com.
  3. János Bolyai, Wikipedia.
  4. E. Kiss, Mathematical Gems from the Bolyai Chests, Akadémiai Kiadó, Budapest, 1999.
  5. P. Stäckel, Wolfgang und Johann Bolyai: Geometrische Untersuchungen, 2 vols., Teubner, Leipzig, 1913.

Mathematics

  1. J. Bolyai, Appendix scientiam spatii absolute veram exhibens, in W. Bolyai, Tentamen, vol. 1, Marosvásárhely, 1832. English translation by G. B. Halsted, The Science Absolute of Space, Austin, 1891; reprinted in Bonola (below).
  2. S. Rose, Notes on Bolyai’s “Appendix”, arXiv:2312.00213.
  3. R. Bonola, Non-Euclidean Geometry: A Critical and Historical Study of its Development, 1912; Dover reprint.
  4. J. J. Gray, János Bolyai, Non-Euclidean Geometry, and the Nature of Space, Burndy Library, MIT Press, 2004 (includes a facsimile of the Appendix).
  5. J. J. Gray, Worlds Out of Nothing: A Course in the History of Geometry in the 19th Century, 2nd ed., Springer, 2010.
  6. J. Davis, Squaring the circle in the hyperbolic disk, Rose-Hulman Undergraduate Mathematics Journal 15(1).
  7. M. J. Greenberg, Euclidean and Non-Euclidean Geometries, 4th ed., W. H. Freeman, 2008.
  8. J. Milnor, Hyperbolic geometry: the first 150 years, Bull. Amer. Math. Soc. 6 (1982) 9–24.