courseshub.world · Lives in Mathematics and Physics

Giovanni Girolamo Saccheri

Giovanni Girolamo (Gerolamo) Saccheri, S.J. · Hieronymus Saccherius

A course in two strands: the life of the Jesuit logician who taught at Pavia for thirty-six years, and the book he published in the year of his death, which set out to vindicate Euclid and instead proved, one proposition after another, the first theorems of hyperbolic geometry.

Born 5 September 1667, San Remo Died 25 October 1733, Milan Euclides ab omni naevo vindicatus Milan, 1733 Level undergraduate → advanced undergraduate
A Saccheri quadrilateral in the Poincaré disk (curvature −1, legs 1.3, base 1.6). Base AB, equal legs AC and BD, summit CD with acute angles of about 42°. The dashed curve is the equidistant line at height 1.3 above AB: under the acute hypothesis it is not straight.
Lesson 00

How to use this course

Giovanni Girolamo Saccheri spent his working life as a Jesuit teacher of philosophy, theology and mathematics in northern Italy. He is remembered for one book, Euclides ab omni naevo vindicatus (Euclid freed from every flaw), printed in Milan in 1733. Its aim was to prove Euclid’s parallel postulate by showing that its denial leads to absurdity. To do so he developed the consequences of the denial in more depth than anyone before him. Most of what he proved is correct, and much of it is now a chapter of hyperbolic geometry. The book was forgotten for a century and a half and then recognized as the most important work on parallels before Nikolai Lobachevsky and János Bolyai.

Prerequisites for Part II
  • Euclid’s Elements, Book I: congruence of triangles (SAS, SSS, ASA), the exterior angle theorem (I.16), and Propositions I.27–I.29 on parallels.
  • Proof by contradiction and simple propositional logic.
  • For Lesson 10: hyperbolic functions, and ideally a first look at the Poincaré disk or upper half-plane. The course on Nikolai Lobachevsky covers the background.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. Proposition numbers (Prop. I, Prop. XXIII and so on) are Saccheri’s own, in Book I of Euclides vindicatus. Lessons 05 to 09 follow the order of his book. Lesson 10 steps outside it and computes his quadrilateral in the models that settled the question he could not.

The course on Omar Khayyam describes the same quadrilateral six centuries earlier. The course on Nikolai Lobachevsky continues the story after 1826.

Reference

Timeline

1667Born 5 September in San Remo, in the Republic of Genoa, son of the lawyer Giovanni Felice Saccheri.
168524 March: enters the Jesuit novitiate in Genoa.
1687–90Teaches at the Jesuit college in Genoa while studying.
1690Sent to the Jesuit Collegio di Brera in Milan to study philosophy and theology. Tommaso Ceva turns him towards mathematics.
1693Quaesita geometrica, his first book, published in Milan.
1694March: ordained priest at Como. Begins teaching philosophy at the Jesuit college in Turin.
1697Logica demonstrativa printed in Turin. Moves to the Jesuit college in Pavia.
1699Appointed to the chair of mathematics at the University of Pavia, which he holds until his death.
1701Second edition of Logica demonstrativa.
1708Neo-statica, Milan.
1713Declines the chair of mathematics at Turin offered by Victor Amadeus II.
173313 July: approval of the Inquisition for Euclides ab omni naevo vindicatus; 16 August: approval of the Jesuit provincial. The book appears in Milan.
1733Dies in Milan on 25 October, aged 66.
1763Georg Simon Klügel’s Göttingen dissertation reviews about thirty attempts to prove the parallel postulate, Saccheri’s among them.
1889Eugenio Beltrami draws attention to Euclides vindicatus as a forerunner of non-Euclidean geometry.
1903Giovanni Vailati rediscovers the Logica demonstrativa.
1920George Bruce Halsted’s English translation, Euclides Vindicatus, Chicago.
Part I · Formation
Lesson 01

San Remo and Genoa, 1667–1690

Giovanni Girolamo Saccheri was born on 5 September 1667 in San Remo, a coastal town of the Republic of Genoa. His father, Giovanni Felice Saccheri, was a lawyer. Early accounts describe the boy as precocious and inquisitive, and little else is known of his childhood.

On 24 March 1685, aged seventeen, he entered the novitiate of the Society of Jesus in Genoa. The Jesuits ran the largest network of schools in Catholic Europe, and their training was long: a novitiate, then years of teaching in the lower classes of a college alternating with study of philosophy and theology. From 1687 Saccheri taught at the Jesuit college in Genoa, whose building, the Palazzo Balbi, now houses the University of Genoa.

Jesuit mathematics

Mathematics had an assured if modest place in the Jesuit curriculum. Its model was Christopher Clavius, whose Latin edition of Euclid’s Elements (first published 1574) was the standard school text across Europe for a century. Clavius had himself discussed the parallel postulate and proposed a proof of it. A young Jesuit with mathematical talent would meet Euclid through Clavius, and Saccheri did.

Lesson 02

Brera and the Ceva brothers, 1690–1694

In 1690 his superiors sent Saccheri to the Collegio di Brera in Milan, the principal Jesuit college of Lombardy, to study philosophy and theology while teaching grammar. Brera’s professor of mathematics was Tommaso Ceva, a Jesuit poet and geometer who taught there for about forty years. Ceva recognized Saccheri’s ability, encouraged him to read Clavius’s Euclid closely, and urged him to devote himself to mathematics.

Through Tommaso he met his brother Giovanni Ceva, mathematician to the Gonzaga court at Mantua, whose 1678 book De lineis rectis contains the theorem on concurrent lines in a triangle that bears his name. Saccheri worked in close contact with both brothers. On Tommaso’s advice he also began a correspondence with Vincenzo Viviani in Florence, a pupil of Galileo.

Quaesita geometrica

His first book, Quaesita geometrica, appeared in Milan in 1693. It solves a set of elementary geometric problems that had been posed by the Sicilian mathematician Ruggero Ventimiglia, using the methods of Giovanni Ceva. Viviani responded by sending Saccheri his own solution of a problem he had circulated, the so-called window of Viviani. In March 1694 Saccheri was ordained priest at Como.

Lesson 03

Turin and Pavia, 1694–1708

From 1694 to 1697 Saccheri taught philosophy at the Jesuit college in Turin. There he came to the notice of Victor Amadeus II, Duke of Savoy, who reportedly called on him when difficult calculations were needed. His philosophy course became his second book, Logica demonstrativa, printed in Turin in 1697. It was reissued in 1701 and again at Cologne in 1735. Lesson 04 describes its content.

In 1697 he moved to the Jesuit college at Pavia, where he taught philosophy and theology for the rest of his life. In 1699 the Senate of Milan, which governed the University of Pavia, appointed him to its chair of mathematics. He stayed in that chair for thirty-four years. In 1713 Victor Amadeus offered him the chair of mathematics at Turin, and later he was offered the chair at Padua that Galileo had held. He declined both.

Neo-statica

In 1708 he published Neo-statica, dedicated to the Senate of Milan. It is a treatise on the equilibrium of weights, partly a critique of Tommaso Ceva’s De natura gravium (1669). It stays within the Aristotelian tradition of statics and is of little interest today. His real mathematical project, slowly developed over the next quarter century, was the foundations of Euclid’s geometry.

DiscussionSaccheri twice turned down prestigious chairs to remain at Pavia. What would a Jesuit professor have gained from remaining within his order’s college, and what might a court or university appointment have offered instead?

Part II · The mathematics
Lesson 04

Definitions and the logic of reductio

For the general reader. Before he wrote about parallels, Saccheri wrote a textbook of logic. Two ideas from it shaped everything he did later. A definition does not prove that the thing defined exists. And one powerful way to prove a statement is to assume it false and show that the assumption leads back to the statement itself.

Goals
  • Distinguish nominal and real definitions, as Saccheri did.
  • State and justify the rule “if not-P implies P, then P”.
  • See how both ideas frame Euclides vindicatus.

Nominal and real definitions

The Logica demonstrativa follows the plan of Aristotle’s logical works in four parts. Its most original pages concern definitions. Saccheri separated a definitio quid nominis, which only fixes the meaning of a word, from a definitio quid rei, which in addition asserts that the thing exists. A real definition is a hidden postulate, and its use in a proof must be justified. He also asked whether the clauses of a complex definition are compatible with one another, a question close to the modern notion of consistency. Thomas Heath later observed that John Stuart Mill’s account of the same distinction had been fully anticipated by Saccheri.

Saccheri’s example was the definition of parallel lines as lines everywhere equidistant, used by Giovanni Alfonso Borelli and others. As a nominal definition it is harmless. Used as a real definition it assumes that the points at a fixed distance from a line form a straight line, and that assumption is equivalent to the parallel postulate (Exercise 4.2).

The rule of Euclid IX.12

Principle (consequentia mirabilis)

If a proposition \(P\) can be derived from the assumption \(\neg P\), then \(P\) is true. In symbols, \((\neg P\to P)\to P\).

Saccheri found the pattern in Euclid’s proof of Elements IX.12, where the denial of what is to be proved yields the proposition itself. It is stronger in feel than ordinary reductio, which derives some contradiction \(Q\wedge\neg Q\), though the two are equivalent in classical logic. This was his plan for the parallel postulate: assume it false, and derive it.

Exercise 4.1Prove \((\neg P\to P)\to P\) using a truth table, and then derive it from ordinary reductio (from \(\neg P\vdash Q\wedge\neg Q\) infer \(P\)).

Show solution
If \(P\) is true, the implication is true. If \(P\) is false, \(\neg P\to P\) is “true implies false”, which is false, so the whole formula is “false implies false”, which is true. For the derivation: assume \(\neg P\to P\) and \(\neg P\). Then \(P\) by modus ponens, so \(P\wedge\neg P\). By reductio, \(P\).

Exercise 4.2Work in absolute geometry (no parallel postulate). Suppose that for some line \(AB\) and distance \(a\), the points on one side of \(AB\) at distance \(a\) from it all lie on one straight line \(\ell\). Show that every Saccheri quadrilateral with base on \(AB\) and legs \(a\) has right summit angles.

Show solution
Take feet \(A,B,F\) in order on the base line with perpendiculars \(AC,BD,FE\) of length \(a\). Then \(C,D,E\) lie on \(\ell\), with \(D\) between \(C\) and \(E\). Let \(\varphi\) be the summit angle of \(ABDC\), so \(\angle BDC=\varphi\). In the Saccheri quadrilateral \(BFED\), \(\angle BDE=\angle FED\). In \(AFEC\), \(\angle ACE=\angle FEC\). Now \(\angle ACE=\angle ACD=\varphi\) and \(\angle FEC=\angle FED=\angle BDE=\pi-\angle BDC=\pi-\varphi\). So \(\varphi=\pi-\varphi\) and \(\varphi=\pi/2\). The right-angle hypothesis follows, and with it the parallel postulate (Lesson 07).

Exercise 4.3Describe the structure of Saccheri’s Proposition XIV (Lesson 07) as an instance of the principle above.

Show solution
Let \(P\) be “the obtuse hypothesis is false”. Saccheri assumes \(\neg P\), the obtuse hypothesis, and derives Euclid’s postulate, which implies the right hypothesis and hence \(P\). So \(\neg P\to P\), and he concludes \(P\). He hoped to do the same with the acute hypothesis.
Lesson 05

The quadrilateral and three hypotheses

For the general reader. Take a line segment, raise two equal perpendiculars at its ends, and join their tops. Without the parallel postulate, nobody can prove that the top angles are right angles. Saccheri’s idea was to split geometry into three cases according to these two angles, and to follow each case to the end.

Goals
  • Prove Saccheri’s first propositions in absolute geometry.
  • Relate summit angles to the lengths of summit and base.
  • Understand why one quadrilateral decides the case for all.

All results in this lesson belong to absolute geometry: Euclid’s Book I up to Proposition 28, without the fifth postulate. Saccheri explicitly allowed himself the first 26 propositions of Euclid, together with the Archimedean axiom and a principle of continuity.

Definitions

A Saccheri quadrilateral \(ABDC\) has base \(AB\), legs \(AC\) and \(BD\) perpendicular to \(AB\) on the same side with \(AC=BD\), and summit \(CD\). Its summit angles are \(\angle ACD\) and \(\angle BDC\). The figure appears in the work of Omar Khayyam in the eleventh century. Whether Saccheri knew Khayyam’s text is not established; he did know the treatment in the school of Nasir al-Din al-Tusi through John Wallis, whose argument he criticizes.

Props. I–II (absolute geometry)

(I) The summit angles of a Saccheri quadrilateral are equal. (II) The line joining the midpoints \(M\) of \(AB\) and \(N\) of \(CD\) is perpendicular to both.

For (I), triangles \(CAB\) and \(DBA\) are congruent by SAS, so \(BC=AD\); then triangles \(ACD\) and \(BDC\) are congruent by SSS. For (II), triangles \(AMC\) and \(BMD\) are congruent (SAS), so \(MC=MD\), and triangles \(CMN\) and \(DMN\) are congruent (SSS), giving right angles at \(N\); a similar argument with \(N\) gives right angles at \(M\).

Each half, such as \(AMNC\), has three right angles. Johann Heinrich Lambert later built his own theory on this figure, now called a Lambert quadrilateral.

Definition: the three hypotheses

The common summit angle is right, obtuse or acute. Saccheri calls these the hypothesis of the right angle, of the obtuse angle and of the acute angle.

Comparing sides and angles

Lemma

Let \(ABDC\) be a quadrilateral with right angles at \(A\) and \(B\) and \(C,D\) on the same side of \(AB\). Then \(AC>BD\) if and only if \(\angle BDC>\angle ACD\).

Proof. If \(AC>BD\), choose \(E\) on \(AC\) with \(AE=BD\). Then \(ABDE\) is a Saccheri quadrilateral, so \(\angle AED=\angle BDE\) by Prop. I. The angle \(\angle AED\) is exterior to triangle \(EDC\), so it exceeds \(\angle ECD=\angle ACD\) by I.16. Since \(E\) lies inside \(\angle BDC\), \(\angle BDC>\angle BDE\). Hence \(\angle BDC>\angle ACD\). The converse follows by exchanging the roles of the legs and using Prop. I when \(AC=BD\). ∎

Prop. III

Under the hypothesis of the right, obtuse or acute angle, the summit \(CD\) is respectively equal to, shorter than, or longer than the base \(AB\).

Apply the lemma to the Lambert quadrilateral \(MNCA\) with base \(MN\) and right angles at \(M\) and \(N\): the sides are \(MA=\tfrac12AB\) and \(NC=\tfrac12CD\), and the opposite angles are \(\angle NCA\) (the summit angle) and \(\angle MAC=\pi/2\). The summit angle is less than \(\pi/2\) exactly when \(NC>MA\), that is when \(CD>AB\). The other cases are identical.

Props. V–VII

If one of the three hypotheses holds in a single Saccheri quadrilateral, it holds in every Saccheri quadrilateral.

Saccheri’s proofs combine Prop. III with a continuity argument: a summit angle that changed from acute to obtuse as the figure was varied would have to pass through a right angle, and a single right summit angle already forces right angles everywhere. The modern reason is simpler. Lesson 10 shows that the summit angle is governed by the curvature of the plane, and its sign is the same everywhere.

Exercise 5.1Complete the proof of Prop. II by showing that \(MN\perp AB\).

Show solution
Triangles \(ANC\) and \(BND\) are congruent by SAS: \(AC=BD\), \(NC=ND\), and \(\angle ACN=\angle BDN\) by Prop. I. Hence \(NA=NB\). Then triangles \(AMN\) and \(BMN\) are congruent by SSS (\(MA=MB\), \(NA=NB\), common \(MN\)), so \(\angle AMN=\angle BMN\). These are supplementary, so each is right.

Exercise 5.2Show that in a Lambert quadrilateral \(AMNC\) (right angles at \(A,M,N\)) under the acute hypothesis, each side adjacent to the acute angle is longer than the opposite side: \(AC>MN\) and \(NC>MA\).

Show solution
Apply the lemma twice. With base \(AM\) (right angles at \(A,M\)), the opposite angles to the legs \(AC\), \(MN\) are at \(N\) (right) and \(C\) (acute). Since \(\angle MNC=\pi/2>\angle ACN\), the lemma gives \(AC>MN\). With base \(MN\) the same reasoning gives \(NC>MA\).

Exercise 5.3Show that under the acute hypothesis the segment \(MN\) is the shortest segment joining the line \(AB\) to the line \(CD\) among those perpendicular to \(AB\). (This is the common perpendicular of Lesson 08.)

Show solution
Let \(P\) on \(AB\) with \(P\ne M\), and let the perpendicular at \(P\) meet line \(CD\) at \(Q\). Then \(MPQN\) is a Lambert quadrilateral with right angles at \(M,N,P\), and the angle at \(Q\) is acute because the acute hypothesis holds everywhere. By Exercise 5.2, \(PQ>MN\).
Lesson 06

Angle sums of triangles

For the general reader. At school we learn that the angles of a triangle add up to 180 degrees. Saccheri showed that this fact is tied to his quadrilateral. Under each of his three hypotheses, every triangle has angle sum equal to, more than, or less than 180 degrees, and the three cases cannot mix.

Goals
  • Prove Prop. IX for right triangles and extend it to all triangles.
  • See the converse (Prop. XV).
  • Apply the result to angles in a semicircle.
Props. IX and XV

Under the hypothesis of the right, obtuse or acute angle, the angle sum of every triangle is respectively equal to, greater than, or less than \(\pi\). Conversely, if the angle sum of a single triangle is equal to, greater than or less than \(\pi\), the corresponding hypothesis holds.

Proof sketch

Take a right triangle \(ABC\) with right angle at \(A\). Erect the perpendicular \(BD\) at \(B\) with \(BD=AC\), on the same side as \(C\), to form a Saccheri quadrilateral \(ABDC\) with summit angle \(\varphi\). Under the right hypothesis triangles \(ABC\) and \(DCB\) are congruent, so the angles of \(ABC\) at \(B\) and \(C\) sum to \(\pi/2\). Under the obtuse hypothesis \(CD<AB\) (Prop. III); comparing triangles \(ABC\) and \(DCB\), which share \(BC\) and have \(AC=BD\), the angle opposite the longer side \(AB\) is larger, so \(\angle ACB>\angle DBC=\pi/2-\angle ABC\), giving an angle sum above \(\pi\). The acute case is the mirror image. Any triangle is split into two right triangles by an altitude from its largest angle, and the angle sums add up correctly.

For the converse, the three hypotheses are exhaustive and their conclusions about triangles exclusive, so each conclusion forces its own hypothesis. Saccheri also proved a statement on angles in a semicircle: they are right, obtuse or acute according to the hypothesis (Exercise 6.1).

Corollary (Saccheri, later Legendre)

If one triangle has angle sum \(\pi\), every triangle has angle sum \(\pi\). If one triangle has angle sum less than \(\pi\), every triangle does.

Adrien-Marie Legendre rediscovered these facts around 1794, and they are often called the Saccheri–Legendre theorems. The course on Lobachevsky (Lesson 04) gives Legendre’s doubling proof that, in absolute geometry, no triangle has angle sum above \(\pi\).

Exercise 6.1Let \(AB\) be a diameter of a circle with centre \(O\), and \(P\) a point on the circle. Show that \(\angle APB\) equals half the angle sum of triangle \(APB\). Deduce Saccheri’s result on angles in a semicircle.

Show solution
\(OA=OP=OB\), so triangles \(OAP\) and \(OBP\) are isosceles: \(\angle OAP=\angle OPA=\alpha\) and \(\angle OBP=\angle OPB=\beta\). Then \(\angle APB=\alpha+\beta\), and the angle sum of \(APB\) is \(\alpha+\beta+(\alpha+\beta)=2\angle APB\). So \(\angle APB\) is right, obtuse or acute exactly when the angle sum equals, exceeds or falls short of \(\pi\), that is, by Prop. IX, according to the hypothesis.

Exercise 6.2Show that the angle sum of a convex quadrilateral is equal to, greater than, or less than \(2\pi\) according to the hypothesis. What is the angle sum of a convex \(n\)-gon under the acute hypothesis?

Show solution
A diagonal splits the quadrilateral into two triangles whose angles together make up the angles of the quadrilateral. Each triangle has sum \(=,>,<\pi\), so the total is \(=,>,<2\pi\). A convex \(n\)-gon splits into \(n-2\) triangles from one vertex, so under the acute hypothesis its angle sum is less than \((n-2)\pi\).
Lesson 07

Refuting the obtuse angle

For the general reader. On a globe, a quadrilateral with two equal sides along meridians standing on the equator has obtuse top angles. So the obtuse hypothesis is not absurd in itself. Saccheri ruled it out using a property of Euclid’s lines that great circles lack: they go on forever without returning.

Goals
  • Follow Saccheri’s route through Props. XI–XIV.
  • Locate the use of infinite lines in Euclid I.16.
  • See why the sphere escapes the argument.
Props. XI–XIII

Under the hypothesis of the right angle or of the obtuse angle, a line perpendicular to a given segment at one end and a line making an acute angle with it at the other end, on the same side, always meet. Consequently Euclid’s fifth postulate holds.

Prop. XIV

The hypothesis of the obtuse angle is entirely false, because it destroys itself.

The structure is the one of Lesson 04. From the obtuse hypothesis, Props. XI–XIII derive the fifth postulate. But the fifth postulate implies that every triangle has angle sum exactly \(\pi\) (Euclid I.32), which by Prop. XV is the right hypothesis. So the obtuse hypothesis implies its own negation. The proofs of XI and XII lay off equal steps along the oblique line and use Archimedes’ axiom to show it must eventually cross the perpendicular.

Where infinity enters

Saccheri’s argument depends on Euclid’s first 26 propositions, and these use the second postulate in a strong form: a segment can be extended indefinitely, and two lines meet at most once. The key case is I.16, the exterior angle theorem. In its proof one extends a median to twice its length, and the argument fails if the line returns on itself. On a sphere it does. In modern terms the obtuse hypothesis describes spherical or elliptic geometry, where lines are finite closed curves and I.16 fails for large triangles. The Dictionary of Scientific Biography remarks that Saccheri’s refutation is not cogent by modern lights for exactly this reason: he refuted the obtuse hypothesis only in the presence of infinite lines.

Modern version

In absolute geometry with Archimedes’ axiom, the angle sum of every triangle is at most \(\pi\) (Saccheri, Legendre). Hence the obtuse hypothesis is false.

Exercise 7.1On the unit sphere take the triangle with one vertex at the north pole and two on the equator, separated by longitude \(\theta\). Compute its angle sum and test the exterior angle theorem I.16.

Show solution
The angles at the equator are both \(\pi/2\) and the angle at the pole is \(\theta\), so the sum is \(\pi+\theta>\pi\). The exterior angle at one equatorial vertex is \(\pi/2\), which equals (and does not exceed) the interior angle at the other equatorial vertex, so I.16 fails for every \(\theta\). In Euclid’s proof one doubles the median from a vertex; on the sphere the doubled segment can run past the antipode, and the point it reaches no longer lies inside the exterior angle.

Exercise 7.2On a sphere of radius \(R\), take a Saccheri quadrilateral with base on the equator of length \(b\) and legs along meridians of length \(a\). Using spherical trigonometry one finds \(\sin\frac{s}{2R}=\cos\frac aR\sin\frac{b}{2R}\) for the summit \(s\). Check Prop. III in the obtuse case: \(s<b\).

Show solution
For \(0<a<\pi R/2\), \(\cos(a/R)<1\), so \(\sin(s/2R)<\sin(b/2R)\). Both angles lie in \((0,\pi/2)\) when \(b<\pi R\), where sine is increasing, so \(s<b\). For example \(R=1\), \(a=0.6\), \(b=1\) gives \(s\approx0.814\).
Lesson 08

The acute angle: common perpendiculars and asymptotes

For the general reader. The acute hypothesis would not die. Saccheri pursued it through two dozen propositions, deducing stranger and stranger consequences in the hope that one would contradict the others. What he found is now a consistent geometry. In it, two lines that never meet either spread apart in both directions or approach each other ever more closely in one direction without touching.

Goals
  • State Saccheri’s results on non-intersecting lines under the acute hypothesis.
  • Classify pairs of lines in the hyperbolic plane.
  • Verify the behaviour in the upper half-plane.
Prop. XVII

Under the hypothesis of the acute angle, a line perpendicular to a segment and a line oblique to it at its other end, making an acute angle on the same side, need not meet.

In the quadrilateral \(ABDC\), the line \(CD\) makes the acute angle \(\varphi\) with \(CA\), and \(AB\) is perpendicular to \(CA\). Yet \(AB\) and \(CD\) never meet, since both are perpendicular to \(MN\) (Euclid I.28). So the fifth postulate fails.

Props. XXII–XXV

Assume the acute hypothesis, and let two coplanar lines not meet. (XXII) If some transversal makes acute interior angles with both on one side and they still do not meet on that side, they have a common perpendicular. (XXIII) Otherwise they approach each other continually in one direction. (XXV) In that case their distance eventually becomes less than any given segment: the lines are asymptotic.

Saccheri also proved that from a point outside a line, the lines through the point that meet the given line are separated from those that do not by two boundary lines, which do not meet it and are asymptotic to it. A century later Lobachevsky called these boundary lines the parallels and organized his geometry around the angle they make with the perpendicular.

Theorem (classification, modern form)

In the hyperbolic plane two distinct lines are of exactly one of three kinds: (i) intersecting; (ii) ultraparallel, with a unique common perpendicular, the distance between them growing without bound on both sides of it; (iii) limiting parallel, with no common perpendicular, the distance tending to 0 in one direction and to infinity in the other.

Exercise 8.1In the upper half-plane with metric \(ds^2=(dx^2+dy^2)/y^2\), the vertical lines \(x=0\) and \(x=1\) are geodesics. The distance between \((0,y)\) and \((1,y)\) is \(\operatorname{arcosh}\!\big(1+\frac{1}{2y^2}\big)\). Show that the lines are asymptotic as \(y\to\infty\) and diverge as \(y\to0\).

Show solution
As \(y\to\infty\), \(1+\frac1{2y^2}\to1\) and the distance tends to \(0\). Since the distance from \((0,y)\) to the whole line \(x=1\) is at most this value, the lines approach each other without limit. As \(y\to0\) the argument tends to infinity, and so does the distance. This is case (iii); the common endpoint at \(\infty\) is the “point at infinity” of Lesson 09.

Exercise 8.2In the same model, the semicircles \(|z|=1\) and \(|z|=e^{d}\) (with \(d>0\)) are ultraparallel. Find their common perpendicular and the distance between them.

Show solution
The imaginary axis meets both circles at right angles, since each circle is centred on the origin. So it is the common perpendicular, and the distance is the length of the segment from \(i\) to \(e^{d}i\), which is \(\int_1^{e^d}dy/y=d\). Moving away from the axis along either circle, the distance grows without bound because the circles end at different points of the real line.

Exercise 8.3Use Exercise 5.3 to show that two ultraparallel lines have only one common perpendicular.

Show solution
If \(MN\) and \(PQ\) were two common perpendiculars, \(MPQN\) would have four right angles. But under the acute hypothesis every quadrilateral has angle sum less than \(2\pi\) (Exercise 6.2). Contradiction.
Lesson 09

“Repugnant to the nature of the straight line”

For the general reader. After thirty-two careful propositions, Saccheri declared the acute hypothesis false. His reason was not a contradiction but an appeal to what straight lines are like. He seems to have known this was weak, for he tried a second, longer refutation. It contains a mistake.

Goals
  • State Prop. XXXIII and identify the illegitimate step.
  • Understand the equidistant curve and why it is not a line.
  • Compute its length in the half-plane model.

The first refutation

Saccheri considered two asymptotic lines, as in Prop. XXV. Extending to the infinitely distant region results proved for finite figures, he argued that the two lines would meet there and have a common perpendicular at that point at infinity. Two lines sharing a point and a perpendicular there would coincide.

Prop. XXXIII

The hypothesis of the acute angle is absolutely false, because it is repugnant to the nature of the straight line.

The flaw is the passage to the limit. Asymptotic lines approach each other, but no point at infinity belongs to either of them, and properties of finite configurations need not survive the limit. Corrado Segre judged the work up to Prop. XXXII logically perfect; the break comes exactly here.

The second refutation

In the remaining propositions of Book I Saccheri tried again with the locus of points at fixed distance from a line, the curve that the advocates of “equidistant” parallels had assumed to be straight (Lesson 04). He sought to show that under the acute hypothesis an arc of this curve would have the same length as the segment of the base below it, which would make the curve a straight line and so force the right hypothesis. His own summary concedes the point: he could not reach his goal without first proving that the equidistant line equals its base. The argument he gave for that rests on a faulty limiting step about infinitesimal pieces of the curve.

Theorem (equidistant curve, modern form)

In the hyperbolic plane of curvature \(-1/k^2\), the points at distance \(h\) from a line \(\ell\) on one side form a smooth curve (a hypercycle) that is not a geodesic. The arc of it lying above a segment of \(\ell\) of length \(b\) has length \(b\cosh(h/k)\).

Exercise 9.1In the upper half-plane (\(k=1\)), take \(\ell\) to be the positive imaginary axis. Show that the ray \(\{re^{i(\pi/2-\theta)}: r>0\}\) with \(0<\theta<\pi/2\) consists of points at constant distance \(h\) from \(\ell\) with \(\cosh h=1/\cos\theta\). Then verify the length formula.

Show solution
The map \(z\mapsto\lambda z\) (\(\lambda>0\)) is an isometry fixing \(\ell\) and the ray, so all points of the ray are at the same distance from \(\ell\). The point \(e^{i(\pi/2-\theta)}=(\sin\theta,\cos\theta)\) is joined to \(i\) by the unit circle, which is perpendicular to \(\ell\); the distance along it is \(\int_0^{\theta}\frac{dt}{\cos t}\), and \(\cosh\) of that equals \(1/\cos\theta\). On the ray, \(y=r\cos\theta\) and \(ds=dr/(r\cos\theta)\), so the arc above the segment of \(\ell\) from \(i\) to \(e^bi\) has length \(b/\cos\theta=b\cosh h\).

Exercise 9.2Show that the equidistant ray of Exercise 9.1 is not a geodesic, and explain the relation with the summit of a Saccheri quadrilateral.

Show solution
Geodesics in the half-plane are vertical half-lines and semicircles centred on the real axis. A Euclidean ray from the origin, not vertical, is neither. In the quadrilateral with base on \(\ell\) and legs \(h\), the summit is the geodesic through the two top vertices, while the equidistant curve passes through them too. Between them the summit dips towards the base (Exercise 5.3), so the equidistant curve lies above it and is longer, exactly as in the hero figure.

DiscussionSome historians read Prop. XXXIII as a deliberate retreat by a cleric unwilling to publish a heterodox geometry; others take it as sincere. Using the text of the book and the fact that he added a second refutation, argue for one reading.

Lesson 10

The quadrilateral in modern geometry

For the general reader. Since Eugenio Beltrami’s work of 1868 we know that each of Saccheri’s hypotheses describes a real geometry, realized on a surface of constant curvature. In these models his quadrilateral can be computed exactly. The summit angle turns out to measure the curvature of the plane.

Goals
  • Derive exact formulas for the summit and summit angle in the hyperbolic plane.
  • Compare with the sphere and take the Euclidean limit.
  • Relate the summit angle to area.
Theorem

Let a Saccheri quadrilateral have legs \(a\), base \(b\), summit \(s\) and summit angle \(\varphi\). In the hyperbolic plane of curvature \(-1/k^2\):

\[ \sinh\frac{s}{2k}=\cosh\frac ak\,\sinh\frac{b}{2k},\qquad \cot\varphi=\sinh\frac ak\,\tanh\frac{b}{2k},\qquad \text{area}=k^2(\pi-2\varphi). \]

On a sphere of radius \(R\) (with \(a<\pi R/2\), \(b<\pi R\)):

\[ \sin\frac{s}{2R}=\cos\frac aR\,\sin\frac{b}{2R},\qquad \cot\varphi=-\sin\frac aR\,\tan\frac{b}{2R},\qquad \text{area}=R^2(2\varphi-\pi). \]

Derivation of the summit length

Use the hyperboloid model with \(k=1\): the plane is \(\{x:\langle x,x\rangle=-1,\ x_0>0\}\) with \(\langle x,y\rangle=-x_0y_0+x_1y_1+x_2y_2\), and distance given by \(\cosh d(x,y)=-\langle x,y\rangle\). Put the base on the geodesic \(x_2=0\) with \(A=(\cosh\frac b2,-\sinh\frac b2,0)\) and \(B=(\cosh\frac b2,\sinh\frac b2,0)\). The unit normal to the base along it is \(n=(0,0,1)\), so the tops of the legs are

\[ C=\Big(\cosh a\cosh\tfrac b2,\ -\cosh a\sinh\tfrac b2,\ \sinh a\Big),\qquad D=\Big(\cosh a\cosh\tfrac b2,\ \cosh a\sinh\tfrac b2,\ \sinh a\Big). \]

Then \(\cosh s=-\langle C,D\rangle=\cosh^2a\,\cosh b-\sinh^2a\). Writing \(\cosh s=1+2\sinh^2\frac s2\) and \(\cosh b=1+2\sinh^2\frac b2\) gives \(\sinh^2\frac s2=\cosh^2a\,\sinh^2\frac b2\). The angle formula follows by applying the right-angled trigonometry of the Lambert quadrilateral \(AMNC\), and the area formula from the Gauss–Bonnet theorem, which says that area equals \(k^2\) times the angle defect \(2\pi-(\pi/2+\pi/2+2\varphi)\).

Every quantity here depends on the size of the figure. In Euclid’s plane the summit angle is \(\pi/2\) whatever the size. In Saccheri’s acute plane, a quadrilateral with given summit angle has a determined size, and this is the “absolute unit of length” that Saccheri found unacceptable and Lambert found intriguing.

Exercise 10.1Take \(k=1\), \(a=1\), \(b=1\). Compute \(s\), \(\varphi\) and the area.

Show solution
\(\sinh(s/2)=\cosh1\,\sinh\tfrac12\approx1.5431\times0.5211=0.8041\), so \(s/2\approx0.7359\) and \(s\approx1.472>b\), as Prop. III requires. \(\cot\varphi=\sinh1\,\tanh\tfrac12\approx1.1752\times0.4621=0.5431\), so \(\varphi\approx1.0733\) rad \(\approx61.5^\circ\). Area \(=\pi-2\varphi\approx0.995\).

Exercise 10.2Show that both families of formulas tend to the Euclidean ones as \(k\to\infty\) or \(R\to\infty\). Find the leading correction to \(\varphi\) in the hyperbolic case.

Show solution
For large \(k\), \(\sinh(s/2k)\approx s/2k\) and \(\cosh(a/k)\approx1\), so \(s\to b\). Also \(\cot\varphi\approx\frac ak\cdot\frac{b}{2k}=\frac{ab}{2k^2}\to0\), so \(\varphi\to\pi/2\). More precisely \(\tfrac\pi2-\varphi\approx\frac{ab}{2k^2}\), so the area \(k^2(\pi-2\varphi)\approx ab\), the area of a Euclidean rectangle. The spherical case is identical with the sign reversed.

Exercise 10.3Fix the base \(b\) and let the legs \(a\to\infty\) in the hyperbolic plane. What happens to \(\varphi\) and to the area? Interpret the limit figure.

Show solution
\(\sinh(a/k)\to\infty\), so \(\cot\varphi\to\infty\) and \(\varphi\to0\). The area \(k^2(\pi-2\varphi)\) increases to \(\pi k^2\). The legs lie on two lines perpendicular to the base, which are ultraparallel. In the limit the summit becomes the line joining their two endpoints at infinity, and the figure becomes a quadrilateral with two right angles and two vertices at infinity, where the angles are 0. Its area is \(k^2(2\pi-\pi)=\pi k^2\).
Part III · Life and legacy
Lesson 11

The book of 1733 and its author’s death

Saccheri seems to have worked on the parallel problem for many years at Pavia, alongside his teaching of philosophy and theology. Euclides ab omni naevo vindicatus was dedicated to the Senate of Milan. As the work of a Jesuit it required the approval of the Inquisition, granted on 13 July 1733, and of his provincial, granted on 16 August. It was printed in Milan that year.

The book has two parts. Book I, in 39 propositions, treats the parallel postulate: Lessons 05 to 09 follow it. Book II defends Euclid’s theory of proportion in Book V of the Elements, in particular the definition of equal ratios, against critics who found it obscure. Halsted, the book’s English translator, praised this second part as evidence of Saccheri’s penetration and modernity. The title promises to remove “every blemish” from Euclid, and Saccheri in fact treated three, of which the parallel postulate is by far the most important.

Saccheri died in Milan on 25 October 1733, about two months after the second approval, aged 66. He did not see the book’s reception, and in his lifetime there was little to see.

DiscussionSaccheri accepted the first 26 propositions of Euclid without question and used them throughout. Which of these propositions would a modern axiomatization (for example Hilbert’s) treat as theorems, and which rest on assumptions that Saccheri never examined?

Lesson 12

Oblivion and rediscovery

The book was not entirely ignored. In 1763 Georg Simon Klügel, a student of Abraham Gotthelf Kästner at Göttingen, wrote a dissertation examining nearly thirty attempted proofs of the parallel postulate, Saccheri’s among them, and concluded that none succeeded. Through Klügel, Saccheri’s approach reached Johann Heinrich Lambert, whose Theorie der Parallellinien (written about 1766) pursued the acute hypothesis further using the three-right-angled quadrilateral. After that the book faded from view. Lobachevsky and Bolyai, who created hyperbolic geometry around 1830, apparently did not know it.

Rediscovery

In 1889 Eugenio Beltrami, who had given the first models of hyperbolic geometry in 1868, published a note drawing attention to Saccheri’s book and to how much of the new geometry it contained. Giovanni Vailati rediscovered the Logica demonstrativa in 1903 and placed Saccheri in the history of logic. George Bruce Halsted’s Latin–English edition appeared in 1920, and Roberto Bonola’s history of non-Euclidean geometry (1906, English 1912) gave Saccheri a central chapter. A modern edition with commentary, edited by Vincenzo De Risi, appeared in 2014.

Legacy

  • Mathematics: the Saccheri quadrilateral, the three hypotheses, and the Saccheri–Legendre theorem are standard in every course on foundations of geometry. His Props. I–XXXII are a correct and systematic development of the first theorems of hyperbolic geometry, including asymptotic and ultraparallel lines.
  • Method: he showed how far one can go in deducing consequences from a hypothesis one believes false, which is the method of independence proofs, even though he drew the wrong conclusion.
  • Logic: his analysis of definitions and existence anticipated later work on consistency and on nominal and real definitions.

Final projectWrite a two-page account for a strong undergraduate of Saccheri’s Props. I–IX, stating each in modern language, marking which axioms of absolute geometry it uses, and checking each in the Poincaré disk.

Reference

Connections

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

Teachers and advisors

  • Tommaso Ceva 1690-1694

Influenced by

Influenced

Colleagues and collaborators

  • Giovanni Ceva
  • Vincenzo Viviani

Also mentioned in this life

Fields

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Giovanni Girolamo Saccheri, MacTutor History of Mathematics.
  2. A. Natucci, Saccheri, Girolamo, Dictionary of Scientific Biography.
  3. Giovanni Girolamo Saccheri, Wikipedia.
  4. J. J. O’Connor and E. F. Robertson, Georg Simon Klügel, MacTutor (on the 1763 dissertation).

Mathematics

  1. G. Saccheri, Euclides ab omni naevo vindicatus, Milan, 1733. English translation by G. B. Halsted, Euclides Vindicatus, Open Court, Chicago, 1920; reprinted Chelsea, 1986.
  2. G. G. Saccheri, Euclid Vindicated from Every Blemish, ed. V. De Risi, transl. G. B. Halsted and L. Allegri, Birkhäuser (Springer), 2014.
  3. R. Bonola, Non-Euclidean Geometry: A Critical and Historical Study of its Development, 1912; Dover reprint. Italian original La geometria non-euclidea, 1906.
  4. Saccheri quadrilateral, Wikipedia.
  5. A. Papadopoulos and G. Théret, Hyperbolic geometry in the work of Johann Heinrich Lambert, arXiv:1503.01865.
  6. M. J. Greenberg, Euclidean and Non-Euclidean Geometries, 4th ed., W. H. Freeman, 2008.
  7. J. Gray, Ideas of Space: Euclidean, Non-Euclidean and Relativistic, 2nd ed., Oxford, 1989.
  8. B. A. Rosenfeld, A History of Non-Euclidean Geometry, Springer, 1988.