Discovery of Non-Euclidean Geometry

April 24, 2013

JánosBolyai (1802-1860), Carl Friedrich Gauss (1777-1855), and Nikolai Ivanovich Lobachevsky (1792-1856) are three founders of non-Euclidean geometry. Hyperbolic geometry is, by definition, the geometry that assume all the axioms for neutral geometry and replace Hilbert’s parallel postulate by its negation, which is called the hyperbolic axiom. Hyperbolic axiom (Negation of Hilbert axiom). There exists a line l and a point P not on l such that at least two distinct lines parallel to l pass through P . Theorem 1.1. In hyperbolic geometry, all triangles have angle sum less than 180◦, and all convex quadrilaterals have angle sum less than 360◦. In particular, there is no rectangle. Proof. Trivial. Theorem 1.2 (Universal Hyperbolic Theorem). In hyperbolic geometry, for every line l and every point P not on l there pass through P at least two distinct lines parallel to l. In fact there are infinitely many lines parallel to l through P .

t t' P m

S n

S' n' l Q R R'

Figure 1: Existence of inﬁnite parallels

Proof. Drop segment PQ perpendicular to l with foot Q on l. Erect line m at P perpendicular to PQ. Then l, m are parallel. Pick a point R on l other than Q, and erect line t at R perpendicular to l. Drop line n through P perpendicular t, intersecting t at S. If S is on m, then S is the intersection m and t, and subsequently ¤P QRS is a rectangle, which is impossible in hyperbolic geometry. So point S is not on m. Hence m, n are distinct lines through P , both are parallel to l. See Figure 1. Let R0 be point on l other than Q, R, and t0 be line through R0 perpendicular to l. There exists line n0 through P perpendicular to t0, intersecting t0 at S0. If PS = PS0, then ¤RR0S0S is a rectangle, which is impossible. So PS, PS0 are distinct lines. Thus for all points R on l other than Q, the lines PS through P perpendicular to l are all distinct.

1 Deﬁnition 1. Two triangles are ∆ABC and ∆A0B0C0 said to be similar if their vertices can be put in on-to-one correspondence so that corresponding angles are congruent, i.e., ∠A, ∠B, ∠C are congruent to angles ∠A0, ∠B0, ∠C0 respectively.

Theorem 1.3 (AAA criterion for congruence of hyperbolic triangles). In hyperbolic geometry, if two triangles are similar then they are congruent.

A A'

B C B'' C''

B' C'

Figure 2: Similar triangles are congruent in hyperbolic geometry

Proof. Given similar triangles ∆ABC and ∆A0B0C0. Suppose the statement is not true, i.e., ∆ABC is not congruent to ∆A0B0C0. Then AB 6∼= A0B0, AC 6∼= A0C0, BC 6∼= B0C0; otherwise ∆ABC ∼= ∆A0B0C0 by ASA. We may assume AB < A0B0 and AC < A0C0. Lay oﬀ segment AB on ray r(A0,B0) to have point B00 such that AB ∼= A0B00 and A0 ∗ B00 ∗ B0, and lay oﬀ segment AC on ray r(A0,C0) to have point C00 such that AC ∼= A0C00 and A0 ∗ C00 ∗ C0. See Figure 2. Then ∆ABC ∼= ∆A0B00C00 by SAS. Hence

∠A0B00C00 ∼= ∠B ∼= ∠B0, ∠A0C00B00 ∼= ∠C ∼= ∠C0.

It follows that lines B00C00, B0C0 are parallel because of Congruent Corresponding Angles. So ¤B0C0C00B00 is a convex quadrilateral. Since angles ∠A0B00C00, ∠B0B00C00 are supplementary, angles ∠A0C00B00, ∠C0C00B00 are supplementary, and ∠B0 ∼= ∠A0B00C00, ∠C0 ∼= ∠A0C00B00, then the angle sum of ¤B0C0C00B00 is 360◦. This is a contradiction. Theorem 1.3 says that in hyperbolic geometry it is impossible to magnify or shrink a triangle without distortion. So in hyperbolic world photography would be inherently surrealistic. Another consequence of Theorem 1.3 is that the length of a segment may be determined by angles in hyperbolic geometry. For example, an angle of an equilateral triangle determines the length of a side uniquely. This fact is sometimes referred to that hyperbolic geometry has an absolute unit length.

2 Parallels that admit a common perpendicular

Given lines l, l0 and points A, B, C, . . . on l. Drop perpendiculars AA0,BB0,CC0,... from A, B, C, . . . to l0 with feet A0,B0,C0,... on l0 respectively. We say that A, B, C, . . . are equidistant from l0 if all these perpendicular segments are congruent to one another. See Figure 3.

Theorem 2.1 (At most two points equidistant). Given two distinct parallels l, l0 in hyperbolic geometry. Then any set of points on l equidistant from l0 contains at most two points.

Proof. Suppose it is not true, i.e., there is a set of three points A, B, C on l equidistant from l0. Then quadrilaterals ¤ABB0A0, ¤ACC0A0, ¤BCC0B0 are Saccheri quadrilaterals (the base

2 ABC l

l' A' B' C'

Figure 3: No more than two equidistant points between two parallels

angles are right angles and the sides are congruent). Then the summit angles of the Saccheri quadrilaterals are congruent, i.e.,

∠BAA0 ∼= ∠ABB0, ∠CAA0 ∼= ∠ACC0, ∠CBB0 ∼= ∠BCC0.

Thus ∠ABB0 ∼= ∠CBB0. Since ∠ABB0, ∠CBB0 are supplementary, they must be right angles. Hence all ¤ABB0A0, ¤ACC0A0, ¤BCC0B0 are rectangles, which is impossible. Lemma 2. Given a Saccheri quadrilateral ¤ABB0A0 with base right angles ∠A0, ∠B0 and equal opposite sides AA0,BB0. Let M,M 0 be the middle points of AB, A0B0 respectively. Then segment MM 0 is perpendicular to both lines AB and A0B0. Proof. Draw segments A0M and B0M. Note that AA0 ∼= BB0, ∠A ∼= ∠B, and AM ∼= BM. Then ∆A0AM ∼= ∆B0BM by SAS. So A0M ∼= B0M. Hence ∆A0MM 0 ∼= ∆B0MM 0 by SSS. We then have ∠A0M 0M ∼= ∠B0M 0M. Subsequently, ∠A0M 0M and ∠B0M 0M are right angles. So MM 0 is perpendicular to the base A0B0. Note that ∠A0MM 0 ∼= ∠B0MM 0 and

AM B

A' M' B'

Figure 4: Perpendicular middle point segment

∠AMA0 ∼= ∠BMB0. Then ∠AMM 0 ∼= ∠BMM 0 by angle addition. Subsequently, ∠AMM 0 and ∠BMM 0 are right angles. So MM 0 is perpendicular to the summit AB. Theorem 2.2 (Divergent and symmetric parallels). Let l, l0 be two lines perpendicular to a segment MM 0 with M ∈ l, M 0 ∈ l0. (a) Then |MM 0| < |XY 0| for all X ∈ l, Y 0 ∈ l0 with XY 0 6= MM 0. (b) If M is the middle point of a segment AB on l, then A, B are equidistant from l0. (c) If M ∗ B ∗ D on l and BB0,DD0 are segments perpendicular to l0 with feet B0,D0 ∈ l0, then BB0 < DD0.

Proof. (a) It is clear that MM 0 < MY 0 for all Y 0 on l0 with Y 0 6= M 0. Let X be a point on l with X 6= M. Let XX0 be segment perpendicular to l0 with foot X0 on l0. Then ¤MM 0XX0 is a Lambert quadrilateral. Thus MM 0 < XX0 by properties of Lambert quadrilaterals. Since XX0 < XY 0 for Y 0 on l0 with Y 0 6= X0. We see that MM 0 < XY 0. (b) Let AA0,BB0 be segments perpendicular to l0 with A0,B0 ∈ l0. Draw segments AM 0 and BM 0. Then ∆AMM 0 ∼= ∆BMM 0 by SAS. So AM 0 ∼= BM 0 and ∠AM 0M ∼= ∠BM 0M.

3 l C D A B M

l' C'A'M' B' D'

Figure 5: Divergent parallels are symmetric

Subsequently, ∠AM 0A0 ∼= ∠BM 0B0 by angle subtraction. Thus ∆AA0M 0 ∼= ∆BB0M 0 by SAA. Hence AA0 ∼= BB0 and A0M 0 ∼= B0M 0. (c) Note that ¤M 0B0BM and ¤M 0D0DM are Lambert quadrilaterals. Then ∠B0BM and ∠D0DM are acute angles. So ∠B0BD is obtuse, for it is supplementary to ∠B0BM. Hence ∠D0DB = ∠D0DM < ∠B0BD. Therefore BB0 < DD0 by the property of ¤B0D0DB.

Proposition 2.3 (Asymptotic and monotonic parallels). Given parallels l, l0 in hyperbolic geometry, no two points of l are equidistant from l0. Let AA0,BB0,CC0 be perpendicular segments to l0 with A ∗ B ∗ C on l and A0,B0,C0 ∈ l0. See Figure 6. (a) If AA0 < BB0, then BB0 < CC0. (b) If BB0 < CC0, then AA0 < BB0.

C l B A

l' A' B' C'

Figure 6: Monotone distance between asymptotic parallels

Proof. Consider quadrilaterals ¤A0B0BA and ¤B0C0CB. (a) Since AA0 < BB0, then ∠ABB0 < ∠BAA0. Since the angle sum of ¤A0B0BA is less than 360◦, it follows that ∠ABB0 is acute. So ∠CBB0 is obtuse. Hence ∠BCC0 must be acute, since the angle sum of ¤B0C0CB is less than 360◦. Of course ∠BCC0 < ∠CBB0, subsequently, BB0 < CC0 by the property of ¤B0C0CB. (b) Fix a point D on l with A ∗ B ∗ C ∗ D. For each X on the open ray ˚r(D,A) we write |DX| = x and deﬁne f(x) = |XX0|, where XX0 is perpendicular to l0 with foot X0 on l0. We claim that f(x) is a continuous function for x > 0. In fact, ﬁx an x0 with point X0 on l 0 0 0 0 such that |BX0| = x0. Let X0X0 be segment perpendicular to l with X0 ∈ l . Note that

0 0 0 0 0 0 |XX | ≤ |XX0| ≤ |X0X0| + |XX0|, |X0X0| ≤ |X0X | ≤ |XX | + |XX0|.

Then ½ 0 0 0 0 |XX | − |X0X0| if |XX | ≥ |XX0| |f(x) − f(x0)| = 0 0 0 0 |X0X0| − |XX | if |XX | < |XX0|

≤ |XX0| = |x − x0|.

Clearly, f(x) is continuous at x0. So f(x) is a continuous function for x > 0. Suppose AA0 > BB0. Note that AA0 6∼= CC0. If |AA0| < |CC0|, by intermediate value theorem there exists a Y with B ∗ Y ∗ C such that |YY 0| = |AA0|. Then A, Y are equidistant

4 from l0, which is impossible. If |AA0| > |CC0|, by intermediate value theorem there exists a point Z with A ∗ Z ∗ B such that |ZZ0| = |CC0|. Then C,Z are equidistant from l0, which is impossible. We then must have AA0 < BB0.

3 Limiting parallel rays

Given a line l in hyperbolic geometry and a point P not on l. Let m be a line through P parallel to l with left ray r(P,R). Drop perpendicular segment PQ to l with foot Q on l. We consider rays between r(P,Q) and r(P,R), and want to ﬁnd the critical ray r(P,X), called the left limiting parallel ray to l through P , that does not meet l but any ray between r(P,X) and r(P,Q) meets l. Likewise, there is a right limiting parallel ray to l through P on the opposite side of PQ. See Figure 7.

Theorem 3.1. Given a line l and a point P not on l in hyperbolic geometry. Let PQ be segment perpendicular to l with foot Q on l. Then there exist two non-opposite rays r(P,X), r(P,X0) on opposite sides of line PQ, satisfying the properties: (a) Each of rays r(P,X), r(P,X0) does not meet l. (b) A ray r(P,Y ) meets l if and only if it is between r(P,X) and r(P,X0). (c) ∠QP X ∼= ∠QP X0.

Proof. Let m be the line through P perpendicular to PQ. Pick a point R on the left side of m and a point R0 on the right side of m separated by P . Draw segments QR and QR0. Then all rays between r(P,Q) and r(P,R) inclusive are represented by r(P,Y ) with Y ∈ QR. See Figure 7.

RP R' m X X' U

l TS V Q V'

Figure 7: Limiting parallel rays

(a) Let Σ1 be the set of points Y ∈ r(Q, R) such that the ray r(P,Y ) does not meet l, and Σ2 the complement of Σ1 in QR. It is easy to see that both Σ1, Σ2 are convex. So Σ1, Σ2 form a Dedekind cut of QR. Then there exists a unique point X ∈ QR such that Σ1, Σ2 are two rays (one of them is an open ray) of QR separated by X. We claim that X ∈ Σ1. Suppose X ∈ Σ2, i.e., r(P,X) meets l at S. Pick a point T on l such that T ∗ S ∗ Q. Then ray r(P,T ) is between r(P,R) and r(P,X). So r(P,T ) meets RQ at U and R ∗ U ∗ X, i.e., 0 U ∈ Σ2, which is a contradiction. The existence of ray r(P,X ) is analogous. (b) Since R ∈ Σ1 and Q ∈ Σ2, we see that R ∗ X ∗ Q. It is obvious that if a ray r(P,Y ) is contained in the open half-plane opposite to H˚(m, Q) then r(P,Y ) does not meet l. We then see that a ray r(P,Y ) meets l if and only if r(P,Y ) is between r(P,X) and r(P,X0). (c) Suppose that ∠XPQ is not congruent to ∠X0PQ, say, ∠XP Q < ∠X0PQ. Find point V 0 on l such that r(P,V 0) is between r(P,Q) and r(P,X0), and ∠QP V 0 ∼= ∠QP X. Mark a point V on l such that V ∗ Q ∗ V 0 and QV ∼= QV 0. Then ∆P QV ∼= ∆P QV 0 by SAS. So ∠QP V ∼= ∠QP V 0 ∼= ∠QP X , i.e., r(P,X) meets l at V , which is a contradiction.

5 The angle ∠QP X is called the angle of parallelism at point P with respect to l, its degree measure is denoted Π(PQ)◦. We have

Π(PQ)◦ < 90◦.

4 Classiﬁcation of parallels

Theorem 4.1. Given parallel lines l, l0 in hyperbolic geometry. (a) If l contains a limiting parallel ray to l0, then l, l0 are asymptotic parallels. (b) If l does not contain limiting parallel ray to l0, then l, l0 are divergent parallels.

D l C R P m B X A S l' T A' B' Q C' D'

Figure 8: The limiting parallel ray is asymptotic and monotonic

Proof. Fix a point P not on l0 and drop a perpendicular PQ to l0 with foot Q ∈ l0. Let m be the line through P perpendicular to PQ. Pick a point R on m other than P . Let r(P,X) be a limiting parallel ray to l0 with X ∈ QR. See Figure 8. (a) Let A, B, C, D be points on l with A ∗ B ∗ P ∗ C ∗ D and A, B ∈ ˚r(P,X). Let AA0,BB0,CC0,DD0 be segments perpendicular to l0 with feet A0,B0,C0,D0 ∈ l0. Note that ∠XPQ is acute, ∠CPQ is obtuse, and the angle sum of ¤P QC0C is less than 360◦. Then ∠PCC0 is acute. Of course ∠PCC0 < ∠CPQ. So P Q < CC0 by property of quadrilaterals with two base right angles. Analogously, ∠PDD0 is acute and ∠DCC0 is obtuse. Of course ∠PDD0 < ∠DCC0. Then CC0 < DD0 by property of quadrilaterals with two base right angles. We claim |AA0| ≤ |PQ| for all A on open ray ˚r(P,X). Suppose |AA0| > |PQ|. Let 0 1 0 0 S be a point on AA such that |AS| = 2 (|AA | − |PQ|). Clearly, |SA | > |PQ|. Then ∠A0SP < ∠XPQ by property of quadrilateral with two base right angles. Of course ∠A0SP is acute. Since r(P,S) is between r(P,Q) and r(P,X), the ray r(P,S) meets l0 at T . Note that ∠A0ST is acute. So ∠A0SP is obtuse, contradicting to that ∠A0SP is acute. We further claim AA0 < BB0 for two points A, B on closed ray r(P,X) with A ∗ B ∗ C. Suppose |AA0| ≥ |BB0|. There exists a point E on BP (maybe B = P ) such that AA0 ∼= EE0 by continuity of distance function. Let M,M 0 be the middle points of AE, A0E0 respectively. Then l, l0 are divergent parallels. Let F be on l such that F ∗ M ∗ C and MF ∼= MC. We have |FF 0| = |CC0| > |PQ|, which is a contradiction. (b) Assume that l does not contain any limiting parallel ray. If l = m, then l, l0 are already divergent parallels. If l 6= m, we may assume that a ray r(P,Y ) of l is between r(P,R) and r(P,X), where R ∗ Y ∗ X. It is easy to see that P Q < CC0 < DD0 by similar arguments. Since ∠XPY is acute, by Aristotle’s axiom there exists a point A on r(P,Y ) such that AE > P Q, where AE is perpendicular to r(P,X) with foot E ∈ r(P,X). Of course AA0 >

6 D l C R P m A Y X E F

l' A' Q C' D'

Figure 9: Non-limiting parallel ray is symmetric

AF > AE. So AA0 > P Q. Thus l, l0 cannot be asymptotic (monotonic) parallels. So l, l0 must be divergent (symmetric) parallels. Let A, A0 be two distinct points on the same side of a line AB such that lines AA0, BB0 are parallel. Then the ﬁgure, consisting of the segment AB (called the base) and the rays r(A, A0) and r(B,B0) (called the sides), is called a biangle with vertices A and B, denoted ∠A0ABB0. See Figure 10. The interior of biangle ∠A0ABB0 is

˚∠A0ABB0 := ˚∠A0AB ∩ ˚∠ABB0.

If P ∈ ˚∠A0ABB0, either of rays r(A, P ), r(B,P ) is called an interior ray of ∠DABC. If

D A A'

E Q P

B B' GF

Figure 10: Biangle and limiting parallel

each interior ray r(A, P ) intersects r(B,B0), we say that r(A, A0) is limiting parallel to r(B,B0) and that biangle ∠A0ABB0 is closed at A, written r(A, A0) · r(B,B0). Lemma 3. Let ∠A0ABB0 be a biangle. See Figure 10. (a) If D ∗ A ∗ A0, then r(D,A0) · r(B,B0) if and only if r(A, A0) · r(B,B0). (b) If r(A, A0) · r(B,B0), so is r(B,B0) · r(A, A0). Proof. (a) Assume r(D,A0) · r(B,B0). Take a point P in the interior of ∠A0ABB0. It is clear that P is an interior point of biangle ∠A0DBB0. Then r(D,P ) meets r(B,B0) at F since ∠A0DBB0 is closed at D. Note that P is an interior point of ∠BAF . Then r(A, P ) is between r(A, B) and r(A, F ). Thus r(A, P ) meets BF at G with B ∗ G ∗ C. By deﬁnition r(A, A0) · r(B,B0). Conversely, assume r(A, A0) · r(B,B0). For each ray r between r(D,B) and r(D,A0), we have r meeting AB at E between A and B. Pick a point P on r such that D∗E ∗P . Note that ∠A0AB > ∠AED ∼= ∠BEP . There is a ray r(A, Q) such that ∠BAQ ∼= ∠BEP . Then r(A, Q)kr(E,P ). Since r(A, Q) meets r(B,B0), we see that r(E,P ) must meet r(B,B0), i.e., r(D,P ) meet r(B,B0). Hence ∠A0DBB0 is closed at D. (b) Given an interior point P ∈ ∠ABB˚ 0 and consider the ray r(B,P ). Suppose r(B,P ) does not meet r(A, A0). By the corollary of Aristotle’s axiom there exists a point Q on r(B,P ) such that ∠AQB < ∠PBB0. See Figure 11. Note that r(A, Q) meets r(B,B0) at C. Then we have triangle ∆BCQ. Thus ∠AQB > ∠QBC = ∠PBB0, which contradicts

7 A A'

Q P

B B' C

Figure 11: Biangle and limiting parallel

∠AQB < ∠PBB0. So r(B,P ) must meet r(A, A0). Hence ∠A0ABB0 is closed at B, i.e., r(B,B0) · r(A, A0).

Proposition 4.2 (Transitivity of limiting parallelism). If both rays r(A, A0) and r(B,B0) are limiting parallel to ray r(C,C0), then r(A, A0) and r(B,B0) are limiting parallel to each other.

A A' P C DE C'

E' B B' F

Figure 12: Limiting parallels

Proof. Case 1. Lines AA0 and BB0 are on opposite sides of line CC0. See Figure 12. It is clear that r(A, A0) and r(B,B0) have no point in common. Let AB meet CC0 at D. We may assume C ∗ D ∗ C0. Now for each point P interior to ∠A0AB, the ray r(A, P ) meets r(C,C0) at E since r(A, A0) · r(C,C0). We may assume C ∗ E ∗ C0. Then r(E,E0) meets r(B,B0) at F since r(C,C0) · r(B,B0), where E ∗ E0 ∗ F . Hence r(A, A0) meets r(B,B0) at F . Therefore by deﬁnition r(A, A0) and r(B,B0) are limiting parallel to each other. Case 2. Lines AA0 and BB0 are on the same side of line CC0.

A B' D B D' A'

C C'

Figure 13: Limiting parallels

We ﬁrst claim that AA0 and BB0 do not meet. Suppose AA0 and BB0 meet at point D. We may assume that D belongs to both rays r(A, A0), r(B,B0), and assume A ∗ D ∗ A0, B ∗ D ∗ B0. Take a point D0 such that A ∗ D ∗ D0. Then r(D,D0) meets r(C,C0) since r(D,B0) · r(C,C0), i.e., r(A, A0) meets r(C,C0), which is a contradiction. See Figure 13. Let AC meet BB0 at point D. We may assume B ∗ D ∗ B0. For each point P interior to ∠A0AC, the ray r(A, P ) meets r(C,C0) at point E. Since r(B,D) (= r(B,B0)) meets the triangle ∆DCE, the ray r(B,B0) meets either AE or CE. Since r(B,B0) does not meet r(C,C0), so r(B,B0) meet AE at F such that A ∗ F ∗ E. For point P interior to ∠BAD, the ray r(A, P ) meets BD between B and D, of course r(A, P ) meets r(B,B0). Hence r(A, A0) and r(B,B0) are limiting parallel to each other.

8 A A'

P B B' D F

C C' E

Figure 14: Limiting parallels

Two rays r, s are said to be limiting parallel, denoted r q s, if r ⊂ s or s ⊂ r or r · s. Then q is an equivalence relation on rays in hyperbolic geometry. An equivalence class of rays is called an ideal point or end, viewing it lying on each ray contained in the equivalence class. Since a point on a line separates the line into two opposite rays, and opposite rays are not equivalent, we see that every line has two ends on it. If A, B are vertices of two rays r, s with r · s. Let R denote the ideal point determined by these rays, i.e., R = [r] = [s]. We write r = AR and s = BR and refer to the closed biangle with side r, s as a singly asymptotic triangle ∆ABR. We shall see that these triangles have some properties in common with ordinary triangles.

Lemma 4. In hyperbolic geometry if two lines l, m are cut by a line t such that the alternate interior angles are congruent, then l, m are divergent parallels.

t Q' P l

m Q P'

Figure 15: Asymptotic triangle

Proof. Let t meet l at P and meet m at Q such that ∠Q0PQ ∼= ∠P 0QP , where PP 0 is perpendicular to m with foot P 0 on m and QQ0 is perpendicular to l with foot Q0 on l. Then ∆P QQ0 ∼= ∆QP P 0. So PP 0 ∼= QQ0. Hence l, m are divergent parallel lines. Proposition 4.3. Let ∆ABR be a singly asymptotic triangle with a single ideal point R. Then the exterior angles at A, B are greater than their respective opposite interior angles, i.e., ∠A < ext ∠B.

A A' R

B' B E D C

Figure 16: Asymptotic triangle

9 Proof. Extend AB to C such that A ∗ B ∗ C. Draw ray r(B,D) such that ∠CBD ∼= ∠BAA0 and extend DB to E such that E ∗ B ∗ D. Then ∠ABE ∼= ∠CBD ∼= ∠BAA0. Thus lines BD, AA0 are divergent parallels. Since r(B,B0) · r(A, A0), we see that r(B,D) 6= r(B,B0). If r(B,D) is between r(B,B0) and r(B,A), then r(B,D) meets r(A, A0), which is a contradiction. So we must have r(B,D) between r(B,C) and r(B,B0). This means that ∠CBD < ∠CBB0, i.e., ∠CBB0 > ∠BAA0.