Chapter 6: Hyperbolic geometry §6.1 Basic Theorems of Hyperbolic Geometry.
MTH 411/511
Foundations of Geometry
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 It’s good to have goals
Goals for today: • Introduce hyperbolic geometry. • Review basic theorems in hyperbolic geometry and properties of quadrilaterals. • Prove the AAA congruence condition.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Universal Hyperbolic Theorem
First we recall a result from neutral geometry.
The Universal Hyperbolic Theorem (Theorem 4.9.1)
If there exists one line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0, then for every line ` and for every external point P there exist at least two lines that pass through P and are parallel to `.
(Corollary 4.9.2) The Universal Hyperbolic Theorem is equivalent to the negation of the Euclidean Parallel Postulate.
(Corollary 4.9.3) In any model for neutral geometry either the Euclidean Parallel Postulate or the Hyperbolic Parallel Postulate will hold.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Hyperbolic geometry
In hyperbolic geometry we assume all of the axioms of neutral geometry and the Hyperbolic Parallel Postulate. Our goal in this section will be to develop enough of an understanding of Hyperbolic geometry so that we can outline a model (the Poincar´e Disk Model).
Hyperbolic Parallel Postulate For every line ` and for every point P that does not lie on `, there are at least two lines m and n such that P lies on both m and n and both m and n are parallel to `.
Hence, by (the corollary to) the Universal Hyperbolic Theorem, any statement equivalent to the Euclidean Parallel Postulate is false in hyperbolic geometry. Thus, the next results are immediate.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Basic theorems in hyperbolic geometry
(Theorem 6.1.1) For every triangle 4ABC, σ(4ABC) < 180◦.
(Corollary 6.1.2) For every triangle 4ABC, 0◦ < δ(4ABC) < 180◦.
(Theorem 6.1.3) ◦ For every convex quadrilateral ABCD, σ(ABCD) < 360 .
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Basic theorems in hyperbolic geometry
(Corollary 6.1.4) The summit angles in a Saccheri quadrilateral are acute.
(Corollary 6.1.4) The fourth angles in a Lamber quadrilateral is acute.
(Theorem 6.1.6) There does not exist a rectangle.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Quadrilaterals
(Theorem 6.1.7) In a Lambert quadrilateral, the length of a side between two right angles is strictly less than the length of the opposite side.
Proof. Let ABCD be a Lambert quadrilateral. Assume that the angles at vertices A, B, and C are right angles. We claim BC < AD. Since BC ≤ AD (Properties of Lambert quadrilaterals), it suffices to prove BC 6= AD.
Suppose BC = AD (RAA hypothesis). Then ABCD is a Saccheri quadrilateral so ∼ ∠BAD = ∠CDA (Properties of Saccheri quadrilaterals). Hence, ABCD is a rectangle, contradicting Theorem 6.1.6. So we reject the RAA hypothesis and conclude that BC 6= AD.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Quadrilaterals
Definition 1
Let ABCD be a Saccheri quadrilateral with base AB. The segment joining the midpoint of AB to the midpoint of CD is called the altitude of the quadrilateral. The length of the altitude is called the height of the quadrilateral.
By properties of Saccheri quadrilaterals, the altitude is perpendicular to the base and the summit.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Quadrilaterals
(Corollary 6.1.9) In a Saccheri quadrilateral, the length of the altitude is less than the length of a side.
(Corollary 6.1.10) In a Saccheri quadrilateral, the length of the summit is greater than the length of the base.
The next result proves that in hyperbolic geometry, similarity implies congruence in triangles.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Similar/Congruent Triangles
AAA (Theorem 6.1.1) If 4ABC is similar to 4DEF , then 4ABC is congruent to 4DEF .
Proof. Let 4ABC and 4DEF be two triangles such that 4ABC ∼ 4DEF . We claim 4ABC =∼ 4DEF .
Suppose one pair of sides are congruent. Then 4ABC =∼ 4DEF (ASA). We now assume there are no pair of congruent sides (RAA hypothesis). WLOG, we assume AB > DE and AC > DF . −→ −→ Choose a point B0 on AB such that AB0 = DE and a point C 0 on AC such that 0 0 0 0 0 ∼ AC = DF (PCP). Then BCC B is convex (Theorem 4.6.7) and 4AB C = 4DEF 0 0 ∼ 0 0 ∼ (SAS). Thus ∠AB C = ∠ABC and ∠AC B = ∠ACB (definition of congruent 0 0 0 0 0 0 0 0 triangle). Since µ(∠CC B ) = 180 − µ(∠AC B ) and µ(∠BB C ) = 180 − µ(∠AB C ) 0 0 ◦ (LPT), then σ(BCC B ) = 360 , contradicting Theorem 6.1.3. Therefore we reject the RAA hypothesis and conclude that 4ABC =∼ 4DEF .
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Saccheri quadrilaterals
(Theorem 6.1.12) 0 0 0 0 If ABCD and A B C D are two Saccheri quadrilaterals such that 0 0 0 0 ∼ 0 0 ∼ 0 0 0 0 δ(ABCD) = δ(A B C D ) and CD = C D , then ABCD = A B C D .
Proof. 0 0 0 0 Let ABCD and A B C D be two Saccheri quadrilaterals such that 0 0 0 0 ∼ 0 0 ∼ 0 0 0 0 δ(ABCD) = δ(A B C D ) and CD = C D . We claim that ABCD = A B C D . −→ WLOG, suppose AD ≤ A0D0, so also BC ≤ B0C 0. Choose a point E on AD such that −→ DE = D0A” and a point F on CB such that CF = C 0B0. We have ∼ 0 0 0 ∼ ∼ 0 0 0 ∠ADC = ∠A D C = ∠DCB = D C B (Properties of Saccheri Triangles). Hence 4EDC =∼ 4A0D0C 0 (SAS). Since Saccheri quadrilaterals are convex, then ∼ 0 0 0 ∼ 0 0 0 ∠ECF = ∠A C B (AAP) so 4ECF = ∠A C B (SAS). Thus ∠EFC is a right angle ∼ 0 0 0 0 and a similar proof shows the same for ∠FED. It follows that EFCD = A B C D . ←→ ←→ Suppose E 6= A (RAA hypothesis). Then AB k EF (AIAT), so F 6= B and ABFE is a rectangle, contradicting Theorem 6.1.3. Thus we reject the RAA hypothesis and conclude that A = E and B = F .
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Next time
Before next class: Read Section 6.2.
In the next lecture we will: • Discuss parallel and perpendicular lines in hyperbolic geometry. • Define common perpendiculars and use this to consider alternate interior angles.
MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020