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MI314 – History of Mathematics: Episodes in Non-Euclidean Geometry The Poincaré Disk Poincaré created a model of a hyperbolic plane in the Euclidean plane and then demonstrated that Hilbert’s axioms were upheld but that the parallel postulate was not. Hilbert’s Axioms Incidence.1: For every two points A, B there exists a line a that contains each of the points A, B. Incidence.2: For every two points A, B there exists no more than one line that contains each of the points A, B. Parallelism.1: Let a be any line and A a point not on it. Then there is at most one line in the plane that contains a and A that passes through A and does not intersect a. Poincaré’s Model Definition. The polar inversion, A1 , of a point, A, that is inside a circle satisfies the formula AO ¨ A1 O “ r2 , where O is the center of the circle and r is the radius. Polar inverse. To find the polar inverse of a point. Let A be a point inside the circle Γ with center O. To find its polar inverse. We join OA and draw P A K AO. We draw P A1 K P O and extend it to meet OA extended at A1 . Then, A1 is the polar inverse of A. 1 Proof. Note that △OP A and △OP A1 are both right and share =O. Hence they are similar, so that OP OA “ . OP OA1 Which we can rewrite as AO ¨ A1 O “ OP 2 “ r2 P A' Γ A O Γ' Definitions. Poincaré’s disk is a model of the hyperbolic plane, H 2 , defined as the inside of circle Γ, excluding the circumference. A point in the hyperbolic plane, a P-point, is any point inside the circle. A line in the hyperbolic plane, a P-line, is either a line through the center of Γ or a circle perpendicular to Γ. Q Γ A O B P Verification of axioms of Incidence 1.1 and 1.2. Any two P-points are contained by one and only P-line. Proof. Let there be two P-points, A and B, inside circle Γ. Then A and B ⑴ either lie on a diameter of Γ, or ⑵ not. 2 (1): If A1 and B1 lie on a diameter, then A1 B1 goes though O and is the only line passing through A1 and B1 . Hence, there is a unique P-line passing through A1 and B1 . (2): If A2 and B2 do not lie on a diameter, we find the polar inverse of A2 at A1 and pass a circle, γ, through the three points A2 , A1 and B2 . Since only one circle passes through three points, circle γ is unique. But circle γ is a P-line passing through A2 and B2 . Hence, there is a unique P-line passing through A2 and B2 . Γ B1 O A1 B2 A2 γ A' Axiom of parallelism fails. Given a P-line and a P-point not on it, a number of P-lines can be drawn through the P-point such that they do no intersect the original P-line. Proof. Let γ be a P-line in Γ and A a P-point not on it. To show that there are more than one P-lines passing through A and not intersecting γ. Take the polar inverse of A at A1 . Then any circle passing through both A and A1 will be a P-line. Hence, it will be possible to construct more than one such circle which does not intersect γ. Hence, there is more than one P-line passing through A and not intersecting γ. 3 A' A γ O Γ Angle of parallelism. The two angles of parallelism, Πpαq, associated with a given P-line and P-point are equal. Proof. Let AM B be a P-line and D a P-point not on it. Let the two P-parallel lines to AM B through D be AD and BD. To show that =ADM “ =BDM . Assume, on the contrary, that =ADM ą =BDM . Then cut off ⑴ =M DE “ =BDM. Cut off GM “ EM , so that △GM D – △EM D (SAS). Combining this with ⑴ we have =M DE “ =BDM “ =M DG. Therefore DB coincides with DG and intersects AB at G. But this is contrary to the assumption that they are P-parallel. Hence, =ADM “ =BDM . B D G M E A 4