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The Line Element Approach for the Geometry of Poincaré Disk

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The Line Element Approach for the Geometry of Poincaré Disk Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 January 2019 doi:10.20944/preprints201901.0151.v1 The line element approach for the geometry of Poincar´edisk Jong Ryul Kim Abstract The geometry of Poincar´edisk itself is interpreted without any mapping to different spaces. Our approach might be one of the shortest and is intended for educational contribution. Mathematics Subject Classification (2010) 53A35, 53C22 97-02 Keywords: Poincar´edisk; inner product; cross ratio; holomorphic function; angle; line element Email: [email protected] Address: Department of Mathematics, Kunsan National University, Kunsan, 573- 701, Republic of Korea 1. Introduction For vectors vp = (v1; v2) and wp = (w1; w2) in 2-dimensional Euclidean space 2 p p 2 2 R , the norm of a vector jvpj = vp · vp = v1 + v2 which is the Pythagorean theorem is defined by the dot product vp · wp = v1w1 + v2w2 2 v·w and the angle θ formed by two vector v; w 2 R is given by cos θ = jvjjwj . The arc length of a differentiable curve α(t) in R2 from α(0) to α(1) is given by Z 1 Z 1 jα0(t)j dt = pα0(t) · α0(t) dt 0 0 and the arc length of a piecewise differentiable curve is the sum of the arc length of differentiable parts. The distance from α(0) to α(1) is defined by the shortest arc length among all curves. We can easily show that the straight line from α(0) to α(1) is the shortest arc length when the dot product is given on R2. Thus by considering an inner product g(v; w) on a vector space V ⊂ R2 and defining the arc length of a curve α(t) by Z 1 Z 1 jα0(t)j dt = pg(α0(t); α0(t)) dt; 0 0 we can have a distance different from Euclidean geometry. A geometry where four Euclidean postulates except for the Parallel one hold is known as absolute geometry ([6]). A non-Euclidean geometry with an inner product g on Poincar´e 1 © 2019 by the author(s). Distributed under a Creative Commons CC BY license. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 January 2019 doi:10.20944/preprints201901.0151.v1 2 2 2 2 disk DP = f(x; y) 2 R j x + y < 1g satisfying the following three observations is going to be determined. • First, if g(vp; wp) = f(p)(vp · wp), then the angle defined by an inner product is equal to the angle defined by the dot product g(vp; wp) cos θ = p p g(vp; vp) g(wp; wp) f(p)(vp · wp) = p p f(p)(vp · vp) f(p)(wp · wp) v · w = p p jvpjjwpj = cos θ : • Second, the geometry of Poincar´edisk DP is assumed to be rotationally sym- metric, that is, the geometry of a neighborhood at p is isometric to that of a neighborhood at any point q related to p by rotation of any angle. It means that a function f(r; θ) = f(x; y) in g(vp; wp) = f(p)(vp · wp) depends only on r for the polar coordinates p = (x; y) = (r cos θ; r sin θ) 2 DP . • Third, the Euclidean norm of a vector vp = (v1; v2) at p = (p1; p2) 2 DP with (p1 + v1; p2 + v2) 2 DP must be scaled to infinity as p goes to the boundary of DP , since the boundary is considered to be a circle of radius 1. Under these three assumptions, one of the simplest candidates for an inner prod- uct g on Poincar´edisk DP could be 2(v · w ) g(v ; w ) = p p p p 1 − (x2 + y2) 2 for all points p = (x; y) 2 DP and scaling constant 2. The line element ds of Poincar´edisk is 4(dx2 + dy2) ds2 = : (1 − (x2 + y2))2 Let α(t) = (x(t); y(t)) be a differentiable curve from α(0) to α(1) in DP . The arc length of α(t) from α(0) = to α(1) is Z 1 Z 1 0 p 0 0 2jα (t)j g(α (t); α (t)) dt = 2 dt; 0 0 1 − jα(t)j where jα0(t)j = pα0(t) · α0(t) and jα(t)j2 = x(t)2 +y(t)2. The distance d(α(0); α(1)) is the shortest arc length among all curve from α(0) to α(1) Z 1 2jα0(t)j d(α(0); α(1)) = infα 2 dt : 0 1 − jα(t)j We are going to find a shortest path between any two points in Poincar´edisk by using a distance-preserving biholomorphic mapping on Poincar´edisk or a linear fractional transformation which preserves the cross ratio and the distance. We also show that Poincar´eDisk is isometric to one connected component of two-sheeted hyperboloid −x2 + y2 + z2 = −1 in 3-dimensional Minkowski space-time and the sum of the interior angles of a triangle, a Saccheri quadrilateral on Poincar´edisk is less than π, 2π, respectively. There are plenty lecture notes, papers([1],[5], [7]) and books ([2] [3],[4], [6]) on the hyperbolic geometry. The picture of the hyperbolic geometry is well-known. Here we suggest intuitive and direct approaches for the effective understanding of the hyperbolic geometry. 2. A distance-preserving biholomorphic mapping on Poincar´edisk Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 January 2019 doi:10.20944/preprints201901.0151.v1 3 A function f : DP −! DP ⊂ C is said to be holomorphic if f(x; y) = u(x; y) + iv(x; y) satisfies the Cauchy Riemann equations @u @v @v @u = ; = − : @x @y @x @y A biholomorphic function is a holomorphic function f which is bijective and whose inverse f −1 is also holomorphic. Let α(t) = x(t) + iy(t) be a differentiable curve from α(0) = z1 to α(1) = z2 in DP . For f(z1) = (f ◦ α)(0) and f(z2) = (f ◦ α)(1), the distance d(f(z1); f(z2)) is Z 1 2j(f ◦ α)0(t)j d(f(z1); f(z2)) = inff(α) 2 dt : 0 1 − j(f ◦ α)(t)j We show that if f is a holomorphic function on DP , then d(f(z1); f(z2)) ≤ d(z1; z2). Schwarz-Pick Theorem For a holomorphic function f : DP −! DP , it holds that jf 0(z)j 1 ≤ : 1 − jf(z)j2 1 − jzj2 Proof. Define z1 − z f(z1) − z g(z) = ; h(z) = z1 2 DP : 1 − z1z 1 − f(z1)z −1 Since g−1(0) = z1, we have h f(g (0)) = 0. Using the Schwarz Lemma, we get −1 h f(g (z)) ≤ jzj: Hence −1 h f(z) = h f(g (g(z))) ≤ jg(z)j f(z ) − f(z) z − z 1 1 ≤ 1 − f(z1)f(z) 1 − z1z f(z )−f(z) 1 1 lim z1−z ≤ lim : z!z z!z 1 1 − f(z1)f(z) 1 1 − z1z So we have jf 0(z)j 1 ≤ : 1 − jf(z)j2 1 − jzj2 For z1 = α(0), z2 = α(1) and f(z1) = (f ◦ α)(0), f(z2) = (f ◦ α)(1), we get Z 1 2jf 0(α(t))jjα0(t)j Z 1 2jα0(t)j inff(α) 2 dt ≤ infα 2 dt = d(z1; z2) 0 1 − jf(α(t))j 0 1 − jα(t)j by Schwarz-Pick Theorem. So we get ([5]) d(f(z1); f(z2)) ≤ d(z1; z2): (1) −1 If f is a holomorphic function on DP , then we have −1 −1 d(z1; z2) = d(f (f(z1)); f (f(z2))) ≤ d(f(z1); f(z2)) by applying the above (1). Hence we obtain d(f(z1); f(z2)) = d(z1; z2) (2) for a biholomorphic function f on DP . Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 January 2019 doi:10.20944/preprints201901.0151.v1 4 3. A shortest path between any two points in Poincar´edisk We show that a straight line connecting the origin and an arbitrary point p 2 DP is a shortest path. Let α(t) = (x(t); y(t)) = (r(t) cos θ(t); r(t) sin θ(t)) be a differentiable curve from α(0) = (0; 0) to α(1) = (x(1); y(1)) = p for the polar coordinates. Calculations show that α0(t) = (r0(t) cos θ(t) − r(t) sin θ(t)θ0(t); r0(t) sin θ(t) + r(t) cos θ(t)θ0(t)) Z 1 Z 1 p 0 2 2 0 2 p 0 0 2 (r (t)) + (r(t)) (θ (t)) g(α (t); α (t)) dt = 2 dt 0 0 1 − r(t) Z 1 2p(r0(t))2 ≥ 2 dt 0 1 − r(t) Z 1 2jr0(t)j = 2 dt; 0 1 − r(t) where we use the inequality p(r0(t))2 + (r(t))2(θ0(t))2 ≥ p(r0(t))2: It means that the length of an arbitrary curve connecting the origin and an arbitrary point p 2 DP is greater than equal to the length of a straight line which is the case of a constant θ0 = θ(t) connecting the origin and an arbitrary point p 2 DP . Fix two constants cos θ(t) = a and sin θ(t) = pb. Let α(t) = (at; bt) be a straight line from α(0) = (0; 0) to α(1) = (a; b) for c = a2 + b2 < 1. Then we have p Z 1 Z 1 2 2 p 0 0 2 a + b g(α (t); α (t)) dt = 2 2 2 dt 0 0 1 − (a + b )t Z 1 2c = 2 2 dt 0 1 − c t Z 1 c −c = − dt 0 1 + ct 1 − ct 1 + c = ln : 1 − c Also let α(t) = (0; ct) be a straight line from α(0) = (0; 0) to α(1) = (0; c) for c < 1.
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