The line element approach for the geometry of Poincaré disk

The geometry of Poincaré disk 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


Introduction
For vectors v p = (v 1 , v 2 ) and w p = (w 1 , w 2 ) in 2-dimensional Euclidean space R 2 , the norm of a vector 2 which is the Pythagorean theorem is defined by the dot product and the angle θ formed by two vector v, w ∈ R 2 is given by cos θ = v•w |v||w| .The arc length of a differentiable curve α(t) in R 2 from α(0) to α(1) is given by 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 R 2 .Thus by considering an inner product g(v, w) on a vector space V ⊂ R 2 and defining the arc length of a curve α(t) by 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 Poincaré disk D P = {(x, y) ∈ R 2 | x 2 + y 2 < 1} satisfying the following three observations is going to be determined.
• First, if g(v p , w p ) = f (p)(v p • w p ), then the angle defined by an inner product is equal to the angle defined by the dot product cos θ = g(v p , w p ) g(v p , v p ) g(w p , w p ) • Second, the geometry of Poincaré disk D P is assumed to be rotationally symmetric, 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(v p , w p ) = f (p)(v p • w p ) depends only on r for the polar coordinates p = (x, y) = (r cos θ, r sin θ) ∈ D P .
• Third, the Euclidean norm of a vector v p = (v 1 , v 2 ) at p = (p 1 , p 2 ) ∈ D P with (p 1 + v 1 , p 2 + v 2 ) ∈ D P must be scaled to infinity as p goes to the boundary of D P , since the boundary is considered to be a circle of radius ∞.
Under these three assumptions, one of the simplest candidates for an inner product g on Poincaré disk D P could be for all points p = (x, y) ∈ D P and scaling constant 2. The line element ds 2 of Poincaré disk is Let α(t) = (x(t), y(t)) be a differentiable curve from α(0) to α(1) in D P .The arc length of α(t) from α(0) = to α(1) is is the shortest arc length among all curve from α(0) to α(1) We are going to find a shortest path between any two points in Poincaré disk by using a distance-preserving biholomorphic mapping on Poincaré disk or a linear fractional transformation which preserves the cross ratio and the distance.We also show that Poincaré Disk is isometric to one connected component of two-sheeted hyperboloid −x 2 + y 2 + z 2 = −1 in 3-dimensional Minkowski space-time and the sum of the interior angles of a triangle, a Saccheri quadrilateral on Poincaré disk 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.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 Schwarz-Pick Theorem For a holomorphic function f : by Schwarz-Pick Theorem.So we get ( [5]) If f −1 is a holomorphic function on D P , then we have by applying the above (1).Hence we obtain for a biholomorphic function f on D P .

A shortest path between any two points in Poincaré disk
We show that a straight line connecting the origin and an arbitrary point p ∈ D P 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 where we use the inequality It means that the length of an arbitrary curve connecting the origin and an arbitrary point p ∈ D P is greater than equal to the length of a straight line which is the case of a constant θ 0 = θ(t) connecting the origin and an arbitrary point p ∈ D P .Fix two constants cos θ(t) = a and sin Also let α(t) = (0, ct) be a straight line from α(0) = (0, 0) to α(1) = (0, c) for c < 1.
Then we have is called a linear fractional transformation.Note that a Möbius transformation f for |a| 2 = aā < 1 and a ∈ C is a linear fractional transformation which is a bijective mapping on D P , since Let C be a circle with center z 0 = (x 0 , y 0 ) and radius r on the complex plane C, that is, We show that a part of C in D P which meets orthogonally at two points of the boundary of Poincaré disk is a shortest path.Rewrite (4) as Let L be a line through the origin on the complex plane C Rewrite (6) as We find the image of y−axis (that is, β = β by ( 7)) in D P by a biholomorphic function f (z) = z + a az + 1 for a real a ∈ D P . and Two circles meet at two points with x = 1 z0 , since As a goes to 0, 1 z0 goes to 0 by (8).The equations ( 8) and (9) imply that the center and radius of a circle depending on a are on the hyperbola |z 0 | 2 − r 2 = 1.As a goes to 1, 1 z0 goes to 1 and the norm of radius of a circle goes to zero by |z 0 | 2 − r 2 = 1.Two circles meet orthogonally at two points by the following two facts.The equation |z 0 | 2 = r 2 + 1 implies that the triangle consisting of three points (the origin, z 0 and one of the two meeting points) is a right triangle.Note that the position vector from the origin to a point p of the boundary of Poincaré disk is always orthogonal to the tangent vector at a point p (Fig. 1).
Finally, a rotation f (z) = e iθ z on D P is also holomorphic, it preserves the distance by (2) (Fig. 2).

The cross ratio and the distance between two points on Poincaré disk
Let α(t) = (t, 0) ⊂ D P be a differentiable curve from α(0) = (0, 0) to α(x).The shortest arc length of α(t) from α(0) to α(x) is The cross ratio [z 0 , z 1 , w 1 , w 0 ] for four points z 0 , z 1 , w 1 , w 0 ∈ C is defined by Then we see T (z 1 ) = 1, T (w 1 ) = 0, T (w 0 ) = ∞.For a real number 0 < x < 1, we have From ( 10), ( 11) and ( 12), we can define the distance from x to the origin O of D P by For 0 < iy < i, we have We find the image of y−axis by a linear fractional transformation for a real a with |a| < 1.We get the same results as in section 3. A linear fractional transformation f preserves the cross ratio So we can define the distance by where w 1 , w 0 ∈ ∂D.It is easy to check that Remark.Let us denote by H the Poincaré upper half plane.Take a bijective mapping h : Therefore we can define a inner product on H as for all p ∈ H.

The line element
Let S be a regular surface in 3-dimensional Euclidean space R 3 .The first fundamental form I p : T p S × T p S −→ R, I p (v, w) = v • w is the inner product on the tangent space T p S at p ∈ S of a surface S induced by the dot product of R 3 .Let X : for a curve c(t) = (u(t), v(t) ⊂ U ⊆ R 2 .The length of the curve from α(t 0 ) to α(t) is Therefore we have the so-called line element Note that the line element ds 2 of 2-dimensional Euclidean space R 2 with the dot product is imply the Pythagorean Theorem.Using the polar coordinates, we get y) = (r cos θ, r sin θ) (dx, dy) = (dr cos θ − r sin θdθ, dr sin θ + r cos θdθ).The line element ds 2 of Poincaré upper half plane, Poincaré disk is respectively.Let us find the line element ds 2 of Poincaré disk with respect to the polar coordinates.By (10), we can put for real 0 ≤ r < 1 r = ln 1 + r 1 − r .
Then we have sinh r = e r − e −r 2 = A rotation f (z) = e iθ z on D P preserves the isometry.Rewrite it on D P ds 2 = dr 2 + sinh 2 rdθ 2 .

Figure 4 .
Figure 4.One connected component of two-sheeted hyperboloid