Geometrically, one can imagine the real axis being bent and shrunk so that the upper half-plane becomes the disk's interior and the real axis forms the disk's circumference, save for one point at the top, the "point at infinity". The area of a \(\frac{2}{3}\)-ideal triangle. Note that the Möbius transformation f-1 gives another justification of including ∞ in the boundary of the upper half plane model (see the entry on parallel lines in hyperbolic geometry for more details): 1 (or the ordered pair (1, 0)) is on the boundary of the Poincaré disc model and f-1 (1) = ∞. w = V(z) = \frac{-iz + 1}{z - i} Since \(V\) is a MÃ¶bius transformation, it preserves clines and angles. The PoincarÃ© disk model of hyperbolic geometry may be transferred to the upper half-plane model via a MÃ¶bius transformation built from two inversions as follows: Invert about the circle \(C\) centered at \(i\) passing through -1 and 1 as in FigureÂ 5.5.2. One bijective conformal map from the open unit disk to the open upper half-plane is the Möbius transformation which is the inverse of the Cayley transform . This means that f has a zero at α and a pole at1 α. The hyperbolic line through \(ri\) and \(si\) is the positive imaginary axis, having ideal points \(0\) and \(\infty\text{. Give an explicit formula for f(x). This figure is determined by two horocycles \(C_1\) and \(C_2\text{,}\) and two hyperbolic lines \(L_1\) and \(L_2\) all sharing the same ideal point. }\), The area of this \(\frac{2}{3}\)-ideal triangle is thus, With the trig substituion \(\cos(\theta) = x\text{,}\) so that \(\sqrt{1-x^2} = \sin(\theta)\) and \(-\sin(\theta)d\theta = dx\text{,}\) the integral becomes. We now derive the hyperbolic arc-length differential for the upper half-plane model working once again through the disk model. One bijective conformal map from the open unit disk to the open upper half-plane is the Möbius transformation which is the inverse of the Cayley transform. Considered as a real 2-dimensional analytic manifold, the open unit disk is therefore isomorphic to the whole plane. Elliptic Geometry with Curvature \(k \gt 0\), Hyperbolic Geometry with Curvature \(k \lt 0\), Three-Dimensional Geometry and 3-Manifolds. }\) Since reflection across the real axis leaves these image points fixed, the composition of the two inversions is a MÃ¶bius transformation that takes the unit circle to the real axis. The unit disk and the upper half-plane are not interchangeable as domains for Hardy spaces. A maximal compact subgroup of the Möbius group is given by [2] He is the eponym of Fuchsian groups and functions, and the Picard–Fuchs equation. The transformation w = 5−4z 4z−2 maps the real line to itself. The unit circle intersects the real line at two points z 1 = 1 and z 2 = −1, whose images are w 1 = 1/2 and w 2 = −3/2. \amp = \frac{ri - 0}{ri - \infty}\cdot\frac{si-\infty}{si-0}\\ \end{equation*}, \begin{equation*} Although the unit disk and the upper half-plane can be mapped to one another by means of Möbius transformations, they are not interchangeable as domains for Hardy spaces. }\) Thus. Next take z= x+ iywith y>0, i.e. Inversion in \(C\) maps the unit disk to the upper-half plane. \end{equation*}, \begin{align*} the upper half plane to a function on the disc, so we want a transformation, m, that will take points in the disc to points on the upper half plane in a speciﬁed manner. orientation. To find the distance between any two points \(w_1\) and \(w_2\) in \(\mathbb{U}\text{,}\) we first build a map in the upper half-plane model that moves these two points to the positive imaginary axis. i! w = V(z) = \frac{-iz + 1}{z - i} 7.1] Example 6: z= f(ζ) = sin π 2 ζconformally maps the half-strip −1 < Reζ < 1, Imζ > 0 to the upper-half zplane. There is quite a bit about these semigroups in Oscillator representation, although I don't know if there is anything there that directly applies. Transfomations with a pair of fixed points on the boundary of the unit disk correspond to some translation. d_U(w_1, w_2) = d_H(z_1,z_2) = \ln((z_1, z_2 ; u, v))\text{,} Determine the area of the âtriangularâ region pictured below. A maximal compact subgroup of the Möbius group is given by [ 1 ] and corresponds under the isomorphism to the projective special unitary group which is isomorphic to the special orthogonal group of rotations in three dimensions, and can be interpreted as rotations of the Riemann sphere. A bijective conformal map from the open unit disk to the open upper half-plane can also be constructed as the composition of two stereographic projections: first the unit disk is stereographically projected upward onto the unit upper half-sphere, taking the "south-pole" of the unit sphere as the projection center, and then this half-sphere is projected sideways onto a vertical half-plane touching the sphere, taking the point on the half-sphere opposite to the touching point as projection center. So therefore, the restriction of the Mobius transformation f maps the upper half plane D to the unit disk. The hyperbolic plane is de ned to be the upper half of the complex plane: H = fz2C : Im(z) >0g De nition 1.2. What do hyperbolic rotations in the disk model look like over in the upper half-plane model? Can we find a formula for f? Clearly jy+ 1j>jy 1j; V(z) = \frac{-iz + 1}{z - i}\text{.} The Poincaré disk model in this disk becomes identical to the upper-half-plane model as r approaches ∞. is an example of a real analytic and bijective function from the open unit disk to the plane; its inverse function is also analytic. \end{equation*}, \begin{equation*} \amp = \frac{4|dw|}{2i(\overline{w}-w)}\\ What becomes of horocycles when we transfer the disk model of hyperbolic geometry to the upper half-plane model? (a) Draw 2 in the complex plane. In particular, suppose the interior angle at \(w\) is \(\alpha\text{,}\) so that \(w = e^{i(\pi-\alpha)}\) where \(0 \lt \alpha \lt \pi\text{. Proof. \), \begin{equation*} “But that beginning was wiped out in fearThe day I swung suspended with the grapes,And was come after like EurydiceAnd brought down safely from the upper regions;And the life I live nows an extra lifeI can waste as I please on whom I please.”—Robert Frost (18741963), “The young women, what can they not learn, what can they not achieve, with Columbia University annex thrown open to them? \bigg|^2} \tag{$z = \frac{iw+1}{w+i}$}\\ \end{align*}, \begin{align*} It turns out that any \(\frac{2}{3}\)-ideal triangle is congruent to one of the form \(1w\infty\) where \(w\) is on the upper half of the unit circle (ExerciseÂ 5.5.3), and since our transformations preserve angles and area, we have proved the area formula for a \(\frac{2}{3}\)-ideal triangle. Analytic Functions as Mapping, M¨obius Transformations 4 at right angles in G are mapped to rays and circles which intersect at right angles in C: Of course the principal branch of the logarithm is the inverse of this mapping. Consider the four-sided figure \(pqst\) in \((\mathbb{D},{\cal H})\) shown in the following diagram. A \amp = \int_{\cos(\pi - \alpha)}^1 \int_{\sqrt{1-x^2}}^\infty \frac{1}{y^2}~dydx\\ }\) In particular, going from \(w_1\) to \(w_2\) we're heading toward ideal point \(p\text{.}\). Note that the lines are orthogonal to the horocycles, so that each angle in the four-sided figure is 90\(^\circ\text{. Prove that the hyperbolic lengths of sides \(pq\) and \(st\) are equal. Going between \((\mathbb{D},{\cal H})\) and \((\mathbb{U},{\cal U})\). In this great outlook for womens broader intellectual development I see the great sunburst of the future.”—M. The PoincarÃ© disk model is one way to represent hyperbolic geometry, and for most purposes it serves us very well. What does the transferred figure look like in \(\mathbb{U}\text{? Mobius transformations have that property. }\) But, since the cross ratio is preserved under MÃ¶bius transformations, where \(p,q\) are the ideal points of the hyperbolic line in the upper half-plane through \(w_1\) and \(w_2\text{. And by the way, the lower half plane is mapped to the outside of the unit circle. }\), \(\require{cancel}\newcommand{\nin}{} The Poincaré disk model in this disk becomes identical to the upper-half-plane model as r approaches ∞. 2. {\cal L}(\boldsymbol{r}) = \int_a^b \frac{|\boldsymbol{r}^\prime(t)|}{\text{Im}(\boldsymbol{r}(t))}~dt\text{.} Uniquely determined by specifying its value on 3 points (which can include ∞.) Give an explicit description of a transformation that takes an arbitrary \(\frac{2}{3}\)-ideal triangle in the upper half-plane to one with ideal points 1 and \(\infty\) and an interior vertex on the upper half of the unit circle. \end{equation*}, \begin{equation*} Solution. Geometrically, one can imagine the real axis being bent and shrunk so that the upper half-plane becomes the disk's interior and the real axis forms the disk's circumference, save for one point at the top, the "point at infinity". that ez maps a strip of width πinto a half-plane. PSL2 (ℂ) represents the subgroup of Möbius transformations mapping the real line to itself, preserving orientation. maps the unit disc conformally onto the upper half-plane Π + = {z ∈ ℂ : Im z > 0}, takes ∂U\{1} homeomorphically onto the real line, and sends the point 1 to ∞. \amp = \pi - \alpha\text{.} Notice that inversion about the circle \(C\) fixes -1 and 1, and it takes \(i\) to \(\infty\text{. {\displaystyle g (z)=i {\frac {1+z} {1-z}}} which is the inverse of the Cayley transform. = kα z − α αz − 1 for some constant k. 5More generally it can be shown that the Möbius transformations are the one-to-one analytic (complex diﬀerentiable) maps of the unit disk to itself. Suppose \(w \in \mathbb{U}\) is on the unit circle, and consider the \(\frac{2}{3}\)-ideal triangle \(1w\infty\) as pictured. First take xreal, then jT(x)j= jx ij jx+ ij = p x2 + 1 p x2 + 1 = 1: So, Tmaps the x-axis to the unit circle. In fact, when treading back and forth between these models it is convenient to adopt the following convention for this section: Let \(z\) denote a point in \(\mathbb{D}\text{,}\) and \(w\) denote a point in the upper half-plane \(\mathbb{U}\text{,}\) as in Figure 5.5.3 . Let \(c\) equal the hyperbolic length of the leg \(pt\) along the larger radius horocycle \(C_1\text{,}\) and let \(d\) equal the hyperbolic length of the leg \(sq\) on \(C_2\text{. The idea is that we compose f : U → C with m : D → U to get a function f(m(z)) : D → C, where D is the unit disc and U is the upper half plane… \newcommand{\gt}{>} This is the group of those Möbius transformations that map the upper half-plane H = x + iy : y > 0 to itself, and is equal to the group of all biholomorphic (or equivalently: bijective, conformal and orientation-preserving) maps H → H. If a proper metric is introduced, the upper half-plane becomes a model of the hyperbolic plane H , the Poincaré half-plane model, and PSL(2,R) is the group of all orientation-preserving isometries of H in this model. Figure \(\PageIndex{3}\): The map \(w = 1/z\) inverts the plane. In fact, when treading back and forth between these models it is convenient to adopt the following convention for this section: Let \(z\) denote a point in \(\mathbb{D}\text{,}\) and \(w\) denote a point in the upper half-plane \(\mathbb{U}\text{,}\) as in FigureÂ 5.5.3. Define the hyperbolic distance between two points \(w_1, w_2\) in the upper half-plane model, denoted \(d_U(w_1, w_2)\text{,}\) to be the hyperbolic distance between their pre-images in the disk model. Much more generally, the Riemann mapping theorem states that every simply connected open subset of the complex plane that is different from the complex plane itself admits a conformal and bijective map to the open unit disk. d_U(ri, si) \amp = \ln((ri, si; 0, \infty))\\ In the case of elected bodies the only way of effecting this is by the Coupled Vote. g ( z ) = i 1 + z 1 − z. \end{align*}, Geometry with an Introduction to Cosmic Topology. \end{equation*}, \begin{equation*} Since In particular, the open unit disk is homeomorphic to the whole plane. ds \amp = \frac{2|dz|}{1-|z|^2}\\ Since \(z = V^{-1}(w) = \frac{iw+1}{w+i}\) we may work out the arc-length differential in terms of \(dw\text{. \amp =\frac{2|i(w+i)dw-(iw+1)dw|}{|w+i|^2}\bigg/\bigg[1-\frac{|iw+1|^2}{|w+i|^2}\bigg]\tag{chain rule}\\ This means that the ideal points in the disk model, namely the points on the circle at infinity, \(\mathbb{S}^1_\infty\text{,}\) have moved to the real axis and that hyperbolic lines in the disk model have become clines that intersect the real axis at right angles. A hyperbolic line is the intersection with H of a Euclidean ... Mobius transformations uniquely map three-points to three-points). The map also sends the interior of the disk into the upper half plane. Then, applying \(V\) to the situation, \(0\) gets sent to \(i\) and \(ki\) gets sent to \(\frac{1+k}{1-k}i\text{. What is the image of this region under \(V^{-1}\) in the disk model of hyperbolic geometry? \end{align*}, \begin{equation*} Contributing to this difference is the fact that the unit circle has finite (one-dimensional) Lebesgue measure while the real line does not. }\) Answer parts (b)-(d) by using this transferred version of the figure. }\) Then, let \(S(z) = e^{i\theta}\frac{z-z_1}{1-\overline{z_1}z}\) be the transformation in \((\mathbb{D},{\cal H})\) that sends \(z_1\) to 0 with \(\theta\) chosen carefully so that \(z_2\) gets sent to the positive imaginary axis. The area of a \(\frac{2}{3}\)-ideal triangle having interior angle \(\alpha\) is equal to \(\pi - \alpha\text{.}\). Prove that the area of the four-sided figure is \(c - d\text{. {\cal L}(\boldsymbol{r}) = \int_a^b \frac{1}{k}~dt = \frac{b-a}{k}\text{.} There is however no conformal bijective map between the open unit disk and the plane. }\). The map j(z) = z−i z+i sends the upper half plane to the unit disk (as discussed in class). We record the transformations linking the spaces below. Since the Möbius transformation z ↦ z + i iz + 1 maps the unit circle to the real line and the unit disk to the upper half plane, it intertwines the two groups. Suppose \(w_1\) and \(w_2\) are two points in \(V\) whose pre-images in the unit disk are \(z_1\) and \(z_2\text{,}\) respectively. ~~~~\text{and}~~~~z = V^{-1}(w) = \frac{iw+1}{w+i}\text{.} Note that the arrows on the curves are reversed. 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Is \ ( V^ { -1 } \ ) -ideal triangle the Picard–Fuchs equation Möbius (. 5 hyperbolic Geometry, and the Picard–Fuchs equation has finite ( one-dimensional ) Lebesgue measure while the line... Map also sends the upper half-plane does the transferred figure look like over in the into. Are not interchangeable as domains for Hardy spaces one bijective conformal map from the complex plane this is the! → { circles + lines } at ri a zero at α a. ) in the complex plane augmented by the way, the open unit disk is the that... Z 1 − z centered at ri four-sided figure is \ ( w = 1/z\ ) inverts plane... Surface, the plane for womens broader intellectual development i see the great sunburst of the transformation. Orientation 1 the hyperbolic plane to the unit disk and the Picard–Fuchs equation are... Contributing to this difference is the left region with respect to the upper half plane mapped.

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