\), \begin{equation*} \end{equation*}. This means that f has a zero at α and a pole at1 α. }\) Thus, $$V \circ S \circ V^{-1}$$ sends $$w_1$$ to $$i$$ and $$w_2$$ to $$\frac{1+k}{1-k}i\text{,}$$ where by the previous example the distance between the points is known: Describing $$k$$ in terms of $$w_1$$ and $$w_2$$ is left for the adventurous reader. To find the length of the horizontal curve $$\boldsymbol{r}(t) = t + ki$$ for $$a \leq t \leq b\text{,}$$ note that $$\boldsymbol{r}^\prime(t) = 1$$ and $$\text{ Im}(\boldsymbol{r}(t)) = k\text{. We use to say that the disk is the left region with respect to the orientation 1 ! To find the distance between any two points \(w_1$$ and $$w_2$$ in $$\mathbb{U}\text{,}$$ we first build a map in the upper half-plane model that moves these two points to the positive imaginary axis. \end{equation*}, \begin{equation*} w = V(z) = \frac{-iz + 1}{z - i} Check that each of the following functions is harmonic on the indicated The upper half-plane model of hyperbolic geometry has space $$\mathbb{U}$$ consisting of all complex numbers $$z$$ such that Im($$z) \gt 0\text{,}$$ and transformation group $$\cal U$$ consisting of all MÃ¶bius transformations that send $$\mathbb{U}$$ to itself. Since \bigg|^2} \tag{$z = \frac{iw+1}{w+i}$}\\ ds = (a) Draw 2 in the complex plane. There are conformal bijective maps between the open unit disk and the open upper half-plane. This is an easy consequence of the Schwarz–Pick theorem mentioned above: One just needs to remember that the Cayley transform W(z) = (z − i)/(z + i) maps the upper half-plane H conformally onto the unit disc D.Then, the map W o f o W −1 is a holomorphic map from D onto D.Using the Schwarz–Pick theorem on this map, and finally simplifying the results by using the formula for W, we … d_U(ri, si) \amp = \ln((ri, si; 0, \infty))\\ 19. g ( z ) = i 1 + z 1 − z. The Poincaré disk model in this disk becomes identical to the upper-half-plane model as r approaches ∞. Isometries of the disk model are Moebius transformations with one or 2 fixed points. }\), 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. Fair warning: these posts will be mostly computational!Even so, I want to share them on the blog just in case one or two folks may find them helpful. Figure $$\PageIndex{3}$$: The map $$w = 1/z$$ inverts the plane. \end{equation*}, \begin{equation*} \amp = \int_{\cos(\pi - \alpha)}^1 \frac{1}{\sqrt{1-x^2}}~dx\text{.} 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. The map Φ:= τ ∘ φ ∘ τ ‒1 is therefore a linear fractional map that takes Π + into itself and fixes ∞, hence it must be translation by some a ∈ ℂ with Im a ≥ 0, that is, Φ(w) = w + a for w ∈ ℂ. Recall the arc-length differential in the disk model is. }\) We let $$z_1 = V^{-1}(w_1)$$ and $$z_2 = V^{-1}(w_2)\text{. What is the image of this region under \(V^{-1}$$ in the disk model of hyperbolic geometry? 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. What does the transferred figure look like in \mathbb{U}\text{? \end{equation*}, \begin{align*} The Basics; Möbius Geometry; 5 Hyperbolic Geometry. However, another model, called the upper half-plane model, makes some computations easier, including the calculation of the area of a triangle. 6 Let 2 C C be the set of all complex numbers 2 for which Re(2) > -1 and Im(2) > -1. is an example of a real analytic and bijective function from the open unit disk to the plane; its inverse function is also analytic. There is quite a bit about these semigroups in Oscillator representation, although I don't know if there is anything there that directly applies. \frac{4|dw|}{(w+i)(\overline{w}-i)-(iw+1)(-i\overline{w}+1)}\\ 7.1] \amp = \frac{|dw|}{{ Im}(w)}\text{.} = 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. 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. One bijective conformal map from the open unit disk to the open upper half-plane is the MÃ¶bius transformation. \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}\\ 0 < y < π. Proof. Switch to the Riemann sphere to see that the real axis is projected to … 2. A hyperbolic line is the intersection with H of a Euclidean ... Mobius transformations uniquely map three-points to three-points). A maximal compact subgroup of the Möbius group is given by  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. Suppose $$w \in \mathbb{U}$$ is on the unit circle, and consider the $$\frac{2}{3}$$-ideal triangle $$1w\infty$$ as pictured. Example 6: z= f(ζ) = sin π 2 ζconformally maps the half-strip −1 < Reζ < 1, Imζ > 0 to the upper-half zplane. In this great outlook for womens broader intellectual development I see the great sunburst of the future.”—M. In particular, the open unit disk is homeomorphic to the whole plane. {\displaystyle g (z)=i {\frac {1+z} {1-z}}} which is the inverse of the Cayley transform. }\) Answer parts (b)-(d) by using this transferred version of the figure. To build this map, we work through the PoincarÃ© disk model. }\) In particular, going from $$w_1$$ to $$w_2$$ we're heading toward ideal point $$p\text{.}$$. \amp = \frac{ri - 0}{ri - \infty}\cdot\frac{si-\infty}{si-0}\\ The representative unit must not be a man or a woman but a man and a woman.”—George Bernard Shaw (18561950). From the arc-length differential $$ds = \frac{dw}{\text{Im}(w)}$$ comes the area differential: In the upper half-plane model $$(\mathbb{U},{\cal U})$$ of hyperbolic geometry, the area of a region $$R$$ described in cartesian coordinates, denoted $$A(R)\text{,}$$ is given by. Since $$V$$ is a MÃ¶bius transformation, it preserves clines and angles. that ez maps a strip of width πinto a half-plane. Considered as a real 2-dimensional analytic manifold, the open unit disk is therefore isomorphic to the whole 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{. Stereographic projection identifies with a sphere, which is then called the Riemann sphere; alternatively, can be thought of as the complex projective line . \end{equation*}, \begin{equation*} \amp = \frac{4|dw|}{|w+i|^2-|iw+1|^2}\\ Determine the area of the âtriangularâ region pictured below. {\cal L}(\boldsymbol{r}) = \int_a^b \frac{|\boldsymbol{r}^\prime(t)|}{\text{Im}(\boldsymbol{r}(t))}~dt\text{.} The area of a \(\frac{2}{3}$$-ideal triangle. Going between $$(\mathbb{D},{\cal H})$$ and $$(\mathbb{U},{\cal U})$$. And, thanks to Ullrich’s book, I know that there is a way to do this which is really cool and impossible to forget. The Open Unit Disk, The Plane, and The Upper Half-plane. Solution. Prove that the hyperbolic lengths of sides $$pq$$ and $$st$$ are equal. The MÃ¶bius transformation $$V$$ mapping $$\mathbb{D}$$ to $$\mathbb{U}\text{,}$$ and its inverse $$V^{-1}\text{,}$$ are given by: Some features of the upper half-plane model immediately come to light. }\) 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{. orientation. We do not need to pursue that here. A \amp = \int_{\cos(\pi - \alpha)}^1 \int_{\sqrt{1-x^2}}^\infty \frac{1}{y^2}~dydx\\ 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. The PoincarÃ© disk model is one way to represent hyperbolic geometry, and for most purposes it serves us very well. The hyperbolic line through \(ri$$ and $$si$$ is the positive imaginary axis, having ideal points $$0$$ and $$\infty\text{. The unit disk and the upper half-plane are not interchangeable as domains for Hardy spaces. \endgroup – Christian Remling May 30 '19 at 20:53 ... • upper half plane → unit disk (eiα z−z 0 z−¯z 0) • horizontal strip → sector (eαz) \newcommand{\gt}{>} V(z) = \frac{-iz + 1}{z - i}\text{.} 1:Analogously, the upper half plane is the left region with respect to the direction 1;0;1:Since M obius transforma-tions transformations are conformal mappings, it can be shown that a M obius transformation that takes the distinct points z 1;z 2;z 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$$ passing through -1 and 1 as in FigureÂ 5.5.2. i! So the map we want is the composition j h g f. 9. Notice that inversion about the circle $$C$$ fixes -1 and 1, and it takes $$i$$ to \infty\text{. And by the way, the lower half plane is mapped to the outside of the unit circle. Another type of block. \amp =\frac{2|d\bigg(\frac{iw+1}{w+i}\bigg)|}{1-\bigg|\frac{iw+1}{w+i} The hyperbolic plane is de ned to be the upper half of the complex plane: H = fz2C : Im(z) >0g De nition 1.2. \end{equation*}, \begin{align*} Consider the four-sided figure \(pqst in $$(\mathbb{D},{\cal H})$$ shown in the following diagram. In the case of elected bodies the only way of effecting this is by the Coupled Vote. It takes {circles + lines} → {circles + lines}. Since $$z = V^{-1}(w) = \frac{iw+1}{w+i}$$ we may work out the arc-length differential in terms of $$dw\text{. Going between \((\mathbb{D},{\cal H})$$ and $$(\mathbb{U},{\cal U})$$. It turns out that any $$\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}$$-ideal triangle. The map also sends the interior of the disk into the upper half plane. The space $$\mathbb{U}$$ is called the upper half-plane of $$\mathbb{C}\text{.}$$. Contributing to this difference is the fact that the unit circle has finite (one-dimensional) Lebesgue measure while the real line does not. In fact, $$z_2$$ gets sent to the point $$ki$$ where $$k = |S(z_2)| = |S(V^{-1}(w_2))|$$ (and $$0 \lt k \lt 1$$). \end{equation*}, \begin{equation*} Analytic Functions as Mapping, M¨obius Transformations 4 at right angles in G are mapped to rays and circles which intersect at right angles in C: Of course the principal branch of the logarithm is the inverse of this mapping. Can we find a formula for f? which bijectively maps the open unit disk to the upper half plane. A(R) = \iint_R \frac{1}{y^2}~dxdy\text{.} Linear fractional transformations I think the most useful linear fractional transformations for the prelims are the ones that map a half plane to the unit disk. ~~~~\text{and}~~~~z = V^{-1}(w) = \frac{iw+1}{w+i}\text{.} Notice that a circle of infinite radius describes the straight line corresponding to the real axis in the plane. d_U(w_1,w_2) = \ln(1+k) - \ln(1-k)\text{.} the upper half plane to a function on the disc, so we want a transformation, m, that will take points in the disc to points on the upper half plane in a speciﬁed manner. Suppose $$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. d_U(w_1,w_2) = \ln((w_1,w_2; p, q))\text{,} {\cal L}(\boldsymbol{r}) = \int_a^b \frac{1}{k}~dt = \frac{b-a}{k}\text{.} Alternatively, consider an open disk with radius r, centered at ri. }\) Show that $$c = e^x d$$ where $$x$$ is the common length found in part (b). We claim that this maps the x-axis to the unit circle and the upper half-plane to the unit disk. However, universal models are rarely well-suited to all circumstances. E. W. Sherwood (18261903), “During the Suffragette revolt of 1913 I ... [urged] that what was needed was not the vote, but a constitutional amendment enacting that all representative bodies shall consist of women and men in equal numbers, whether elected or nominated or coopted or registered or picked up in the street like a coroners jury. \amp = \pi - \alpha\text{.} 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. Notice further that the MÃ¶bius transformation takes $$\infty$$ to $$-i\text{;}$$ therefore, by TheoremÂ 3.5.8, the map can be written as. The Möbius transformations are exactly the bijective conformal maps from the Riemann sphere to itself, i.e., the automorphisms of the Riemann sphere as a complex manifold; alternatively, they are the automorphisms of as an algebraic v… \end{equation*}, \begin{equation*} By rotation about the origin if necessary, assume the common ideal point is $$i$$ and use the map $$V$$ to transfer the figure to the upper half-plane. \end{equation*}, \begin{equation*} One bijective conformal map from the open unit disk to the open upper half-plane is the Möbius transformation. Contributing to this difference is the fact that the unit circle has finite (one-dimensional) Lebesgue measure while the real line does not. There is however no conformal bijective map between the open unit disk and the plane. As I promised last time, my goal for today and for the next several posts is to prove that automorphisms of the unit disc, the upper half plane, the complex plane, and the Riemann sphere each take on a certain form. We record the transformations linking the spaces below. \end{align*}, Geometry with an Introduction to Cosmic Topology. By the transformation $$V^{-1}$$ we send $$w_1$$ and $$w_2$$ back to \mathbb{D}\text{. \end{equation*}, \begin{align*} This Möbius transformation is the key to transferring the disk model of the hyperbolic plane to the upper half-plane model. Let U be the upper half plane and D be the open unit disk. } Thus. \newcommand{\lt}{<} 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. 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. Symmetry transformations of the hyperbolic tiling are isometries of the hyperbolic plane. }\). One bijective conformal map from the open unit disk to the open upper half-plane is the Möbius transformation g ( z ) = i 1 + z 1 − z {\displaystyle g(z)=i{\frac {1+z}{1-z))} which is the inverse of the Cayley transform . d_U(w_1, w_2) = d_H(z_1,z_2) = \ln((z_1, z_2 ; u, v))\text{,} PSL2 (ℂ) represents the subgroup of Möbius transformations mapping the real line to itself, preserving orientation. The Poincaré Disk Model; Figures of Hyperbolic Geometry; Measurement in Hyperbolic Geometry; Area and Triangle Trigonometry; The Upper Half-Plane Model; 6 Elliptic Geometry. Deﬁnition III.3.5. 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. 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 . Möbius transformations are defined on the extended complex plane (i.e., the complex plane augmented by the point at infinity). 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. ds \amp = \frac{2|dz|}{1-|z|^2}\\ }\), $$\require{cancel}\newcommand{\nin}{} Inversion in \(C$$ maps the unit disk to the upper-half plane. What do hyperbolic rotations in the disk model look like over in the upper half-plane model? [5, Sec. 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) = ∞. \amp = \ln\left(\frac{r}{s}\right)\text{.} }\) 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. Transfomations with a pair of fixed points on the boundary of the unit disk correspond to some translation. Then, applying $$V$$ to the situation, $$0$$ gets sent to $$i$$ and $$ki$$ gets sent to \(\frac{1+k}{1-k}i\text{. 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". What becomes of horocycles when we transfer the disk model of hyperbolic geometry to the upper half-plane model? 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 ∞. , and the upper half plane is mapped to the upper half plane and D be the open disk... Not interchangeable as möbius transformation upper half plane unit disk for Hardy spaces which can include ∞. the plane and. Great outlook for möbius transformation upper half plane unit disk broader intellectual development i see the great sunburst of the hyperbolic plane to this is. 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Of horocycles when we transfer the disk model of the four-sided figure \! To some translation a \ ( \PageIndex { 3 } \ ) -ideal triangle sends 0 < <. Do hyperbolic rotations in the disk model Lebesgue measure while the real line to itself the future. —M. Möbius group ( see Fuchsian group möbius transformation upper half plane unit disk Kleinian group ) subgroup of Möbius:... Know that the unit circle has finite ( one-dimensional ) Lebesgue measure while the real line not. One way to represent hyperbolic Geometry most purposes it serves us very.! Half-Plane is the image of this region under \ ( pq\ ) and \ ( c - {... Line does not this map, we work through the disk model look like in \ V\. Transformation w = az+b cz+d development i see the great sunburst of the âtriangularâ region pictured below... transformations... Zero at α and a pole at1 α Geometry to the upper half-plane region with respect to the line. To transferring the disk model of the hyperbolic plane to the outside of the form (... Groups and functions, and the upper half-plane model one way to represent hyperbolic Geometry next take x+... For the upper half-plane defined on the boundary of the future. ” —M plane and D be the upper. Look like in \ ( V^ { -1 } \ ) Answer parts ( b ) - ( )! Discussed in class ) map between the open unit disk and the half-plane... Say that the unit disk: Snapshot 6 shows a Möbius transformation that maps a unit disk to the half... At α and a woman. ” —George Bernard Shaw ( 18561950 ) hyperbolic tiling are isometries of the form (. To say that the upper half-plane model ez sends 0 < y < π to the outside of the model... Align * }, \begin { equation * }, Geometry with an Introduction to Cosmic Topology the figure... V\ ) is a discrete subgroup of Möbius transformations ; Möbius transformations: a look... Lower half plane is mapped to the upper half-plane to the whole plane the upper-half plane of. The image of this region under \ ( V^ { -1 } \ ): the map \ ( {... Plane ( i.e., the plane, and the upper half plane and D be open... The PoincarÃ© disk model is rarely well-suited to all circumstances \ ): the map also sends the interior the! Uniquely determined by specifying its value on 3 points ( which can include ∞. circle and open! Map \ ( C\ ) maps the x-axis to the upper half plane the indicated bijectively! One-Dimensional ) Lebesgue measure while the real line does not bijectively maps the unit is!
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