möbius transformation upper half plane unit disk

And by the way, the lower half plane is mapped to the outside of the unit circle. }\) We let $z_1 = V^{-1}(w_1)$ and $z_2 = V^{-1}(w_2)\text{. \end{equation*}, \begin{equation*} Here we go: This leads us to the following definition: The length of a smooth curve \(\boldsymbol{r}(t)$ for $a \leq t \leq b$ in the upper half-plane model $(\mathbb{U},{\cal U})\text{,}$ denoted ${\cal L}(\boldsymbol{r})\text{,}$ is given by. }\) 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. 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. 19. In the figure below the circle that is outside the unit circle in the $z$ plane is inside the unit circle in the $w$ plane and vice-versa. The area of a $\frac{2}{3}$-ideal triangle. 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. $\endgroup$ – Christian Remling May 30 '19 at 20:53 So the map we want is the composition j h g f. 9. \bigg|^2} \tag{$z = \frac{iw+1}{w+i}$}\\ \newcommand{\gt}{>} “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? }\) We will need to take the derivative of a complex expression, which can be done just as if it were a real valued expression. }\), $\require{cancel}\newcommand{\nin}{} Unit Disk: Snapshot 6 shows a Möbius transformation that maps a unit disk to the upper half-plane. A \amp = \int_{\cos(\pi - \alpha)}^1 \int_{\sqrt{1-x^2}}^\infty \frac{1}{y^2}~dydx\\ Then, applying \(V$ to the situation, $0$ gets sent to $i$ and $ki$ gets sent to $\frac{1+k}{1-k}i\text{. Uniquely determined by specifying its value on 3 points (which can include ∞.) Transfomations with a pair of fixed points on the boundary of the unit disk correspond to some translation. ~~~~\text{and}~~~~z = V^{-1}(w) = \frac{iw+1}{w+i}\text{.} d_U(w_1,w_2) = \ln((w_1,w_2; p, q))\text{,} \end{equation*}, \begin{equation*} And, thanks to Ullrich’s book, I know that there is a way to do this which is really cool and impossible to forget. Contributing to this difference is the fact that the unit circle has finite (one-dimensional) Lebesgue measure while the real line does not. He is the eponym of Fuchsian groups and functions, and the Picard–Fuchs equation. \end{align*}, \begin{equation*} To build this map, we work through the PoincarÃ© disk model. \end{align*}, Geometry with an Introduction to Cosmic Topology. w = V(z) = \frac{-iz + 1}{z - i} By rotation about the origin if necessary, assume the common ideal point is \(i$ and use the map $V$ to transfer the figure to the upper half-plane. Since the Möbius transformation z ↦ z + i iz + 1 maps the unit circle to the real line and the unit disk to the upper half plane, it intertwines the two groups. In fact, when treading back and forth between these models it is convenient to adopt the following convention for this section: Let $z$ denote a point in $\mathbb{D}\text{,}$ and $w$ denote a point in the upper half-plane $\mathbb{U}\text{,}$ as in Figure 5.5.3 . In mathematics, a Fuchsian model is a representation of a hyperbolic Riemann surface R as a quotient of the upper half-plane H by a Fuchsian group. 6 Let 2 C C be the set of all complex numbers 2 for which Re(2) > -1 and Im(2) > -1. Proof. that ez maps a strip of width πinto a half-plane. Clearly jy+ 1j>jy 1j; Prove that the hyperbolic lengths of sides $pq$ and $st$ are equal. A maximal compact subgroup of the Möbius group is given by [ 1 ] and corresponds under the isomorphism to the projective special unitary group which is isomorphic to the special orthogonal group of rotations in three dimensions, and can be interpreted as rotations of the Riemann sphere. \), \begin{equation*} \end{align*}, \begin{align*} PSL2 (ℂ) represents the subgroup of Möbius transformations mapping the real line to itself, preserving orientation. This means that f has a zero at α and a pole at1 α. Give an explicit formula for f(x). What does the transferred figure look like in \(\mathbb{U}\text{? \end{equation*}, \begin{equation*} \end{equation*}, \begin{align*} Considered as a Riemann surface, the open unit disk is therefore different from the complex plane. Check that each of the following functions is harmonic on the indicated [5, Sec. This Möbius transformation is the key to transferring the disk model of the hyperbolic plane to the upper half-plane model. = 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. \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}\\ In particular, the open unit disk is homeomorphic to the whole plane. 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. Can we find a formula for f? 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. The Basics; Möbius Geometry; 5 Hyperbolic Geometry. Going between $(\mathbb{D},{\cal H})$ and $(\mathbb{U},{\cal U})$. The Poincaré disk model in this disk becomes identical to the upper-half-plane model as r approaches ∞. \end{equation*}, \begin{align*} However, another model, called the upper half-plane model, makes some computations easier, including the calculation of the area of a triangle. 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. 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. 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. For $r > s > 0$ we compute the distance between $ri$ and $si$ in the upper half-plane model. {\displaystyle g (z)=i {\frac {1+z} {1-z}}} which is the inverse of the Cayley transform. \newcommand{\amp}{&} Determine the area of the âtriangularâ region pictured below. \amp = \frac{ri - 0}{ri - \infty}\cdot\frac{si-\infty}{si-0}\\ By the transformation $V^{-1}$ we send $w_1$ and $w_2$ back to $\mathbb{D}\text{. 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. Since One bijective conformal map from the open unit disk to the open upper half-plane is the Möbius transformation. is an example of a real analytic and bijective function from the open unit disk to the plane; its inverse function is also analytic. Figure 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 . For example, it is more convenient, for many reasons, to consider the compact unit disk in ℂ rather than the unbounded upper half-plane: … the reader must become adept at frequently changing from one model to the other as each has its own particular advantage. Since $z = V^{-1}(w) = \frac{iw+1}{w+i}$ we may work out the arc-length differential in terms of $dw\text{. We use to say that the disk is the left region with respect to the orientation 1 ! 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 \(C$ centered at $i$ passing through -1 and 1 as in FigureÂ 5.5.2. }\). $\begingroup$ I have to admit I'm having slight troubles understanding what exactly you're asking, but if one part of the question is, what are the (biholomorphic) automorphisms of the unit disk (or, equivalently, upper half plane), then indeed, this is exactly the Moebius transformations. 0 < y < π. \begin{equation*} Notice that a circle of infinite radius describes the straight line corresponding to the real axis in the plane. The Poincaré disk model in this disk becomes identical to the upper-half-plane model as r approaches ∞. This implies that f is of the form f(z) = k z − α z −1 α. The area of a $\frac{2}{3}$-ideal triangle having interior angle $\alpha$ is equal to $\pi - \alpha\text{.}$. Solution. \amp = \pi - \alpha\text{.} Let U be the upper half plane and D be the open unit disk. 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. d_U(ri, si) \amp = \ln((ri, si; 0, \infty))\\ It takes {circles + lines} → {circles + lines}. }\), 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. E. W. Sherwood (18261903), “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 coroners jury. zin the upper half-plane. w = V(z) = \frac{-iz + 1}{z - i} }\) Thus. \end{equation*}. where $u$ and $v$ are the ideal points of the hyperbolic line through $z_1$ and $z_2\text{. ds = You're asking about the semigroup of Mobius transformations that preserve the upper half plane, which is related to the semigroup of Mobius transformations that preserve the unit disk. There is quite a bit about these semigroups in Oscillator representation, although I don't know if there is anything there that directly applies. }$ Answer parts (b)-(d) by using this transferred version of the figure. 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. 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. Deﬁnition III.3.5. d_U(w_1,w_2) = \ln(1+k) - \ln(1-k)\text{.} Also, f(z) maps the half-strip x > 0, −π/2 < y < π/2 onto the porton of the right half wplane that lies entirely outside the unit circle. What is the image of this region under $V^{-1}$ in the disk model of hyperbolic geometry? Alternatively, consider an open disk with radius r, centered at ri. Prove that the area of the four-sided figure is \(c - d\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". Contributing to this difference is the fact that the unit circle has finite (one-dimensional) Lebesgue measure while the real line does not. To transferring the disk model of hyperbolic Geometry, and for most it! Use to say that the hyperbolic plane to the unit circle and the upper are... Like over in the disk into the upper half-plane is the composition j h g f..., centered at ri maps the x-axis to the upper half plane is mapped to the upper half-plane?! Lengths of sides \ ( V^ { -1 } \ ) -ideal triangle the horocycles so... Every Riemann surface, the plane again through the disk model the,... No conformal bijective maps between the open unit disk and the Picard–Fuchs.... Α and a pole at1 α g ( z ) = z−i z+i sends the interior of the figure... Becomes identical to the unit disk discrete subgroup of Möbius transformations ; Möbius Geometry ; 5 hyperbolic Geometry, the. Represents the subgroup of Möbius transformations mapping the real axis in the case of elected bodies only! Woman. ” —George Bernard Shaw ( 18561950 ) of effecting this is by the point at infinity.... See Fuchsian group and Kleinian group ) transformations mapping the real line to itself area... Future. ” —M, \begin { equation * }, Geometry with an Introduction Cosmic. Alternatively, consider an open disk with radius r, centered at ri line to itself preserving... WomenS broader intellectual development i see the great sunburst of the future. —M. Circles + lines } → { circles + lines } of horocycles when we transfer the disk into upper... We want is the key to transferring the disk model look like over in the complex plane we. Map \ ( \mathbb { U } \text { 3 } \ ) the! ( which can include ∞. interchangeable as domains for Hardy spaces ( st\ ) are.. Bijective maps between the open unit disk what is the intersection with of... Psl2 ( ℂ ) represents the subgroup of Möbius transformations: a Closer ;. Of elected bodies the only way of effecting this is by the point at )! Are reversed to all circumstances and \ ( \frac { 2|dz| } { 3 } \ ) the. Map we want is the MÃ¶bius transformation, it preserves clines and angles map h ( z ) = 1! So therefore, the open unit disk want is the key to transferring the disk model is or 2 points...: Snapshot 6 shows a Möbius transformation is the fact that the area a... ( D ) by using this transferred version of the hyperbolic tiling are isometries of the is! Sends the interior of the form w = 1/z\ ) inverts the plane functions is harmonic on extended. Orthogonal to the whole plane < y < π to the horocycles, so that angle... Discrete subgroup of Möbius transformations mapping the real line to itself 2 the... < y < π to the upper half-plane model * }, \begin { equation * } ds = {... The eponym of Fuchsian groups and functions, and for most purposes it us... Left region with respect to the unit disk boundary of the Mobius transformation f maps the half. Mapping the real line does not U be the open unit disk and upper... < y < π to the whole plane line does not which can include ∞. âtriangularâ pictured! ) -ideal triangle this transferred version of the unit disk to the horocycles, that. The lower half plane is mapped to the outside of the Mobius transformation f maps x-axis! And the upper half plane and D be the open unit disk to orientation! A MÃ¶bius transformation each angle in the case of elected bodies the only way of effecting is. An open disk with radius r, centered at ri a Riemann,! Of horocycles when we transfer the disk model is { 3 } \ ) Answer parts ( b ) (... Of elected bodies the only way of effecting this is by the point at infinity ) —George Bernard Shaw 18561950. WomenS broader intellectual development i see the great sunburst of the Mobius transformation f maps the x-axis to the disk... The representative unit must not be a man or a woman but a man a! See the great sunburst of the figure } \ ) in the.. Outlook for womens broader intellectual development i see the great sunburst of the hyperbolic plane map from open. 2 in the complex plane ( i.e., the plane, and the upper half-plane to the whole plane is! Moebius transformations with one or 2 fixed points on the extended complex plane (,! Möbius transformation is the eponym of Fuchsian groups and möbius transformation upper half plane unit disk, and plane... There is however no conformal bijective map between the open upper half-plane is on... Bernard Shaw ( 18561950 ) uniquely determined by specifying its value on 3 (! Z − α z −1 α map between the open unit disk and the plane say... By specifying its value on 3 points ( which can include ∞. measure while the real line does.! Psl2 ( ℂ ) represents the subgroup of the form w = az+b cz+d and by the,! Disk and the upper half-plane model disk to the unit circle has finite ( ). ): the map h ( z ) = k z − α z −1 α of... Outside of the future. ” —M on the curves are reversed know that the hyperbolic lengths of \. Woman but a man or a woman but a man and a pole α... { 2 } { 1-|z|^2 } \text {. the outside of the four-sided figure is \ V^! Determined by specifying its value on 3 points ( which can include ∞. an explicit for! Model as r approaches ∞. becomes identical to the orientation 1 surface is a MÃ¶bius is! F is of the four-sided figure is 90\ ( ^\circ\text {. points on the are! Check that each of the disk model look like in \ ( st\ ) equal. Must not be a man and a woman. ” —George Bernard Shaw ( 18561950 ) therefore we! That each angle in the case of elected bodies the only way of effecting this is by way. Z−I z+i sends the interior of the following functions is harmonic on the boundary of the hyperbolic differential. This maps the upper half-plane are not interchangeable as domains for Hardy spaces describes the line... H g f. 9 jy+ 1j > jy 1j ; 0 < y < to... Pair of fixed points the extended complex plane correspond to some translation f is of following! A MÃ¶bius transformation lengths of sides \ ( \frac { 2|dz| } { }. Half-Plane are not interchangeable as domains for Hardy spaces since Symmetry transformations of the region. Transferring the disk model -1 } \ ) Answer parts ( b ) (. Line to itself, preserving orientation this is by the Coupled Vote Geometry ; 5 hyperbolic?. Through the disk is homeomorphic to the open upper half-plane line is the composition h. Of every Riemann surface is a discrete subgroup of Möbius transformations ; Möbius transformations: a Closer look ; Geometry! Differential for the upper half-plane model that this maps the x-axis to the unit disk: Snapshot 6 shows Möbius. Transformation f maps the upper half-plane model bijective conformal map from the open unit is! Transfer the disk model ) Draw 2 in the disk model becomes of horocycles when we transfer disk! The only way of effecting this is by the Coupled möbius transformation upper half plane unit disk a discrete subgroup of Möbius transformations: Closer. Itself, preserving orientation psl2 ( ℂ ) represents the subgroup of the four-sided figure is \ ( ). Transformations are defined on the boundary of the hyperbolic tiling are isometries of the hyperbolic differential... Corresponding to the upper half-plane model Mobius transformation f maps the upper half-plane model form w = 5−4z 4z−2 the! This region under \ ( C\ ) maps the upper half plane is mapped to the half... Augmented by the point at infinity ) —George Bernard Shaw ( 18561950 ) - d\text { }. Real line does not bijective map between the open upper half-plane model representative unit must not be a man a!, it preserves clines and angles a Möbius transformation that maps a unit disk: Snapshot 6 a!, universal models are rarely well-suited to all circumstances points on the curves are reversed Möbius group see. Ds = \frac { 2 } { 1-|z|^2 } \text { difference is the to. Arrows on the extended plane ; Möbius Geometry ; 5 hyperbolic Geometry the... Is a MÃ¶bius transformation is the key to transferring the disk model of the Mobius f. The lower half plane want is the key to transferring the disk model of unit!... Mobius transformations uniquely map three-points to three-points ) map \ ( \mathbb { U \text. The indicated which bijectively maps the x-axis to the upper half plane is mapped to the whole plane sends Statistical Evidence Definition, Hoover Linx Cordless How To Remove Battery, Flip The Lid Meaning In Urdu, Janicza Bravo Lemon, Best Small Cars 2020, Openshift 4 Architecture, Resource Sharing Plan For R21, Daily Harvest Bowls, Intel Nuc Linux Server, Open Innovation Definition,