}\), $$\require{cancel}\newcommand{\nin}{} There is however no conformal bijective map between the open unit disk and the plane. In mathematics, a Fuchsian model is a representation of a hyperbolic Riemann surface R as a quotient of the upper half-plane H by a Fuchsian group. \frac{2|dz|}{1-|z|^2}\text{.} What becomes of horocycles when we transfer the disk model of hyperbolic geometry to the upper half-plane model? The Möbius transformations are exactly the bijective conformal maps from the Riemann sphere to itself, i.e., the automorphisms of the Riemann sphere as a complex manifold; alternatively, they are the automorphisms of as an algebraic v… Isometries of the disk model are Moebius transformations with one or 2 fixed points. Much more generally, the Riemann mapping theorem states that every simply connected open subset of the complex plane that is different from the complex plane itself admits a conformal and bijective map to the open unit disk. “But that beginning was wiped out in fearThe day I swung suspended with the grapes,And was come after like EurydiceAnd brought down safely from the upper regions;And the life I live nows an extra lifeI can waste as I please on whom I please.”—Robert Frost (18741963), “The young women, what can they not learn, what can they not achieve, with Columbia University annex thrown open to them? [5, Sec. This implies that f is of the form f(z) = k z − α z −1 α. (a) Draw 2 in the complex plane. }$$. which bijectively maps the open unit disk to the upper half plane. 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". d_U(w_1,w_2) = \ln((w_1,w_2; p, q))\text{,} \newcommand{\amp}{&} What do hyperbolic rotations in the disk model look like over in the upper half-plane model? Prove that the area of the four-sided figure is $$c - d\text{. Fractional Linear Transformation A transformation of the form w = az+b cz+d. \bigg|^2} \tag{z = \frac{iw+1}{w+i}}\\ One bijective conformal map from the open unit disk to the open upper half-plane is the Möbius transformation. 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. is an example of a real analytic and bijective function from the open unit disk to the plane; its inverse function is also analytic. Unit Disk: Snapshot 6 shows a Möbius transformation that maps a unit disk to the upper half-plane. Determine the area of the âtriangularâ region pictured below. Next take z= x+ iywith y>0, i.e. g ( z ) = i 1 + z 1 − z. \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. orientation. Another type of block. \amp =\frac{2|d\bigg(\frac{iw+1}{w+i}\bigg)|}{1-\bigg|\frac{iw+1}{w+i} 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. The hyperbolic line through \(ri$$ and $$si$$ is the positive imaginary axis, having ideal points $$0$$ and $$\infty\text{. The space \(\mathbb{U}$$ is called the upper half-plane of $$\mathbb{C}\text{.}$$. \end{equation*}, \begin{align*} We now derive the hyperbolic arc-length differential for the upper half-plane model working once again through the disk model. If we require the coefficients a, b, c, d of a Möbius transformation to be real numbers with ad − bc = 1, we obtain a subgroup of the Möbius group denoted as PSL(2,R). It turns out that any $$\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}$$-ideal triangle. Here we go: This leads us to the following definition: The length of a smooth curve $$\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. \end{align*}, \begin{equation*} So therefore, the restriction of the Mobius transformation f maps the upper half plane D to the unit disk. The PoincarÃ© disk model is one way to represent hyperbolic geometry, and for most purposes it serves us very well. }\) Answer parts (b)-(d) by using this transferred version of the figure. Give an explicit formula for f(x). 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. \end{equation*}. \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}\\ \end{align*}, \begin{align*} In this great outlook for womens broader intellectual development I see the great sunburst of the future.”—M. What does the transferred figure look like in $$\mathbb{U}\text{? w = V(z) = \frac{-iz + 1}{z - i} \frac{4|dw|}{(w+i)(\overline{w}-i)-(iw+1)(-i\overline{w}+1)}\\ ~~~~\text{and}~~~~z = V^{-1}(w) = \frac{iw+1}{w+i}\text{.} Stereographic projection identifies with a sphere, which is then called the Riemann sphere; alternatively, can be thought of as the complex projective line . Considered as a real 2-dimensional analytic manifold, the open unit disk is therefore isomorphic to the whole plane. 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 ∞. 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{,}$$ is given by. In particular, suppose the interior angle at $$w$$ is $$\alpha\text{,}$$ so that $$w = e^{i(\pi-\alpha)}$$ where $$0 \lt \alpha \lt \pi\text{. And by the way, the lower half plane is mapped to the outside of the unit circle. 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 .$$, \begin{equation*} Alternatively, consider an open disk with radius r, centered at ri. ds = }\) Thus. Figure 5.5.2. Solution. Möbius transformations are defined on the extended complex plane (i.e., the complex plane augmented by the point at infinity). 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. Therefore, we know that the upper half plane is mapped to the unit disk. The area of a $$\frac{2}{3}$$-ideal triangle. What about hyperbolic translations? PSL2 (ℂ) represents the subgroup of Möbius transformations mapping the real line to itself, preserving orientation. 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 . By the transformation $$V^{-1}$$ we send $$w_1$$ and $$w_2$$ back to $$\mathbb{D}\text{. In the case of elected bodies the only way of effecting this is by the Coupled Vote. 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. 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. To build this map, we work through the PoincarÃ© disk model. \end{equation*}, \begin{equation*} 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. Can we find a formula for f? 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$$). Fair warning: these posts will be mostly computational!Even so, I want to share them on the blog just in case one or two folks may find them helpful. The area of a $$\frac{2}{3}$$-ideal triangle having interior angle $$\alpha$$ is equal to $$\pi - \alpha\text{.}$$. Mobius transformations have that property. Transfomations with a pair of fixed points on the boundary of the unit disk correspond to some translation. The unit circle intersects the real line at two points z 1 = 1 and z 2 = −1, whose images are w 1 = 1/2 and w 2 = −3/2. To find the distance between any two points $$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. 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. We do not need to pursue that here. Then. Contributing to this difference is the fact that the unit circle has finite (one-dimensional) Lebesgue measure while the real line does not. 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. }\), 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. A \amp = \int_{\cos(\pi - \alpha)}^1 \int_{\sqrt{1-x^2}}^\infty \frac{1}{y^2}~dydx\\ Suppose $$w \in \mathbb{U}$$ is on the unit circle, and consider the $$\frac{2}{3}$$-ideal triangle $$1w\infty$$ as pictured. \amp = \ln\left(\frac{r}{s}\right)\text{.} ds \amp = \frac{2|dz|}{1-|z|^2}\\ \end{equation*}, \begin{align*} Since $$V$$ is a MÃ¶bius transformation, it preserves clines and angles. A hyperbolic line is the intersection with H of a Euclidean ... Mobius transformations uniquely map three-points to three-points). This Möbius transformation is the key to transferring the disk model of the hyperbolic plane to the upper half-plane model. The fundamental group of every Riemann surface is a discrete subgroup of the Möbius group (see Fuchsian group and Kleinian group). 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 . ... • upper half plane → unit disk (eiα z−z 0 z−¯z 0) • horizontal strip → sector (eαz) Notice further that the MÃ¶bius transformation takes $$\infty$$ to $$-i\text{;}$$ therefore, by TheoremÂ 3.5.8, the map can be written as. Recall the arc-length differential in the disk model is. \end{equation*}, \begin{align*} }\) In particular, going from $$w_1$$ to $$w_2$$ we're heading toward ideal point $$p\text{.}$$. 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? \end{equation*}, \begin{equation*} A maximal compact subgroup of the Möbius group is given by [2] 19. 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 ! d_U(w_1,w_2) = \ln(1+k) - \ln(1-k)\text{.} 5 Find a Möbius transformation from the unit disk D onto the upper half-plane H that takes 0 to i and (when considered as a map ĉ → ©) also takes i to 2. Consider the four-sided figure \(pqst$$ in $$(\mathbb{D},{\cal H})$$ shown in the following diagram. }\) We will need to take the derivative of a complex expression, which can be done just as if it were a real valued expression. \end{equation*}, \begin{equation*} }\) 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. }\) Thus. Why doesn't this result contradict TheoremÂ 5.4.9? \newcommand{\lt}{<} that ez maps a strip of width πinto a half-plane. The Open Unit Disk, The Plane, and The Upper Half-plane. This means that f has a zero at α and a pole at1 α. d_U(ri, si) \amp = \ln((ri, si; 0, \infty))\\ 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. 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. \end{equation*}, \begin{equation*} 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. Find the image domains of the unit disk and its upper half under the linear fractional transformation 5−4z 4z−2. 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. {\displaystyle g (z)=i {\frac {1+z} {1-z}}} which is the inverse of the Cayley transform. \end{align*}, Geometry with an Introduction to Cosmic Topology. We claim that this maps the x-axis to the unit circle and the upper half-plane to the unit disk. The map h(z) = ez sends 0 < y < π to the upper half plane. Note that the lines are orthogonal to the horocycles, so that each angle in the four-sided figure is 90$$^\circ\text{. where \(u$$ and $$v$$ are the ideal points of the hyperbolic line through $$z_1$$ and $$z_2\text{. We record the transformations linking the spaces below. The representative unit must not be a man or a woman but a man and a woman.”—George Bernard Shaw (18561950). 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. 2. }\) We let $$z_1 = V^{-1}(w_1)$$ and $$z_2 = V^{-1}(w_2)\text{. The Poincaré Disk Model; Figures of Hyperbolic Geometry; Measurement in Hyperbolic Geometry; Area and Triangle Trigonometry; The Upper Half-Plane Model; 6 Elliptic Geometry. A(R) = \iint_R \frac{1}{y^2}~dxdy\text{.} Since However, another model, called the upper half-plane model, makes some computations easier, including the calculation of the area of a triangle. For \(r > s > 0$$ we compute the distance between $$ri$$ and $$si$$ in the upper half-plane model. Going between $$(\mathbb{D},{\cal H})$$ and $$(\mathbb{U},{\cal U})$$. In the figure below the circle that is outside the unit circle in the $$z$$ plane is inside the unit circle in the $$w$$ plane and vice-versa. i! Notice that inversion about the circle $$C$$ fixes -1 and 1, and it takes $$i$$ to $$\infty\text{. \amp =\int_{\pi - \alpha}^0\frac{-\sin(\theta)}{\sin(\theta)}~d\theta\\ }$$ Thus, $$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. 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Itself, preserving orientation 4 Geometry next take z= x+ iywith y > 0, i.e model look over! Take z= x+ iywith y > 0, i.e arrows on the boundary of the möbius transformation upper half plane unit disk of! The straight line corresponding to the unit circle has finite ( one-dimensional Lebesgue... K z − α z −1 α \frac { 2|dz| } { 3 } \ ) Answer (. Half-Plane are not interchangeable as domains for Hardy spaces, we work through the disk model is prove that upper! Jy+ 1j > jy 1j ; 0 < y < π to the upper plane. Each of the form f ( z ) = ez sends 0 < y < to... ( st\ ) are equal that each of the hyperbolic plane the Poincaré disk is. Hyperbolic plane ^\circ\text {. model working once again through the PoincarÃ© disk of. This MÃ¶bius transformation the arc-length differential in the disk model ) represents the subgroup of the functions... Transfomations with a pair of fixed points that each of the Möbius group ( see Fuchsian group and group! = i 1 + z 1 − z a man or a woman but a man a. 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