Cross-Ratio: Z −T −1(0) T −1(1) −T −1(∞) S(Z) = Z −T −1(∞) T −1(1) −T −1(0) Which Takes the Same Points to (0,1,∞) As T

Total Page:16

File Type:pdf, Size:1020Kb

Cross-Ratio: Z −T −1(0) T −1(1) −T −1(∞) S(Z) = Z −T −1(∞) T −1(1) −T −1(0) Which Takes the Same Points to (0,1,∞) As T Differential Geometry: Willmore Flow [Discrete Willmore Flow. Bobenko and Schröder, 2005] Quaternions Quaternions are extensions of complex numbers, with 3 imaginary values instead of 1: a + ib + jc + kd Quaternions Quaternions are extensions of complex numbers, with 3 imaginary values instead of 1: a + ib + jc + kd Like the complex numbers, we can add quaternions together by summing the individual components: (a1 + ib1 + jc 1 + kd 1 ) + (a 2 + ib2 + jc 2 + kd 2 ) = (a1 +a2 )+ i (b1 + b2 )+ j (c 1 +c 2 )+ k (d1 +d 2 ) Quaternions Quaternions are extensions of complex numbers, with 3 imaginary values instead of 1: a + ib + jc + kd Like the imaginary component of complex numbers, squaring the components gives: i 2 = j 2 = k 2 = −1 Quaternions Quaternions are extensions of complex numbers, with 3 imaginary values instead of 1: a + ib + jc + kd Like the imaginary component of complex numbers, squaring the components gives: i 2 = j 2 = k 2 = −1 However, the multiplication rules are more complex: ij = k ik = − j jk = i ji = −k ki = j kj = −i Quaternions Quaternions are extensions of complex numbers, with 3 imaginary values instead of 1: a + ib + jc + kd Like the imaginary component of complex numbers, squaring the components gives: 2 2 2 Notei that= j multiplication= k = −1 of However,quaternions the multiplicationis not commutative. rules are more complex: ij = k ik = − j jk = i ji = −k ki = j kj = −i Quaternions As with complex numbers, we define the conjugate of a quaternion q=a+ib+jc+kd as: q = a −ib − jc −kd Quaternions As with complex numbers, we define the conjugate of a quaternion q=a+ib+jc+kd as: q = a −ib − jc −kd We define the norm of a quaternion q=a+ib+jc+kd as: q = a 2 + b 2 +c 2 +d 2 = qq Quaternions As with complex numbers, we define the conjugate of a quaternion q=a+ib+jc+kd as: q = a −ib − jc −kd We define the norm of a quaternion q=a+ib+jc+kd as: q = a 2 + b 2 +c 2 +d 2 = qq And, we define the reciprocal by dividing the conjugate by the square norm: 1 q = q q 2 Quaternions One way to express a quaternion is as a pair consisting of the real value and the 3D vector consisting of the imaginary components: q = (α,w ) with α = a, w = (b,c ,d ) Quaternions One way to express a quaternion is as a pair consisting of the real value and the 3D vector consisting of the imaginary components: q = (α,w ) with α = a, w = (b,c ,d ) The advantage of this representation is that it is easier to express quaternion multiplication: q1q 2 = (α1,w 1 )(α 2 ,w 2 ) = (α1α 2 − w 1,w 2 ,α1w 2 +α 2w 1 +w 1 ×w 2 ) Möbius Transformations If we think of the plane as the set of complex numbers, any Möbius transformation can be expressed as a fractional linear transformation: az + b f (z) = cz + d with ad-bc≠0. Cross Ratio A more useful description of a conformal map is in terms of the map: z − z z − z S(z) = 3 2 4 z − z4 z2 − z3 for points z2, z3, z4 in the complex plane. Cross Ratio A more useful description of a conformal map is in terms of the map: z − z z − z S(z) = 3 2 4 z − z4 z2 − z3 for points z2, z3, z4 in the complex plane. The map is a Möbius transformation and has the property that: S(z2 ) =1 S(z3 ) = 0 S(z4 ) = ∞ Cross Ratio Claim: Every Möbius transformation can be written in this form. Cross Ratio Claim: Every Möbius transformation can be written in this form. Proof: Given a Möbius transformation T, we can define the cross-ratio: z −T −1(0) T −1(1) −T −1(∞) S(z) = z −T −1(∞) T −1(1) −T −1(0) which takes the same points to (0,1,∞) as T. Cross Ratio Claim: Every Möbius transformation can be written in this form. Proof: z −T −1(0) T −1(1) −T −1(∞) S(z) = z −T −1(∞) T −1(1) −T −1(0) But then ST-1 is a transformation that takes (0,1,∞) back to (0,1,∞). Cross Ratio Claim: Every Möbius transformation can be written in this form. Proof: But if we consider the expression for the Möbius transformation: az + b f (z) = cz + d we see that the only way to have (0,1,∞) map back to itself is to have f(z) be the identity. Cross Ratio Claim: Every Möbius transformation can be written in this form. Proof: But if we consider the expression for the Möbius transformation: az + b f (z) = cz + d we see that the only way to have (0,1,∞) map back to itself is to have f(z) be the identity. Thus ST-1 is the identity so S=T. Cross Ratio Definition: Given four (distinct) points in the complex plane z1, z2, z3, z4∈C, the cross-ratio is value: z1 − z3 z2 − z4 (z1, z2 , z3 , z4 ) = z1 − z4 z2 − z3 Cross Ratio Definition: Given four (distinct) points in the complex plane z1, z2, z3, z4∈C, the cross-ratio is value: z1 − z3 z2 − z4 (z1, z2 , z3 , z4 ) = z1 − z4 z2 − z3 That is, the cross-ratio is the image of z1 under the Möbius transformation sending (z2,z3,z4) to (0,1,∞). Cross Ratio Claim 1: The cross ratio of four points z1, z2, z3, z4∈C is invariant under Möbius transformations: (z1, z2 , z3 , z4 ) = (Tz1,Tz2 ,Tz3 ,Tz4 ) Cross Ratio Claim 1: The cross ratio of four points z1, z2, z3, z4∈C is invariant under Möbius transformations: (z1, z2 , z3 , z4 ) = (Tz1,Tz2 ,Tz3 ,Tz4 ) Proof: If S is the Möbius transformation sending z to -1 (z,z2,z3,z4) then ST takes (Tz2,Tz3,Tz4) to (0,1,∞). Cross Ratio Claim 1: The cross ratio of four points z1, z2, z3, z4∈C is invariant under Möbius transformations: (z1, z2 , z3 , z4 ) = (Tz1,Tz2 ,Tz3 ,Tz4 ) Proof: If S is the Möbius transformation sending z to -1 (z,z2,z3,z4) then ST takes (Tz2,Tz3,Tz4) to (0,1,∞). But that means that: −1 ST (Tz1) = (Tz1,Tz2 ,Tz3 ,Tz4 ) Cross Ratio Claim 1: The cross ratio of four points z1, z2, z3, z4∈C is invariant under Möbius transformations: (z1, z2 , z3 , z4 ) = (Tz1,Tz2 ,Tz3 ,Tz4 ) Proof: If S is the Möbius transformation sending z to -1 (z,z2,z3,z4) then ST takes (Tz2,Tz3,Tz4) to (0,1,∞). But that means that: −1 ST (Tz1) = (Tz1,Tz2 ,Tz3 ,Tz4 ) (z1, z2 , z3 , z4 ) = (Tz1,Tz2 ,Tz3 ,Tz4 ) Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. z4 z1 z2 β z3 Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. Proof: z The angle of the product of two 4 complex numbers is the sum of their angles: z1 arg(c ⋅c ) = arg(c ) + arg(c ) 1 2 1 2 z2 iθ1 iθ2 i(θ1 +θ2 ) β r1e r2e = r1r2e z3 Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. Proof: z Thus, the angle of the cross-ratio: 4 z1 − z3 z2 − z4 (z1, z2 , z3 , z4 ) = z1 − z4 z2 − z3 is the sum: z1 z − z z − z = 1 3 + 2 4 z arg(z1, z2 , z3 , z4 ) arg arg 2 z1 − z4 z2 − z3 β z3 Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. Proof: z − z z − z 1 3 2 4 z4 (z1, z2 , z3 , z4 ) = z1 − z4 z2 − z3 But the angle of (z1-z3)/(z1-z4) is z the difference between the 1 z angles of (z3-z1) and (z4-z1). 2 β z3 Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. Proof: z − z z − z 1 3 2 4 z4 (z1, z2 , z3 , z4 ) = z1 − z4 z2 − z3 Similarly, we know that the z1 angle of (z2-z4)/(z2-z3) is… z2 β z3 Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. Proof: z − z z − z 1 3 2 4 z4 (z1, z2 , z3 , z4 ) = z1 − z4 z2 − z3 So, the angle of the cross-ratio z is the sum of the angles that 1 are opposite the edge, which is z2 β exactly π-β. z3 Cross Ratio Claim 2: Given the four points z1, z2, z3, z4∈C, the angle of the cross ratio is π-β, where β is the angle between the circum-circles. Note: For a complex number c∈C, the cosine of the angle of c is just the real part of c c=a+ib divided by its length, so we have: |c| b a Re(z , z , z , z ) Re(z , z , z , z ) cos(π − β ) = 1 2 3 4 cos β = − 1 2 3 4 (z1, z2 , z3 , z4 ) (z1, z2 , z3 , z4 ) “Extended” Cross Ratio We can extend the notion of cross-ratio to 3D by using the imaginary parts of quaternions.
Recommended publications
  • The Riemann Mapping Theorem Christopher J. Bishop
    The Riemann Mapping Theorem Christopher J. Bishop C.J. Bishop, Mathematics Department, SUNY at Stony Brook, Stony Brook, NY 11794-3651 E-mail address: [email protected] 1991 Mathematics Subject Classification. Primary: 30C35, Secondary: 30C85, 30C62 Key words and phrases. numerical conformal mappings, Schwarz-Christoffel formula, hyperbolic 3-manifolds, Sullivan’s theorem, convex hulls, quasiconformal mappings, quasisymmetric mappings, medial axis, CRDT algorithm The author is partially supported by NSF Grant DMS 04-05578. Abstract. These are informal notes based on lectures I am giving in MAT 626 (Topics in Complex Analysis: the Riemann mapping theorem) during Fall 2008 at Stony Brook. We will start with brief introduction to conformal mapping focusing on the Schwarz-Christoffel formula and how to compute the unknown parameters. In later chapters we will fill in some of the details of results and proofs in geometric function theory and survey various numerical methods for computing conformal maps, including a method of my own using ideas from hyperbolic and computational geometry. Contents Chapter 1. Introduction to conformal mapping 1 1. Conformal and holomorphic maps 1 2. M¨obius transformations 16 3. The Schwarz-Christoffel Formula 20 4. Crowding 27 5. Power series of Schwarz-Christoffel maps 29 6. Harmonic measure and Brownian motion 39 7. The quasiconformal distance between polygons 48 8. Schwarz-Christoffel iterations and Davis’s method 56 Chapter 2. The Riemann mapping theorem 67 1. The hyperbolic metric 67 2. Schwarz’s lemma 69 3. The Poisson integral formula 71 4. A proof of Riemann’s theorem 73 5. Koebe’s method 74 6.
    [Show full text]
  • The Chazy XII Equation and Schwarz Triangle Functions
    Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (2017), 095, 24 pages The Chazy XII Equation and Schwarz Triangle Functions Oksana BIHUN and Sarbarish CHAKRAVARTY Department of Mathematics, University of Colorado, Colorado Springs, CO 80918, USA E-mail: [email protected], [email protected] Received June 21, 2017, in final form December 12, 2017; Published online December 25, 2017 https://doi.org/10.3842/SIGMA.2017.095 Abstract. Dubrovin [Lecture Notes in Math., Vol. 1620, Springer, Berlin, 1996, 120{348] showed that the Chazy XII equation y000 − 2yy00 + 3y02 = K(6y0 − y2)2, K 2 C, is equivalent to a projective-invariant equation for an affine connection on a one-dimensional complex manifold with projective structure. By exploiting this geometric connection it is shown that the Chazy XII solution, for certain values of K, can be expressed as y = a1w1 +a2w2 +a3w3 where wi solve the generalized Darboux{Halphen system. This relationship holds only for certain values of the coefficients (a1; a2; a3) and the Darboux{Halphen parameters (α; β; γ), which are enumerated in Table2. Consequently, the Chazy XII solution y(z) is parametrized by a particular class of Schwarz triangle functions S(α; β; γ; z) which are used to represent the solutions wi of the Darboux{Halphen system. The paper only considers the case where α + β +γ < 1. The associated triangle functions are related among themselves via rational maps that are derived from the classical algebraic transformations of hypergeometric functions. The Chazy XII equation is also shown to be equivalent to a Ramanujan-type differential system for a triple (P;^ Q;^ R^).
    [Show full text]
  • Conjugacy Classification of Quaternionic M¨Obius Transformations
    CONJUGACY CLASSIFICATION OF QUATERNIONIC MOBIUS¨ TRANSFORMATIONS JOHN R. PARKER AND IAN SHORT Abstract. It is well known that the dynamics and conjugacy class of a complex M¨obius transformation can be determined from a simple rational function of the coef- ficients of the transformation. We study the group of quaternionic M¨obius transforma- tions and identify simple rational functions of the coefficients of the transformations that determine dynamics and conjugacy. 1. Introduction The motivation behind this paper is the classical theory of complex M¨obius trans- formations. A complex M¨obiustransformation f is a conformal map of the extended complex plane of the form f(z) = (az + b)(cz + d)−1, where a, b, c and d are complex numbers such that the quantity σ = ad − bc is not zero (we sometimes write σ = σf in order to avoid ambiguity). There is a homomorphism from SL(2, C) to the group a b of complex M¨obiustransformations which takes the matrix ( c d ) to the map f. This homomorphism is surjective because the coefficients of f can always be scaled so that σ = 1. The collection of all complex M¨obius transformations for which σ takes the value 1 forms a group which can be identified with PSL(2, C). A member f of PSL(2, C) is simple if it is conjugate in PSL(2, C) to an element of PSL(2, R). The map f is k–simple if it may be expressed as the composite of k simple transformations but no fewer. Let τ = a + d (and likewise we write τ = τf where necessary).
    [Show full text]
  • Conformal Mapping Solution of Laplace's Equation on a Polygon
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Journal of Computational and Applied Mathematics 14 (1986) 227-249 227 North-Holland Conformal mapping solution of Laplace’s equation on a polygon with oblique derivative boundary conditions Lloyd N. TREFETHEN * Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, U.S.A. Ruth J. WILLIAMS ** Department of Mathematics, University of California at San Diego, La Jolla, CA 92093, U.S.A. Received 6 July 1984 Abstract: We consider Laplace’s equation in a polygonal domain together with the boundary conditions that along each side, the derivative in the direction at a specified oblique angle from the normal should be zero. First we prove that solutions to this problem can always be constructed by taking the real part of an analytic function that maps the domain onto another region with straight sides oriented according to the angles given in the boundary conditions. Then we show that this procedure can be carried out successfully in practice by the numerical calculation of Schwarz-Christoffel transformations. The method is illustrated by application to a Hall effect problem in electronics, and to a reflected Brownian motion problem motivated by queueing theory. Keywor& Laplace equation, conformal mapping, Schwarz-Christoffel map, oblique derivative, Hall effect, Brownian motion, queueing theory. Mathematics Subject CIassi/ication: 3OC30, 35525, 60K25, 65EO5, 65N99. 1. The oblique derivative problem and conformal mapping Let 52 by a polygonal domain in the complex plane C, by which we mean a possibly unbounded simply connected open subset of C whose boundary aa consists of a finite number of straight lines, rays, and line segments.
    [Show full text]
  • Schwarz-Christoffel Transformations by Philip P. Bergonio (Under the Direction of Edward Azoff) Abstract the Riemann Mapping
    Schwarz-Christoffel Transformations by Philip P. Bergonio (Under the direction of Edward Azoff) Abstract The Riemann Mapping Theorem guarantees that the upper half plane is conformally equivalent to the interior domain determined by any polygon. Schwarz-Christoffel transfor- mations provide explicit formulas for the maps that work. Popular textbook treatments of the topic range from motivational and contructive to proof-oriented. The aim of this paper is to combine the strengths of these expositions, filling in details and adding more information when necessary. In particular, careful attention is paid to the crucial fact, taken for granted in most elementary texts, that all conformal equivalences between the domains in question extend continuously to their closures. Index words: Complex Analysis, Schwarz-Christoffel Transformations, Polygons in the Complex Plane Schwarz-Christoffel Transformations by Philip P. Bergonio B.S., Georgia Southwestern State University, 2003 A Thesis Submitted to the Graduate Faculty of The University of Georgia in Partial Fulfillment of the Requirements for the Degree Master of Arts Athens, Georgia 2007 c 2007 Philip P. Bergonio All Rights Reserved Schwarz-Christoffel Transformations by Philip P. Bergonio Approved: Major Professor: Edward Azoff Committee: Daniel Nakano Shuzhou Wang Electronic Version Approved: Maureen Grasso Dean of the Graduate School The University of Georgia December 2007 Table of Contents Page Chapter 1 Introduction . 1 2 Background Information . 5 2.1 Preliminaries . 5 2.2 Linear Curves and Polygons . 8 3 Two Examples and Motivation for the Formula . 13 3.1 Prototypical Examples . 13 3.2 Motivation for the Formula . 16 4 Properties of Schwarz-Christoffel Candidates . 19 4.1 Well-Definedness of f .....................
    [Show full text]
  • Use of Conformal Mappings for Solution of 2D PDE Problems 1. Use
    Use of conformal mappings for solution of 2D PDE problems 1. Use of conformal mappings onto a circle, for 2D Dirichlet problems. Consider a Dirichlet problem for the Laplace equation: 8 < ¢u(x; y) = 0; (x; y) 2 D; (1) : u(x; y)j@D = g(x; y): We will solve this problem using Conformal Mappings. iÁ Remark 1. Mean Value Theorem for Analytic Functions. If Cr = fz = z0 + re ; 0 · Á < 2¼g is a circle centered at z0 with radius r, and f(z); z = x + iy 2 C, is analytic inside Cr and continuous inside and on Cr, then the function value in the center of the circle is expressed via the boundary integral: Z 2¼ I 1 iÁ 1 f(z0) = f(z0 + re )dÁ = f(z)jdzj; (2) 2¼ 0 2¼r Cr where jdzj = rdÁ is a differential arc length. Corollary. Mean Value Theorem for Harmonic Functions. If u(x; y) is a harmonic function iÁ inside Cr = fz = z0 + re ; 0 · Á < 2¼g, i.e. ¢u = uxx + uyy = 0, then it is a real part of some analytic function f(z) = u(x; y) + iv(x; y) (this is true since a circle is simply connected.) Hence, taking the real part of (2), we get the value of u(x; y) at z0 = x0 + iy0 (in the center of the circle) is also expressed as the boundary integral: I 1 u(x0; y0) = u(x; y)jdzj; (3) 2¼r Cr where in the integrand (x; y) 2 Cr is a point on the circle.
    [Show full text]
  • Arxiv:1602.04855V4 [Math.NA] 2 Nov 2018 1 Z Dσ(X) = − Σ(Y)N ˆ · ∇Y Log |Y − X| Ds(Y), X∈ / Γ
    CONFORMAL MAPPING VIA A DENSITY CORRESPONDENCE FOR THE DOUBLE-LAYER POTENTIAL ¨ MATT WALA∗ AND ANDREAS KLOCKNER∗ Abstract. We derive a representation formula for harmonic polynomials and Laurent polynomials in terms of densities of the double-layer potential on bounded piecewise smooth and simply connected domains. From this result, we obtain a method for the numerical computation of conformal maps that applies to both exterior and interior regions. We present analysis and numerical experiments supporting the accuracy and broad applicability of the method. Key words. conformal map, integral equations, Faber polynomial, high-order methods AMS subject classifications. 65E05, 30C30, 65R20 1. Introduction. This paper presents an integral equation method for numerical conformal mapping, using an integral equation based on the Faber polynomials (on the interior) and their counterpart, Faber-Laurent polynomials (on the exterior). Our method is applicable to computing the conformal map from the interior and exterior of domains bounded by a piecewise smooth Jordan curve Γ onto the interior/exterior of the unit disk. Like most techniques for conformal mapping, this method relies on computing a boundary correspondence function between Γ and the boundary of the target domain. From the boundary correspondence, the mapping function can be derived via a Cauchy integral [25, p. 381]. The numerical construction of a function that maps the exterior of a simply connected region conformally onto the exterior of some other region arises in a number of applications including fluid mechanics [8, Ch. 4.5], the generation of finite element meshes for problems in fracture mechanics [45], the design of optical media [31], the analysis of iterative methods [15,16,40], and the solution of initial value problems [32].
    [Show full text]
  • Quaternionic Analysis on Riemann Surfaces and Differential Geometry
    Doc Math J DMV Quaternionic Analysis on Riemann Surfaces and Differential Geometry 1 2 Franz Pedit and Ulrich Pinkall Abstract We present a new approach to the dierential geometry of 3 4 surfaces in R and R that treats this theory as a quaternionied ver sion of the complex analysis and algebraic geometry of Riemann surfaces Mathematics Sub ject Classication C H Introduction Meromorphic functions Let M b e a Riemann surface Thus M is a twodimensional dierentiable manifold equipp ed with an almost complex structure J ie on each tangent space T M we p 2 have an endomorphism J satisfying J making T M into a onedimensional p complex vector space J induces an op eration on forms dened as X JX A map f M C is called holomorphic if df i df 1 A map f M C fg is called meromorphic if at each p oint either f or f is holomorphic Geometrically a meromorphic function on M is just an orientation 2 preserving p ossibly branched conformal immersion into the plane C R or 1 2 rather the sphere CP S 4 Now consider C as emb edded in the quaternions H R Every immersed 4 4 surface in R can b e descib ed by a conformal immersion f M R where M is a suitable Riemann surface In Section we will show that conformality can again b e expressed by an equation like the CauchyRiemann equations df N df 2 3 where now N M S in R Im H is a map into the purely imaginary quaternions of norm In the imp ortant sp ecial case where f takes values in 1 Research supp orted by NSF grants DMS DMS and SFB at TUBerlin 2 Research supp orted by SFB
    [Show full text]
  • The Conformal Center of a Triangle Or Quadrilateral
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Keck Graduate Institute Claremont Colleges Scholarship @ Claremont HMC Senior Theses HMC Student Scholarship 2003 The onforC mal Center of a Triangle or Quadrilateral Andrew Iannaccone Harvey Mudd College Recommended Citation Iannaccone, Andrew, "The onforC mal Center of a Triangle or Quadrilateral" (2003). HMC Senior Theses. 149. https://scholarship.claremont.edu/hmc_theses/149 This Open Access Senior Thesis is brought to you for free and open access by the HMC Student Scholarship at Scholarship @ Claremont. It has been accepted for inclusion in HMC Senior Theses by an authorized administrator of Scholarship @ Claremont. For more information, please contact [email protected]. The Conformal Center of a Triangle or a Quadrilateral by Andrew Iannaccone Byron Walden (Santa Clara University), Advisor Advisor: Second Reader: (Lesley Ward) May 2003 Department of Mathematics Abstract The Conformal Center of a Triangle or a Quadrilateral by Andrew Iannaccone May 2003 Every triangle has a unique point, called the conformal center, from which a ran- dom (Brownian motion) path is equally likely to first exit the triangle through each of its three sides. We use concepts from complex analysis, including harmonic measure and the Schwarz-Christoffel map, to locate this point. We could not ob- tain an elementary closed-form expression for the conformal center, but we show some series expressions for its coordinates. These expressions yield some new hy- pergeometric series identities. Using Maple in conjunction with a homemade Java program, we numerically evaluated these series expressions and compared the conformal center to the known geometric triangle centers.
    [Show full text]
  • Conformal Mappings and Shape Derivatives for the Transmission Problem with a Single Measurement
    Conformal mappings and shape derivatives for the transmission problem with a single measurement. Lekbir Afraites, Marc Dambrine, Djalil Kateb To cite this version: Lekbir Afraites, Marc Dambrine, Djalil Kateb. Conformal mappings and shape derivatives for the transmission problem with a single measurement.. 2006. hal-00020177 HAL Id: hal-00020177 https://hal.archives-ouvertes.fr/hal-00020177 Preprint submitted on 7 Mar 2006 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Conformal mappings and shape derivatives for the transmission problem with a single measurement. Lekbir Afraites∗ Marc Dambrine∗ Djalil Kateb∗ March 7, 2006 Abstract In the present work, we consider the inverse conductivity problem of recovering inclusion with one measurement. First, we use conformal mapping techniques for determining the location of the anomaly and estimating its size. We then get a good initial guess for quasi-Newton type method. The inverse problem is treated from the shape optimization point of view. We give a rigorous proof for the existence of the shape derivative of the state function and of shape functionals. We consider both Least Squares fitting and Kohn and Vogelius functionals. For the numerical implementation, we use a parametrization of shapes coupled with a boundary element method.
    [Show full text]
  • Conformal Metrics
    Differential Geometry: Conformal Metrics [Conformal Flattening by Curvature Prescription and Metric Scaling. Ben-Chen et al., 2008] [Conformal Equivalence of Triangle Meshes. Springborn et al., 2009] Conformal Maps Recall: A conformal map is an angle-preserving transformation. Liouville's Theorem: In dimensions greater than 2, the Möbius transformations (translations, rotations, scales, and inversions) are the only conformal maps. Möbius Transformations Given points x,y∈Rn and given a Möbius transformation φ, we have: 1. If φ is a translation: φ(x) −φ(y) = x + c − y − c = x − y Möbius Transformations Given points x,y∈Rn and given a Möbius transformation φ, we have: 1. If φ is a translation: φ(x) −φ(y) = x + c − y − c = x − y 2. If φ is a rotation: φ(x) −φ(y) = φ(x − y) = x − y Möbius Transformations Given points x,y∈Rn and given a Möbius transformation φ, we have: 1. If φ is a translation: φ(x) −φ(y) = x + c − y − c = x − y 2. If φ is a rotation: φ(x) −φ(y) = φ(x − y) = x − y 3. If φ is a scale: φ(x) −φ(y) = sx − sy = s x − y Möbius Transformations Given points x,y∈Rn and given a Möbius transformation φ, we have: 1. If φ is a translation: φ(x) −φ(y) = x + c − y − c = x − y 2. If φ is a rotation: φ(x) −φ(y) = φ(x − y) = x − y 3. If φ is a scale: φ(x) −φ(y) = sx − sy = s x − y 4. If φ is an inversion: x y 1 φ(x) −φ(y) = 2 − 2 = x − y x y x y Möbius Transformations Thus, given points x,y∈Rn and given a Möbius transformation φ, we can express the distance between the transformed points as the original distance, modulated by the product of functions that only depends on the individual points’ positions.
    [Show full text]
  • Conformal Geometry Processing
    CONFORMAL GEOMETRY PROCESSING Keenan Crane • Symposium on Geometry Processing Grad School • Summer 2017 Outline PART I: OVERVIEW PART II: SMOOTH THEORY PART III: DISCRETIZATION PART IV: ALGORITHMS These slides are a work in progress and there will be errors. As always, your brain is the best tool for determining whether statements are actually true! Please do not hesitate to stop and ask for clarifications/corrections, or send email to [email protected]. PART I: OVERVIEW PART I: PART II: SMOOTH THEORY OVERVIEW PART III: DISCRETIZATION PART IV: ALGORITHMS CONFORMAL GEOMETRY PROCESSING Keenan Crane • Symposium on Geometry Processing Grad School • Summer 2017 Motivation: Mapmaking Problem • How do you make a flat map of the round globe? • Hard to do! Like trying to flatten an orange peel… Impossible without some kind of distortion and/or cutting. Conformal Mapmaking • Amazing fact: can always make a map that exactly preserves angles. E (Very useful for navigation!) N N E Conformal Mapmaking • However, areas may be badly distorted… (Greenland is not bigger than Australia!) Conformal Geometry More broadly, conformal geometry is the study of shape when one can measure only angle (not length). Conformal Geometry—Visualized Applications of Conformal Geometry Processing Basic building block for many applications… CARTOGRAPHY TEXTURE MAPPING REMESHING SENSOR NETWORKS SIMULATION 3D FABRICATION SHAPE ANALYSIS Why Conformal? • Why so much interest in maps that preserve angle? • QUALITY: Every conformal map is already “really nice” • SIMPLICITY: Makes
    [Show full text]