Hyperbolic Geometry In Parameterization Of Surfaces

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Hyperbolic Geometry in Parameterization of SurfacesDirected Reading Program ProjectZhecheng Wangsupervised by Neža Žager KorenjakApril 20201IntroductionParameterization is the process to find explicit equations of a curve (R2 ) or a surface (R3 ),or a manifold (Rn ) in general. Intuitively, a parameterization of a geoemtry shape is torepresent the shape in terms of “parameters”. In this paper, I will refresh the concept ofparameterization from calculus, introduce a specific problem exists in the parameterizationin R3 , and provide a solution.22.1Parameterization, and Reparameterization in R2ParameterizationI believe readers have seen the parameterization of a circle before in multi-variable calculus.Therefore, let’s review the common definition of parameterization.Definition: A parameterized curve in Rn is a map γ : (α, β) Rn , for some α, β with α β .The symbol (α, β) denotes the open interval(α, β) {t R α t β}Therefore, a parameterized curve in R2 isγ(t) σ(u(t), v(t))1

2.2ReparameterizationNow, as we can imagine, there exists a process to reparameterize the curve, which is toexpress the curve with a different set of “parameters”.Definition: A parameterized curve curve γ̃ : (α̃, β̃) Rn is a reparamatrization of a parameterized curve γ : (α, β) Rn if there is a smooth bijective map θ : (α̃, β̃) (α, β) (thereparameterization map) such that the inverse map φ 1 : (α, β) (α̃, β̃) is also smooth andγ̃(t̃) γ(φ(t̃)) for all t̃ (α̃, β̃)Note that, since φ has a smooth inverse, γ is a reparameterization of γ̃:γ̃(φ 1 (t)) γ(φ(φ 1 (t))) γ(t) for all t (α, β)Two curves that are reparameterizations of each other have the same image, so they shouldhave the same geometric properties. Informally, we can perceive this as a changing of the”speed” of t.2.3Example: Parameterization of a CircleI believe the form of parameterization function must be familiar to the readers, we recall theimplicit formula of a circle is in the form ofx2 y 2 1The unit circle has a parameterizationγ̃(t) (sin t, cos t)Then we could find a reparameterization of σ, we have to find a reparameterization map φsuch that(cos φ(t), sin φ(t)) (sin t, cos t)One solution could be θ(t) π/2 t. The images before reparameterization and after thetransition maintain the same shapes and geometric properties.2

Figure 1: Parameterization of a Circle33.1Parameterization in R3Surface and Surface Patch (Parameterization)Definition: A subset S of R3 is a surface if, for every point p S, there is an open set Uin R2 and an open set W in R3 containing p such that S W is homemorphic to U .Similarly, we could find a parameterization, or surface patch for R3 as we did for R2 .Definition: A homemorphism σ : U S W as in this definiton is called a surface patchor parameterization of the open subset S W of S.Definition: A collection of such surface patches whose images cover the whole of S is calledan atlas of S.As we can see, there is analogy between parameterization in R2 and R3 . In fact, the mostimportant application in R3 is UV mapping, commonly used in Computer Graphics andVisual Effects. As you can see, this is also a bijective mapping.3

Figure 2: Parameterization of a 3D Model, credit to Keenan Crane3.2Genus 1 Surface and the Parameterization of a TorusThe most common form of a Genus 1 Surface is a torus. Given the parameterization equations:x(θ, ϕ) (a b cos θ) cos ϕy(θ, ϕ) (a b cos θ) sin ϕz(θ, ϕ) b sin θTherefore,σ(θ, ϕ) ((a b cos θ) cos ϕ, (a b cos θ) sin ϕ, b sin θ)Figure 3: Parameterization of a Torus, credit to Rob WomersleyTo get this parameterization, we could imagine the first two terms define the circle, then thethird term construct another circle whose radius is perpendicular to the first circle at anygiven point on the first circle.4

44.1The Problem in the Parameterization of A Genus 2SurfaceGenus 2 Surface and its Common ConstructionSince Genus 1 Surface is a Torus, Genus 2 surface, occasionally called Double Torus, is a asurface formed by the connected sum of 2 tori. A famous exemple of a non-orientable surfacegenus two is the Klein bottle.Figure 4: Klein Bottle and Its Fundamental Polygon, credit to Wikipedia user InductiveloadThe common construction of a double torus is a octagon with opposite identified.Figure 5: A Octagon to Double-TorusBy our definition of surface patch, this process is a parameterization of the genus surface inR3 .4.2the Importance of Euclidean IntuitionAs we all know, we live in a world where Euclidean space is the “norm”. However, it isimpossible to always obtain an understandable Euclidean graph when study geometry. Inthe case mentioned in the last sub-section, we could obtain 4 lines intersect one point, whichimplies the sum of 4 angles produced is equal to 360 since genus 2 surface is locally Euclidean.Apparently, the octagon is in R2 and the sum of its interior angles is 1080 . Let us assumethe octagon is an equilateral octagon, then each angle has the degree of 135 . Apparently,the angles deformed during the surface patch since the sum of angles changed.5

However, as we have mentioned before, we want to preserve this Euclidean intuition to helpus solving more questions. Then it is clear we want a better way to that preserve the angleand length of the R2 open set in surface patch.5The Solution to the Problem5.1Conformal Mapping and Its LimitationFirstly, let us think within the box: start with R2 , there exists a mapping that can preserveangles.Definition: If S1 and S2 are surfaces, a conformal map f : S1 S2 is a local diffeomorphisms.t. if γ1 and γ̂1 are any two curves on S1 that intersect, say at a point p S1 , and if γ2 andγˆ2 are their images under f , the angle of intersection of γ1 and γ̂1 at p are equal to the angleof intersection of γ2 and γˆ2 at f (p).However, conformal map does not preserve the length of our equilateral octagon, here is anexample of in Computer Graphics: as you can see, the disproportional stretch of polygons isnot ideal to our mapping.Figure 6: An Example of Conformal Mapping in Computer Graphics, credit to CCGL5.2Poincaré Disk Model and Hyperbolic GeometryThe only difference between Hyperbolic Geometry and Euclidean Geometry is the one postulate: the parallel postulate (Playfair’s axiom). In the following of this paper, I will use thePoincaré Disk Model to demonstrate some properties of hyperbolic geometry.Definition: Here is the five axioms of Hyperbolic Geometry:(1) To draw a straight line from any point to any point.(2) To produce (extend) a finite straight line continuously in a straight line.6

(3) To describe a circle with any centre and distance (radius).(4) That all right angles are equal to one another.(5) For any given line R and point P not on R, in the plane containing both lineR and point P there are at least two distinct lines through P that do notintersect R. An alternative way to say it is there exists two or more parallel line to aline through a given point that is not on the line.Figure 7: A Straight Line in Hyperbolic Plane H2Now, let us denote hyperbolic space as H. Based on Poincaré disk model, H2 is an opendisk and H2 is a subspace of R2 . This implies we could use Euclidean intuition to constructhyperbolic lines/shape (H2 ) in R2 .5.3Properties of Hyperbolic Geometry in H2Back to the equilateral octagon we used in double torus parameterization, we could reconstruct the equilateral octagon in hyperbolic plane H2 that preserve the same length yet havea different sum of interior angles.7

Figure 8: An equilateral octagon with each interior angles equal to 90 in H2Now, let me use triangles to demonstrate some “magical” properties on hyperbolic plane. ItFigure 9: Three equilateral triangles in H2 with different degree of interior angleis easy to notice that all triangles are equilateral, but the closer the triangle to the center ofhyperbolic plane, the more geometrically similar it is to its counterpart in Euclidean plane.Vice versa: if the triangle is far from the center, its shape will “deformed” (in an Euclideanintuition) and the sum of interior angles will approach to 0. To put it simply: while all threetriangles are equilateral triangles, they have different interior angle sum.5.4The Solution: An Octagon on A Hyperbolic PlaneIn summary, instead of traditional R2 based shapes, we choose an equilateral octagon onH2 . Using the similar idea we used in last sub-section, we could construct arbitrary equilateral octagon with ideal interior angle degree. Therefore, even after the surface patch, the8

sum of interior angles of our hyperbolic shape can be preserved. Furthermore, with carefulconstruction, we could potentially preserve the length as well.6ConclusionParamatrization of a surface in R3 can be challenging. In this paper, I provided an interestingsolution to the problem of angle deformations in R2 R3 surface patch with introducing hyperbolic geometry. Lastly and most importantly, thanks to my DRP mentor Neža’s patienceand effort in helping me compiling this note.9

AppendicesABasic Topology DefinitionDefinitionQuantifier StatementNotationp is an interior point of G ( r 0)(Nr (p) G)p G p is in the closure of E( r 0)(Nr (p) E 6 )p EG is openN/AG GF is closedN/AF 6 F or F FFigure 10: Basic Topology Definition ChartReferences[1] Andrew Pressley, Elementary Differential Geometry (Springer, 2001. ISBN 1-85233-1526)10

hyperbolic lines/shape (H 2) in R . 5.3 Properties of Hyperbolic Geometry in H2 Back to the equilateral octagon we used in double torus parameterization, we could recon-struct the equilateral octagon in hyperbolic plane H2 that preserve the same length yet have a di erent sum of interior angles. 7

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