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The Rigidity of the Sphere THE RIGIDITY OF THE SPHERE April 25, 2016 A thesis submitted To Kent State University in partial Fulllment of the requirements for the Degree of Master of Science By Paul C. Havens April 2016 c Copyright All rights reserved Except for previously published materials Thesis Written By Paul C. Havens B.S., Kent State University, 2009 M.S., Kent State University, 2016 Approved by Dmitry Ryabogin , Masters Advisor Dr. Andrew Tonge , Chair, Departmentof Mathematical Sciences Dr. James Blank , Dean, College of Arts and Sciences Contents 1 Introduction 1 2 Basic Dierential Geometry 4 2.1 First Fundamental Form . .4 2.2 Second Fundamental Form . .4 2.3 Curvatures . .5 2.4 Conformal Maps . .7 2.5 Gauss Map in Local Coordinates . 11 2.6 Christoel Symbols and the Mainardi-Codazzi Equations . 12 3 The Rigidity of the Sphere 15 3.1 Preliminaries . 15 3.2 Conclusion . 20 iii ACKNOWLEDGEMENTS I would like to thank Dmitry Ryabogin for being my advisor and guiding me through the research process and working patiently with me. His knowledge and experience were of immense value to this writing. I also extend my thanks to Benjamin Jaye and Artem Zvavitch for serving on my thesis comittee. A special thanks goes to Carl Stitz for getting me started in my pursuit of my mathematics major and guiding me during my early undergraduate years. I would like to thank my mother for her support throughout the process. Last but not least, I thank my friend Joy for her encouragement and support over the years. iv 1 Introduction Our thesis concerns the area of surfaces of constant mean curvatures. Such surfaces have been applied to the shapes of soap lms and soap bubbles. For a soap lm, we may imagine taking a bubble wand and dipping it into our bubble solution. The surface stretched across the opening to the bubble wand is an example of a constant mean curvature surface, particularly a surface of constant mean curvature zero. If we likewise take a wire frame of a polyhedron and dip it into our solution, the resulting shape is also a minimal surfaces. These surfaces have equal pressure on both sides of a sheet of soap lm, and have a mean curvature of zero. Such surfaces are called minimal surfaces, so called because surface tension minimizes the surface energy. We will in fact not be working with these directly, as we wish to be more general and study surfaces of constant mean curvature. For examples of these, we turn our eyes towards soap bubbles. Soap bubbles also have their surfaces minimized by surface tension, but they have a dierence in pressure between their interior and exterior. This translates to a constant yet nonzero mean curvature. Beyond soap lms, surfaces of constant mean curvature have applications to the shape of a water droplet on a superhydrophobic surface as well as the design of air-supported structures such as various domes used in sports complexes. We focus in this thesis on rigid surfaces. Intuitively, we are considering surfaces made of exible, inelastic materials. We call these surfaces rigid if it is not possible to deform them. A classical result of Cauchy says that a convex polyhedron in R3 that is made of rigid plates but exible hinges is a rigid surface. This thesis concerns an extension of this idea, a theorem due to Hopf: a regular surface with constant mean curvature which is homeomorphic to a sphere is a sphere. We will show that a surface with this constant mean curvature is composed of so called umbilic points, which are exactly the points where the principle curvatures are equal. We will also show that such a surface must be either a sphere or a plane. Since we have assumed homeomorphism to a sphere, our surface must then be a sphere. To proceed heuristically, we begin by introducing isothermal parameters, ds2 = E du2 + dv2, which not only are required conditions, but are nice to work with and simplify many of our calculations. We arrive at these from taking the coecients of the rst fundamental form, E, F , and G, such that E = G and F = 0. Among the rst items that we will need are the Gaussian curvature K and the mean curvature H. The Gaussian curvature is the product of the two principal curvatures, while the mean curvature is the average. Under our isothermal parameters, these take the forms eg − f 2 K = = k k ; E2 1 2 e + g k1 + k2 H = = : 2E 2 Where k1 and k2 are the principle curvatures. 1 The next major items that we will need are the Codazzi equations: ! e − g + f = EH ; 2 v u u ! e − g − f = −EH : 2 u v v 2 To see how the two sets of items are related, we follow the example of Hopf and consider the function e − g φ(ζ) = φ(u; v) = − if. We have that 2 v u !2 u e − g t + f 2 jφj 2 = ; E E s e2 − 2eg + g2 + 4f 2 = ; 4E2 s e2 + 2eg + g2 − 4eg + 4f 2 = ; 4E2 s (e + g)2 eg − f 2 = − ; 4E2 E2 v u !2 u k1 + k2 = t − k k : 2 1 2 Solving for zeros, we nd v u !2 u k1 + k2 0 = t − k k ; 2 1 2 2 jk1 − k2j 0 = : 2 jφj Thus, = 0 when k = k . E 1 2 Chapter 2 is dedicated to the basics of dierential geometry that we will need for our purposes, including the denitions and constructions of the rst and second fundamental forms, currvature, and the Codazzi equations mentioned above. We also prove the existence of isothermal parameters. A reader comfortable with these topics may move on to Chapter 3. In this chapter, we take the concepts we have gathered and develop our Codazzi equations into the form used above. We also introduce two parametrizations on the sphere which will work together to help us prove Hopf's theorem, which is handled in the nal section. 3 2 Basic Dierential Geometry 2.1 First Fundamental Form Denition 1. [1] A subset S ⊂ R3 is a regular surface if for each p 2 S, 9 a nbd V in R3 and a map r : U ! V T S of an open set U ⊂ R2 onto V T S ⊂ R3 such that 1. r is dierentiable; so r(u; v) = (x(u; v); y(u; v); z(u; v)), (u; v) 2 U, then x(u; v), y(u; v), z(u; v) have continuous partials of all orders in U. 2. r is a homeomorphism. r is continuous by condition 1, so r has an inverse r−1 : V \ S ! U which is continuous; i.e. r−1 is the restriction of a continuous map F : W ⊂ R3 ! R2 dened on an open set W containing V \ S. 2 3 3. (regularity) For each q 2 U, the dierential drq : R ! R is 1-1. r is called a parametrization of coordinates in p. V \ S is called a coordinate neighborhood. On each tangent plane Tp(S) of a regular surface S, we can get an inner product from the natural inner 3 product on R . We denote this as < w1; w2 >p, w1 · w2, and (when the meaning is clear) w1w2. Denition 2. Let R ⊂ S be a bounded region of a regular surface contained in the coordinate neighborhood of the parametrization 2 . The positive number RR , −1 , is r : U ⊂ R ! S Q kru × rvk du dv = A(R) Q = r (R) the area of R. 2 Denition 3. [2] The quadratic form Ip(w) =< w; w >p= jwj ≥ 0 on Tp(S) is called the rst fundamental form of the regular surface S ⊂ R3 at p 2 S. In the basis frurvg associated to a parametrization r(u; v) at p, since a tangent vector w 2 Tp(S) is the tangent to the curve α(t) = r(u(t); v(t)), t 2 (−, ), with p = α(0) = r(u0; v0), we get 0 0 0 Ip(α (0)) =< α (0); α (0) >p; 0 0 0 0 =< ruu + rvv ; ruu + rvv >p; 0 2 0 0 0 2 = ruru(u ) + 2rurvu v + rvrv(v ) ; = E(u0)2 + 2F (u0v0)2 + G(v0)2: (1) 2.2 Second Fundamental Form ru × rv Denition 4. [3] Let N(q) = , q 2 r(U). Let S ⊂ R3 be a normal surface. The map N : S ! R3 jru × rvj lies on the unit sphere S2 = (x; y; z) 2 R3 : x2 + y2 + z2 = 1 . The map N : S ! S2 is the Gauss Map of S. Proposition 1. [4] The dierential dNp : Tp(S) ! Tp(S) of the Gauss map is a self-adjoint linear map; that is, for an operator A, hAv; wi = hv; Awi for all u; v 2 V . 4 Proof. dNp is linear, so it suces to show hdNp (w1) ; w2i = hw1; dNp (w2)i in the basis fw1; w2g of Tp(S). Let r(u; v) be a parametrization of S at p and fru; rvg be the basis of Tp(S). If α(t) = r (u(t); v(t)) is a parametrized curve in S with α(0) = p, we have 0 0 0 dNp (α (0)) = dNp (ruu (0) + rvv (0)) ; d = N (u(t); v(t)) j ; dt t=0 0 0 = Nuu (0) + Nvv (0): In particular, dNp(ru) = Nu and dNp(rv) = Nv. We need only show that hNu; rvi = hru;Nvi. By taking the derivatives of hN; rui = 0 and hN; rvi = 0 relative to v and u respectively, we get hNv; rui + hN; ruvi = 0; hNu; rvi + hN; rvui = 0: Therefore, hNu; rvi = − hN; ruvi =hNv; rui.
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