Hilbert's theorem (differential geometry)

Hilbert's theorem (differential geometry) In differential geometry, Hilbert's theorem (1901) states that there exists no complete regular surface {displaystyle S} of constant negative gaussian curvature {displaystyle K} immersed in {displaystyle mathbb {R} ^{3}} . This theorem answers the question for the negative case of which surfaces in {displaystyle mathbb {R} ^{3}} can be obtained by isometrically immersing complete manifolds with constant curvature.

Contents 1 History 2 Proof 3 See also 4 References History Hilbert's theorem was first treated by David Hilbert in "Über Flächen von konstanter Krümmung" (Trans. Amer. Math. Soc. 2 (1901), 87–99). A different proof was given shortly after by E. Holmgren in "Sur les surfaces à courbure constante négative" (1902). A far-leading generalization was obtained by Nikolai Efimov in 1975.[1] Proof The proof of Hilbert's theorem is elaborate and requires several lemmas. The idea is to show the nonexistence of an isometric immersion {displaystyle varphi =psi circ exp _{p}:S'longrightarrow mathbb {R} ^{3}} of a plane {displaystyle S'} to the real space {displaystyle mathbb {R} ^{3}} . This proof is basically the same as in Hilbert's paper, although based in the books of Do Carmo and Spivak.

Observations: In order to have a more manageable treatment, but without loss of generality, the curvature may be considered equal to minus one, {displaystyle K=-1} . There is no loss of generality, since it is being dealt with constant curvatures, and similarities of {displaystyle mathbb {R} ^{3}} multiply {displaystyle K} by a constant. The exponential map {displaystyle exp _{p}:T_{p}(S)longrightarrow S} is a local diffeomorphism (in fact a covering map, by Cartan-Hadamard theorem), therefore, it induces an inner product in the tangent space of {displaystyle S} at {displaystyle p} : {displaystyle T_{p}(S)} . Furthermore, {displaystyle S'} denotes the geometric surface {displaystyle T_{p}(S)} with this inner product. If {displaystyle psi :Slongrightarrow mathbb {R} ^{3}} is an isometric immersion, the same holds for {displaystyle varphi =psi circ exp _{o}:S'longrightarrow mathbb {R} ^{3}} .

The first lemma is independent from the other ones, and will be used at the end as the counter statement to reject the results from the other lemmas.

Lemma 1: The area of {displaystyle S'} is infinite. Proof's Sketch: The idea of the proof is to create a global isometry between {displaystyle H} and {displaystyle S'} . Then, since {displaystyle H} has an infinite area, {displaystyle S'} will have it too.

The fact that the hyperbolic plane {displaystyle H} has an infinite area comes by computing the surface integral with the corresponding coefficients of the First fundamental form. To obtain these ones, the hyperbolic plane can be defined as the plane with the following inner product around a point {displaystyle qin mathbb {R} ^{2}} with coordinates {displaystyle (u,v)} {displaystyle E=leftlangle {frac {partial }{partial u}},{frac {partial }{partial u}}rightrangle =1qquad F=leftlangle {frac {partial }{partial u}},{frac {partial }{partial v}}rightrangle =leftlangle {frac {partial }{partial v}},{frac {partial }{partial u}}rightrangle =0qquad G=leftlangle {frac {partial }{partial v}},{frac {partial }{partial v}}rightrangle =e^{u}} Since the hyperbolic plane is unbounded, the limits of the integral are infinite, and the area can be calculated through {displaystyle int _{-infty }^{infty }int _{-infty }^{infty }e^{u}dudv=infty } Next it is needed to create a map, which will show that the global information from the hyperbolic plane can be transfer to the surface {displaystyle S'} , i.e. a global isometry. {displaystyle varphi :Hrightarrow S'} will be the map, whose domain is the hyperbolic plane and image the 2-dimensional manifold {displaystyle S'} , which carries the inner product from the surface {displaystyle S} with negative curvature. {displaystyle varphi } will be defined via the exponential map, its inverse, and a linear isometry between their tangent spaces, {displaystyle psi :T_{p}(H)rightarrow T_{p'}(S')} .

That is {displaystyle varphi =exp _{p'}circ psi circ exp _{p}^{-1}} , where {displaystyle pin H,p'in S'} . That is to say, the starting point {displaystyle pin H} goes to the tangent plane from {displaystyle H} through the inverse of the exponential map. Then travels from one tangent plane to the other through the isometry {displaystyle psi } , and then down to the surface {displaystyle S'} with another exponential map.

The following step involves the use of polar coordinates, {displaystyle (rho ,theta )} and {displaystyle (rho ',theta ')} , around {displaystyle p} and {displaystyle p'} respectively. The requirement will be that the axis are mapped to each other, that is {displaystyle theta =0} goes to {displaystyle theta '=0} . Then {displaystyle varphi } preserves the first fundamental form.

In a geodesic polar system, the Gaussian curvature {displaystyle K} can be expressed as {displaystyle K=-{frac {({sqrt {G}})_{rho rho }}{sqrt {G}}}} .

In addition K is constant and fulfills the following differential equation {displaystyle ({sqrt {G}})_{rho rho }+Kcdot {sqrt {G}}=0} Since {displaystyle H} and {displaystyle S'} have the same constant Gaussian curvature, then they are locally isometric (Minding's Theorem). That means that {displaystyle varphi } is a local isometry between {displaystyle H} and {displaystyle S'} . Furthermore, from the Hadamard's theorem it follows that {displaystyle varphi } is also a covering map.

Since {displaystyle S'} is simply connected, {displaystyle varphi } is a homeomorphism, and hence, a (global) isometry. Therefore, {displaystyle H} and {displaystyle S'} are globally isometric, and because {displaystyle H} has an infinite area, then {displaystyle S'=T_{p}(S)} has an infinite area, as well. {displaystyle square } Lemma 2: For each {displaystyle pin S'} exists a parametrization {displaystyle x:Usubset mathbb {R} ^{2}longrightarrow S',qquad pin x(U)} , such that the coordinate curves of {displaystyle x} are asymptotic curves of {displaystyle x(U)=V'} and form a Tchebyshef net.

Lemma 3: Let {displaystyle V'subset S'} be a coordinate neighborhood of {displaystyle S'} such that the coordinate curves are asymptotic curves in {displaystyle V'} . Then the area A of any quadrilateral formed by the coordinate curves is smaller than {displaystyle 2pi } .

The next goal is to show that {displaystyle x} is a parametrization of {displaystyle S'} .

Lemma 4: For a fixed {displaystyle t} , the curve {displaystyle x(s,t),-infty

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