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Chapter 5: Problem 5

Let \(S\) be a complete surface with \(K \geq K_{1}>0\), where \(K\) is the Gaussian curvature of \(S\) and \(K_{1}\) is a constant. Prove that every geodesic \(\gamma:[0, \infty) \rightarrow S\) has a point conjugate to \(\gamma(0)\) in the interval \(\left(0, \pi / \sqrt{K_{1}}\right]\).

Short Answer

Expert verified
Using Rauch comparison theorem, geodesics on S have conjugate points within (0, 蟺 / 鈭欿_1].

Step by step solution

01

- Understand the given conditions

The surface is complete with Gaussian curvature (K 鈮� K_1 > 0). This means that everywhere on the surface, the curvature is at least K_1, which is a positive constant.
02

- Recall properties of geodesics on surfaces with curvature bounds

For a surface with a lower bound on Gaussian curvature K_1, any geodesic 纬: [0, 鈭�) 鈫� S must exhibit certain behavior related to its curvature properties. Specifically, due to the curvature lower bound, we can leverage results from comparison theorems in Riemannian geometry.
03

- Use the Rauch comparison theorem

The Rauch comparison theorem helps us relate the behavior of geodesics on our surface S to those on a model surface with constant curvature. In this case, since K 鈮� K_1 > 0, compare it to a sphere with constant curvature K_1.
04

- Solution for constant curvature K_1 (sphere)

On a sphere with constant curvature equal to K_1, every geodesic will have conjugate points. Specifically, the first conjugate point along a geodesic occurs at a distance of 蟺 / 鈭欿_1 from the start of the geodesic.
05

- Apply the comparison

The Rauch comparison theorem implies that the geodesics on our surface S will also have conjugate points with distances that can be compared to those on the model space (sphere). Because our surface's curvature is at least K_1, geodesics on this surface must have conjugate points at distances no greater than 蟺 / 鈭欿_1.
06

- Conclusion

Thus, every geodesic 纬: [0, 鈭�) 鈫� S has a conjugate point to 纬(0) in the interval ( (0, 蟺 / 鈭欿_1].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

geodesics
To understand this problem, we first need to grasp the concept of geodesics. Geodesics are the shortest paths between two points on a curved surface, much like straight lines on a flat plane. Imagine walking along the shortest route across a hilly terrain or a sphere without deviating left or right. This path is called a geodesic. Geodesics are crucial in Riemannian geometry as they reflect the intrinsic geometry of the surface. They serve as the 'straight lines' in this curved world and help in understanding the curvature and shape of the surface.
  • They minimize distance.
  • On a sphere, geodesics are great circles.
  • Geodesics exhibit certain behaviors based on the surface's curvature.

In conclusion, geodesics help us traverse a surface with the least effort and they are pivotal in connecting various curvature properties of the surface.
Rauch comparison theorem
The Rauch comparison theorem is a powerful tool in Riemannian geometry. It helps compare the behavior of geodesics on different surfaces by considering curvature. Essentially, it states that if you have two surfaces with different curvature bounds, you can predict how geodesics behave relative to each other. This is incredibly useful in this problem because we can compare our given surface with a model surface like a sphere.
The theorem can be summarized as:
  • If the curvature of one surface is greater than or equal to the other, geodesics on the first surface will be 'tighter' or have shorter distances between conjugate points compared to the second surface.
  • In our problem, since our surface has curvature at least K_1, we compare it to a sphere with constant curvature K_1.

This comparison allows us to predict that the geodesics on our surface will have conjugate points within the same or shorter intervals as those on the sphere.
conjugate points
Conjugate points are critical in understanding geodesics and surface properties. Imagine two points on a geodesic. If infinitesimally close geodesics converge again at another point, these points are conjugate. Conjugate points essentially measure how geodesics spread or converge, reflecting the curvature of the surface.
  • They signify the beginning and end of the shortest path on a surface.
  • In our problem, the first conjugate point appears at a distance \( 蟺 / \sqrt{K_1} \).
  • They are important in the study of surface stability and curvature properties.
For our given surface, because it has curvature at least K_1, there must be conjugate points within the interval \( (0, \pi / \sqrt{K_1}) \), reflecting the intrinsic nature of its geometry.
Riemannian geometry
Riemannian geometry forms the backbone of our discussion. It is the study of smooth surfaces and their curvature. Unlike Euclidean geometry, which deals with flat spaces, Riemannian geometry explores curved spaces like our surface S and examines properties like distance, angles, and curvature.
  • It generalizes geometry to spaces of any dimension.
  • Enables the study of more complex and realistic shapes and surfaces.
  • Fields like physics and engineering use Riemannian geometry to model complex structures and phenomena.
In our specific problem, we look at a surface with a lower curvature bound. This helps us use tools like the Rauch comparison theorem and understand the placement of conjugate points along geodesics. Riemannian geometry brings a powerful framework to analyze and solve these complex geometric problems.

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Most popular questions from this chapter

(Calculus of Variations.) Geodesics are particular cases of solutions to variational problems. In this exercise, we shall discuss some points of a simple, although quite representative, variational problem. In the next exercise we shall make some applications of the ideas presented here. Let \(y=y(x), x \in\left[x_{1}, x_{2}\right]\) be a differentiable curve in the \(x y\) plane and let a variation of \(y\) be given by a differentiable map \(y=y(x, t)\), \(t \in(-\epsilon, \epsilon)\). Here \(y(x, 0)=y(x)\) for all \(x \in\left[x_{1}, x_{2}\right]\), and \(y\left(x_{1}, t\right)=\) \(y\left(x_{1}\right), y\left(x_{2}, t\right)=y\left(x_{2}\right)\) for all \(t \in(-\epsilon, \epsilon)\) (i.e., the end points of the variation are fixed). Consider the integral 4\. First and Second Variations of Are Length, Bonnet's Theorem $$ I(t)=\int_{x_{1}}^{x_{2}} F\left(x, y(x, t), y^{\prime}(x, t)\right) d x, \quad t \in(-\epsilon, \epsilon) $$ where \(F\left(x, y, y^{\prime}\right)\) is a differentiable function of three variables and \(y^{\prime}=\partial y / \partial x\). The problem of finding the critical points of \(I(t)\) is called a variational problem with integrand \(F\). a. Assume that the curve \(y=y(x)\) is a critical point of \(I(t)\) (i.e., \(d I / d t=0\) for \(t=0\) ). Use integration by parts to conclude that \((I=d I / d t)\) $$ \begin{aligned} I(t) &=\int_{x_{1}}^{x_{2}}\left(F_{y} \frac{\partial y}{\partial t}+F_{y} \frac{\partial y^{\prime}}{\partial t}\right) d x \\ &=\left[\frac{\partial y}{\partial t} F_{y}\right]_{x_{1}}^{x_{2}}+\int_{x_{1}}^{x_{2}} \frac{\partial y}{\partial t}\left(F_{y}-\frac{d}{d x} F_{y^{\prime}}\right) d x \end{aligned} $$

Let \(U \subset R^{3}\) be an open connected subset of \(R^{2}\) and let \(\mathbf{x}: U \rightarrow S\) be an isothermal parametrization (i.e., \(E=G, F=0\); cf. Sec. 4-2) of a regular surface \(S\). We identify \(R^{2}\) with the complex plane \(\mathbb{C}\) by setting \(u+i v=\zeta\), \((u, v) \in R^{2}, \zeta \in \mathbb{C}\). \(\zeta\) is called the complex parameter corresponding to \(\mathbf{x}\). Let \(\phi: \mathbf{x}(U) \rightarrow \mathbb{C}\) be the complex- valued function given by $$ \phi(\zeta)=\phi(u, v)=\frac{e-g}{2}-i f=\phi_{1}+i \phi_{2} $$ where \(e, f, g\) are the coefficients of the second fundamental form of \(S\). a. Show that the Mainardi-Codazzi equations (cf. Sec. 4-3) can be written, in the isothermal parametrization \(\mathrm{x}\), as $$ \left(\frac{e-g}{2}\right)_{a}+f_{*}=E H_{a} \quad\left(\frac{e-g}{2}\right)_{*}-f_{*}=-E H_{*} $$ and conclude that the mean curvature \(H\) of \(\mathbf{x}(U) \subset S\) is constant if and only if \(\phi\) is an analytic function of \(\zeta\) (i.e., \(\left(\phi_{1}\right)_{a}=\left(\phi_{2}\right)_{2},\left(\phi_{1}\right)_{\nu}=\) \(\left.-\left(\phi_{2}\right)_{2}\right) .\) h. Define the "complex derivative" $$ \frac{\partial}{\partial \bar{\zeta}}=\frac{1}{2}\left(\frac{a}{\partial x}-i \frac{\partial}{\partial x}\right) \text {. } $$ and prove that \(\phi(\zeta)=-2\left\langle\mathbf{x}_{\gamma}, N_{\zeta}\right\rangle\), where by \(\mathbf{x}_{c}\), for instance, we mean the vector with complex coordinates $$ \mathbf{x}_{p}=\left(\frac{\partial x}{\partial \zeta}, \frac{\partial y}{\partial \zeta}, \frac{\partial z}{\partial \zeta}\right) $$ c. Let \(f: U \subset C \rightarrow V \subset C\) be a one-to-one complex function given by \(f(u+i v)=x+i y=\eta\). Show that \((x, y)\) are isothermal parameters on \(S\) (i.e., \(\eta\) is a complex parameter on \(S\) ) if and only if \(f\) is analytic and \(f^{\prime}(\zeta) \neq 0, \zeta\) \& \(U\). Let \(\mathbf{y}=\mathbf{x} \circ f^{-1}\) be the correspond ing parametrization and define \(\phi(\eta)=-2\left\\{\mathbf{y}_{4}, N_{4}\right\\}\). Show that on \(x(U) \cap y(V) .\) $$ \phi(\zeta)=\psi(\eta)\left(\frac{3 \eta}{\partial \zeta}\right)^{2} $$ d. Let \(S^{2}\) be the unit sphere of \(R^{3}\). Use the stereographic projection (cf. Exercise 16, Sec. 2.2) from the poles \(N=(0,0,1)\) and \(S=(0,0,-1)\) to cover \(S^{2}\) by the coordinate neighborhoods of two (isothermal) complex parameters, \(\zeta\) and \(\eta\), with \(\zeta(S)=0\) and \(n(N)=0\), in such a way that in the intersection \(W\) of these coordinate neighborhoods (the sphere minus the two poles) \(\eta=\zeta^{-1}\). Assume that there exists on each coontinate neighborhood analytic functicns \(\psi(\zeta), \psi(n)\) such that \((+)\) holds in W. Use Liouville's theorem to prove that \(\varphi(\zeta)=0\) (hence, \(\psi(\eta)=0\) ). c. Let \(S \subset R^{3}\) be a regular surface with constant mean curvature homeomorphic to a sphere. Assume that there exists a conformal diffeomorphism \(\varphi: S \rightarrow S^{2}\) of \(S\) onto the unit sphere \(S^{2}\) (this is a consequence of the uniformization theorem for Riemann surfaces and will be assumed here). Let \(\bar{\zeta}\) and \(\bar{\eta}\) be the complex parameters corresponding ander \(\varphi\) to the parameters \(\zeta\) and \(\eta\) of \(S^{2}\) given in part \(\mathrm{d}\). By part a, the function \(\phi(\bar{\zeta})=((e-g) / 2)-\) if is analytic. The similar function \(\psi(\bar{p})\) is also 5.3. Complete Surfaces. Theorem of Hopf.Aioow \(93 t\) analytic, and by part c they are related by \((+)\). Use part d to show that \(\phi(\bar{\zeta})=0\) (hence, \(\psi(\bar{i})=0\) ). Conclude that \(S\) is made up of umbilical points and hence is a sphere. This proves Hopf's theorem.

Fix a point \(p_{0} \in R^{2}\) and define a family of maps \(\varphi_{t}: R^{2} \rightarrow R^{2}, t \in[0,1]\), by \(\varphi_{t}(p)=t p_{0}+(1-t) p, p \in R^{2}\). Notice that \(\varphi_{0}(p)=p, \varphi_{1}(p)=p_{0}\). Thus, \(\varphi_{t}\) is a continuous family of maps which starts with the identity map and ends with the constant map \(p_{0}\). Apply these considerations to prove that \(R^{2}\) is simply connected.

Let \(\alpha:[0, l] \rightarrow R^{3}\) be a regular closed curve parametrized by arc length. Assume that \(0 \neq|k(s)| \leq 1\) for all \(s \in[0, l]\). Prove that \(l \geq 2 \pi\) and that \(l=2 \pi\) if and only if \(\alpha\) is a plane convex curve.

(Stoke's Remark.) Let \(S\) be a complete geometric surface. Assume that the Gaussian curvature \(K\) satisfies \(K \leq \delta<0\). Show that there is no isometric immersion \(\varphi: S \rightarrow R^{3}\) such that the absolute value of the mean curvature \(H\) is bounded. This proves Efimov's theorem quoted in Remark 2 with the additional condition on the mean curvature. The following outline may be useful: a. Assume such a \(\varphi\) exists and consider the Gauss map \(N: \varphi(S) \subset R^{3} \rightarrow\) \(S^{2}\), where \(S^{2}\) is the unit sphere. Since \(K \neq 0\) everywhere, \(N\) induces a new metric (, ) on \(S\) by requiring that \(N \circ \varphi: S \rightarrow S^{2}\) be a local isometry. Choose coordinates on \(S\) so that the images by \(\varphi\) of the coordinate curves are lines of curvature of \(\varphi(S)\). Show that the coefficients of the new metric in this coordinate system are $$ g_{11}=\left(k_{1}\right)^{2} E, \quad g_{12}=0, \quad g_{22}=\left(k_{2}\right)^{2} G $$ where \(E, F(=0)\), and \(G\) are the coefficients of the initial metric in the same system. b. Show that there exists a constant \(M>0\) such that \(k_{1}^{2}
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