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Chapter 5: Problem 7

A divergent curve on \(S\) is a differentiable map \(\alpha:[0, \infty) \rightarrow S\) such that for every compact subset \(K \subset S\) there exists a \(t_{0} \in(0, \infty)\) with \(\alpha(t) \notin K\) for \(t>t_{0}\) (i.e., \(\alpha\) "leaves" every compact subset of \(S\) ). The length of a divergent curve is defined as $$ \lim _{t \rightarrow \infty} \int_{0}^{t}\left|\alpha^{\prime}(t)\right| d t . $$ Prove that \(S \subset R^{3}\) is complete if and only if the length of every divergent curve is unbounded.

Short Answer

Expert verified
S is complete if and only if the length of every divergent curve on S is unbounded.

Step by step solution

01

- Understand the Definitions

First, understand the definitions given. A divergent curve on a surface S is a differentiable map \(\alpha:[0, \infty) \rightarrow S\) such that for every compact subset K of S, there exists a \(t_0 \in (0, \infty)\) with \(\alpha(t) \otin K\) for \(t > t_0\). The length of this curve is defined as \[\text{length} = \lim_{t \rightarrow \infty} \int_{0}^{t} \left| \alpha^{\prime}(t) \right| \ d t. \]
02

- Prove the 'if' direction

Assume that S is complete. We need to show that the length of every divergent curve on S is unbounded. Suppose the length of a divergent curve \(\alpha\) is bounded, i.e., \[\text{length} < \infty. \] This implies that \(\alpha(t)\) is a Cauchy sequence in S. Since S is complete, \alpha(t)\ converges to a point in S. This contradicts the definition of a divergent curve, which leaves every compact subset of S. Hence, the length of every divergent curve must be unbounded.
03

- Prove the 'only if' direction

Assume the length of every divergent curve on S is unbounded. We need to show that S is complete. Suppose for contradiction that S is not complete. Then there exists a Cauchy sequence \(p_n\) in S that does not converge to any point in S. Construct a curve \(\alpha(t)\) that passes through points of \(p_n\) with increasing t. Because \(p_n\) is Cauchy but does not converge in S, \(\alpha(t)\) is a divergent curve. Since the length of \alpha(t)\ is unbounded, it contradicts the assumption of S being incomplete. Hence, S must be complete.

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Key Concepts

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

divergent curve
A divergent curve on a surface \(S\) plays a crucial role in understanding the geometry of that surface.
Essentially, it is a differentiable map \(\alpha: [0, \infty) \rightarrow S\) that consistently veers off from any compact subset of the surface.
To break it down, a compact subset is a limited and closed part of the surface.
If we imagine a curve \(\alpha(t)\), for every confined area \(K\) we pick on the surface, there exists a point in time \(t_0\) after which the curve stays outside of \(K\).
This concept implies that the curve continually moves away from any restricted zone, hence the term 鈥榙ivergent鈥�.
complete surface
A complete surface is one where all Cauchy sequences converge to a point on the surface.
This concept is fundamental in differential geometry and relates closely to the properties of divergent curves.
To grasp this, envision a sequence of points getting increasingly close to one another 鈥� this is a Cauchy sequence.
On a complete surface, these points will gather and converge to a definite spot on the surface itself.
If the surface facilitates this, it is complete; if not, the surface is considered incomplete.
This completeness is significant as it ensures that various properties and theorems in geometry hold true.
Cauchy sequence
A Cauchy sequence is an ordered set of points that get closer and closer to each other as the sequence progresses.
Mathematically, this means for any small distance \(\epsilon > 0\), there exists an integer \(N\) such that for all integers \(m, n > N\), the distance between the points \(p_m\) and \(p_n\) is less than \(\epsilon\).
In the context of a complete surface, any Cauchy sequence must converge to a point within the surface.
If a surface is not complete, a Cauchy sequence might fail to converge, implying some form of 鈥榠ncompleteness鈥� in the surface鈥檚 structure.
Understanding Cauchy sequences helps in identifying and proving the properties of geometric surfaces.
unbounded length
The length of a curve on a surface can be quantified using calculus.
Specifically, the length of a divergent curve \(\alpha(t)\) is defined as follows: \ \[\text{length} = \lim_{t \rightarrow \infty} \int_{0}^{t} \left| \alpha^{\prime}(t) \right| \ dt. \] When we say this length is 鈥榰nbounded,鈥� it means that as \(t\) approaches infinity, the integral does not reach a finite value.
This indicates that the curve \(\alpha(t)\) stretches out endlessly, without ever looping back or converging to a specific point.
In proving the completeness of a surface, the unbounded nature of divergent curves is crucial, as it directly ties into the behavior of distances and lengths on the surface.

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Most popular questions from this chapter

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 \(S\) and \(\bar{S}\) be regular surfaces and let \(\varphi: S \rightarrow \bar{S}\) be a diffeomorphism. Assume that \(\bar{S}\) is complete and that a constant \(c>0\) exists such that $$ I_{p}(v) \geq c \bar{I}_{\psi(p)}\left(d \varphi_{p}(v)\right) $$ for all \(p \in S\) and all \(v \in T_{p}(S)\), where \(I\) and \(\bar{I}\) denote the first fundamental forms of \(S\) and \(\bar{S}\), respectively. Prove that \(S\) is complete.

a. Use Klingenberg's lemma to prove that if \(S\) is a complete, simply connected surface with \(K \geq 0\), then \(\exp _{p}: T_{p}(S) \rightarrow S\) is one-to-one. b. Use part a to give a simple proof of Hadamard's theorem (Theorem 1).

Let \(S_{1} \subset R^{3}\) be a (connected) complete surface and \(S_{2} \subset R^{3}\) be a connected surface such that any two points of \(S_{2}\) can be joined by a unique geodesic. Let \(\varphi: S_{1} \rightarrow S_{2}\) be a local isometry. Prove that \(\varphi\) is a global isometry.

Show that the map \(\pi: R \rightarrow S^{1}=\left\\{(x, y) \in R^{2} ; x^{2}+y^{2}=1\right\\}\) that is given by \(\pi(t)=(\cos t, \sin t), t \in R\), is a covering map.
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