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To state the context, I am familiar with the Hopf-Rinow theorem.

My request is three fold,

  • I would like to know of general classes of geodesically incomplete spaces. I basically want to see lots of examples for this.

  • I want to know of techniques of proving and testing for geodesic incompleteness or completeness.

  • I want to know of methods of proving and testing for existence of maximal extension of curves.

{Confusingly in quite a bit of literature I have seen the adjective "inextensible" being used when actually they seem to want to mean "maximally extended". Is there some subtle point here that I am missing?}

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I have been told (but don't know how to prove) that every noncompact surface admits a metric of positive sectional curvature. However, the only noncompact surface admitting a complete metric of positive curvature is $\mathbb{R}^2$. Hence, most of the above metrics will be incomplete. –  Jason DeVito Dec 9 '10 at 14:08
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For a silly example, take a proper open subset of a complete connected Riemannian manifold. By Hopf-Rinow, this can't be geodesically complete. –  Akhil Mathew Dec 9 '10 at 15:37
    
I think that @Akhil's example is a very good one to think about. –  Matt E Dec 9 '10 at 18:45
    
@Akhil Thanks for this simple class of examples (should have thought of that!) –  Anirbit Dec 10 '10 at 6:50
    
@Jason Apologies for the rudimentary state of my understanding. Can you kindly explain what is the connection between positive sectional curvature and metric incompleteness? (If you can give a reference etc) –  Anirbit Dec 10 '10 at 6:52

2 Answers 2

Geodesic incompleteness occurs when a space has "holes" in it, or if it is possible to "fall off the edge" of the space. Here are some examples:

  1. The punctured plane $\mathbb{R}^2 - \{(0,0)\}$ is geodesically incomplete in the standard metric. For example, a horizontal geodesic starting at $(-1,0)$ and traveling in the positive $x$-direction can only be extended for one unit.

    Note that this space is homeomorphic to an infinite cylinder $S^1\times\mathbb{R}$, which is complete under the standard metric. The difference is that it isn't possible to get to infinity in a finite distance on the cylinder, while it is on the punctured plane.

    Covers of a geodesically incomplete space are also geodesically incomplete. Thus all of the covers of the punctured plane are geodesically incomplete, including the universal cover.

  2. The open unit disc $\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2 < 1\}$ is geodesically incomplete in the standard metric. No geodesic can be extended for more than $\pi$ units.

  3. The open upper half plane $\{(x,y)\in\mathbb{R}^2\mid y >0\}$ is geodesically incomplete in the standard metric. As with the disc, the idea is that geodesics can fall off the edge of the space.

    Note that the upper half plane is complete under the hyperbolic metric. The difference is that it takes infinite time to reach the $x$-axis under the hyperbolic metric, while it can be reached in finite time under the Euclidean metric.

  4. Most of the examples so far are proper open subsets of complete Riemannian manifolds, but not every geodesically incomplete space has this form. For example, consider the infinite cone $C=\{(x,y,z)\in\mathbb{R}^3 \mid x^2+y^2=z^2\text{ and }z>0\}$ under the standard metric, and note that the vertex $(0,0,0)$ does not lie in $C$. This space is incomplete because geodesics can approach the point $(0,0,0)$, but it doesn't work to add this point because the closed cone $C\cup\{(0,0,0)\}$ isn't a Riemannian manifold.

    The covers of the punctured plane mentioned in example #1 also aren't open subsets of complete Reimannian manifolds.

The most basic way to prove that a Riemannian manifold $M$ is incomplete is to identify a geodesic $\gamma\colon[0,a)\to M$ that cannot be extended any further. The way to prove that $\gamma$ cannot be extended is to show that $\lim_{t\to a} \gamma(t)$ does not exist in $M$.

By the Hopf-Rinow theorem, you can also prove that $M$ is geodesically incomplete by proving that $M$ is incomplete as a metric space. Thus, you can also show that $M$ is geodesically incomplete by demonstrating a Cauchy sequence that doesn't converge, or by demonstrating that some closed and bounded subset of $M$ in not compact.

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The equation for a geodesic is just an ODE with an initial value. The classical Picard-Lindeloef theorem gives the existence of a maximal solution (defined from 0 to t_+). Thus, for instance, if you can show that some "energy" (for instance the L^2 integral of the derivative from 0 to t_+) is finite, then t_+=\infty.

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I'm not sure I understand your last sentence. If the Riemannian manifold is (0,1) (with metric inherited from the inclusion into $\mathbb{R}$, with standard metric), then the usual energy functional is finite, but geodesics cannot be extended for all time. –  Jason DeVito Dec 9 '10 at 15:31
    
Well, stupid me. I implicitly assumed completeness already and was referring to other ODE's. For instance you could have two different Riemannian metrics, one is complete and for the other you don't know (and you are considering geodesics w.r.t. the second metric). Sorry! –  OrbiculaR Dec 9 '10 at 19:01

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