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I know that every compact riemannian manifold $M$ is complete. And by Hopf-Rinow $M$ is geodesically complete.

But now I'm confused by the closed unit disc $B^2\subset\mathbb{R}^2$. (I think $B^2$ is compact, but not geodesically complete.

Is Hopf-Rinow just working for manifolds without boundary?

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Is $B^2$ endowed with the metric it inherits from $\Bbb R^2$? If so, then you know who the geodesics are, and thus wether or not it is complete! –  Olivier Bégassat Oct 4 '12 at 21:42
    
Yes I meant the canonical metric. My question was not, whether B^2 is complete or not, but why it isn't a counterexample to the wellknown fact of Hopf-Ronow. Again: $B^2$ is compact and complete, hence complete as a metric space <=> geod. complete) What is wrong here? –  Braten Oct 5 '12 at 5:29
    
What do you mean by geodesically complete? Complete and has geodesics? Because the closed disc certainly has this property. Note that Hopf-Rinow works for any complete, locally compact length space. –  user641 Oct 5 '12 at 7:21
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Geodesics are straight lines and so they will meet the boundary after "finite time". Hence there are no geodesics defined on the entire real line!! –  Braten Oct 5 '12 at 21:54
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The problem is is purely terminological: For most Riemannian geometers a "compact manifold" means "a compact manifold without boundary". With this definition, every compact Riemannian manifold is indeed metrically complete and, hence, by Hopf-Rinow, is geodesically complete. Other people allow compact manifolds to have boundary (they would call compact boundaryless manifolds "closed"). Hopf-Rinow theorem does not apply to manifolds with boundary: They are never geodesically complete. –  studiosus Mar 19 at 0:57

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