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In every lecture on Riemannian geometry it is standard to prove that geodesic curves are locally length minimizing. The only thing I find confusing about this is, that here length minimizing means: compared to all piecewise smooth curves in contrast to, say, all continuous curves. So my question is:

Are geodesics locally length minimizing in the continuous curves?

If generally they are not: Under which conditions can we obtain such a result? Can you give any counterexamples?

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    $\begingroup$ Continuous curves don't neccessarily have a length. Let's add the hypothesis that the curve be rectifiable, also lets assume we are in a complete Riemannian manifold, then the answer is yes. $\endgroup$
    – Charlie Frohman
    Commented Jun 27, 2011 at 12:43
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    $\begingroup$ I'd think a non-rectifiable continuous curve could naturally be assigned infinite "length" (by using the same supremum of distances as in the definition for rectifiable curves), so that you wouldn't have to worry about them as candidates for minimizing lengths. $\endgroup$
    – Andreas Blass
    Commented Jun 27, 2011 at 13:09
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    $\begingroup$ To amplify @Charlie's comment: how do you define the length of a curve? If as the limit of piecewise-smooth approximations, then the statement is immediate. If not, you should tell us your definition of length... $\endgroup$
    – Igor Rivin
    Commented Jun 27, 2011 at 15:31

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Your question will be trivial once you give a definition of the length of curve in a Riemannian manifold.

For example, you may define distance as infimum of lengths piecewise smooth curves connecting given points. Then you define length of general curve as you do it in a metric space...

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  • $\begingroup$ Thank you, Anton! I made a mistake in formulating the question, because ow I realize that actually I was not really thinking of a Riemannian manifold, but of something like an embedded surface getting its metric from the surrounding space. Does the "program" you suggested also work in this case? $\endgroup$
    – Sebastian Scholtes
    Commented Jun 27, 2011 at 16:52
  • $\begingroup$ Isn't an embedded surface getting a metric from the ambient space also a Riemannian manifold?? $\endgroup$
    – Willie Wong
    Commented Jun 27, 2011 at 17:20
  • $\begingroup$ For an embedded surface one can show that a small neighborhood of the surface admits a retraction to the surface with small Lipschitz constant... $\endgroup$
    – Anton Petrunin
    Commented Jun 27, 2011 at 17:20
  • $\begingroup$ As it answers the question perfectly and as I also realize that this was probably a stupid question to begin with, I accept Anton's answer. Just, so that you know what got me started on this and what I had in mind when asking: if I have a subarc of a great arc on S^2 that has minimal length under all piecewise smooth curves [0,1]->R^3 connecting p and q on S^2, is it clear from differential geomtric methods that this arc is also minimzing under all continous curves [0,1]->R^3 with values in S^2 connecting p and q? $\endgroup$
    – Sebastian Scholtes
    Commented Jun 27, 2011 at 18:00

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