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Let $M$ be smooth compact riemannian manifold with boundary and $\varphi_0: S^1\to M$ be a rectifiable curve (or a smooth one). I would like to find a reference to the following statement:

Statement. There exists a length decreasing homotopy $\varphi_t:S^1\to M$, $t\in [0,1]$, such that the $\varphi_1(S^1)$ is locally length minimizing curve (i.e. can not be slightly perturbed to a shorter curve.) or a point.

$\varphi$ is length decreasing if for any $t_1<t_2$ the length of $\varphi_{t_2}(S^1)$ is smaller than the length $\varphi_{t_1}(S^1)$.

Remark. In the case when $M$ has no boundary, of course, $\varphi_1(S^1)$ should be a closed geodesic.

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    $\begingroup$ Does a point qualify as locally length-minimizing? $\endgroup$
    – Igor Rivin
    Commented Jan 25, 2018 at 0:19
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    $\begingroup$ It seems false to me. How about $M=\mathbb R_s\times S^1_t$ with metric $ds^2+(2-e^{-1/s^2}\cos(\frac 1s))dt^2$ as a probable counterexample. The central circle $\{0\}\times S^1$ is not locally length minimizing ($\{\frac 1{2\pi k}\}\times S^1$ has smaller length for any positive integer $k$). I would suspect there is no length decreasing homotopy starting at $\{0\}\times S^1$ (let alone one converging to a locally length minimizing curve). $\endgroup$
    – John Pardon
    Commented Jan 25, 2018 at 17:09
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    $\begingroup$ Thanks a lot John, indeed it looks as a counterexample...I have not thought carefully of what "locally minimizing" means for a geodesic... $\endgroup$
    – aglearner
    Commented Jan 25, 2018 at 18:13
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    $\begingroup$ Igor, in this example the problem is located close to the geodesic $0\times S^1$, so one can easily make this example compact... $\endgroup$
    – aglearner
    Commented Jan 25, 2018 at 19:04
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    $\begingroup$ @IgorBelegradek it's just $\mathbb R$. The subscript indicates the name of the variable used as the coordinate of that factor. $\endgroup$
    – John Pardon
    Commented Jan 25, 2018 at 19:21

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The argument is basically that given by Milnor in his total curvature of knots paper: For your curve take an "inscribed polygon" (that is, take a collection points on $\gamma$ so that there each of them is on the smooth part of your curve lies in the "Gauss disk" (as in Gauss Lemma) of its neighbors, join each pair of neighbors by a geodesic, repeat. Clearly, this is a length-decreasing homotopy. A standard Arzela-Ascoli argument shows that there is a limit curve, and the only way the homotopy cannot be continued is if the curve is actually geodesic (or a point).

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    $\begingroup$ Igor, thanks. If you are speaking of Brikhoff shortening process, it is known that it does not need to converge: homepages.warwick.ac.uk/~masgak/papers/bhb-catone.pdf so I don't see why this will give you a continuous homotopy. $\endgroup$
    – aglearner
    Commented Jan 25, 2018 at 0:55
  • $\begingroup$ It does not need to converge. There will be a convergent subsequence whose length will go to the infimum. $\endgroup$
    – Igor Rivin
    Commented Jan 25, 2018 at 1:15
  • $\begingroup$ Exactly, and my question is about a continuous homotopy, not about a sequence. I don't see how to get such a homotopy from what you propose $\endgroup$
    – aglearner
    Commented Jan 25, 2018 at 1:23
  • $\begingroup$ @aglearner each curve in the sequence is homotopic to any other one, same is true for any subsequence. $\endgroup$
    – Igor Rivin
    Commented Jan 25, 2018 at 2:06
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    $\begingroup$ Igor, sorry I realized that the confusion probably happened because I had not given the definition of a "length decreasing homotopy". This is a homotopy such that for any $t_1<t_2$ the length of $\varphi_{t_2}(S^1)$ is smaller than the length $\varphi_{t_1}(S^1)$. You agree that in the present form your answer does not settle this question? $\endgroup$
    – aglearner
    Commented Jan 25, 2018 at 9:10

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