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Let M be a smooth Riemannian manifold. Let $f:S^1\to M$ be a locally-flat injective loop. Must there exist such $\varepsilon>0$ that if we connect points $f(0),f(\varepsilon), f(2\varepsilon)... , f([1/\varepsilon]\varepsilon), f(1)=f(0) $ by shortest geodisic path then we obtain piecewise-smooth loop without self-intersection?

If it is true, does this $\varepsilon$ depend on $f$ continuously?

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I'm not sure what you intend by the terminology "locally flat injective loop". Do you intend this in the sense of topology of manifolds, that it's a continuous injective map where each point in the image has a neighborhood making the curve homeomorphic to a line in $\mathbb R^n$?

If this is the correct interpretation, then it isn't true even for curves in the plane. You can start with a circle, and make a countable sequence of embellishments in various intervals by putting in an inward spiral and matching outward spiral of total angle $4\pi$ each. Parametrize it so that the inward spiral for the $i$th embellishment takes a very short time $< \epsilon_i$, then it lingers in the middle for some time say $> 10 \epsilon_i$, and then it spirals outward in a way that steps of length $< 10 \epsilon_i$ traverse angles $< \pi/2$.

Then for any series of geodesic steps with parameter length between $\epsilon_i$ and $10 \epsilon_i$ are self-intersecting in this embellishment, because the turning number of the inward journey does not match the turning number of the outward journey.

With a sequence of embellishments of this form where the $\epsilon_i$ form a geometric sequence, you can make polygonal approximations with steps of sufficiently small $\epsilon$ all have self-intersection. In the plane, of course, if you choose $\epsilon$ large enough that there are only 3 points, you'll get a triangle, and it will be simple. If you want to prevent even large polygons from being simple, I'm confident you can do it with a different Riemannian metric on the plane (and therefore for some Riemannian metric on any other surface and on any manifold of any dimenson).

As noted in comments, every injective image of a circle in the plane is locally flat. Here is a plot of a suitable sample embellishment. The blue curve is the Jordan curve, parametrized in a way that the inward spiral is traversed nearly 20 times as fast as the outward spiral. For some range of stepsizes, the inward journey (starting from the left) is forced to intersect the outward journey, because the total angle of turning in the inward journey is 0 (since there is just one intermediate step), while in the outward journey it turns counterclockwise two full revolutions.

alt text

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Yes, interpretation is right. Sorry, I don't understand why it works. Secondly, there is limit point $x\in S^1$ of embellishment preimages. Why loop is locally-flat in $x$? – Nikita Kalinin Jan 24 '11 at 5:26
@Nikita Kalinin: In the plane, the image of every injective map of $S^1$ is locally flat. This is a consequence of the Riemann mapping theorem, which extends continuously to the boundary for any simply-connected domain whose complement is locally connected. (A theorem of Caratheodory), but this theorem may have earlier history). Do this for the outside and inside on $S^2$, and make an isotopy on the boundary, to construct a homeomorphism of $S^2$ carrying it to the equator. I may try making some pictures to make the idea more clear. – Bill Thurston Jan 24 '11 at 18:20
@Bill Thurston: If you do it at least for one embellishment I will be very happy :) – Nikita Kalinin Jan 24 '11 at 19:24
@Bill Thurston: Thanks for knowledge about plane! I suspect it but don't believ in it. It seems that difference between topological and locally-flat curves appears in codimension 2 and disappears in large codimension... – Nikita Kalinin Jan 24 '11 at 19:28
Thank you very much! – Nikita Kalinin Jan 25 '11 at 6:54

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