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How flat (flat in pancake-style, not in curvature 0-style), in some extrinsic intuitive measure, can a Zoll surface of revolution (embedded in Euclidean three-space) be?

I don't want to impose a measure of "flatness", but, for example, we can (1) normalize the area of the surface to be $4\pi$ and take the extrinsic distance between the poles as a measure of "flatness" or (2) consider the Hausdorff distance between the surface and a flat disc.

If the surface is convex, then it cannot be too flat: fixing the length of the closed geodesics to $2\pi$ (or the area to $4\pi$), the distance between the poles must be greater than $\pi -2$ (really easy exercise, but this quantity is not optimal).

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    $\begingroup$ Now, instead, does it make sense?... ^_o $\endgroup$
    – Qfwfq
    Dec 2, 2014 at 10:28
  • $\begingroup$ In a loose way, yes ... Oh, I forgot to add that the surface is a surface of revolution embedded in Euclidean three-space. $\endgroup$ Dec 2, 2014 at 10:53
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    $\begingroup$ (An aside: An image of a Zoll surface of revolution can be found at the question, "How to draw a Zoll surface?.") $\endgroup$ Dec 2, 2014 at 12:42
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    $\begingroup$ It isn't clear to me if the extrinsic distance from a point to its opposite pole is positive. Keep in mind that Zoll surfaces can have negative curvature on open sets, so not convex. $\endgroup$
    – Ben McKay
    Dec 2, 2014 at 12:50
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    $\begingroup$ @JosephO'Rourke: Yes you can have negative curvature for a Zoll surface of revolution. Have a look at the examples described in Chapter 4, Section C (particularly, see Fig 4.25d-e) of A. Besse, Manifolds all of whose geodesics are closed. (Also, take a look at the comment at the top of page 107.) $\endgroup$ Dec 3, 2014 at 18:29

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