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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).

alvarezpaiva
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    Now, instead, does it make sense?... ^_o – Qfwfq Dec 02 '14 at 10:28
  • In a loose way, yes ... Oh, I forgot to add that the surface is a surface of revolution embedded in Euclidean three-space. – alvarezpaiva Dec 02 '14 at 10:53
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    (An aside: An image of a Zoll surface of revolution can be found at the question, "How to draw a Zoll surface?.") – Joseph O'Rourke Dec 02 '14 at 12:42
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    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. – Ben McKay Dec 02 '14 at 12:50
  • Ben, you change from point to pole in the wording of your comment. Could you please clarify? Do you mean that the Zoll surface could have self-intersections and we could think of the distance between the poles (fixed points of the circle action) as negative? – alvarezpaiva Dec 02 '14 at 12:54
  • As I said in the question, I'd gladly consider other ways of measuring flatness (Hausdorff distance to a plane disc ?). – alvarezpaiva Dec 02 '14 at 12:57
  • @BenMcKay: "can have negative curvature on open sets"---May I ask: Do you know if this is still possible for a Zoll surface of revolution? – Joseph O'Rourke Dec 02 '14 at 19:08
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    @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.) – Robert Bryant Dec 03 '14 at 18:29
  • I can't tell you the answer but I think it should exist;) By this I mean that Zoll surfaces of revolution are given by explicit formulas, using any odd function on the interval [-1,1] you have a Zoll metric. So probably the curvature can be calculated then. What do you think? – Olga May 19 '16 at 17:19
  • @Olga, very roughly speaking the way one constructs Zoll surfaces of revolution is that one is free to choose one of the hemispheres (let's say the north hemisphere, the equator being a closed planar geodesic) and you compensate accordingly with the other hemisphere to make the surface Zoll. One would really have understand this "north-south" construction better (or at least better than I do). – alvarezpaiva Jun 17 '16 at 13:15

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