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Given a simple Riemannian metric $(D,g)$ on the two-disc---its geodesics have no conjugate point and the boundary of the disc is strictly convex---, is it possible to embed $(D,g)$ isometrically into a Zoll two-sphere?

I'm also interested in the same question on the $n$-disc. A related question is whether one can distinguish a Zoll metric locally (or at least say: this metric cannot be Zoll because in the neighborhood of this point it does not behave in such and such way). I'm guessing the answer to this last question is no and this prompted the first question.

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    $\begingroup$ I don't have an answer, but I think it is likely that the answer to your second question is 'no'. Remember that Guillemin proved that any odd function on the $2$-sphere is the first derivative of a conformal deformation of the round $2$-sphere through Zoll metrics. I believe that his proof can be used to show that, for any given smooth metric $g$ on the $2$-disk, a small enough neighborhood of the origin is isometric to a disk on a Zoll $2$-sphere. This would rule out any local obstruction. I haven't gone through the details to check this, but it seems reasonable and probably could be done. $\endgroup$ Jan 19, 2013 at 15:05
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    $\begingroup$ Oh, and I believe that your question does have an affirmative answer in the case of a rotationally symmetric simple metric. I think that, in fact, Zoll's original construction of the rotationally symmetric metrics on the $2$-sphere with all geodesics closed shows this. $\endgroup$ Jan 19, 2013 at 15:12
  • $\begingroup$ Thanks Robert, I was thinking the same thing but I was wondering if in higher dimensions there would be some local obstructions since the infinitesimal obstructions to be a deformation are also trickier (Kiyohara condition and so forth). $\endgroup$ Jan 19, 2013 at 15:13
  • $\begingroup$ @RobertBryant I am not sure about "rotational case": at least, what I was thinking about Zoll's original construction (written up by Berger in his lectures for example) is exactly the contrary: Zoll's surface of rotation has an equator and if you give me any function above the equator, I can continue it with another function below the equator to close this little "rotated cap" to some Zoll surface of rotation. Or I do not understand something?.... $\endgroup$
    – Olga
    Apr 1, 2017 at 10:53
  • $\begingroup$ @alvarezpaiva Did you learn something new about the answer to this question since then?... I am very interested! $\endgroup$
    – Olga
    Apr 1, 2017 at 10:53

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