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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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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. –  Robert Bryant Jan 19 '13 at 15:05
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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. –  Robert Bryant Jan 19 '13 at 15:12
    
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). –  alvarezpaiva Jan 19 '13 at 15:13
    
Thanks also for the second comment. –  alvarezpaiva Jan 19 '13 at 15:14
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