The Zoll surfaces have the property that all of their geodesics are closed.
If one futher stipulates that all geodesics are also *simple*, i.e., non-self-intersecting,
does this leave only the sphere?


Apologies for the simplicity of this question, but I am not finding an answer in the literature,
and I suspect many just know this off the top of their head. Thanks!



Surfaces all of whose geodesics are both closed and simple
From Guillemin's "The Radon transform on Zoll surfaces", it follows that there are deformations of $S^2$ which keep all geodesics closed AND simple.


