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A theorem of Bott states that if a manifold admits a metric with all geodesics closed, then its homology is isomorphic to the homology of one of the manifolds from the list: $S^n, \mathbb{RP}^n, \mathbb{CP}^n, \mathbb{HP}^n$ or $\mathbb{C}a\mathbb{P}^2$.

The problem of constructing such metrics is known to be extremely hard.

So my particular interest is: what else is known about this problem in the case of dimension $3$?

For example, are all homology spheres known to admit such metrics? If a homology sphere (or a homology $\mathbb{RP}^3$ --- by the way, are there any examples which are not $\mathbb{RP}^3$?) has finite $\pi_1$, its universal cover is $S^3$, and I believe that all geodesics will be closed in the metric induced by the standard metric on $S^3$.

But I've just learned that there are examples of homology spheres with infinite fundamental group. Their Thurston geometry is modeled on the universal cover of $\operatorname{SL}(2, \mathbb{R})$. It would be very interesting to know if such homology spheres admit metrics with all geodesics closed.

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    $\begingroup$ Just to be sure: I assume you are aware of Arthur Besse's (old) extensive monograph on this subject? (Manifolds all of whose geodesics are closed, Ergeb. Math., vol. 91, 1976, 250 pp.) $\endgroup$ – Vesselin Dimitrov Jan 9 '18 at 18:20
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    $\begingroup$ A nice quotient of the round three-sphere $S^3$ (say, a lens space) will have all geodesics closed, but need not have the homology of $S^3$ or of $P^3$. So I don't understand the statement that you attribute to Bott. $\endgroup$ – Sam Nead Jan 9 '18 at 18:48
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    $\begingroup$ Ah - indeed, you have left out a hypothesis. Bott is assuming that all geodesics emanating from some special point $p$ are closed and simple. $\endgroup$ – Sam Nead Jan 9 '18 at 18:54
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    $\begingroup$ The connected sum of $\mathbb{RP}^3$ with a non-trivial homology sphere is a non-trivial homology $\mathbb{RP}^3$. $\endgroup$ – Michael Albanese Jan 9 '18 at 21:13
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There are integral homology spheres in the following Thurston geometries: $S^3$, $\mathrm{PSL}(2,\mathbb{R})$, and $H^3$. No manifold of the latter two types (when equipped with any metric) can have all of its geodesics being closed. This follows from observing that the fundamental group contains a non-trivial free group, and then a Gromov-Hausdorff limiting argument.

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