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It is well known that R. Hamilton (JDG 1982) used Ricci flow to show that a closed $3$-manifold with positive Ricci curvature must be diffeomorphic to a spherical space form $S^3/\Gamma$, since such metrics evolve to a limit which has constant curvature.

Together with some colleagues, I was under the suspicion that it was previously known that closed $3$-manifolds with positive sectional curvature satisfied this conclusion, but cannot seem to find such an earlier proof independent of Ricci flow. More precisely, I suspected it was possible to prove this statement by establishing that the Heegaard genus of such a manifold must be $\leq2$. So here is my:

Question. Is there a proof that if a closed manifold $M^3$ has $\sec>0$ then $M^3\cong S^3/\Gamma$ using Heegaard splittings instead of Ricci flow?

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  • $\begingroup$ Is there an analogous result for 2 manifolds which is done without Ricci flow? $\endgroup$
    – Juan Sebastian Lozano
    Commented Mar 2, 2017 at 1:13
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    $\begingroup$ For 2-manifolds you can simply use the uniformization theorem, which longly predates any Ricci flow techniques. $\endgroup$
    – Renato G. Bettiol
    Commented Mar 2, 2017 at 3:37
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    $\begingroup$ It was known (after Hamilton's theorem) by Thurston's orbifold theorem) that genus 2 manifolds satisfy geometrization (one can now prove this independent of Rocci flow). But I don't know how to prove these manifolds are genus 2. $\endgroup$
    – Ian Agol
    Commented Mar 5, 2017 at 3:32
  • $\begingroup$ In dim 3, obviously they are homology 3-sphere. then you need prove that positive sec/Ricci implies being sphere in simply connected case that is same as RH theorem. $\endgroup$
    – C.F.G
    Commented Mar 29, 2021 at 21:08

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No, this was not known before Hamilton's paper. I think a Ricci-flow-free argument would still be of interest, so if you know how to do it, by all means...

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