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I know that the Poincaré conjecture was first proved in dimension ≥ 5, then dimension 4, and finally 3. I'm just curious, does the Ricci flow approach by Perelman shed any light on the high dimension case? Or for some reason one cannot hope for a uniform proof because these dimensions are essentially different?

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    $\begingroup$ One may think of Ricci flow with surgeries as an algorithm. Since there is no algorithm to say that more-than-three-dimensional manifold is simply connected, you do not have much chance. (It will not help even if an old man told you that manifold is simply connected and you trust him.) $\endgroup$
    – Anton Petrunin
    Commented Oct 26, 2012 at 15:43
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    $\begingroup$ An other reason, is that in dimensions 5 and up, the neck surgeries do not lead to a simpler manifold. At least in principle, they can cancel each other. $\endgroup$
    – Anton Petrunin
    Commented Oct 26, 2012 at 15:44
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    $\begingroup$ There's Brendle-Schoen: ams.org/journals/bull/2011-48-01/S0273-0979-2010-01312-4/… You could also consider Ricci flow on cohomogeneity one metrics, which exist on Milnor spheres: arxiv.org/abs/math/0601765 $\endgroup$
    – Ian Agol
    Commented Oct 26, 2012 at 15:45
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    $\begingroup$ This is in the opposite direction to the original question, but the Ricci approach also works in dimension 2: arxiv.org/abs/math/0505163 . (And in dimension 1 as well, I suppose, though this is a vacuous truth.) $\endgroup$
    – Terry Tao
    Commented Oct 26, 2012 at 17:43
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    $\begingroup$ This is somehow hopeless (at this moment) if you want to get general topological conclusions. Ricci flow encounters singularities, and we expect rescaled limit of singularities are Ricci solitons. The problem is we don't have a classification of solitons - in dimension 4 or higher we don't "even" have a classification of Einstein manifolds. In dimension 3, Einstein manifolds are of constant curvature and compact solitons are just Einstein, and one knows a lot on noncompact solitons (Hamilton, Ivey, Perelman, Brendle, ...) That is a necessary condition that Ricci flow is fruitful in 3d. $\endgroup$
    – Caramba
    Commented Nov 21, 2012 at 5:17

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