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Is it true that any simply connected compact $n$-dimensional Riemannian manifold with positive isotropic curvature is diffeomorphic to the standard sphere $S^n$? I know that it is true for the case $n=4$ (R. Hamilton proved it in the 1990s), how about $n\geq 5$ ?

I also see that it is true for $n\geq 4$ under a stronger curvature assumption i.e. $(M,g)\times R$ has positive isotropic curvature (S. Brendle proved it in the 2000s).

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    $\begingroup$ I'll point out the trivial observation that since Micallef-Moore proved that such a manifold is homeomorphic to the sphere, one may conclude in certain dimensions that it is diffeomorphic to the standard sphere (e.g. dims. 5,6,12) since there is a unique smooth structure. So your question is interesting for $n\geq 7$. en.wikipedia.org/wiki/Exotic_sphere $\endgroup$ – Ian Agol Jan 13 '16 at 21:39
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This is an important open problem, see a discussion of some of the chief difficulties in Sec 1.4 of the book "The Ricci flow in Riemannian Geometry" of Andrews and Hopper.

As an unsolicited remark, I favor the shorter way (equivalent, by Micallef-Moore) of phrasing the question as "Does there exist an exotic sphere with positive isotropic curvature?"

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If you are interested there is a paper of A. Fraser and J. Wolfson (http://users.math.msu.edu/users/wolfson/fraser-wolfson-final.pdf) where they prove that for a compact Riemannian manifold $(M^n,g)$ of dimension $n\geq 5$, with positive isotropic curvature, the fundamental group does not contain a subgroup isomorphic to the fundamental group of a compact Riemann surface!

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The recent paper of Simon Brendle seems to answer your question "Ricci flow with surgery on manifolds with positive isotropic curvature"

https://arxiv.org/abs/1711.05167

See Corollary 11.3 for details.

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The result of Brendle, Corollary 11.3 of https://arxiv.org/abs/1711.05167, needs an additional assumption that the manifold, with dimension n, does not contain non-trivial incompressible (n − 1)-dimensional space forms. A recent preprint on arxiv claims that the additional assumption is not necessary, see https://arxiv.org/abs/1909.12265v1.

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  • $\begingroup$ The answer to the question asked was already answered by Brendle's work. In fact if $M$ is PIC, compact and simply connected, then it is homemorphic to $S^n$. Now $S^n$ does not contain incompressible space forms (it is a topological assumption), so we can apply Brendle's result. $\endgroup$ – Thomas Richard Sep 30 '19 at 12:04
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    $\begingroup$ Thanks, I didn't know this. Also, the preprint arxiv.org/abs/1909.12265v1 was revised. The result is not as strong as in version 1. $\endgroup$ – Entao Zhao Oct 15 '19 at 3:18
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    $\begingroup$ So, as of today, is this conjecture considered solved? It seems that the paper of Brendle only deals with dimensions $\geq12$. $\endgroup$ – Colescu Jun 15 '20 at 11:35

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