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In an online webinar, I heard (not directly) the statement that (closed) manifolds of nonnegative curvature operator $\mathcal{R}\geq 0$ are symmetric spaces. Is this a valid theorem? Any reference that contains its proof? I am not sure that in the above statement whether "positive curvature" is a part of assumptions or not.

So by the above claim, it seems that $\mathcal{R}\geq 0\iff$ $M$ is symmetric space!

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    $\begingroup$ mathoverflow.net/a/264899/3948 seems relevant. $\endgroup$
    – Willie Wong
    Commented Apr 19, 2021 at 16:58
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    $\begingroup$ If you slightly deform the standard metric on the unit sphere, you get a metric if positive curvature operator which is not symmetric. The same will happen if you slightly deform a spherical space form. Are you asking if all closed manifolds of nonnegative curvature operator are diffeomorphic to locally symmetric spaces? $\endgroup$
    – Igor Belegradek
    Commented Apr 19, 2021 at 17:46
  • $\begingroup$ @WillieWong: my question is different from that linked MO. $\endgroup$
    – C.F.G
    Commented Apr 19, 2021 at 17:47
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    $\begingroup$ Yes, I think closed irreductible manifolds of nonnegative curvature operator are diffeomorphic to locally symmetric spaces, see e.g. section 2.4 of arxiv.org/pdf/1511.07899.pdf. $\endgroup$
    – Igor Belegradek
    Commented Apr 19, 2021 at 18:00
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    $\begingroup$ @C.F.G ... hence a comment, just the mention that it is relevant, and not a vote to close as duplicate. $\endgroup$
    – Willie Wong
    Commented Apr 19, 2021 at 18:18

1 Answer 1

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As Igor Belegradek commented, the correct statement is as follows:

Theorem (classification of closed simply connected manifold with nonnegative curvature operator): A closed simply connected manifold with nonnegative curvature operator is isometric to a Riemannian product of

  1. standard spheres with metrics of nonnegative curvature operator

  2. closed Kahler manifolds biholomorphic to complex projective spaces whose Kahler metric has nonnegative curvature operator on real (1, 1)-forms

  3. compact irreducible Riemannian symmetric spaces with their natural metrics of nonnegative curvature operator.

See for instance Page 270, theorem 7.34 of

Chow, Bennett; Lu, Peng; Ni, Lei, Hamilton’s Ricci flow, Graduate Studies in Mathematics 77. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-4231-5/hbk). xxxvi, 608 p. (2006). ZBL1118.53001.

So closed simply connected irreducible manifolds of nonnegative curvature operator are isometric to locally symmetric spaces.

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    $\begingroup$ Your conclusion should be diffeomorphic, not isometric, because of Igor's comment (small deformations of the standard metric on the sphere give metric with positive curvature operator which are not locally symmetric). $\endgroup$
    – Jeffrey Case
    Commented Apr 20, 2021 at 10:53
  • $\begingroup$ @JeffreyCase: I see, but above theorem says that "is isometric to" . $\endgroup$
    – C.F.G
    Commented Apr 20, 2021 at 10:54
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    $\begingroup$ The first case (with the irreducible assumption) is a manifold diffeomorphic to a sphere equipped with a metric whose curvature operator is nonnegative. These do not have to be locally symmetric (i.e. $\nabla R$ need not vanish). $\endgroup$
    – Jeffrey Case
    Commented Apr 20, 2021 at 10:57
  • $\begingroup$ @JeffreyCase: I am not sure about the statement and I just quoted it from the books and papers. Fell free to correct the post. $\endgroup$
    – C.F.G
    Commented Apr 20, 2021 at 11:01
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    $\begingroup$ @JeffreyCase I believe it is correct to write "isometric to" because the ambiguity in the metric restricts to the factors and is covered by the phrasing of 1 and 2 ("standard sphere" meaning "non-exotic"). $\endgroup$
    – Sebastian Goette
    Commented Apr 25, 2021 at 9:42

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