As far as I know, in a simply connected compact manifold, still there exists no well-known obstruction for a manifold with a quasi-positive curvature to be a manifold with positive curvature.

But Hopf's conjecture is unsolved, i.e., the conjecture: $S^2\times S^2$ has a positive curvature.

So I think that the quasi-positive curvature-condition can be weakened by conditions : non-negative curvature and non-flatness.

Here my concrete question is : Is there a simply connected compact simple Lie group which does not have a metric with a positive curvature and has a metric with a non-negative curvature ?


Here we say that a Riemannian manifold $M$ has positive (resp. non-negative) curvature if all sectional curvatures are positive (resp. non-negative) at all points of $M$ And a manifold with quasi-positive curvature is a manifold with a non-negative curvature and a point at which all tangent 2-planes have positive curvature.

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    $\begingroup$ Could you please clarify the question? (I suppose that the last sentence of your post is in doubt and you would like to know if it's a true statement?) $\endgroup$ – SashaKolpakov Oct 10 '13 at 4:23
  • $\begingroup$ Okay I will clarify my question. $\endgroup$ – Hee Kwon Lee Oct 10 '13 at 8:17
  • $\begingroup$ what you call "to be positively curved" should be termed "to admit a positively curved Riemannian metric" $\endgroup$ – YCor Oct 10 '13 at 8:35
  • $\begingroup$ That's right. And I eddited. $\endgroup$ – Hee Kwon Lee Oct 10 '13 at 8:42
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    $\begingroup$ What you are asking about is a well-known open problem. See the survey by Ziller: math.upenn.edu/~wziller/papers/SurveyMexico.pdf for most recent summary of obstructions and examples. Also the biinvariant metric on any simply-connected compact Lie group has nonnegative curvature (this is an easy exercise in Do Carmo's "Riemannian geometry''). $\endgroup$ – Igor Belegradek Oct 10 '13 at 11:32

For high rank Lie group, this is in fact the generalized Hopf conjecture which is still open.


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