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In the study of manifolds with positive sectional curvature, I guess the following statement is true:

Let $M$ be a manifold with positive (but not necessarily constant) sectional curvature. Then $M\times M$ admits a Riemannian metric with non-negative sectional curvature such that there exists a direction with $K=0$ and there exists a direction with $K=1$. ($K$= Sectional curvature)

I cannot find any counterexample for this statement. Can anybody give a counterexample for this statement?

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  • $\begingroup$ Take the sphere of very large radius as $M$. Then on $M \times M$ there is no direction with $K=1$, because the largest $K$ is that of the individual spheres $M$. $\endgroup$
    – Ben McKay
    Commented Dec 28, 2016 at 9:34
  • $\begingroup$ When you write "constant (Not necessarily) positive", do you mean positive (not necessarily constant)"? $\endgroup$
    – Ben McKay
    Commented Dec 28, 2016 at 9:34
  • $\begingroup$ Yes. Does this a different meaning? your example is not clear for me? can you explain? $\endgroup$
    – C.F.G
    Commented Dec 28, 2016 at 9:39
  • $\begingroup$ Maybe I made a mistake. My means is that there exists a direction with $K=0$ and there exists a direction with $K>0$. $\endgroup$
    – C.F.G
    Commented Dec 28, 2016 at 9:49

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Perhaps you are thinking of a conjecture of Heinz Hopf: no Riemannian metric on $S^2 \times S^2$ has positive sectional curvature. This conjecture is still open, as far as I can tell.

For more information see Manifold with a quasi-positive curvature and also see Wolfgang Ziller's survey paper: https://www.math.upenn.edu/~wziller/papers/SurveyMexico.pdf

If you take the product metric on $M \times M$, and take any two geodesics $C_1, C_2 \subset M$ then the sectional curvature along $C_1 \times C_2$ is zero, while, for any surface $S \subset M$ and point $p \in M$, the sectional curvature along $S \times \{p\}$ is positive by hypothesis. The sectional curvature is not negative along any surface in $M \times M$.

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