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Whether a complete non-compact non-flat Riemannian $n$-manifold $M$ with non-negative sectional curvature has Euclidean volume growth?

That is, whether there is a constant $C>0$ such that $\mathrm{Vol}(B_x(r))\geq Cr^n$ for all $r>0$ and $x\in M$? Here $\mathrm{Vol}(B_x(r))$ is the volume of the $r$-ball in $M$.

Since manifolds with non-negative Ricci curvature and Euclidean volume growth are studied a lot, I am curious about the non-negative sectional curvature case.

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It is certainly not true that every complete nonflat open manifold of nonnegative curvature has Euclidean volume growth. Counterexamples are trivial to construct. Say, a capped cylinder. More generally any nonnegatively curved manifold $M^n$ with nontrivial soul has slower than Euclidean volume growth. Because its asymptotic cone at infinity has dimension strictly smaller than $n$ if the soul is not a point while manifolds with Euclidean volume growth have asymptotic cones of dimension $n$.

This also implies that any nonnegatively curved manifold with Euclidean volume growth is diffeomorphic to $\mathbb R^n$.

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  • $\begingroup$ Futhermore, does it imply that the metric is isometric to the Euclidean metric up to scaling? Whether it also holds for complete non-compact Alexandroff spaces (with n-dimensional Hausdorff dimension or n-dimensional topological dimension) with curvature bounded below by 0? $\endgroup$
    – Jialong Deng
    Apr 11, 2022 at 3:02
  • $\begingroup$ @JialongDeng I don't understand the question. Does what imply that the metric is Euclidean? and what statement about Alexandrov spaces are you asking? $\endgroup$
    – Vitali Kapovitch
    Apr 11, 2022 at 3:15
  • $\begingroup$ The question askes for the statement in the last part of your answer. $\endgroup$
    – Jialong Deng
    Apr 11, 2022 at 3:20
  • $\begingroup$ if you are asking whether a nonnegatively curved manifold with Euclidean volume growth (meaning $vol B_r(p)\ge cr^n$) must be isometric to $\mathbb R^n$ that is certainly false. if you are asking something else please clarify. $\endgroup$
    – Vitali Kapovitch
    Apr 11, 2022 at 3:24
  • $\begingroup$ Yes, there is my question. And whether a complete non-compact Alexandroff spaces (with n-dimensional Hausdorff dimension or n-dimensional topological dimension) with curvature bounded below by 0 and Euclidean volume growth is homeomorphic to $\mathbb{R} ^n$? $\endgroup$
    – Jialong Deng
    Apr 11, 2022 at 3:31

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