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Is the following statement true? Assume that $(M^n, g, p)$ is a pointed complete manifold with metric $g$, $Rc(g)\geq 0$, $\{s_i\}$ is a positive sequence decreasing to $0$, and $(M^n, s_i\cdot g, p)$ converges to metric space $(N, h, p)$, where $h$ is the metric on $N$. Then we always have that $h= dr^2+ f(r)^2 dX$, where $r$ is the distance function from point $p$, $X$ is another compact metric space.

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  • $\begingroup$ This question might related to Milnor's conjecture on fundamental group of $M$. In fact C. Sormani showed in JDG 53 (1999) "Nonnegative Ricci curvature, small linear diameter growth and finite generation of fundamental groups" that Theorem 11 (Pole Group Theorem). If a complete noncompact manifold, $M^n$, with nonnegative Ricci curvature has a fundamental group which is not finitely generated, then it has a tangent cone at infinity, $(Y,y_o)$, which does not have a pole at its base point. $\endgroup$
    – J. GE
    May 12, 2013 at 19:03

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The answer is no. There are manifold with positive Ricci curvature and tangent cone at infinity is not polar. The example is constructed by Menguy "Examples of nonpolar limit spaces. Amer. J. Math. 122 (2000), no. 5, 927–937".

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  • $\begingroup$ @J. GE, as I know, the example in Menguy's paper is about tangent cone, but maybe the strategy in his argument can be applied to generate the tangent cone at infinity which is not polar. $\endgroup$
    – guoyi xu
    May 17, 2013 at 20:27
  • $\begingroup$ @guoyi xu, It's Theorem 0.2 in Menguy's paper and you can also find it in the section 5. There is no much difference between these two types of constructions. The intuition is the 'peaks' destroys the pole, and the peaks are scaling invariant. $\endgroup$
    – J. GE
    May 17, 2013 at 21:53

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