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I am wondering that the following statement is true or not:

Let $(M,g)$ be a complete non-compact Riemannian manifold with $0 < Sect \leq C\cdot dist(O,x)^{-2a}$, $a\in(0,1]$. ($O$ is a point in $M$ and $Sect$ is the sectional curvature.) Then the injectivity radius $inj(x)$ has a lower bound $c\cdot dist(O,x)^{a}$ for some constant $c>0$.

I believe this is true at least for simply-connected manifolds, but I cannot prove it even when $M$ is diffeomorphic to $\mathbb{R}^n$.

Thanks for any comments.

Edit: According to Igor's comment and answer, this question is only undetermined for $n\geq 3$.

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  • $\begingroup$ This fails for nonnegatively curved metrics on the plane that are cylinders $S^1\times [r,\infty)$ outside of a compact subset. The curvature is zero there but going around the cylinder gives an upper bound on the injectivity radius. Products with flat tori give counterexamples in higher dimensions. Did you mean to ask something else? $\endgroup$ May 18, 2014 at 19:17
  • $\begingroup$ 0 < sec implies M^n IS diffeomorphic to R^n ... $\endgroup$
    – valeri
    May 18, 2014 at 20:24
  • $\begingroup$ Oh, I missed the assumption that the curvature is positive. Still the example on $\mathbb R^2$ stands. All you need to do is to find a strictly concave function $f$ that is asymptotically sufficiently $C^2$ close to a constant and satisfies the usual conditions at $0$, and use it for the metric $dr^2+f^2(r) d\phi^2$. $\endgroup$ May 18, 2014 at 20:49
  • $\begingroup$ To valeri: Thanks. To Igor: yes, you are right. A cylindrical end indeed has zero curvature and $inj(x)\to const$. I forgot to say that $n\geq 3$. Sorry for missing this condition. $\endgroup$ May 19, 2014 at 13:33

1 Answer 1

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As mentioned in comments, positive curvature implies (by Gromoll-Meyer) that the soul is a point. Here is a counterexample on $\mathbb R^2$. The metric will be of the form $dr^2+f^2(r) d\phi^2$ in polar coordinates. Assume $f(r)=\sin r$ for small nonnegative $r$, and $f(r)=1-e^{-r}$ for large $r$. Also assume that $f(r)>0$ for $r>0$ and $f^{\prime\prime}<0$ everywhere. It is easy to see that these conditions are consistent. The point is that we need $f^\prime$ to decrease from slope $1$ at $(0,0)$ to almost horisontal slope $e^{-r}$ at $(r, 1-e^{-r})$ and visibly one can interpolate between the two slopes. The metric is asymptotic to a cylinder, so the injectivity radius is bounded above by $\pi$ for large $r$. The curvature is $-\frac{f^{\prime\prime}}{f}>0$ and it is asymtotically zero.

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  • $\begingroup$ Thank you, Igor. I could accept your answer only later because I wish to see more suggestions (from you or others) for the higher dimensional case. Do you have any comment about that case? $\endgroup$ May 19, 2014 at 13:43
  • $\begingroup$ You certainly should not accept the answer if you really meant to ask about $n>2$, and in any case accepting an answer right away is not a good idea: other people may wish to contribute in a few days. I never thought about manifolds with subquadratic curvature decay, so have no immediate suggestions. $\endgroup$ May 19, 2014 at 14:02

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