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  1. Let $M$ be a non-compact connected Riemannian manifold with $\mathrm{sec}_g=0$ and $\operatorname{vol} B(x,r)\geq c(n)r^n$ for any $r$, where $c(n)>0$. How to prove that $(M,g)$ is isometric to $(R^n,\mathrm{can})$?

  2. Let $M$ be a non-compact connected Riemannian manifold with $\mathrm{ric}_g=0$, $\mathrm{conj.rad}=\infty$, and $\operatorname{vol}B(x,r)\geq c(n)r^n$ for any $r$, where $c(n)>0$. How to prove that $(M,g)$ is isometric to $(R^n,\mathrm{can})$?

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    $\begingroup$ Such manifold should have n linearly independent straight lines issuing from the same point, and then Cheeger-Gromoll splitting theorem gives isometry in question. Does it? $\endgroup$
    – valeri
    Apr 1, 2014 at 16:08
  • $\begingroup$ Could you give some background, motivation, and things you tried? $\endgroup$ Apr 1, 2014 at 17:36
  • $\begingroup$ @valeri we don't know the information about $inj.rad$, but in Shing-Tung Yau. one can proof that, under the condition $Ric\geq 0$, noncompact riemannian manifold, we have $\omega_nr^n\geq volB(x,r)\geq k(n)r$. but in here, we have r$^n$ growth volume, I think this must be linked with the topology. here the riemannian manifold is connected. $\endgroup$
    – qiuguozhou
    Apr 2, 2014 at 4:01
  • $\begingroup$ @Benoît Kloeckner:cheeger and Gromov consider the classes of the remannian manifolds with $|sec|\leq\lambda, volB(x,1)\geq v,diam\leq d$ which convergence in the $C^{1,\alpha}$ topology.in sense,they have finite classes of diffeomorphism types.those ideas was generalied by Anderson.Andersons' ideas is proof that these have uniformly harmonic radius from below,but all by argue by contradiction. which rescal in a factor,that we want to proof that in the pointed convergence of Gromov-hausdorff convergence the limit space is the Euclidean metric (R^n,can),that is we want to proof of those. $\endgroup$
    – qiuguozhou
    Apr 2, 2014 at 4:22
  • $\begingroup$ @Benoît Kloeckner:in here,for the second question:that is we consider the riemnian manifolds with $|ric|\leq\lambda,conj.rad\geq i_o,diam\leq d,volB(x,1)\geq v$, the first question is $|ric|\leq\lambda,|R|_p\leq\lambda,volB(x,r)\geq vr^n$,here $p>\frac{n}{2}$, all we rescal by argue by contradiction,all we want to the limit spaces is $(R^n,can)$ $\endgroup$
    – qiuguozhou
    Apr 2, 2014 at 4:32

1 Answer 1

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Answer to the first question :

a complete flat Riemannian manifold of dimension $n$ is a quotient $\mathbb{R}^n$ by a discrete subgroup of isometry $\Gamma$. The Euclidean volume growth implies that $\Gamma$ is finite but as $\Gamma$ acts freely on $\mathbb{R}^n$, we must have $\Gamma=\{1\}$.

About the second question :

According to an inequality of Calabi and Yau, the volume growth of a complete Riemannian manifold of non negative Ricci curvature is contoled by $$\forall r\ge 1 :{\mathrm vol}B(x,r)\ge k(n)r {\mathrm vol} B(x,1).$$

According to M. Anderson and P. Li a complete Riemannian manifold with non negative curvature and Euclidean growth is simply connected. (M. Anderson, On the topology of complete manifolds of nonnegative Ricci curvature, Topology 29 (1990) 41--55 and P. Li, Large time behavior of the heat equation on complete manifolds with nonnegative Ricci curvature, Ann. of Math. 124 (1986) 1-21.)

In your case, the manifold is necessary diffeomorphic to $\mathbb{R}^n$. Because the hypothesis on the conjugate radius implies that the exponential map is a covering. As the Euclidean volume growth implies that the fundamental group is finite, this will imply that this covering is trivial.

There are metric on $\mathbb{R}^n$ that are rotationnally symmetric, with non negative ricci curvature and Euclidean growth for instance in dimension $2$ take the metric (with $a\in (0,1)$) $$g=(dr)^2+\left(a r+(1-a)\tanh(r) \right)^2(d\theta)^2$$ The conjugate radius at the $0$ is infinite.

If you assume that the conjugate radius is infinite at every point of the manifold. This implies that all geodesic must be a ray : let $\gamma\colon \mathbb{R}\mapsto M$ be a infinite geodesic. $$\gamma(t)=\exp_{\gamma(0)} (tv)$$ We know that $\gamma$ is minimizing on $[0,+\infty)$ and on $(-\infty,0]$ because the exponential map is a diffeomorphism.

But we also have for all $T$ : $$\gamma(t)=\exp_{\gamma(T)} ((t-T)\dot \gamma(T))$$ Hence $\gamma$ is minimizing on $[T,+\infty)$ and on $(-\infty,T]$ hence $\gamma$ is a ray and the Cheeger-Gromoll splitting theorem implies that the manifold is isometric to the Euclidean space.

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  • $\begingroup$ thanks your good advices,here $conj.rad=\infty$ is for any $x\in M$, But it is hard to say just by the $conj.rad=\infty$ without using $Ric=0$,example for the $sec=-1$,simple connected riemanian manifolds. $\endgroup$
    – user49024
    Apr 2, 2014 at 5:46

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