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Consider the Riemannian manifold $\mathbb{R}^n$ and a smooth Riemannian metric $G:\mathbb{R}^n\rightarrow\mathbb{R}^{n\times{n}}$. What is the minimum assumption on $G$ such that the manifold $\mathbb{R}^n$ is complete as a metric space with respect to the Riemannian distance determined by $G$. The Riemannian distance $d_G(x,y)$ between points $x,y\in\mathbb{R}^n$ is defined as follows: $d_G(x,y)=\inf_{\chi\in\Omega(x,y)}\int_0^1{\sqrt{\left(\frac{d\chi(s)}{ds}\right)^TG(\chi(s))\frac{d\chi(s)}{ds}}ds}$, where $\Omega(x,y)$ is the set of all the piecewise smooth paths connecting $x$ to $y$. It can be easily checked that if $G$ satisfies: $\omega_1\Vert{y}\Vert_2^2\leq y^TG(x)y\leq\omega_2\Vert{y}\Vert_2^2$ for any $x,y\in\mathbb{R}^n$ and for some positive constants $\omega_1$ and $\omega_2$, then $\mathbb{R}^n$ is a complete metric space with respect to $d_G$. Moreover, if we can find a change of coordinate such that the metric in the new coordinate satisfies the preceding assumption, again the result is true. I am just wondering if there exist weaker assumptions on $G$ to show the result or there exists counter examples to show that this assumptions are necessary.

edited Sep 23 2010 at 0:02

jc
3,915●9●23

asked Sep 22 2010 at 23:59

Majid
41●1

1

These conditions are certainly not necessary. The bounds on $G$ imply that the volume of a geodesic ball grows like $r^n$ where $r$ is the radius. But there are plenty of complete Riemannian manifolds whose geodesic balls grow at either a slower or faster rate. In fact, these conditions are extremely strong and far stronger than needed for completeness. – Deane Yang Sep 23 2010 at 0:21

Assuming that $G \ge c g$, where $c$ is a positive constant and $g$ is the standard flat metric, suffices. – Deane Yang Sep 23 2010 at 0:47

How can I show the result using this condition? Do I need to construct a Cauchy sequence on the original metric and show that it has a limit? – Majid Sep 23 2010 at 1:27

Same proof as before should work. I don't see why $\omega_2$ is needed. – Deane Yang Sep 23 2010 at 1:34

By having that lower bound, I can show that any Cauchy sequence in $\mathbb{R}^n$ with respect to $d_G$ is a Cauchy sequence with respect to $d_g$, where $g$ is Euclidean metric or the flat metric. As we know, $\mathbb{R}^n$ with respect to $d_g$ is a complete metric space. Hence, the sequence converge to a point in $\mathbb{R}^n$ with respect to $d_g$. Now, how can I show that the sequence converges to the same point with respect to $d_G$? – Majid Sep 23 2010 at 2:17

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When does the Riemannian manifold $\mathbb{R}^n$ is complete as a metric space with respect to the Riemannian distance?

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