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 exist counterexamples to show that these assumptions are necessary.
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3$\begingroup$ 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. $\endgroup$– Deane YangSep 23, 2010 at 0:21
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1$\begingroup$ Assuming that $G \ge c g$, where $c$ is a positive constant and $g$ is the standard flat metric, suffices. $\endgroup$– Deane YangSep 23, 2010 at 0:47
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$\begingroup$ 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? $\endgroup$– MajidSep 23, 2010 at 1:27
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$\begingroup$ Same proof as before should work. I don't see why $\omega_2$ is needed. $\endgroup$– Deane YangSep 23, 2010 at 1:34
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$\begingroup$ 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 $d_G$? $\endgroup$– MajidSep 23, 2010 at 2:17
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2 Answers
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You are assuming that the identity map between the Riemannian metric and the Euclidean metric is bi-Lipschitz, but this is certainly far too strong. Here are two counter-examples:
Take some self-diffeomorphism $f$ of $\mathbb{R^n}$ with derivative unbounded below and above and pull back the Euclidean metric.
Take hyperbolic $n$-space, which is complete and diffeomorphic to $\mathbb{R}^n$ but not bi-Lipschitz to it.
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$\begingroup$ Is this condition also necessary? At the moment I don't see why that sould be the case, but I cannot construct a counter-example. An intuitive approach would be: Let $p_n=(n,0,0,...)$. Choose the metric such that $G$ equals the usual Euclidian metric everywhere except $\bigcup K_n$ where $K_n$ is the ball with radius $1/n$ around $p_n$. Now choose $G$ such that $G\leq 1/n g$ somewhere in $K_n$. Then certainly $G$ does not fulfill the condition stated, but probably $R^n$ will be complete wrt this metric as well. $\endgroup$– user34711Jun 6, 2013 at 11:01
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A standard necessary and sufficient condition is that every bounded set for your metric is bounded in $\mathbb{R}^n$ with the standard Euclidean norm.
answered Oct 29, 2014 at 9:11
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$\begingroup$ I'm sure this is well known to people experienced with manifolds, but do you know of a text book mention that this fact? $\endgroup$– ttbJun 30, 2017 at 22:08