The metric $g_{ij} = \frac{\delta_{ij}}{(1 + \frac{K}{4}\|x\|^2)^2}$ is not complete on $\mathbb{R}^n$, if $K > 0$.

Question

Suppose that in $\mathbb{R}^n$ is given the metric:

$$g_{ij} = \frac{\delta_{ij}}{(1 + \frac{K}{4}\|x\|^2)^2},$$ where $\|x\|$ is the standard metric of $\mathbb{R}^n$ and $K>0$ a constant and $\delta_{ij}$ is the Kronecker delta. How can I show that the this metric, that is defined on the entire $\mathbb{R}^n$ is complete?

I appreciate any comments, hints or suggestions? I tried to work with divergent curves but I do not think that it is a good idea.

score 1 · Accepted Answer · answered Oct 23 '16 at 15:23

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$c(t)=(t,0,\cdots, 0)$ so that $$| c'(t)|=\frac{1}{1+\frac{K}{4} t^2},\ t\geq 0 $$

$$ {\rm length}\ c|[0,\infty) \leq 1 +\int_1^\infty \frac{4}{K} \frac{1}{t^2} < C<\infty $$ for some $C>0$ That is $$ d_g(O,c(t_n))\leq C$$ If $M:=\mathbb{R}^n$ is complete then bounded sequence has a convergent subsequence and its limit is in $M$ But its limit does not exist

answered Oct 23 '16 at 15:23

HK Lee

19,964

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Just a few questions. What do you mean by $\rm O$? Why does the sequence $c(t_n)$ does not converge? – L.F. Cavenaghi Oct 23 '16 at 15:28
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Ah I mean that $O$ is origin in $\mathbb{R}^n$ – HK Lee Oct 23 '16 at 15:29
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If $t_n=n$ then $c(t_n)$ goes to $\infty$ – HK Lee Oct 23 '16 at 15:29

score 1 · Answer 2 · answered Oct 23 '16 at 15:24

The curve $c(t) = (t,t,\cdots,t)$, $t \geq 0$ is clearly divergent. But $$L[c] = \int_0^\infty \sqrt{g_{c(t)}(c'(t),c'(t))}\,{\rm d}t = \int_0^\infty \frac{\sqrt{n}}{1+\frac{nKt^2}{4}}\,{\rm d}t < \infty.$$If you already know that a Riemannian manifold $(M,g)$ is complete if and only if every divergent curve in $M$ has infinite length, we're done.

The metric $g_{ij} = \frac{\delta_{ij}}{(1 + \frac{K}{4}\|x\|^2)^2}$ is not complete on $\mathbb{R}^n$, if $K > 0$.

2 Answers2