Showing that the riemannian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete

Question

I would like to show that a certain Riemannian metric defined on $\mathbb{R^3}$ is complete. The metric is given in the following sentence from this article (pg 160):

... a Riemannian metric on $\mathbb{R^3}$ which, outside the unit ball, is $\frac{g_{(x, y, z)}}{\sqrt{x^2+y^2+z^2}}$, where $g_{(x, y, z)}$ denotes the standard Euclidean metric. This defines a complete Riemannian metric...

If we name this metric h, I understand it to be: $$ h_{(x, y, z)}(u, v) = \begin{cases} g_{(x, y, z)}(u, v) & ||x|| < 1 \\ \\ \frac{g_{(x, y, z)}(u, v)}{\sqrt{x^2 + y^2 + z^2}} & \text{otherwise} \\ \end{cases} $$

for $u, v \in T_{(x, y, z)}\mathbb{R^3}$.

As far as I understand, a Riemannian metric on a manifold is complete if the geodesics for that metric are defined on all of $\mathbb{R}$. How may one proceed to show such a thing? The best I could come up with is to try to find a diffeomorphism from $\mathbb{R^3}$ to itself that will induces this metric, but I have not managed to find any diffeomorphism whose differential at each point would yield the above metric.

All help appreciated!

Answer 1

There is a very usefull criterion for proving the completeness of a Riemannian metric. First of all we introduce the following definition: a smooth curve $ \gamma:[0,+\infty) \rightarrow M $ is called divergent if for every compact subset $ K \subset M $ there exists $ t_0 $ such that for every $ t \geq t_0 $ $ \gamma(t) \notin K $.

A riemmanian manifold $ M $ is complete if and only if the length of any divergent path is unbounded.

Now for your example let $ \gamma:[0,b) \rightarrow R^3 $ be a path such that $ |\gamma(0)|=1 $ and $ |\gamma(b)|=R $ where $ |\cdot | $ is the standard euclidean metric. Therefore if $ l(\gamma) $ is the length of $ \gamma $ in your conformal metric, using Schwartz inequality, you can easily prove that

$ l(\gamma) = \int_{0}^{b}\frac{|\dot{\gamma}(t)|}{|\gamma(t)|} dt \geq \log R $

Therefore if you take $ R \rightarrow \infty $ you can prove that divergent path have unbounded length.

Answer 2

Here's a different approach. By the Hopf-Rinow theorem, if, at some particular point $p$, all geodesics emanating from $p$ are define for all time, then $M$ is complete.

So, we may as well focus on geodesics emanating from the origin. Second, by expressing the metric in spherical coordinates, we see the metric is spherically symmetric. This leads to two facts. First, the isometry group of $M$ will act transitively on the set of geodesics emanating from the origin, so it's enough to prove that a single geodesic emanating from the origin is defined for all time. Second, geodesics through the origin just look like regular straight lines (but with a different parameterization).

So, let's pick a geodesic, say the one in the $x$ direction. We can write it as $\gamma(t) = (x(t), 0, 0)$ where $x(t)$ is some smooth (except, perhaps, at $t = 1$) increasing function satisfying $x(0) = 0$. Further, we have $x(t)=t$ for $t\leq 1$ because the metric is Euclidean there. We'll now solve the geodesic equation $$x'' + \Gamma^x_{xx}x'x' = 0$$ to find out what the rest of $x$ is. (Note the only possible Christoffel symbol we'll care about is $\Gamma^x_{xx}$ since $\gamma'$ and $\gamma''$ are both $0$ in the $y$ and $z$ slots.)

We have $$\Gamma^x_{xx} = \frac{1}{2}g^{xx}\left(\frac{\partial g_{xx}}{\partial x} \right) = \frac{1}{2} x \left(\frac{-1}{x^2}\right) = -\frac{1}{2x}.$$

Plugging this into our geodesic equation, we get $$x'' -\frac{1}{2x} (x')^2 = 0$$ which is solved, according to Wolfram, by $$x(t) = \frac{(c_1^2 t^2)}{(4 c_2)}+c_1 t+c_2.$$

The initial conditions $x(1) = 1$ and $x'(1) = 1$ force $c_1 = \frac{1}{2}$ and $c_2 = \frac{1}{4}$, so over all, our solution is $$x(t) = \frac{t^2}{4} + \frac{1}{2}t + \frac{1}{4}.$$

It's now clear from inspection that $x(t)$ is defined for all (positive) time. This implies $M$ is complete.