4
$\begingroup$

Let $(M,g)$ be a connected Riemannian manifold which admits a universal cover $(\tilde{M}, \tilde{g})$, where $\tilde{g}$ is the Riemannian metric such that the covering is a Riemannian covering. I want to know under what conditions the universal cover $\tilde{M}$ is complete. The reason for this questions is that I want to know under what conditions on $M$ the Hopf-Rinow theorem can be applied to the universal cover.

On Wolfram (http://mathworld.wolfram.com/CompleteRiemannianMetric.html) it says that if $M$ is compact, its universal cover is complete. Would someone be able to give a proof of this?

And what deductions can we make if $M$ is complete (and possibly fulfills some other conditions)? (I'm not really looking for curvature conditions like corollaries of the Bonnet-Myers theorem).

Thanks in advance for any help!

$\endgroup$
  • 5
    $\begingroup$ A Riemannian manifold is complete if and only if its universal cover is complete. To see this, use Hopf Rinow in order to conclude that it suffices to prove that geodesics are indefinitely extendable, together with the lifting property and the fact that geodesics lift to geodesics. $\endgroup$ – Aloizio Macedo Jul 18 '16 at 17:09
7
$\begingroup$

It's actually true that $M$ is complete if and only if its universal cover $\widetilde{M}$ is complete. Let $p: \widetilde{M} \to M$ be the universal covering map, $q \in M$ and $\tilde{q} \in p^{-1}(q)$. As has already been stated, by Hopf-Rinow, all we need to if we want to conclude that $\widetilde{M}$ implies $M$ is complete is prove the corresponding statement for the exponential maps based at $q$ and $\tilde{q}$.

Now if $\widetilde{M}$ is complete and $\widetilde{E}: T_{\widetilde{q}}\widetilde{M} \to \widetilde{M}$ is its exponential map based at $\widetilde{q}$, define a map

$$ E = p \circ \widetilde{E} \circ (dp)^{-1}: T_qM \to M$$

You can show that $p$ sends geodesics to geodesics by showing their images are locally length minimizing. Since $(dp)^{-1}$ is linear it sends radial lines to radial lines, and you can use this to show that $E$ is exactly the exponential map for $M$ based at $p$. Then from the above it follows immediately that $E$ is defined on the whole tangent space.

The other answer shows the reverse implication.

$\endgroup$
1
$\begingroup$

If $M$ is compact,Hopf Rinow implies that $M$ is complete, let $\hat M$ be the universal cover of $M$, and $p:\hat M\rightarrow M$.Lift the metric defined on $M$ with $p$ and $p:\hat M\rightarrow M$ preserves the metric.Suppose $\hat M$ is incomplete. Let $c:I=(a,b)\rightarrow \hat M$ an incomplete geodesic maximal that you can't extend. $p(c)$ can be extended to $c':(a,b+\epsilon)\rightarrow M$ since $M$ is complete, and you can lift $c'$ to a geodesic $(a,b+\epsilon)\rightarrow \hat M$ since $p$ is a covering map. Contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.