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This question is from Lectures on the Geometry of Manifolds by Nicolaescu. (Exercise 6.2.8 b)

Let $(M,g)$ be a complete (connected) Riemannian manifold and let $(\tilde{M},\tilde{g})$ be its universal cover, $\pi:\tilde{M}\to M$ the covering transformation with $\tilde{g}=\pi^{\ast}g$. Prove that $(\tilde{M},\tilde{g})$ is complete.

My thoughts: $\pi:\tilde{M}\to M$ is a local isometry. It suffices to prove that $\tilde{M}$ satisfies one of the hypotheses of Hopf-Rinow, and then we're done. My idea was to show that all geodesics on $\tilde{M}$ can be defined forever using a path-lifting argument. So if we have a geodesic for some $p\in M, V\in T_pM$, namely $\gamma:\mathbb{R}\to M$, take a fixed $q\in \pi^{-1}(p)$ and $X\in T_q\tilde{M}$ such that $d\pi_q(X)=V$ (since it is a local diffeomorphism), and let $\delta:(-\epsilon,\epsilon)\to \tilde{M}$ be a geodesic for $q,X$. Moreover, $\pi\circ \delta=\gamma$ at least on a small interval, because they have the same initial point and same initial derivative.

Now, we can lift $\gamma$ to $\tilde{\gamma}:\mathbb{R}\to \tilde{M}$ starting at $q$. We have that $\pi\circ \tilde{\gamma}=\gamma$.By uniqueness, on a small interval, $\tilde{\gamma}=\delta$. Thus the geodesic $\delta$ can be extended to a map $\tilde{\gamma}:\mathbb{R}\to\tilde{M}$. Since this was arbitrary, $\tilde{M}$ is geodesically complete.

Does this work? If not, how should I approach this problem?

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    $\begingroup$ This is completely correct. $\endgroup$ – Andreas Cap Oct 29 '15 at 16:55

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