Universal cover of a riemannian manifold inherits the metric of space

Question

I read on http://mathworld.wolfram.com/UniversalCover.html that any property of X can be lifted to its universal cover, as long as it is defined locally, the riemannian metric of a manifold $(X)$ is defined locally, so the I want to think that the unviersal cover have a metric deffined by the metric of $X$, but I don't know how to prove this or where to find it. If you can give me some book that help me with this would be very helpfull. Thanks.

Answer 1

If $(M,g)$ is a Riemmanian manifold and $p: \tilde M \to M$ is a covering space projection, then in particular $p$ is a local diffeomorphism. (Note: $\tilde M$ does not need to be the universal cover. It can be any cover, since all we'll use is that $p$ is a local diffeomorphism.)

Then for $x \in \tilde M$, the differential of $p$ is an isomorphism $p_*: T_x \tilde M \tilde\rightarrow T_{p(x)} M$, which allows us to define an inner product $\tilde g_x$ on $T_x \tilde M$ from the inner product $g_{p(x)}$ on $T_{p(x)} M$.

Explicitly: $\tilde g_x(Y,Z) := g_{p(x)}(p_*Y,p_*Z)$.

Doing so for each $x\in \tilde M$ defines a metric $\tilde g$ on $\tilde M$, whose smoothness follows from smoothness of $g$ and the fact that $p$ is a local diffeomorphism.