What does it mean for a covering to be trivial and one other quetsion from Lawson and Micheldohn's "Spin Geometry"

Question

On page 78 of Lawson and Michelsohn's Spin Geometry they say the following:

consider the bundle $P_O(E)$ of orthonormal frames in $E$ (here $\pi:E\to X$ is a vector bundle equipped with a metric). This is the principal $O(n)$-bundle whose fibre at a point $x\in X$ is the set of orthonormal bases of $E_x \cong \pi^{-1}(x)$. The bundle of orientations in $E$ is then just the quotient $Or(E) = P_O(E)/SO(n)$, where two bases of $E_x$ are identified if the orthogonal matrix transforming one to the other has determinant $+1$. Note that $Or(E)$ is a $2$-sheeted covering space of $X$ and that $E$ is orientable if and only if this covering space is the trivial one.

Questions:

1. Why is $Or(E)$ a two sheeted covering? In fact I'm not even sure how it a covering space. I suppose I have to prove that the trivializing open sets of $\pi:E\to X$ are the evenly covered open sets of $X$. But I am not sure how to begin.

2. What do they mean by trivial covering space here? As I understand a trivial covering space is just a space covering itself via the identity map. Does that mean $Or(E)$ must be $X$, in which case how would it be a two sheeted covering?

Thank you.

Answer 1

1. It is a double covering because every vector space has two orientations.

2. A non-trivial covering is a covering space which is connected. A trivial $k$-sheeted covering of a space $X$ is $X\times\{1, \dots, k\} \to X$ given by $(x, a) \mapsto x$.