Orientation double cover Let $M$ be a manifold and let $\bigwedge^\text{top}TM$ be the top exterior product of the tangent bundle. Then this becomes a line bundle. Let $g$ be any metric on $\bigwedge^\text{top}TM$ and define $\hat{M}:=\{x\in\bigwedge^\text{top}TM:g(x,x)=1\}$, thus $\hat{M}$ is the space of all unit vectors. My question is: Why is $\pi:\hat{M}\rightarrow M$ a smooth double cover which is independent of the choice of the metric and why $\hat{M}$ has a natural orientation. I thought that the independence is implied by the fact that we have a line bundle.
Moreover: $M$ is orientable iff $\hat{M}=M\coprod M$. Can someone help me, with this because i have no idea, how to start. Thanks.
 A: Here are some hints:


*

*The projection $\pi:\hat M \to M;(p,\omega)\mapsto p$ is the restriction of the smooth projection $\wedge^{top}TM \to M$.

*For the degree of the cover, think about the preimage $\pi^{-1}(p)$. How many "unit-length" elements $\omega$ are there in $\wedge^{top} T_pM$?

*In terms of it being a local diffeomorphism, think about what a neighborhood of $(p,\omega)$ in $\wedge^{top} TM$ looks like. The smoothness of the metric $g$ (as $p$ varies) is crucial here.

*How are you defining orientability? If it is a "smooth assignment" of equivalence classes of bases for $T_p \hat M$ (positive vs negative bases), then you can use a unit-length top-dimensional form $\omega$ to give each basis of $T_p \hat M$ a sign. The same thing goes for $T_{(p,\omega)}\hat M$.

*As for linking the orientability (or lack thereof) of $M$ to the connectivity of $\hat M$, think about showing that a certain subset of $\hat M$ is open and closed, hence a connected component.

*To see that $\hat M$ is independent of $g$, first convince yourself that you can define a set bijection between $\hat M_g$ and $\hat M_{g'}$ (respectively constructed using $g$ and $g'$). Show that the obvious map for a set bijection is smooth.
