[Math] Is the orientation double cover unique

algebraic-topologygeneral-topology

My question comes from the following. We usually say "the universal cover" because, as we know, it is unique up to isomorphism on the adequate category (in particular, every universal cover is homeomorphic to each other).

I never dwelled in thinking about the orientation double cover. I just read its construction and knew that, if a manifold is orientable, the construction would yield two copies of the manifold. If it is not orientable, it will yield a connected orientable manifold. And unconsciously I absorbed the fact that it should be "unique". But it seems like it isn't the case.

For instance, take $\mathbb{R}P^3$. It is orientable and, as such, its orientation double cover should be two disjoint $\mathbb{R}P^3$. But $S^3$, through the projection, is also a double cover, and orientable, and obviously $S^3$ is not homeomorphic to two disjoint $\mathbb{R} P^3$.

Therefore, I have these questions:

1) Am I getting something wrong? (For instance, is my implicit definition of orientation double cover as a two-fold orientable cover wrong?)
EDIT: Thanks @GrumpyParsnip for telling me that my definition is wrong. Based on the answers, I think I didn't get my intention with this question clear. Now that my definition is "wrong", I'll ask: Why defining it the way it is? What makes this special cover… special? Why not allow any two-fold orientable cover to be a orientation double cover?

2) Does this issue not happen when the base manifold is non-orientable?

Best Answer

The orientation cover of a path connected manifold $M$ is indeed unique. To see why, there is a homomorphism $OR : \pi_1(M) \to \mathbb{Z}/2\mathbb{Z}$ which takes a closed path $\gamma$ to the nontrivial element of $\mathbb{Z}/2\mathbb{Z}$ if and only if $\gamma$ is orientation reversing. If $OR$ is nontrivial then the orientation cover is the covering space corresponding to $\text{kernel}(OR)$. Otherwise, if $OR$ is trivial then the orientation cover is two copies of $M$.

Regarding your example of $\mathbb{R}P^3$, it is not too surprising that some orientable manifolds have orientable double covering spaces. Any closed orientable surface (except the 2-sphere) is an example.

Related Question