Is it possible for a manifold to have a normal vector that is zero everywhere, if so, would this indicate that the manifold is non-orientable? Basically I've been thinking about defining a non-orientable three-dimensional metric space via defining the normal vector and looking to see if there is two possible vectors for the same point.  I'm not at all sure if this is the right approach (haven't had any luck with it so far) but I thought that maybe if the normal vector was zero everywhere that it would indicate that it's impossible to have a normal vector which would therefore mean that the manifold is non-orientable...?  Anyway I hope my question is straight forward enough and as a side note, I was using the Levi-Civita symbols and Shift tensor to find the normal vector.  Thanks for your help.
 A: Non-orientablility is a global property. It can't be confirmed based on looking at one point or a small neighborhood of one point because any point has a neighborhood diffeomorphic to euclidean space, which is orientable. That is to say, locally any manifold is orientable.
In terms of normal vectors, these are non-zero at any given point, but the question is whether there's a globally consistent (i.e. continuous) choice of non-zero normal vector. That is, an $n$-dimensional submanifold of $\mathbb{R}^{n+1}$ is orientable if there's a field of non-zero normal vectors. In other words, a manifold is non-orientable if any field of normal vectors is zero somewhere (not everywhere).
For example, if you consider the Möbius band, it has non-zero normal vectors everywhere, but if you try to give a non-zero field of normal vectors, you run into trouble when you go around the loop (this is a global failure we're seeing) and they're pointing the opposite way --- to make it a continuous field of normal vectors we'd need for it to be zero somewhere.
