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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.

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  • $\begingroup$ Sorry, I am having a hard time understanding your question. Are you talking about a 3-dimensional manifold embedded in $R^4$? Or are you talking about a 2-dimensional manifold embedded in $R^3$? What do you mean by "a manifold to have a normal vector that is zero everywhere"? I mean, any manifold can have a vector field that is identically equal to zero, but I don't think that's what you mean. $\endgroup$
    – Braindead
    Jan 5 '15 at 8:06
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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.

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  • $\begingroup$ Thanks this was pretty helpful. Just to clarify, if I used Levi-Civita symbols and the shift tensor to find a formula for a vector normal to a given 3-manifold at any given point, and this vector was zero at some point, that would show that the manifold is non-orientable? $\endgroup$
    – Ebanflo
    Jan 6 '15 at 4:13
  • $\begingroup$ To make sure I'm being clear I'll use the two-dimensional example of the Mobius band. I once used the parametrization found on wikipedia to define a coordinate system, shift tensor, and then a formula for a normal vector. If I understand correctly, showing that this formula gives a normal vector that is zero at some point on the surface would be a proof that the Mobius band is non-orientable. And the same logic would apply to a three-dimensional manifold. Thanks again for your answer. $\endgroup$
    – Ebanflo
    Jan 6 '15 at 4:13
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    $\begingroup$ @Ebanflo That's not quite right, because maybe you just wrote down a bad formula that happens to be zero at some point. The statement is that it's non-orientable if no matter how you write down a field of normal vectors, it's zero at some point. The way to prove it's non-orientable: write down non-zero normal vector fields over different patches of your manifold, then show there's no way to patch them together into a global non-zero field. $\endgroup$
    – aes
    Jan 6 '15 at 4:21
  • $\begingroup$ @Ebanflo For example, for the Möbius band, you can pretty easily write down non-zero normal vector fields over two different patches, but on one part of the overlap they're positive multiples of each other, and on the other part of the overlap they're negative multiples of each other. So there's no way to get them to agree without being zero somewhere (if you multiply one by $-1$ they still won't agree up to a positive multiple). $\endgroup$
    – aes
    Jan 6 '15 at 4:27
  • $\begingroup$ okay I'm beginning to see what you're saying but I'm having a hard time visualizing the example. So two patches on the Mobius band, are they overlapping? I'm pretty sure you imply it but I just want to be clear. And is there a limit on their size? So I assume that they are overlapping and that on one side of the overlap the normal vector field points towards the center of the strip and on the other side they point away from the center? (is this what you mean by positive and negative mutiples of each other?) I actually can see how that would prove its non-orientability, if thats right $\endgroup$
    – Ebanflo
    Jan 6 '15 at 6:51

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