Why is the Möbius strip not orientable?

I am trying to understand the notion of an orientable manifold.
Let M be a smooth n-manifold. We say that M is orientable if and only if there exists an atlas A={(Uα,ϕα)} such that det(J(ϕαϕ1β))>0 (where defined). My question is:
Using this definition of orientation, how can one prove that the Möbius strip is not orientable?

Thank you!

Answer

If you had an orientation, you’d be able to define at each point p a unit vector np normal to the strip at p, in a way that the map pnp is continuous. Moreover, this map is completely determined once you fix the value of np for some specific p. (You have two possibilities, this uses a tangent plane at p, which is definable using a (Uα,ϕα) that covers p.)

The point is that the positivity condition you wrote gives you that the normal at any p is independent of the specific (Uα,ϕα) you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of np changes sign when you return to p, which of course is a contradiction.

(This is just a formalization of the intuitive argument.)

Attribution
Source : Link , Question Author : Richard G , Answer Author : Andrés E. Caicedo

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