Orientable double-cover of the Mobius strip

Question

Let C₀ be the cylinder, M₁ the Mobius band (a cylinder with 1 twist), and C₂ a cylinder with 2 twists (each embedded in R³).

Although C₀ and C₂ are diffeomorphic, intuitively one should think that C₂ is a better choice for the oriented double-cover of M₁, as it should preserve the notion of the "twisting" of M₁. Clearly C₀ is diffeomorphic to C₂, but the two are not ambient isotopic within R³.

How specifically can we state "C₂ is the better choice for the oriented double-cover of M₁, as opposed to C₀"? Is there an intrinsic way of doing so? Is there some way to phrase this in terms of cobordism?

Answer 1

I don't know whether there's a way to phrase it in terms of (co)bordism, but you might observe that the standard embedding of $C_1$ has a normal bundle; the tubular nhd theorem says that that bundle, too, can be embedded so that that the 0-section is exactly $C_1$. Taking the image of $[-1, 1]$ in each fiber, you get a 3-manifold with boundary; if you take the boundary in the fiber direction, you get an embedding of the double cover. Here's a picture, before and after taking the double-cover (crudely drawn, but it's the best I can do at 5 AM):

Before:

After:

Now you can take that second surface, and claim (or try to claim) that it's ambient isotopic to the twice-twisted cylinder, but not to the untwisted cylinder. Experimenting with a piece of paper confirms this. (The isotopy to the twice-twisted cylinder is easy. To prove there's no isotopy to the untwisted cylinder, look at the linking number between the top and bottom edge of the cylinder. That's invariant under isotopy, but in the untwisted case it's zero, and in the double-twist case, it's two.)