Distinguishing the Cylinder from a "full-twist" Möbius strip

Playing around with the definition of a fiber bundle, I found that while a Möbius strip (with its usual "half-twist") is a nontrivial fiber bundle, it seems that a Möbius strip with a "full-twist" is again trivial. In turn, a full-twist strip is homeomorphic to a simple cylinder.

I was wondering if it was well-known "how much more structure" is required before you can distinguish these two spaces. Intuitively, I would guess at most a reimann manifold structure allows you to do so, since there is more "curving" going on in the double-twist. Is it known if it can be done with less?

Solution 1:

A cylinder and a full-twist Möbius strip are distinguishable when they are embedded in $\mathbb R^3$ and we consider equivalence only up to ambient isotopies of $\mathbb R^3$. This is very easy to see because the boundary of the cylinder is a pair of unlinked circles, while for the full-twist Möbius strip the boundary circles are linked.

Intuitively this is what I'm thinking about if I try to differentiate the two; that is to say that the term "full-twist Möbius strip" already implies that it is embedded in $\mathbb R^3$. The fact that these are diffeomorphic means that they abstractly admit exactly the same structures, so something like a Riemannian metric can't even hope to distinguish between the two unless e.g. we require that metric to come from some embedding (I'm not actually sure if this is sufficient or not).

Solution 2:

The twisted cylinder and the cylinder are not isomorphic as bundles with structure group Z2. They are isomorphic as bundles with structure group SO(2). This is a classic example in Steenrod's Topology of Fiber Bundles.