The two are related but not as closely as you might hope. The first thing happens because Möbius strips are not orientable. One way to describe what it means for a manifold to be orientable is that its tangent bundle admits a reduction of the structure group from $O(n)$ to $SO(n)$. There is a $\mathbb{Z}_2$ floating around here, one manifestation of which is the short exact sequence

$$1 \to SO(n) \to O(n) \xrightarrow{\det} \mathbb{Z}_2 \to 1$$

and another of which is the first Stiefel-Whitney class $w_1 \in H^1(BO(n), \mathbb{Z}_2)$.

Spinors instead happen on manifolds which are not only orientable but which also have a spin structure, which means a reduction of the structure group to a group called $Spin(n)$. There is a different $\mathbb{Z}_2$ floating around here, one manifestation of which is the short exact sequence

$$1 \to \mathbb{Z}_2 \to Spin(n) \to SO(n) \to 1$$

and another of which is the second Stiefel-Whitney class $w_2 \in H^2(BSO(n), \mathbb{Z}_2)$.

One thing to say here is that when people say spinors have to be rotated twice to be returned to the same position, this rotation is not happening in space. It is a "gauge transformation," or an "internal" symmetry. Mathematically this has to do with the reduction of the structure group to $Spin(n)$ and then the construction of associated vector bundles of the corresponding principal $Spin(n)$-bundle; spinors are sections of these.

Spin isn't called spin because anything is spinning in space, it's called spin because it behaves the way angular momentum behaves in quantum mechanics. Mathematically this is because both things are controlled by the representation theory of $Spin(3)$.

This post imported from StackExchange Mathematics at 2015-06-03 19:35 (UTC), posted by SE-user Qiaochu Yuan