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Ryan Budney
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There's the traditional obstruction-theoretic perspective. Orientability means the tangent bundle trivializes over a 1-skeleton. Dually you could think of that as saying the complement of a co-dimension $2$ subcomplex has a trivial tangent bundle.

So admitting a spin structure is the same, but it will be the tangent bundle trivializes over a 2-skeleton, dually the complement of a co-dimension three subcomplex admits a trivial tangent bundle.

A surface is orientable if and only if it contains no Moebius bands -- a regular neighbourhood of any simple closed curve must be a cylinder. In higher dimensions this translates into a manifold being orientable if and only if it contains no twisted bundles $D^{n-1} \rtimes S^1$, i.e. regular neighbourhoods of simple closed curves are diffeomorphic to $D^{n-1} \times S^1$.

For spin structures there's something very similar. Of course, a surface admits a spin structure if and only if it is orientable. It's a more interesting notion in higher dimensions. The statement there is the manifold is orientable, and if you take a regular neighbourhood of any surface in the manifold, then it has a trivial tangent bundle. So manifolds like $\mathbb RP^3$ are perfectly valid spin manifolds -- $\mathbb RP^3$ contains $\mathbb RP^2$ but the total space of its normal bundle has a perfectly trivializable tangent bundle. Technically, the condition is a little stronger than that -- you can trivialize the tangent bundle of the complement of a co-dimension $3$ subset. No not only can you trivialize the total spaces of normal bundles of surfaces, but even the regular neighbourhoods of unions of surfaces.

So if you want a manifold that isn't spin, the archetype would be a vector bundle over a surface so that the total space does not have a trivializable tangent bundle. Take the $D^2$-bundle over $S^2$ with Euler Class $\chi$. I think this happens if and only if $\chi$ is even. I suppose you have more entertaining examples when dealing with the regular neighbourhood of a 2-complex that isn't itself a manifold.

Ryan Budney
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