I'm trying to understand why there are only 2 spin structures on the circle $S^1$. From my understanding, in dimension 1 at least, a spin structure on a manifold $M$ is just a double cover of the frame tangent bundle.
According to this MathOverflow post, the frame bundle of $S^1$ is homeomorphic to $S^1$ - then, since the circle only has 2 distinct double covers, the result would follow.
But I don't see why the frame bundle of the circle is homeomorphic to $S^1$ - Isn't the frame bundle a principal GL$(1,\mathbb{R})$ bundle in this case, meaning that each fiber has dimension 1? How could the whole bundle also have dimension 1?
On any manifold $M$ we can pick a Riemannian metric, which allows us to deformation retract the frame bundle onto the orthonormal frame bundle; if $M$ is an $n$-manifold this is a principal $O(n)$-bundle, compared to the frame bundle itself which is a principal $GL_n(\mathbb{R})$-bundle. If $M$ is oriented then we can similarly deformation retract the oriented frame bundle onto the oriented orthonormal frame bundle (as Andrew says in the comments, this is presumably what the linked post is really referring to by "frame bundle"). This is a principal $SO(n)$-bundle, so for $n = 1$ it's a principal $SO(1)$-bundle.
Since $SO(1)$ is trivial, the fibers of this bundle are trivial (this says that at each tangent space to $S^1$ there is a unique unit tangent vector compatible with a given choice of orientation), so the oriented orthonormal frame bundle of $S^1$ is $S^1$ again.