# Spin structures on $S^1$ and Spin cobordism

I'm trying to understand the 2 spin structures on the circle. Since the frame bundle for the circle is just the circle itself, Spin structures on $S^1$ correspond to double covers of $S^1$. There are two choices: the connected double cover and the disconnected double cover.

From the point of view of Spin cobordism, we can view the circle as the boundary of the disk in the plane. The disk has a unique spin structure, and we can ask which spin structure this induces on the boundary.

Lawson/Michelson's "Spin Geometry" claims that this induces the spin structure coming from the double cover, but I'm having trouble seeing that. The frame bundle for the disk $D^2$ must be trivial, and thus isomorphic to $D^2\times SO(2) = D^2 \times S^1.$ There is a natural double cover given again by $D^2 \times S^1,$ and the map is just the identity on $D^2$ and $z \rightarrow z^2$ on the $S^1$ factor.

To see what the induced spin structure on the boundary is, we must view the frame bundle of the boundary as sitting inside the frame bundle of $D^2\times S^1$ by fixing an outward normal vector field and then using it to complete any frame on $S^1$ to a frame on $D^2.$ To me, this seems to say that we view the frame bundle of $S^1$ (which is itself $S^1)$ as $S^1\times \{1\} \subset D^2 \times S^1,$ since once we fix one vector of a frame (in this case given by the normal) the other is entirely determined since we are in 2 dimensions.

But now if we look at the inverse image of that in the double cover, we appear to get two disjoint copies of $S^1,$ i.e. the disconnected double cover. What am I doing wrong?

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I do not think the "natural double cover" you wrote is natural. You may be talking about $z\rightarrow z^{2}$ on $D^{2}$? This should give you the desired double cover, and on the boundary it is a non-trivial double cover of $S^{1}$. – Bombyx mori Nov 27 '12 at 5:10
But we are not looking for a double cover of $D^2,$ we are looking for a double cover of it's bundle of frames coming with an action of Spin(2). Furthermore, what you mentioned isn't a double cover of $D^2;$ the preimage of 0 is a single point. – mck Nov 27 '12 at 12:34

The frame bundle on the disk is indeed $D^2 \times S^1$. But the problem with your argument is that the outward normal vector field does not extend to a non-vanishing vector field on the whole disk, so you cannot find a trivialization $D^2 \times S^1$ of the frame bundle of $D^2$ such that the outward pointing normal vector field is given by $(s,1)$ on the boundary of $D^2$ in your trivialization.
So we have to be more careful in identifying the map $S^1 \rightarrow D^2 \times S^1$ given by extending a frame on $S^1$ to a frame on the boundary of $D^2$ by adding the outward pointing normal vector field. Since everything happens in $\mathbb{R}^2$, we have a canonical trivialization of all tangent bundles. Let $\lambda: S^1 \rightarrow S^1$ be multiplication by $i$ or, in other words, rotation by $90°$. Then $(s, \lambda(s))$ is a frame in the tangent space of $D^2$ at the point $s$ in the boundary. Your map $S^1 \rightarrow D^2 \times S^1$ is then $s \mapsto (s, \lambda(s))$ (think of $\lambda(s))$ as a tangent vector to $s$ in $S^1$). Now it is easy to see that the restriction of the cover of $D^2 \times S^1$ is the connected double cover of $S^1$.
Depending on conventions, the map $S^1 \rightarrow D^2 \times S^1$ may also be $s \mapsto (s,s)$, but this doesn't change the argument. – Fabian Lenhardt Nov 27 '12 at 15:28