# 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?

(This is crossposted on stack exchange as http://math.stackexchange.com/questions/245480/spin-structures-on-s1-and-spin-cobordism).

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The tangent bundle of the disk is indeed trivial, but the outward pointing normal vector field does not extend over the whole disk, so you cannot assume that it is part of a trivialization. I think this throws your argument off. – Fabian Lenhardt Nov 27 '12 at 15:04

As Fabian pointed out in the comments, you have to be more careful about how you trivialize $SO(D^2)$. I'm going to use the standard coordinates $(x,y)$ on $\mathbb{R}^2$ (note that these are not global coordinates on $D^2$, but they still trivialize the frame bundle). Thus we have a global section of $SO(D^2)$ which assigns to each point in $D^2$ the standard orthonormal basis $(e_x,e_y)$, and the action of $S^1 = SO(2)$ on $SO(D^2) \cong D^2 \times S^1$ is by counterclockwise rotation of this basis. It is fairly clear from this picture that the principal spin bundle is $Spin(D^2) \cong D^2 \times S^1$, and the double cover $Spin(D^2) \to SO(D^2)$ is the identity on $D^2$ and the doubling map on $S^1$.
Now let's figure out how $SO(S^1) \cong S^1$ sits inside $SO(D^2)$ using this trivialization. The boundary circle is the set of points $(\cos \theta, \sin \theta)$ in $D^2$. The (oriented) unit tangent vector to $S^1$ at the point $(\cos(0),\sin(0)) = (1,0)$ is just $e_y$ in the frame used to trivialize $SO(D^2)$ above, and at any other point $(\cos \theta, \sin \theta)$ it is just $R_\theta e_y$ where $R_\theta$ denotes counterclockwise rotation by the angle $\theta$. So if we denote the oriented unit vector tangent to $\theta \in S^1$ by $e_\theta$ then the embedding $SO(S^1) \to SO(D^2) \cong D^2 \times S^1$ is given by $e_\theta \mapsto ((\cos \theta, \sin \theta), \theta)$. Finally, in this picture it is clear that the inverse image of the set $\lbrace ((\cos \theta, \sin \theta), \theta) \rbrace$ under the map $Spin(D^2) \to SO(D^2)$ is the connected double cover of $S^1$.