Spin structures on $S^1$ and Spin cobordism - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:49:19Z http://mathoverflow.net/feeds/question/114660 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114660/spin-structures-on-s1-and-spin-cobordism Spin structures on $S^1$ and Spin cobordism mkreisel 2012-11-27T14:49:46Z 2012-11-27T16:34:34Z <p>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.</p> <p>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.</p> <p>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. </p> <p>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. </p> <p>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?</p> <p>(This is crossposted on stack exchange as <a href="http://math.stackexchange.com/questions/245480/spin-structures-on-s1-and-spin-cobordism" rel="nofollow">http://math.stackexchange.com/questions/245480/spin-structures-on-s1-and-spin-cobordism</a>).</p> http://mathoverflow.net/questions/114660/spin-structures-on-s1-and-spin-cobordism/114669#114669 Answer by Paul Siegel for Spin structures on $S^1$ and Spin cobordism Paul Siegel 2012-11-27T16:23:27Z 2012-11-27T16:34:34Z <p>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 $S^1$.</p> <p>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 $S^1$.</p>