Relation between the two spin structures of $S^1$

Question

I know that $S^1$ has two spin structures $s$ and $t$, corresponding to its two double covers. I am trying to understand which diffeomorphism $f \colon S^1 \rightarrow S^1$ sends $s$ to $t$, i.e. $f^*(s)=t$.

I am thinking about spin structures as cohomology classes in $H^1(E(M);\mathbb{Z}/2\mathbb{Z})$, where $E(M)$ is the tangent frame bundle on $M$. Any help would be appreciated: I guess that my problem depends on my poor understanding of spin structures.

Answer 1

Usually a spin structure is defined for an oriented manifold. (A given spin structure on a orientable manifold should induce an orientaton.) As shown in Theorem 2.1 in Lawson & Michelsohn's Spin Geometry the spin structures of a spinnable manifold are then in 1-1 correspondence with elements in $H^1(M;\mathbb{Z}/2\mathbb{Z})$.

This is related to your statement. But one has to consider the oriented frame bundle $P_{SO(n)}$ on $M$. There is an exact sequence ( see page 81 of the book) \begin{align*} 0 \to H^1(M;\mathbb{Z}/2\mathbb{Z}) \overset{\pi^*}{\to} H^1(P_{SO(n)};\mathbb{Z}/2\mathbb{Z}) \overset{\iota^*}{\to} H^1(SO(n);\mathbb{Z}/2\mathbb{Z}) \overset{w_2}{\to} H^2(M;\mathbb{Z}/2\mathbb{Z}) \end{align*} where $\iota$ is the inclusion of a fibre.

Now lets say $M=S^1$ (oriented). We have $H^1(M;\mathbb{Z}/2\mathbb{Z}) \overset{\sim}{=} \mathbb{Z}/2\mathbb{Z}$, that is, there are two spin structures on $M$. For any continous map $f:S^1 \to S^1$ the induced map on cohomology $f^*$ is multiplication with its mapping degree.