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

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.

• $SO(n)$ is connected but not simply connected if $n \geq 2$. More precisely, $\pi_1(SO(2)) \simeq \mathbb{Z}$ and $\pi_1(SO(n)) \simeq \mathbb{Z}/2\mathbb{Z}$ if $n \geq 3$. On the other hand $SO(1)$ is just the trivial group containing only the identity. I don't think it makes much sense to talk about its spin group. So I don't think it makes much sense to talk about "spin curves". If one does talk about "spin curves", maybe they just mean a double covering of the curve. My point is, the "spin" notion becomes too degenerate for curves. Oct 15, 2021 at 1:54

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.