I've seen it written (for example in Spin Geometry by Lawson and Michelsohn) that there are only two Spin structures on $S^1$, corresponding to the covers of $S^1$ given by $f_1: S^1\rightarrow S^1,\,z\mapsto z^2$ and $f_2: S^1\sqcup S^1\rightarrow S^1$ the trivial two-to-one map. Is it easy to see that these are the only two double covers of the circle there are?

  • 2
    $\begingroup$ $\mathbb{Z}$ only has one subgroup of index $2$, so it only admits one connected double cover. In a disconnected one, each component would account for at least one sheet, so the only way to do that is two have two one-sheeted components, i.e. the trivial double cover. $\endgroup$
    – Pedro
    Dec 8, 2016 at 18:18

1 Answer 1


It depends on what you find simple. With a bit of covering theory, it is simple.

First note that if $p \colon E \to S^1$ is a covering map, and $C$ a connected component of $E$, then $p\lvert_C \colon C \to S^1$ is a covering map. Since the restriction of a local homeomorphism to an open subset is again a local homeomorphism, and all covering spaces of $S^1$ are locally connected, it follows that $p\lvert_C$ is a local homeomorphism, and if $U\subset S^1$ is a connected open set that is evenly covered by $p$, then it is also evenly covered by $p\lvert_C$, since each connected component of $p^{-1}(U)$ is either contained in $C$ or doesn't intersect $C$, so $p\lvert_C$ is a covering map.

If we have a two-sheeted covering $p \colon E \to S^1$ with disconnected $E$, it follows that each connected component of $E$ gives a single-sheeted covering, so there are precisely two connected components, and each is mapped homeomorphically to $S^1$ by $p$. That is, $p$ is equivalent to the trivial two-sheeted covering.

And if we have a two-sheeted covering $p \colon E \to S^1$ with connected covering space $E$, then using the universal covering $\pi \colon \mathbb{R} \to S^1$ we find that $p$ is equivalent to the covering $z \mapsto z^2$ by factoring $\pi$ through $E$, $\pi = p \circ q$, where $q \colon \mathbb{R} \to E$ is a covering. Since $p$ is two-sheeted, the deck transformation group of $q$ has index $2$ in the deck transformation group of $\pi$, so

$$E \cong \mathbb{R}/\operatorname{Deck}(q) = \mathbb{R}/(2\cdot\operatorname{Deck}(\pi)),$$

and $p$ is equivalent to the induced covering $\overline{\pi} \colon \mathbb{R}/(2\operatorname{Deck}(\pi)) \to S^1$, which is - with the appropriate identifications - just $z \mapsto z^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.