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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?

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    $\begingroup$ $\mathbb{Z}$ 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

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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 $z \mapsto z^2$.

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