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The two Spin structures on the oriented circle

Example

Write $S^{1} = R / Z$ with its standard orientation and global unit tangent frame. The oriented frame bundle has structure group $S O (1) = 1$ . The two Spin structures are the disconnected double cover $S^{1} ⊔ S^{1} \to S^{1}$ , with the two components exchanged by $Z / 2$ , and the connected double cover $R / 2 Z \to R / Z$ , $[t]_{2 Z} \mapsto [t]_{Z}$ , with deck action $[t] \mapsto [t + 1]$ .

Facts & Assumptions

Given: Full AC, the oriented circle, the trivialization of its oriented frame bundle and the covering maps in the Example.

[F1]

Given Spin and Spin-c structures and their spinor bundles: for rank one, a Spin structure is a principal $Spin (1) = Z / 2$ bundle with an equivariant map to the trivial oriented frame bundle.

[F2]

Spin and Spin-c obstruction and torsor classification: when $w_{2} = 0$ , Spin structures form a torsor for $H^{1} (S^{1}; Z / 2)$ .

[F3]

Singular cohomology with coefficients: singular cohomology is computed from the singular chain complex by the cochain Hom complex.

[F4]

Homology of spheres: $H_{0} (S^{1}; Z) = Z$ and $H_{1} (S^{1}; Z) = Z$ .

[F5]

Topological universal coefficient short exact sequence for cohomology: $0 \to Ext (H_{0} (S^{1}; Z), Z / 2) \to H^{1} (S^{1}; Z / 2) \to Hom (H_{1} (S^{1}; Z), Z / 2) \to 0$ is exact.

Verification

Given: The data above and [F1]–[F5].

1.1F1givenconstruct

The disconnected cover is a principal $Z / 2$ bundle because each fiber has two points and the group interchanges the two sheets. The connected map $[t]_{2 Z} \mapsto [t]_{Z}$ is a covering: each sufficiently short interval in $R / Z$ lifts to two disjoint intervals, and translation by $1$ interchanges them freely and transitively. Both total spaces therefore give principal $Spin (1)$ bundles; since the oriented frame bundle is trivial, their projections supply the equivariant Spin lifts required by [F1].

2.1F2F3F4F5step 1.1algebra∎

The disconnected cover has two components, whereas $R / 2 Z$ is connected, so the covers are not isomorphic. By the cochain definition [F3], cohomology is the middle term in the UCT sequence [F5]; [F4] gives $H_{0} (S^{1}; Z) = H_{1} (S^{1}; Z) = Z$ , so $Ext (Z, Z / 2) = 0$ and $Hom (Z, Z / 2) ≅ Z / 2$ . Hence $H^{1} (S^{1}; Z / 2) ≅ Z / 2$ . The tangent line is trivial, so $w_{2} = 0$ ; [F2] says the set of Spin structures is a nonempty torsor for this two-element group. The two explicit nonisomorphic covers therefore exhaust it.

The two Spin structures on the oriented circle

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources