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It is well-known that $SO(2)$-principal bundles over a manifold $M$ are topologically characterized by their first Chern class. I was wondering what was the characterization of $O(2)$-bundles in terms of characteristic classes. I guess the first and second Setiefel-Whitney classes are necessary for the topological characterization of $O(2)$-bundles, but they can't be enough, because if $w_{1} = 0$ then one should recover the classification of $SO(2)$-bundles, which is given by the first Chern class and not by the second Stiefel-Whitney class.

Thanks.

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  • $\begingroup$ Unfortunately characteristic classes are not enough to completely determine $O(2)$-bundles. You only have $p_1$, $w_1$, and $w_2$, and these aren't enough to classify rank 2 bundles over $S^2$, even! $\endgroup$
    – user98602
    Mar 25, 2016 at 3:41
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    $\begingroup$ I'm confused... it seems like an unoriented bundle $E$ is the same as the knowledge of (1) $\text{det}(E)$ and (2) $E \otimes \text{det}(E)$ (or maybe duals thereof, etc.) In other words, an oriented rank 2 bundle is the same as line bundle and an oriented rank 2 bundle with the correspondence given by $(V, L) \mapsto V \otimes L^{-1}$. I think that's what Chanler's answer is saying, but it feels like there's a contradiction with what Mike is saying... what's going on? $\endgroup$
    – Dylan Wilson
    Mar 25, 2016 at 16:23
  • $\begingroup$ (I guess one possibility is that this correspondence is surjective but not bijective.) $\endgroup$
    – Dylan Wilson
    Mar 25, 2016 at 16:24
  • $\begingroup$ The Cech cocycle corresponding to the $O(2)$-bundle defines an element $w_{1}\in H^{1}(M,\mathbb{Z}_{2})$ together with an element $w_{2}\in H^{2}(M,\mathbb{Z}_{2})$. They look like the first and second Stiefel-Whitney classes of the $O(2)$ bundle. However, I don't see the implication in the other direction. $\endgroup$
    – Bilateral
    Mar 25, 2016 at 16:33
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    $\begingroup$ @Dylan: $E \otimes \det(E)$ need not be orientable when $E$ is even-dimensional. $\endgroup$
    – Oscar Randal-Williams
    Mar 26, 2016 at 12:46

1 Answer 1

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We have a fiber sequeunce $BSO(2)\to BO(2)\to B\mathbb{Z}/(2)$, and so we have that a $O(2)$-bundle, or a map $X\to BO(2)$ factors $X\to BSO(2)$ if and only if the composite $X\to B\mathbb{Z}/(2)$ is null-homotopic. $Hom_{\mathcal{h}Top}(X, B\mathbb{Z}/(2))=H^1(X; \mathbb{Z}/(2))$, so we have a class $x\in H^1(X; \mathbb{Z}/(2))$ representing the isomorphism class of the bundle. This class is clearly the pullback of the the universal class $x\in H^1(BO(2); \mathbb{Z}/(2))=(\mathbb{Z}/(2)[w_1, w_2])_{deg=1}=\mathbb{Z}/(2)w_1$. We note that since $O(2)\neq SO(2)\times \mathbb{Z}/(2)$, this class cannot vanish identically, so that $x=w_1$. Now we are left with an $SO(2)$-bundle, which you already know about!

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  • $\begingroup$ Thanks for the nice answer. This means that a $O(2)$-bundles is completely characterized by its first Stiefel-Whitney class and a class in $H^{2}(M,\mathbb{Z})$? In other words, an element of $H^{1}(M,\mathbb{Z}_{2})$ together with an element of $H^{2}(M,\mathbb{Z})$ defines a unique $O(2)$-bundle? $\endgroup$
    – Bilateral
    Mar 25, 2016 at 15:45

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