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I had asked this question in stackexchange but there seems to be no consensus in the answer

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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    $\begingroup$ Mike's comment is wrong. An unoriented bundle is the same as a line bundle and an oriented bundle, so $BO(2) = BSO(2) × BZ/2 = B^2 Z × BZ/2$. Thus such bundles are completely classified by $w_1 \in H^1(Z/2)$ and a class in $H^2(Z)$ which maps to $w_2$ under $Z \to Z/2$. I suppose we should call it an integral Stiefel-Whitney class $\hat w_2$. If we look at cohomology then for $B^2 Z = \Bbb C P^\infty$ we have $H^*(Z) = Z[[x]]$ with $\mathrm{deg} x = 2$, $x = \hat w_2$. In particular $x^2 = p_1$. $\endgroup$ – Anton Fetisov Mar 26 '16 at 2:31
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    $\begingroup$ Anton's comment is wrong. $BO(2)$ does not split as claimed, as can be seen by e.g. calculating its rational cohmology. As $\pi_2(BO(2))= \mathbb{Z}$ but $H_2(BO(2);\mathbb{Z})$ is torsion, the Hurewicz map in degree 2 cannot be injective, and so $O(2)$-bundles cannot be classified by characteristic classes. $\endgroup$ – Oscar Randal-Williams Mar 26 '16 at 8:35
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The $O(2)$ bundles $\xi$ over a manifold $M$ are classified by their first Stiefel-Whitney class $w_1(\xi)\in H^1(M;\mathbb{Z}/2)$ and their twisted Euler class $e(\xi)\in H^2(M;\mathbb{Z}_{w_1(\xi)})$.

This is because the space $BO(2)$ is a generalized Eilenberg--Mac Lane space $L_{w_1}(\mathbb{Z},2)$ in the sense of

Samuel Gitler, Cohomology operations with local coefficients, Amer. J. Math. 85 (1963), 156--188.

In (slightly) more detail, there is a fibration $$ K(\mathbb{Z},2)\to E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)\to B\mathbb{Z}/2 $$ given by the twisting of $w_1$ on the universal $SO(2)$ bundle, and this fibration agrees up to homotopy with the fibration $$BSO(2)\to BO(2)\to BO(1).$$

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    $\begingroup$ In general if you want to study $G$-bundles you should take a look at the Postnikov tower of $BG$. Unfortunately this tower is often infinite (so you'll need an infinite number of twisted characteristic classes) $\endgroup$ – Denis Nardin Mar 26 '16 at 13:02
  • $\begingroup$ @MarkGrant: Thanks. I think your description fits with what I get by considering the Cech cocycle defined by $O(2)$, which, when decomposed in terms of $O(2) = SO(2)\rtimes\mathbb{Z}_{2}$ gives rise to the Stiefel-Whitney class $w_{1}$ of $O(2)$ and some 2-cohomology class which locally takes values in $\mathbb{Z}_{2}$. $\endgroup$ – Bilateral Mar 26 '16 at 13:37
  • $\begingroup$ @DenisNardin: Thanks for the pointer. Could you please elaborate a little bit on your answer? The references dealing with the Postnikov tower seem to be quite specialized. $\endgroup$ – Bilateral Mar 26 '16 at 16:53
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    $\begingroup$ @Bilateral I've written an answer expanding the point, since a proper response didn't fit in this margin :) $\endgroup$ – Denis Nardin Mar 26 '16 at 17:42
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To complement Mark Grant's excellent answer, I'll say something more about the general case. This topic goes under the name of obstruction theory.

The first observation is that a $G$-bundle on $X$ is the same thing as an homotopy class of maps $X\to BG$. To study them we will use the Postnikov tower of $BG$. This is a tower assembled by spaces $P_n(BG)$ together with a map $BG\to P_n(BG)$ such that

  • The map $\pi_i(BG)\to \pi_i(P_n(BG))$ is an isomorphism for $i\le n$;
  • $\pi_i(P_n(BG))=0$ for $i>n$.

We can assemble this spaces together so to form a tower as follows:

$\require{AMScd}$ \begin{CD} @. \vdots\\ @. @VVV \\ @. P_2(BG)\\ @. @VVV \\ X @>>d> P_1(BG) \end{CD}

and moreover the limit of the tower is $BG$. So we can study the homotopy classes $[X,BG]$ by studying the collections of arrows $[X,P_i(BG)]$ making the diagram commute.

Now let's start at the bottom of the diagram. By definition we have that $P_1(BG)$ is a $K(\pi_1BG,1)=K(\pi_0G,1)$, so we have

$[X,P_1(BG)] = [X,K(\pi_0G,1)] = H^1(X;\pi_0G)$

This is our first cohomology class, corresponding to $w_1$ in the case of $BO(n)$.

Now let us suppose that we have lifted our map all the way to $P_n(BG)$ and we want to see what algebraic information corresponds to a lift to $P_{n+1}(BG)$. It turns out that there is a cartesian diagram $\require{AMScd}$ \begin{CD} @. P_{n+1}(BG) @>>> K(\pi_0G,1)\\ @. @VVV @VVV\\ X @>>> P_n(BG) @>>> K(\pi_{n+1}G,n+2)_{h\pi_0G} \end{CD} (don't be scared by all those homotopy quotients you see: they're just the homotopy theorist's way of saying that we're dealing with twisted cohomology classes). So lifting a map from $P_n(BG)$ to $P_{n+1}(BG)$ is the same thing as lifting a map from $K(\pi_{n+1}G,n+2)_{h\pi_0G}$ to $K(\pi_0G,1)$. This is saying that the lift exists if and only if some class in $H^{n+2}(X,\pi_{n+1}G)$ vanishes (not all choices of characteristic classes will correspond to a $G$-bundle!) but, more importantly for us, this is exactly the same situation as in Mark Grant's answer and so the possible choices are parametrized by a class in $H^{n+3}(X,\pi_{n+1}G)$.

So, to sum up we will have

  • A class $\alpha$ in $H^1(X;\pi_0G)$
  • An infinite sequence of classes in $H^{n+1}(X;\pi_nG)$ for $n\ge1$ where the coefficients are twisted by $\alpha$.
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    $\begingroup$ As an added remark, the classes of maps $[X,P_n(BG)]$ correspond to the failure of supporting some additional structure. For example if $G=O(k)$ we have that the map $X\to P_1(BO(k))$ is nullhomotopic iff the bundle is orientable, while the map $X\to P_2(BO(k))$ is nullhomotopic iff the bundle is spin etc.. $\endgroup$ – Denis Nardin Mar 26 '16 at 17:48
  • $\begingroup$ Thanks for the detailed answer. Could you recommend a reference to further explore what you are explaining? $\endgroup$ – Bilateral Apr 27 '16 at 7:07
  • $\begingroup$ What does exactly mean "coefficients twisted by $\alpha$"? $\endgroup$ – Bilateral Apr 27 '16 at 7:26
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    $\begingroup$ @Bilateral I don't have a good reference for this particular story, but being familiar with Hatcher's book on algebraic topology will certainly help to see many similar constructions, so that this is not too surprising. I learnt the general structure of the Postnikov tower from Blanc, Dwyer, Goerss The realization space of a Pi-algebra but that might be a bit advanced. (cont.) $\endgroup$ – Denis Nardin Apr 27 '16 at 12:43
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    $\begingroup$ (cont.) When I say "twisted by $\alpha$" I mean in the sense of cohomology with local coefficients (that is $\pi_nG$ is secretly a local system on $X$ via the $\pi_1X$-action). Also beware that the groups where the higher classes lie are not canonically identified (that is you'll have to fix a G-bundle with first class $\alpha$ and use it to compare the other G-bundles) $\endgroup$ – Denis Nardin Apr 27 '16 at 12:43

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