# Classification of $O(2)$-bundles in terms of characteristic classes

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.

• 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 $x^2 = p_1$. – Anton Fetisov Mar 26 '16 at 2:31
• 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. – Oscar Randal-Williams Mar 26 '16 at 8:35

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 $$BSO(2)\to BO(2)\to BO(1).$$

• 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) – Denis Nardin Mar 26 '16 at 13:02
• @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 $\mathbb{Z}_{2}$. – Bilateral Mar 26 '16 at 13:37
• @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. – Bilateral Mar 26 '16 at 16:53
• @Bilateral I've written an answer expanding the point, since a proper response didn't fit in this margin :) – Denis Nardin Mar 26 '16 at 17:42

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 $$\alpha$$.
• 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.. – Denis Nardin Mar 26 '16 at 17:48
• Thanks for the detailed answer. Could you recommend a reference to further explore what you are explaining? – Bilateral Apr 27 '16 at 7:07
• What does exactly mean "coefficients twisted by $\alpha$"? – Bilateral Apr 27 '16 at 7:26
• @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.) – Denis Nardin Apr 27 '16 at 12:43
• (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) – Denis Nardin Apr 27 '16 at 12:43