This question was originally posted on mathoverfow: below the question there were some useful comments however no canonical answer was given. Normally I would offer a bounty for this question but it was voted +3-2 so probably was not well suited for mathoverflow. This is the reason why I'm posting it here. So here is the question:

Let $E$ be an $n$-plane bundle over CW complex $X$. Then $E$ is a pullback of tautological bundle $\gamma_n$ over $BO(n)$ i.e. $E=f^*(\gamma_n)$. This $f$ is called classyfing map. One can show that the universal covering of $BO(n)$ is $BSO(n)$ and $f$ can be lifted to $BSO(n)$ (meaning that there is $\tilde{f}:X \to BSO(n)$ such that $p \circ \tilde{f}=f$ where $p$ is the projection) if and only if $f_*:\pi_1(X) \to \pi_1(BO(n)) \cong \mathbb{Z}_2$ is zero. Such a map from $\pi_1(X)$ to $\mathbb{Z}_2$ gives rise to the cohomology class in $H^1(X,\mathbb{Z}_2)$.

One can also show that the $2$-connected covering of $BSO(n)$ is $BSpin(n)$ and one can ask for further extensions of the classyfing map to $BSpin(n)$.

I would like to understand how the existence of such extension is equivalent to the vanishing of some class in $H^2(X,\mathbb{Z}_2)$.

I know that this class should be exactly the second Stiefel-Whitney class and I would also like to understand:

Why this class is the second Stiefel-Whitney class?

  • $\begingroup$ I think that if you look at Chapter 10 of Milnor and Stasheff's "Characteristic Classes", you may find yourself able to answer this question. (I can't promise it, but if you can formulate it this clearly, then M&S's discussion of obstructions and their reduction to S-W classes is probably just the thing you need.) $\endgroup$ – John Hughes Sep 15 '17 at 16:32

The keyword to look up is obstruction theory, and the story goes like this.

First let me describe a simpler but analogous story about groups. Suppose $f : G \to H$ is a map of groups, and you want to understand when this map can be lifted to a map $G \to K$, where $K \subseteq H$ is a subgroup. (It might sound a bit strange to call such a thing a lift, but bear with me.) If $K$ is a normal subgroup, there's a nice answer: there is a short exact sequence

$$0 \to K \to H \xrightarrow{q} H/K \to 0$$

exhibiting $K$ as the kernel of a map, namely the quotient map $H \xrightarrow{q} H/K$, and by the universal property of kernels, $f$ lifts to $K$ iff the composite map

$$G \xrightarrow{f} H \xrightarrow{q} H//K$$

is zero.

The story in homotopy theory is a categorification of this story. Now suppose $f : X \to E$ is a map of spaces, and you want to understand when this map can be lifted to a map $X \to F$, where $F$ is some space equipped with a map $F \to E$, up to homotopy. When $E = BG$ for a group $G$ and $F = B \widetilde{G}$ for some group $\widetilde{G}$ equipped with a map $\widetilde{G} \to G$ giving such a lift is called reducing the structure group of the $G$-bundle $f : X \to BG$ from $G$ to $\widetilde{G}$, and includes as a special case finding orientations, spin structures, almost complex structures, etc.

The answer to this question is nicest if the map $F \to E$ is a "homotopy kernel" in the sense that it is part of a fiber sequence

$$F \to E \xrightarrow{q} B.$$

Here $F, E, B$ are pointed spaces and this is a sequence of pointed maps, and being part of a fiber sequence means $F$ is the homotopy fiber of the map $E \to B$, which means precisely that it satisfies the following universal property: a map $f : X \to E$ lifts to a map $X \to F$ up to homotopy if and only if the composite map

$$X \xrightarrow{f} E \xrightarrow{q} B$$

is nullhomotopic (homotopic to the map $X \to B$ sending every point in $X$ to the basepoint of $B$).

General fact: Let $E$ be $(n-1)$-connected and let $\widetilde{E}$ be its $n$-connected cover. There is a unique map (up to homotopy) $q : E \to B^n \pi_n(E) \cong K(\pi_n(E), n)$ inducing the identity on $\pi_n$, and it fits into a fiber sequence

$$\widetilde{E} \to E \xrightarrow{q} B^n \pi_n(E).$$

In other words, a map $f : X \to E$ lifts to a map $f : X \to \widetilde{E}$ up to homotopy if and only if the composite $X \xrightarrow{f} E \xrightarrow{q} B^n \pi_n(E)$ is nullhomotopic, or equivalently if and only if the cohomology class $f^{\ast}(q) \in H^n(X, \pi_n(E))$ is zero.

When $E = BO(n)$ this recovers the usual story about $w_1$ and orientations, basically because $w_1$ is the unique nontrivial map $BO(n) \to B \mathbb{Z}_2$ up to homotopy. Similarly when $E = BSO(n)$ this recovers the usual story about $w_2$ and spin structures, because $w_2$ is the unique nontrivial map $BSO(n) \to B^2 \mathbb{Z}_2$ up to homotopy (both of these are straightforward corollaries of the Hurewicz theorem + universal coefficients).

What is a bit surprising here is that $w_2$ is already defined on $BO(n)$, but the story doesn't continue this way: the next step is about string structures and is controlled by a class $\frac{p_1}{2} \in H^4(BSpin(n), \mathbb{Z})$ called the first fractional Pontryagin class which is not the pullback of a cohomology class on $BSO(n)$ (here I need $n \ge 4$ or maybe $n \ge 5$, not sure).

  • $\begingroup$ Dear Qiaochu Yuan, thank you for the informative answer! Your answer gives a broader perspective on the issue from the point of view of homotopy theory. However I'm also interested in the direct argument answering my questions, not invoking the General Fact which you quoted. As I explained, in order to understand $w_1$ from this point of view you don't need the whole machinery since the classyfing map lifts to its universal cover iff the induced map in first homotopy group is zero. But since $\pi_1(BO(n))=\pi_0(O(n)) \cong \mathbb{Z}_2$, this gives a homomorphism $\pi_1(M) \to \mathbb{Z}_2$... $\endgroup$ – truebaran Sep 23 '17 at 22:55
  • $\begingroup$ ...which gives rise to homomorphism $H_1(M) \to \mathbb{Z}_2$ and using universal coefficients, a class in $H^1(M,\mathbb{Z}_2)$. One can show that this class is natural and that for $M=BO(n)$ this class is nonzero element in $H^1(M,\mathbb{Z}_2)$. This is enough to identify this class as $w_1$. For $w_2$ one has to show the naturality and that for $M=BSO(n)$ one gets nonzero element in $H^2(M,\mathbb{Z}_2)$ (for each $n$). $\endgroup$ – truebaran Sep 23 '17 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.