Let $E \to X$ be an oriented vector bundle over a CW complex $X$. Suppose that the second Stiefel-Whitney class $w_2(E)=0$, then $E$ admits a spin structure. In terms of homotopy theory this means that the classifying map $X \to BSO(n)$ lifts to $X \to BSpin(n)$. Does this mean that the homotopy classes $[X,BSpin(n)]$ classify all oriented vector bundles with $w_2(E)=0$? And can it be that there is more than one lift?

What puzzles me also is that the universal bundle over $BSpin(n)$ has not the same rank as that of $BSO(n)$. And therefore the pulled back universal bundle cannot be of same rank as $E$, therefore not isomorphic. How is it then possible that $[X,BSpin(n)]$ classifies vector bundles in $[M,BSO(n)]$ with $w_2(E)=0$?

  • 3
    $\begingroup$ It classifies oriented bundle together with a witness to w_2 vanishing. There can be lots of lifts. $\endgroup$ Apr 1, 2018 at 14:17
  • $\begingroup$ Observe the fibration sequence $BSpin(n)\rightarrow BSO(n)\xrightarrow{w_2}K(\mathbb{Z}_2,2)$. Hence the set of lifts of a map $f:X\rightarrow BSO(n)$ with $f^*w_2=0$ is a torsor for $H^1(X;\mathbb{Z}_2)$ (although the action need not be free). $\endgroup$
    – Tyrone
    Apr 1, 2018 at 16:57
  • $\begingroup$ @Tyrone which action? $\endgroup$ Apr 1, 2018 at 18:30

1 Answer 1


Let me just expand what people written in the comments. Given a fibration $F \to E \to B$ of pointed spaces and $X$ be a space with a map $f: X \to E$. Then, the space of lifts $f' : X \to F$ is equivalent to the space of null-homotopies of the composition $X \to E \to B$.
Choosing $F = BSpin_n$, $E = BSO_n$, $B = K(\mathbb{Z}/2,2)$ we see that the space of lifts of a map $f : X \to BSO_n$ to $BSpin_n$ is precisely the space of null-homotopies of the map $w_2(f) : X \to K(\mathbb{Z}/2,2)$. If this space is not empty, then the map $w_2(f)$ is null homotopic, and so we can choose a given null-homotopy $H$. Then, $H$ can be used to identify the space of null-homotopies with the space of null-homotopies of the constant map, which is $Hom(X,\Omega(K(\mathbb{Z}/2,2))) \cong Hom(X,K(\mathbb{Z}/2,1))$. Algebraically, this correspond to choosing a given $\alpha \in C^1(X,\mathbb{Z}/2)$ such that $d\alpha = w_2(f)$. Then, every other such $\alpha'$ gives a 1-cocycle $\alpha - \alpha' \in Z^1(X,\mathbb{Z}/2)$.
Anyway, without choosing the homotopy $H$, we just have an action of the "group" ($\mathbb{E}_1$-space) $Hom(X,\Omega K(\mathbb{Z}/2,1))$ on the space of null-homotopies of $f$, and this action is a torsor in the sense that the action map $$Hom(X,\Omega K(\mathbb{Z}/2,2)) \times NH(f) \to NH(f) \times NH(f)$$, $(T,H) \mapsto (T \square H,T)$ is a homotopy equivalence, where here $NH(f)$ stands for the space of null-homotopies of $f$.
In particular, for every choice of point $H \in NH(f)$ we get a homotopy equivalence as stated above.

Now all the results about cohomology are obtained by taking $\pi_0$ from the topological fact above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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