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.


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