Let $E \to B$ be real vector bundle. Consider the homomorphism $\pi_1(B) \to \mathbb{Z}_2$ which assign $0$ or $1$ to each loop according to whether orientations of fibers are preserved or not. Since $\mathbb{Z}_2$ is abelian this homomorphism factors through $H_1(B)$ therefore it gives a class in $H^1(B,\mathbb{Z}_2)$.

How to prove that this class is exactly the first Stiefel_Whitney class?

EDIT: My preferred definition of S-W classes is via axiomatic approach: naturality, Whitney formula and normalization

  • $\begingroup$ math.stackexchange.com/questions/1820394/… $\endgroup$ – Lorenzo Najt Jul 5 '17 at 15:59
  • $\begingroup$ What is your initial definition of Stiefel Whitney first class ? $\endgroup$ – R. Alexandre Jul 5 '17 at 16:02
  • 1
    $\begingroup$ Consider how the given class behaves under direct sum and pullbacks, then reduce the problem (e.g., via the splitting principle) to the case of the tautological bundle over $\mathbb{RP}^\infty$. $\endgroup$ – anomaly Jul 5 '17 at 16:40
  • 1
    $\begingroup$ It has the same functorality so it suffices to check the universal examples, namely the universal vector bundles on $BO(n)$. Here you can just check that there's only one nonzero class in that cohomology group so it must be $w_1$ if it's nonzero. $\endgroup$ – Qiaochu Yuan Jul 5 '17 at 16:40

What is your preferred definition of the first Stiefel Whitney class?

I'm more comfortable with classifying spaces (this is the only definition I can produce for Stiefel Whitney classes right now), and so I'd prefer to think of your description as whether or not the classifying map $f : B \to BO(n)$ factors through $BSO(n)$, i.e. whether or not the bundle is orientable (i.e. whether or not the transition maps can be taken to be determinant one).

Because of the short exact sequence $0 \to SO(n) \to O(n) \to \mathbb{Z}/2Z \to 0$, where the last map is the determinant, we get a fibration $B(SO(n)) \to B(O(n)) \to B(Z/2Z) = \mathbb{RP}^{\infty}$. The bundle is orientable iff the classifying map can be lifted to $SO(n)$ (up to homotopy), iff it can be homotoped to live in a fiber, iff the induced map to $\mathbb{RP}^{\infty}$ is null homotopic (homotopy lifting property for Serre fibrations), iff the determinant line bundle is orientable, iff the principle $O(1)$ bundle (2 sheeted cover) associated to the determinant line bundle is trivial. This is exactly keeping track of the $\pi_1(B)$ action on the orientation of a fiber.

The last bunch of equivalences can be used to solve the problem. The point is the the map from X to $RP^{\infty}$ is nullhomotopic iff the induced map on $H_1$ is trivial iff the induced map on $\pi_1$ is trivial, iff it is lifts to the cover which is $S^{\infty}$, and in particular contractible. (We use that $\pi_1(RP^{\infty})) = \mathbb{Z}/2$.) The induced map on $H_1$ is dual to the induced map on $H^1$ (everything is over a field*), so you can conclude from this, since the first Stiefel Whitney class is the pullback of the generator $w_1$ of $H^1(B(O(n))$ under the classifying map.

(*This gave me pause, so let me explain. This is pure homological algebra. We have two chain complexes of k-vector spaces, $C_*$ and $D_*$, an a map $f : C \to D$. We have a functor $F : k-Vect \to k-Vect$, $Hom(\_, k)$, which is exact. Then the induced map of $H_* (F(C_*))) \to H_*(F(D_*))$ is $F$ of the induced map $H_*(C_*) \to H_*(D_*)$. The point is that exact functors "commute" with taking homology of a complex.)

I'm assuming that $X$ is a CW-complex, because at some point I use a result about Serre fibrations, but probably this can be done away with. Let me know if you have questions, my writing here is hardly masterly exposition - I'm trying to understand this material as well.

This is very close to the content of The first Stiefel-Whitney class is zero if and only if the bundle is orientable but I think I address your question (about the action of $\pi_1$) more directly, and add some more detail.

If this is all nonsense to you (or other reader), I learned these ideas (except the mistakes) from these notes, which I cannot advertise enough: https://www3.nd.edu/~mbehren1/18.906/prin.pdf

| cite | improve this answer | |

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.