5
$\begingroup$

How to prove that the first Stiefel-Whitney class $w_1 (E)$ of a real rank $n$ vector bundle over a manifold M is equal to $w_1(\det E)$, where $\det E$ is the $n$-th wedge power of $E$?

(I want to assume the "axiomatic" definition of Stiefel-Whitney classes, as given e.g. in the book by Milnor and Stasheff).

I have just been asked an analogous question by a younger guy, but I think I could only find a proof starting from a different definition of the $w_i$'s. Perhaps I'm just missing something? Of course, feel free to close it if you find it's to homework-ish for MO standards.

Improve this question
$\endgroup$
4
  • 7
    $\begingroup$ splitting principle, reduce to line bundles. $\endgroup$
    – Martin Brandenburg
    Commented Apr 17, 2010 at 7:13
  • $\begingroup$ @Martin: doing so, it's my impression that one is led to compare $w1(L1)+ \cdots + w1(Ln)$ with $w1(L1 \otimes \cdots \otimes Ln)$. How does $w1$ behave with respect to tensor products of line bundles? $\endgroup$
    – Qfwfq
    Commented Apr 17, 2010 at 8:55
  • $\begingroup$ Ok, Allen Hatcher proves that $w_1(L\otimes L')=w_1(L)+w_1(L')$ for line bundles. But he uses a "Leray-Hirsch" definition of characteristic classes, not the axiomatic one. Also, the proof that $w_1(L\otimes L')=w_1(L)+w_1(L')$ uses the property of the classifying space -not just the axioms- and is relatively complicated. I would've thought there was a one-line-proof directly from the axioms... $\endgroup$
    – Qfwfq
    Commented Apr 17, 2010 at 9:19
  • $\begingroup$ You can prove the identity for line bundles by proving it for the universal bundles over RP^inf, which follows from the Kunneth isomorphism. $\endgroup$
    – Tobias Shin
    Commented Oct 11, 2018 at 0:07

2 Answers 2

Reset to default
12
$\begingroup$

$E \oplus det E$ is orientable (its structure group $O(n)$ is represented in $SO(n+1)$), so its $w_1$ vanishes; and $w_1(E \oplus det E) = w_1(E) + w_1(det E)$.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ I would accept this answer if it didn't make use of the fact that $w_1=0$ iff "orientable": in fact the guy -I didn't specify it in the OP- wanted me to explain why a (differentiable) manifold has $w_1=0$ iff it is orientable. I'd have used $w_1(E)=w_1(detE)$ plus Proposition 4 from Milnor & Stasheff... $\endgroup$
    – Qfwfq
    Commented Apr 17, 2010 at 9:56
  • $\begingroup$ (continued) ..., the latter stating "a rank $n$ vector bundle has a nowhere vanishing section iff $w_n=0$". In our case that section would of course be a volume form on the manifold. $\endgroup$
    – Qfwfq
    Commented Apr 17, 2010 at 9:58
  • 2
    $\begingroup$ The orientability of a manifold can be tested on loops, and on loops, there are only two kinds of bundles: oriented ones - those are trivial, and non-oriented ones - those are the sum of the $\gamma^1$ and a trivial (one can split trivial subbundles off as long as the rank of the bundle is more than the dimension of the base, which is 1). So the axioms tell us what the pullbacks of $w_1$ to the loops are - 0 if it is an orientation-preserving one, 1 if it is not. $\endgroup$
    – Thorny
    Commented Apr 17, 2010 at 10:50
  • $\begingroup$ Is $w_i$ zero on $M$ iff it has zero pullback along any loop? $\endgroup$
    – Qfwfq
    Commented Apr 17, 2010 at 11:19
  • 1
    $\begingroup$ @Qfwfq: for $w_1$ yes $\endgroup$
    – Thomas Rot
    Commented Jul 3, 2018 at 7:38
11
$\begingroup$

This is really a long comment regarding some of the above discussion.

Hatcher (in his Vector Bundles notes) certainly proves that the characteristic classes defined using Leray-Hirsch satisfy the axioms from Milnor-Stasheff. But $w_1$ is a more basic object: the axioms specify its value on the tautological bundle over ${\Bbb R}P^1$ (= $S^1$) and this immediately determines its values on all line bundles (see p. 81 of Hatcher's notes). One can then see that as an element of

$H^1(X; {\Bbb Z}/2) = Hom(H_1 X, {\Bbb Z}/2) = Hom(\pi_1 X, {\Bbb Z}/2)$,

$w_1(L)$ simply answers the question: "Along which loops is L trivial?" (Actually, this is true for all bundles, not just lines.) From this point of view, multiplicativity ($w_1 (L\otimes K) = w_1(L) + w_1(K)$) is a quick exercise (hmmm... what should a homomorphism from a multiplicative group to an additive group be called? Anyway, I just mean it's a homomorphism from the Picard group of line bundles to $H^1$.). Alternatively, it follows from the H-space structure on ${\Bbb R}P^\infty$ defined via the map ${\Bbb R}P^\infty\times {\Bbb R}P^\infty \to {\Bbb R}P^\infty$ classifying $\gamma^1_\infty \otimes \gamma^1_\infty$. This is spelled out in my notes on vector bundles (it's written in the complex case but works the same way in the real case). I couldn't quickly see where Hatcher discusses this point.

Incidentally, my notes also discuss the relationship between orientability and $w_1$. I've never been crazy about discussions of this point in the literature (e.g. Hatcher states this relationship only for spaces of the homotopy type of a CW complex, although he doesn't seem to use that assumption in his proof).

Improve this answer
$\endgroup$
4
  • $\begingroup$ For vector bundles, shouldn't the question answered by $w_1(E)$ be something like "along which loops is $E$ orientable?". For line bundles over the circle orientability is the same as nontriviality, but for higher rank bundles? (b.t.w. I havent read all the answer carefully yet: I will do it later!) $\endgroup$
    – Qfwfq
    Commented Apr 17, 2010 at 12:08
  • 2
    $\begingroup$ Bundles over the circle are pretty simple things... They're classified by maps into the Grassmanian, so you need to think about the fundamental group of the Grassmanian. This is just $Z/2 =\pi_1 BO(n)= \pi_0 O(n)$, so there are only two of them in each dimension. This is also discussed in my notes :) $\endgroup$
    – Dan Ramras
    Commented Apr 17, 2010 at 12:16
  • $\begingroup$ @DanRamras: "This is spelled out in my notes on vector bundles" - could you please give us a link to them? Thank you. $\endgroup$
    – Alex M.
    Commented Oct 5, 2018 at 8:49
  • 1
    $\begingroup$ @AlexM. I've updated the link (when I went to edit the post, it seemed the word "notes" was supposed to be a link, but for some reason it hadn't looked like it in the original post). The full notes are here: math.iupui.edu/~dramras/697.html $\endgroup$
    – Dan Ramras
    Commented Oct 10, 2018 at 23:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .