Timeline for How to prove that $w_1(E)=w_1(\det E)$?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 3, 2018 at 7:38 | comment | added | Thomas Rot | @Qfwfq: for $w_1$ yes | |
| Apr 17, 2010 at 12:00 | comment | added | Dan Ramras | Yup, see my answer. | |
| Apr 17, 2010 at 11:19 | comment | added | Qfwfq | Is $w_i$ zero on $M$ iff it has zero pullback along any loop? | |
| Apr 17, 2010 at 10:50 | comment | added | Thorny | 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. | |
| Apr 17, 2010 at 9:58 | comment | added | Qfwfq | (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. | |
| Apr 17, 2010 at 9:56 | comment | added | Qfwfq | 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... | |
| Apr 17, 2010 at 9:50 | vote | accept | Qfwfq | ||
| Apr 17, 2010 at 9:50 | |||||
| Apr 17, 2010 at 9:50 | vote | accept | Qfwfq | ||
| Apr 17, 2010 at 9:50 | |||||
| Apr 17, 2010 at 9:36 | history | answered | Thorny | CC BY-SA 2.5 |