0
$\begingroup$

I am studying Stiefel-Whitney classes at the moment. In his book

http://www.math.cornell.edu/~hatcher/VBKT/VB.pdf

Hatcher defines orientability of a vector bundle $p:E\to B$ (with path-connected base space $B$) by a map $\pi_1(B)\to \mathbb{Z}_2$ by assigning $0$ if a loop preserves orientation and $1$ if it reverses orientation. If the map is zero for every loop then the bundle is orientable.

(This map can also be seen as an element in $H^1(B,\mathbb{Z}_2)$ and is named the first Stiefel-Whitney class $w_1(p:E\to B)$. Thus we can neatly write $p:E\to B$ iff $w_1(p:E\to B)=0$)

I am concerned now about the orientation definition of vector bundles. We define it usually as Wikipedia does

https://en.wikipedia.org/wiki/Orientation_of_a_vector_bundle

i.e. an orientation of a vector bundle coincides locally with the $\mathbb{R}^n$-standard orientation.

Now why are these two notions equivalent? It is not obvious to me because on the one hand we have loops and on the other hand open sets.

A corollary of this would also be that if the base space $B$ is a CW-complex then $B$ is orientable iff its 1-skeleton is orientable, because every loop can be homotopically deformed to lie in the 1-skeleton.

$\endgroup$
  • $\begingroup$ For the converse direction ie orientation imply $w_1=0$ for that have a look at chp 12 of characteristics class by Milnor (Gysin seq). $\endgroup$ – Anubhav Mukherjee Oct 4 '16 at 8:32
  • 1
    $\begingroup$ @Anubhav Isn't that killing a fly with a sledgehammer? If $E$ is orientable, every loop preserves orientation (because $E$ restricts to an orientable vector bundle on the circle, so trivial) so one easily has $w_1 =0$. For the converse direction - the map $w_1$ seems precisely to be given by sending a vector bundle to the classifying element in $H^1(B; \Bbb Z/2)$ of the determinant bundle. In which case, $w_1 = 0$ tells the determinant bundle of $E$ is trivial, so $E$ is orientable. $\endgroup$ – Balarka Sen Oct 4 '16 at 14:56
  • $\begingroup$ @Balarka the point is I didn't want to think anything... So the closest reference what I remembered at that time, I wrote that down. I know easy solution, but too lazy to write :p . $\endgroup$ – Anubhav Mukherjee Oct 4 '16 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.