Let $F,E,B$ be manifolds (we may assume them to be compact, without boundary if necessary) and $$F\to E\to B$$ be a fibre bundle. Suppose $$w(F), w(B),$$ the Stiefel-Whitney class of $F$ and $B$, are known. I notice that in particular, if the bundle is trivial, then $E=B\times F$ and $w(E)=w(B)w(F)$.

Question: In general, are there any formulas to compute the Stiefel-Whitney class of the total space $E$ $$ w(E) $$ in terms of $w(B)$ and $w(F)$? Or even in terms of the cohomology ring $$ H^*(B;\mathbb{Z}_2), H^*(F;\mathbb{Z}_2) $$ And other factors?

  • 9
    $\begingroup$ Certainly not just in terms of $w(B)$ and $w(F)$; the Klein bottle is a circle bundle over the circle. $\endgroup$
    – mme
    Sep 25, 2015 at 3:24

1 Answer 1


Just to get started (a bit long for a comment):

For a fiber bundle $F\rightarrow E\overset{\pi}{\rightarrow}B$ one has an exact sequence of bundles over $E$

$$ 0\rightarrow T_vE\rightarrow TE\rightarrow \pi^{*}TB\rightarrow 0. $$

Here $T_vE:=\ker T\pi$ is the vertical part of the tangent space to $E$, i.e. these are the tangent vectors to the fibers. The choice of a complementary bundle amounts to the choice of a connection, but can be identified with $\pi^*TB$. So one sees that

$$ w(TE)=w(T_vE)\pi^*(w(TB)). $$

I think one needs more information if one wants to continue the computation. For example one can go on if $E$ is the projectivization of a vector bundle over $B$.

  • $\begingroup$ How can we proceed further, say if one has that the fibers are actually projective spaces? @ThomasRot $\endgroup$ Nov 11, 2021 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.