# Stiefel-Whitney class of fibre bundles

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?

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

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 $$B$$.

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