Vanishing of Euler class

If $B$ is triangulated, $e(E)\in H^n(B,Z)$ is only the obstruction to have a non-vanishing section on the $n$-skeleton of $B$, but if $\dim B>n$, it is possible that none of these sections extends to the $n+1$ skeleton : the obstruction lies in $H^{n+1}(B,\pi_{n}(S^{n-1}))$, and may be non-zero if $n>2$. This obstruction theory is exposed in Steenrod's classic "Topology of fibre bundles".

Hi Dima,

I think that the answer to your question is no: as it is pointed out in the book "Differential Forms in Algebraic Topology" of R. Bott and W. Tu, cohomological invariants are too coarse to ensure the existence of geometrical objects.

More precisely, Example 23.16 of the book of Bott and Tu shows that $S^4$ has a nontrivial rank 3 vector bundle with vanishing Euler class.

Best,

Matheus

If the base space is a smooth, oriented manifold then the Euler class corresponds to the zero locus of a section in the following way. Let $\sigma \colon B \to E$ be a $generic$ section, and let $[Z] \in H_n(B)$ be the homology class of the zero locus $Z$ of $\sigma$. Then the Euler class $e(E) \in H^n(B)$ of $E$ is the Poincaré dual of $[Z]$. In particular, $e(E)=0$ if and only if the general section of $E$ vanishes nowhere.

The book of Milnor and Stasheff "Characteristic Classes" is a classical introduction to the subject (see in particular p. 98).