Vanishing of Euler class Given a real oriented vector bundle E over the base space B of rank n, such that the Euler characteristic class in the n-th cohomology group of B vanishes, is it true that there exists a global nowhere-vanishing section of the bundle? 
Any idea where to find a proof or a counterexample?
Thanks!
 A: 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).
A: Let $n$ be odd. Recall that $S^n$ is parallelisable if and only if $n = 1, 3, 7$. For every other $n$, there exists $k < n$ such that $S^n$ admits $k$ linearly independent vector fields, but not $k + 1$. As $n$ is odd, $\chi(S^n) = 0$ so $S^n$ admits a nowhere-zero vector field by Poincaré-Hopf; that is, $k > 0$. Therefore $TS^n = E\oplus\varepsilon^k$ where $E$ has rank $n - k$ and does not admit a nowhere-zero section. Note however that $e(E) \in H^{n-k}(S^n; \mathbb{Z}) = \{0\}$ as $0 < n-k < n$.
It should be noted that the precise value of $k$ is known for every $n$ by work of Radon, Hurwitz, and Adams. Namely $k = \rho(n+1) - 1$ where $\rho(n+1)$ denotes the $(n+1)^{\text{st}}$ Radon-Hurwitz number: if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $\rho(n+1) = 8a + 2^b$.
A: One important case when vanishing of the Euler class does imply triviality of the bundle (and hence existence of nowhere zero section) is for oriented rank 2 bundles over paracompact bases. In fact, Euler class gives one-to-one correspondence between the set of isomorphism classes of such bundles and  the second cohomology group [Husemoller's "Fiber bundles" book, 20.2.6].
If rank is $>2$, then a vector bundle is determined by Euler and Pontryagin classes up to finite ambiguity (provided the base is a finite cell complex). In rare cases Euler and Pontryagin determine the bundle completely.
For example, rank 4 bundles over $S^4$ are in one-to-one correspondence with $\pi_3(SO(4))\cong\mathbb Z+\mathbb Z$ where the latter isomorphism can be chosen so that the bundle $(n,m)$ has Euler class $n$ and first Pontryagin class $2m$. If memory serves me, this can be found in Milnor's original paper on exotic 7-sphere, but there are more recent detailed sources, e.g. see

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*Diarmuid Crowley, Christine M. Escher, A classification of $S^3$-bundles over $S^4$, Differential Geometry and its Applications 18, Issue 3 (2003) pp 363–380, https://doi.org/10.1016/S0926-2245(03)00012-3, arXiv:math/0004147
A: 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".
A: You might want to take a look at Milnor and Stasheff. In section 9, oriented bundles and the euler class, they prove that if a bundle has a nowhere zero section then the euler class of that bundle is trivial. I think that this is typically the direction of invariants in algebraic topology. However, the vanishing of various stiefel whitney classes will allow you to put an orientation on the bundle or a spin structure.
A: 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
