TITLE: How to interpret the Euler class?
QUESTION [43 upvotes]: Although I (hardly) understand the formal definition of the Euler class, I have very little intuition of it. I understand that the Euler class of $E\to X$ is zero if and only if there is a section, but what does it mean that the Euler class is non-zero?
For example, when $X$ is a 3-manifold and $E\to X$ a plane bundle with non-trivial Euler class, I would like to have a geometric/topological interpretation of the Euler class. In particular, I would like to have an interpretation of $e(S)$, where $S$ is an element of $H_2(X, \mathbb Z)$.

REPLY [15 votes]: It isn't true that if the Euler class vanishes, then there is a nowhere vanishing section (though the converse is true). 
For instance, there is a three-dimensional, oriented vector  bundle  $E \to S^4$  which admits no nowhere vanishing section, as pointed out on this discussion. In fact this vector bundle can be classified by the nontrivial element of $\pi_3(SO(3)) = \mathbb{Z}$ by the "clutching construction." The Euler class is zero because the cohomology of $S^4$ vanishes in dimension three, but on the other hand, if the vector bundle admitted a nowhere vanishing section, then it would be trivial. (Proof: then the vector bundle would decompose $E = F \oplus \mathbb{1}$ where $F$ is two-dimensional and classified by an element of $\pi_3(SO(2)) = \ast$, so $F$ is trivial and so is $E$.)
The Euler class is the first obstruction to the existence of a nowhere vanishing section, but it isn't the only one.