How to interpret the Euler class?

Question

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 $H_2(X, \mathbb Z)$.

Answer 1

Here is a different perspective from Matthew's: Suppose that we are given a section $s:X\to E$ of our rank $r$ vector bundle $E\to X$ such that $s(X)$ and the zero section intersect transversely. This intersection $I$ is a dimension $\dim(X)-r$ submanifold of $E$ which can naturally be thought of as a submanifold of $X$ by inclusion into the zero section. The Euler class is then $PD[I]\in H^r(X;\mathbb{Z})$, where $PD$ is Poincare duality in $X$.

For a plane field on a $3$-manifold, the Euler class lives in $H^2(X;\mathbb{Z})$, not $H_2(X;\mathbb{Z})$. (These are non-isomorphic if the first homology has torsion.) It is the Poincare dual of an element of $H_1(X;\mathbb{Z})$. For example, suppose that $X=\Sigma_g \times S^1$ and that $E$ is $(T\Sigma_g)\times S^1$. Then the Euler class is $(2-2g)PD[S^1]$.

Answer 2

Well, the Euler class exists as an obstruction, as with most of these cohomology classes. It measures "how twisted" the vector bundle is, which is detected by a failure to be able to coherently choose "polar coordinates" on trivializations of the vector bundle.

In the case where $E = \mathbb{R}^2 \times B$ is a trivial vector bundle with projection map $\pi: E \rightarrow B$, then we can take the form $\phi$ to be the pullback of the form $\frac{1}{2\pi} d \theta$ under the projection $$E - E_0 = (\mathbb{R}^2 - 0) \times B \rightarrow \mathbb{R}^2 - 0 $$ where $d\theta$ is the standard angular form on $\mathbb{R}^2 - 0$. Then we have that the Euler class $\chi$ is defined by $$d \phi = - \pi^* \chi$$ a formula that is generally true, if you think about defining local polar coordinates and having $\phi$ measuring how they fail to piece together over triple intersections.

In this case, $\phi$ is closed and therefore $e$ is 0. This stems from the fact that we can choose global angular coordinates on the vector bundle, which is what this thing measures the failure of in general.

For a thorough discussion of the Euler class from this perspective (and a chance to read a great book that can give you geometric intuition for these algebro-topological things) check out Differential Forms in Algebraic Topology by Bott & Tu.

Answer 3

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.