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 $H_2(X, \mathbb Z)$.


3 Answers 3


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 $(2-2g)PD[S^1]$.

  • $\begingroup$ Excellent; +1 .Sorry for the necropost. In the bottom paragraph, since you're referring to a 3-manifold, I guess you mean $\Sigma_2 \times S^1$? and $E=(T\Sigma_2)\times S^1 $ And , is the Euler class in general, for a plane field (2-distribution D_2) $ χ (D_2)PD[M]$ , i.e., χ is the Euler characteristic, and where the base is $D_2 \times M $ ? $\endgroup$
    – user99680
    May 9, 2014 at 21:48
  • $\begingroup$ @user99680 There is no requirement that the genus of the surface should be 2. Here $\Sigma_g$ was a closed surface of genus $g$ and $E$ was the pullback of its tangent bundle to $\Sigma_g\times S^1$ via the project-out-the-$S^1$ map. I'm having trouble parsing the rest of your question, but I can say that Euler classes behave functorially under pullback. $\endgroup$
    – Adam
    Jun 8, 2015 at 19:05
  • $\begingroup$ Is there a quick way to prove that $e(X)=PD[I]$ from the definition of $e(X)$ as the restriction of the Thom class? $\endgroup$
    – nkm
    Mar 25, 2016 at 3:48
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    $\begingroup$ @NathanielMayer Yes, taking a section is just a transverse version of the standard construction. That is, the usual set of inclusions are $X \hookrightarrow E \hookrightarrow (E,E\setminus E_0)$ where $X \hookrightarrow E$ is the zero section. Replace that map with a section. (It's a vector bundle so that's an isotopic map.) $\endgroup$
    – Adam
    Apr 10, 2016 at 15:20

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.


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.

  • $\begingroup$ I'm sure something similar to this works, but not quite this. We have $\pi_3(SO(3)) = \pi_3(S^3) = \mathbb{Z}$, not $\mathbb{Z}/2\mathbb{Z}$. $\endgroup$ Jun 10, 2012 at 23:13
  • $\begingroup$ On second thought, that doesn't affect the answer at all. All you need is $\pi_3(SO(3))\neq 0$ for this to work. (Also, minor typo: "the cohomology os $S^3$" should be "the cohomology of $S^4$" in the middle of your middle paragraph.) Finally, when the rank of the bundle is greater than or equal to the dimension of the base, I think the same link you have shows that $e = 0 $ iff there is a section. $\endgroup$ Jun 10, 2012 at 23:16
  • $\begingroup$ @Jason: Whoops. I'm not sure what I was thinking there. $\endgroup$ Jun 11, 2012 at 0:03
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    $\begingroup$ @AkhilMathew Regarding the last sentence of your answer, I was wondering what are the other obstructions aside from the Euler class to the existence of a nowhere vanishing section $\endgroup$
    – u184
    Feb 1, 2019 at 13:02

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