Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $M$ be a manifold and $S$ its sphere bundle with fiber $\mathbb{S}^n$. As we know, the notion of the Euler class is raised from the problem of finding a global form on $S$ which restricts on each fiber to a generator of $H_{\textrm{dR}}^n(\mathbb{S}^n)$. One overcomes two obstructions, namely the orientability of the sphere bundle and the Euler class $e(S)$ to find such a form. Now let $N$ and $M$ be two manifolds and $f:N\rightarrow M$ be smooth, $E$ is a vector bundle over $M$. The following functorial property of $e(E)$ is well-known:

$e(f^\ast E)=f^\ast e(E)$

On the other hand, the splitting principle for cmplex vector bundles is wedely-known for its power on computing characteristic classes. In the real case, although the splitting principle does not hold in full generality, a weak version of this type of theorems can still be obtained: (cf. Lecture Notes in Mathematics, 638)

In fact, let $E$ be an even dimensional oriented real vector bundle over $M$. Then there exist a manifold $N$ and a map $g:N\rightarrow M$ satisfying the following two conditions:

(1) the homomorphism $g^\ast: H_{\textrm{dR}}^\ast(M)\rightarrow H_{\textrm{dR}}^\ast(N)$ is injective;

(2) $g^\ast(E)$ is a direct sum of plane bundles.

Therefore one may expect that the Euler class can be defined for an arbitrary vector bundle $E$ by first introducing the Euler class for plane bundles, then extending it to $E$ using the splitting principle stated above. In fact, suppose that $e(P_i)$ has been introduced for every plane bundle $P_i$, and $g^\ast(E)$ is a direct sum of $P_i$, then $e\big(g^\ast(E)\big)$ is well defined, therefore $e(E)$ may be defined by

$e\big(g^\ast(E)\big)=g^\ast e(E)$

However, one still needs to show the existence of $e(E)$ and its independence of the choices of $N$ and $g$. It doen’t seem easy.

share|improve this question
I'm not quite sure what the question is asking. An Euler class $e(E)$ of the vector bundle $E$ is, by definition, the pullback of a Thom class $t(E)\in H^n(D(E),S(E))$ under the zero section. A Thom class is, by definition, a cohomology class which restricts to a generator of $H^n(D^n,S^{n-1})$ in each fibre. Such a class exists (with real coefficinents) if and only if the bundle is orientable. So an Euler class with real coefficients exists iff $E$ is orientable. If you are prepared to use twisted coefficients, its a different story, and every bundle has a twisted Euler class. –  Mark Grant Nov 1 '11 at 20:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.