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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!

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5 Answers 5

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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

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Thanks, I checked in Bott&Tu and understood their example. I was confused by the book of Bredon, where I didn't notice that the dimension of base space and rank of bundle are equal. Still, this is a case where topological invariants produce geometric objects, no? –  Dima Jul 11 '10 at 11:44

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".

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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).

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Hi Francesco, I believe that your argument with generic sections implicitly assumes that the rank of the vector bundle coincides with the dimension of the underlying base manifold (indeed, in this case, the vanishing of the Euler class means that the sum of the local degrees of $Z$ is zero, so that we can "cancel" them), right? Otherwise, if you allow the rank of the bundle to be different from the dimension of the base manifold, you can get counter-examples as in my answer below. Best, Matheus –  Matheus Jul 11 '10 at 10:49
    
Yes, I was implicitly making this assumption. Thank you for the remark. By the way, I do not have the book of Bott and Tu with me now, so I cannot directly check your counterexample. But how can you conclude from the fact that a rank 3 vector bundle is non trivial the fact that it has no nowhere vanishing section? (think of a bundle of the form E+O, where E is a nontrivial rank 2 vector bundle and O is the rank 1 trivial line bundle) –  Francesco Polizzi Jul 11 '10 at 11:31
1  
@Francesco : there is no nontrivial $E$, as it is rank two and orientable, and $H^2(S^4,Z)=0$. Note that this is the $n=2$ case of the question, which happens to be true because $BSO(2)$ coincides the Eilenberg-MacLane space $K(Z,2)$. –  BS. Jul 11 '10 at 11:40
    
Ok, I got the point. Thank you. –  Francesco Polizzi Jul 11 '10 at 13:30

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 here.

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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.

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