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Aug 6, 2019 at 9:42 comment added Thomas Rot @fredy: The map that sends each fiber $v$ to $-v$ is orientation reversing if the vector space is odd dimensional. Hence $e(E)=-e(E)$, so $2e(E)=0$, which in the de Rham theory implies that $e(E)=0$ (but not when the Euler class is defined with integral coefficients).
Jul 23, 2019 at 2:40 comment added Fredy @ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
Jul 22, 2019 at 15:06 comment added Thomas Rot @Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
Jul 22, 2019 at 15:04 vote accept Fredy
Jul 22, 2019 at 13:37 comment added Fredy Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
Jul 22, 2019 at 10:28 comment added BS. Moreover an important example (often denoted $\Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
Jul 22, 2019 at 9:02 history edited Thomas Rot CC BY-SA 4.0
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Jul 22, 2019 at 8:55 history answered Thomas Rot CC BY-SA 4.0