Timeline for Oriented vector bundle with odd-dimensional fibers
Current License: CC BY-SA 4.0
8 events
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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 |
added 208 characters in body
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Jul 22, 2019 at 8:55 | history | answered | Thomas Rot | CC BY-SA 4.0 |