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Oct 14, 2015 at 10:46 history edited Alexey Ustinov CC BY-SA 3.0
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Oct 14, 2015 at 9:05 vote accept QSR
Oct 14, 2015 at 7:30 answer added Andreas Cap timeline score: 5
Oct 10, 2015 at 6:04 comment added Asghar Ghorbanpour The four dimensional vector bundle has a trivial one dimensional sub bundle induced by the invariant vector (1,1,1,1). That means you have to look at the 3 dimensional sub bundle $SO(3)×_{A_4}V$.
Oct 10, 2015 at 0:56 vote accept QSR
Oct 14, 2015 at 9:05
Oct 10, 2015 at 0:42 vote accept QSR
Oct 10, 2015 at 0:43
Oct 9, 2015 at 21:27 comment added PVAL 3-dimensional real vector bundles over CW-complexes of dimension $\leq 3$ are completely classified by the first two Stiefel-Whitney classes. The obstruction for triviality over the 1-skeleton is $w_1$, and the obstruction for the existence of a 2-frame over the 2-skeleton reduces to $w_2$, so there exists a trivial 2-plane bundle inside of your vector bundle restricted to the 2-skeleton. This 2-plane bundle extends over the 3-skeleton as $\pi_2(V_2(\Bbb R^3))=0$. Now endow your bundle with a metric to find a complement to this 2-plane bundle.
Oct 9, 2015 at 20:28 comment added Igor Belegradek @MatthiasWendt: the comment is not quite right. You forgot the Euler class even though it is zero in the case at hand. The reference is "Classification of Oriented Sphere Bundles Over A 4-Complex" by A. Dold and H. Whitney, see maths.ed.ac.uk/~aar/papers/doldwhit.pdf
Oct 9, 2015 at 17:30 answer added Matthias Wendt timeline score: 5
Oct 9, 2015 at 16:10 history edited QSR CC BY-SA 3.0
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Oct 9, 2015 at 13:40 comment added Matthias Wendt Real vector bundles over CW-complexes of dimension $\leq 3$ are completely classified by the Stiefel-Whitney classes (seems to be an old result of Whitney). So if the classes are trivial, then so is the bundle.
Oct 9, 2015 at 13:31 history asked QSR CC BY-SA 3.0