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This question came up yesterday during our index theory seminar.

Let $M$ be a 1-connected smooth manifold and let $E \to M$ be a finite-rank complex vector bundle over $M$. If all the Chern classes of $E$ vanish, what else can one say about $E$? In other words, is there an alternative characterisation of such $E$?

(For a similar question in the case of $M$ having nontrivial fundamental group, see this previous question.)

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A somewhat tautological but perhaps useful answer: A vector bundle given by $M \to BU$ has trivial Chern classes precisely when it lifts to the homotopy fibre of $BU \to \prod_n K(Z,2n)$ given by the Chern classes. –  Torsten Ekedahl Feb 15 '11 at 18:45
    
In the holomorphic world the general answer is "not much" without some kind of stability condition (e.g., take direct sums of line bundles.) But if your $E$ is stable, with respect to some polarization (I'm assuming $M$ projective also), then, since $M$ is $1$-connected, it will be trivial (Donaldson-Uhlenbeck, IIRC). (But maybe this is orthogonal to your real question, sorry.) –  inkspot Feb 15 '11 at 20:41
    
What do you mean by "take direct sums of line bundles"? It should be an example of what? –  diverietti Feb 15 '11 at 22:29
    
The map in Torstens comment is $6$-connected ($\pi_*$-iso in degrees $\leq 5$, epi in degree $6$). Therefore the homotopy fibre is $5$-connected. That means that for manifolds of dimension $\leq 5$, such a vector bundle will be trivial. –  Johannes Ebert Feb 16 '11 at 0:13
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Correction: $c_n$ on $S^{2n}$ has to be divisible by $(n-1)!$ (Bott). So the map isn't surjective on $\pi_6$. But it is still $5$-connected, and bundles with zero Chern classes are trivial in dimensions $\leq 4$. –  Johannes Ebert Feb 16 '11 at 0:18

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Since no one has mentioned it yet, let me point out one possibly interesting observation. If the base manifold $M$ is compact and has no torsion in its integral cohomology, then a vector bundle $E$ with vanishing Chern classes is stably trivial. This was pointed out to me by Robert Lipshitz. The reason is as follows: from looking at the Atiyah-Hirzebruch spectral sequence, one can see that there can't be any torsion in the complex K-theory of $M$. Looking at the Chern character, one concludes that $[E]$ must be trivial in $\widetilde{K}^0(M)$, i.e. $E$ is stably trivial.

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By the way, I've been told (and it fits with the evidence) that Bott was always very interested in determining whether the (co)homology of various spaces was torsion-free. I never really knew why that was such an interesting thing to prove, but the statement above strikes me as one very nice consequence of torsion-free-ness. –  Dan Ramras Apr 1 '11 at 2:19

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