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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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    $\begingroup$ 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. $\endgroup$
    – Torsten Ekedahl
    Feb 15, 2011 at 18:45
  • $\begingroup$ 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.) $\endgroup$
    – inkspot
    Feb 15, 2011 at 20:41
  • $\begingroup$ What do you mean by "take direct sums of line bundles"? It should be an example of what? $\endgroup$
    – diverietti
    Feb 15, 2011 at 22:29
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    $\begingroup$ 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 $\leq 4$. $\endgroup$
    – Johannes Ebert
    Feb 16, 2011 at 0:18
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    $\begingroup$ Let me try to redeem myself by pointing to a paper by Boyer et al (Algebraic cycles and infinite loop spaces, Invent. Math. 113) where they show that there is an infinite loop space structure on $\prod_nK/Z,2n)$ making $BU \to \prod_nK/Z,2n)$ an infinite loop space map so that in particular its homotopy fibre gives a cohomology theory (or better yet the homotopy fibre of the map from connective K-theory). $\endgroup$
    – Torsten Ekedahl
    Feb 17, 2011 at 19:04

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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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    $\begingroup$ 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. $\endgroup$
    – Dan Ramras
    Apr 1, 2011 at 2:19
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    $\begingroup$ Would you mind saying a few more words about how you deduce that $K(M)$ is torsion-free? I can see that the torsion-free condition on cohomology implies that all the terms on the $E_2$ page of the Atiyah-Hirzebruch spectral sequence are free abelian groups of finite rank, but I guess I don't know enough about the differentials. A priori, there could be some non-trivial differential $d^r : E^{p,q}_r \to E^{p+r,q-r+1}_r$ with $r$ odd whose image has finite index in $\operatorname{ker}(d^r : E^{p+r,q-r+1}_r \to E^{p+2r,q-2r+2}_r)$. $\endgroup$
    – Michael Albanese
    May 4, 2022 at 1:18
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    $\begingroup$ Basically the point is that we know a priori that ranks of $K^0 (M)$ and $K^1 (M)$ equal the total ranks of the even and odd degree cohomology groups (respectively), and these total ranks also agree with the total rank of the $E^2$ terms on the relevant anti-diagonals. A non-trivial differential anywhere in the spectral sequence would decrease the total rank on one of these lines, hence would decrease the total rank of $K^*(M)$. Kristen Hendricks wrote out the argument in detail: arxiv.org/abs/1107.2154, Prop. 6.10. See also Section 2.5 of the original Atiyah-Hirzebruch article. $\endgroup$
    – Dan Ramras
    May 6, 2022 at 3:35
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    $\begingroup$ Thanks. I was missing the fact that we could use the Chern character to deduce the ranks on the $E_{\infty}$ page. $\endgroup$
    – Michael Albanese
    May 6, 2022 at 11:17

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