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As far as I understand, one can make a monoid from the space of vector bundles on a (compact) manifold $M$, with respect to the direct sum operation $\oplus$. In order to make this into a group, one needs to add 'inverses' and this leads to the consideration of formal differences $V-W$ of vector bundles. One can again compute Euler classes, Chern classes, etc... for these objects called 'virtual bundles'.

Now my question is:

When is such an object an actual vector bundle? Is there an easy criterion to check? For instance, if I have a complex virtual vector bundle and I know that its Chern classes are integral i.e. in $H^*(M,\mathbb{Z})$, rather than $H^*(M,\mathbb{Q})$, is that sufficient?

Disclaimer: I am only familiar with basic aspects of $K$-theory, so the above question might be completely trivial, but I couldn't find an answer.

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    $\begingroup$ Let $c$ denote the total Chern class. Then the Whitney formula is $c(E\oplus F)=c(E)c(F)$. So, to be consistent, I think you should have $c(E-F)=c(E)c(F)^{-1}=c(E)\cdot 1/(1+c_1(F)+c_2(F)+\ldots + c_r(F))=c(E)\sum_{k\geq 0}(-1)^k(\sum_{i=1}^r c_i(F))^k$ which seems integral. $\endgroup$ – Qfwfq Feb 9 at 14:37
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    $\begingroup$ @Qfwfq 's argument shows that it is necessary to have $c_i(E)=0$ for $i$ large enough. But it is clearly not sufficient because any bundle of the form $V-nC$, where $nC$ is the trivial $n$-dimensional complex bundle verifies this. $\endgroup$ – user43326 Feb 9 at 15:42
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    $\begingroup$ Over a compact space every map $f: X \to BO$ factors through a finite-dimensional Grassmannian. In particular every virtual v.b. is of the form $V - n$ for some vector bundle $V$, and $n$ is determined by the rank of the virtual vb. $\endgroup$ – Mike Miller Feb 9 at 16:19
  • $\begingroup$ And once you've used Mike Miller's observation, a necessary condition is that the last 𝑛 chern classes of 𝑉 are zero. If the rank and the dimension of the manifold interact well you can appeal to some obstruction theory, but I think it's a pretty hard problem in general, since it's the same as asking for $n$ linearly independent sections of $V$. $\endgroup$ – John Greenwood Feb 9 at 21:23

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