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Is there something non-trivial to be said about the subring of $K(X)$ spanned by one-dimensional bundles?

If I am not mistaken, it is still a functor into commutative rings from the category of schemes. Moreover, it seems that there is some interest in the splitting problem for vector bundles, which goes back to Grothendieck's result that for $\mathbb{P}^1$ all bundles split. Also, by the splitting principle, any element of $K(X)$ lands in such a subring after a suitable pullback.

But maybe my question is stupid for some obvious reason.

Edit: Some concrete questions:

  1. Is the existence of an indecomposable bundle on $X$ imply that the subring generated by line bundles is proper?
  2. Is there an analog of higher $K$-groups for this definition? (I don't see any problem with taking the usual motivic construction, but what do I know)
  3. Are there known computations of this ring?
  4. Is there a general relationship between the usual $K$-ring and this subring?
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    $\begingroup$ It is not clear what do you mean by "something non-trivial". $\endgroup$
    – Sasha
    Jan 31, 2023 at 13:46
  • $\begingroup$ @Sasha, Thank you! I added some concrete questions I had in mind. $\endgroup$ Jan 31, 2023 at 16:40

1 Answer 1

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No for question 1.

Indeed, $K(\mathbb{P}^2)$ is generated by line bundles (in fact, it is freely generated by $\mathcal{O}$, $\mathcal{O}(1)$, $\mathcal{O}(2)$), but there are plenty of indecomposable vector bundles (e.g., the tangent bundle).

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  • $\begingroup$ Do we need exceptional collections to see this, or is it an elementary fact? $\endgroup$ Jan 31, 2023 at 18:07
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    $\begingroup$ This can also be proved without using exceptional collections, but not as easy. $\endgroup$
    – Sasha
    Jan 31, 2023 at 19:38

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