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Let $K$ be a field and $n \geq 1$. Then the set of isomorphism classes of vector bundles over $\mathbb{P}^n_K$ is a semiring (i.e. almost a ring, but no additive inverses are possible). By introducing additive inverses and quotienting out exact sequences, we get the $K$-theory of $\mathbb{P}^n_K$, which is known to be $\mathbb{Z}^{n+1}$. But is it also possible to compute exactly the semiring?

For $n=1$, there is a result by Dedekind-Weber (1892) which proves that the semiring is $\mathbb{N}[x,x^{-1}]$, where $x=\mathcal{O}(1)$ (related topic). Some months ago, I was told that the structure is far more complicated for $n>1$. Can anybody elaborate this or even give a presentation of the semiring?

If necessary, you may assume $K = \mathbb{C}$.

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    $\begingroup$ Note that just adding additive inverses to the semiring does <em>not</em> give the usual Grothendieck group. In fact the semiring is, by the Krull-Remak-Schmidt theorem, the free abeliam semigroup on the (isomorphism classes of) indecomposable vector bundles. Hence its group completion is just the free abelian group on the same vector bundles. Hence it is, as Angelo pointed out for the semiring, truly huge. Except in the $1$-dimensional case when the only indecomposable vector bundles are the line bundles. $\endgroup$
    – Torsten Ekedahl
    Commented Apr 6, 2010 at 7:05
  • $\begingroup$ thanks. I've edited the description of the $K$ ring in my question. $\endgroup$
    – Martin Brandenburg
    Commented Apr 6, 2010 at 7:33

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This semiring carries an enourmous amount of information about vector bundles on $\mathbb{P}^n$, including stuff we don't yet know. For example, you can read from it whether there are indecomposable vector bundles of any given rank; and for small rank we know very little about it (this is discussed, for example, in C. Okonek, M. Schneider, H. Spindler, "Vector bundles on complex projective spaces" , Birkhäuser (1987); I don't have access to the book here, and can't give you a more precise reference). I doubt you can can even get close to what you want.

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