# Can this reduced version of algebraic K-theory be identified with this direct limit?

Let $X$ be a connected smooth scheme over a field and let $VB(X)$ be the exact category of vector bundles (i.e. locally free $\mathcal{O}_X$-module whose rank is finite at every point) over $X$.

Let the abelian group $\tilde K_0(X)$ denote the kernel of the rank morphism from the zeroth algebraic K-theory group $K_0(X)$ to $\mathbb{Z}$.

Let $Vec_k(X)$ be the set of isomorphism classes $[A]$ of rank $k$ vector bundles $A$ over $X$. One gets maps $i_k:Vec_k(X)\to Vec_{k+1}(X)$ defined by $[A]\mapsto [A\oplus \mathbb{A}^1_X]$, i.e. by adding a trivial bundle. The directed colimit $cVec(X)=\operatorname{colim}_k Vec_k(X)$ of this system has the structure of a commutative monoid by $[A]+[B]=[A\oplus B]$.

Is there an isomorphism $cVec(X)\cong\tilde K_0(X)$ of monoids?

This holds for topological K-theory over a connected CW-complex. I am a little unconvinced that this isomorphism holds since $VB(X)$ is not split-exact. However, I cannot find an argument.

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## 1 Answer

It's already false for $\mathbb P^1$. Let $L$ be a line bundle with no global sections except zero. Then $L\oplus L^{-1}$ is not a trivial bundle because it is not generated by global sections, and this remains true when you add a trivial bundle to it, but it is trivial in your reduced $K^0$.

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Dear @Tom Goodwillie, could you give me a hint why $L\oplus L^{-1}$ is ''trivial'' in $K^0$, please? –  Ronald Bernard Jul 3 '13 at 5:38
Let $H$ be the hyperplane bundle, the rank one bundle that has a global section possessing a single zero. Every rank one bundle on $\mathbb P^1$ is $H^d$ for a unique $d$. Every rank two bundle is $H^d\oplus H^0$ for a unique $d$. You can tell what $d$ is by considering the second exterior power. It follows that $H^a\oplus H^b$ is $H^{a+b}\oplus H^0$. –  Tom Goodwillie Jul 3 '13 at 13:43