MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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 $K^0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.