Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There are several ways for defining the K-Theory of a category depending on which structure it admits. The K-Theory of schemes is commonly defined as the "group completion" of the category of algebraic vector bundles with respect to its exact category structure rather than with respect to its symmetric monoidal category structure.

In general, these two are not the same and the former is a quotient of the latter. In particular, for schemes these two constructions are not equivalent since exact sequences of algebraic vector bundles do not always split (e.g. projective line $\mathbb{P}_R^1$).

My question: why is the "correct" definition to choose the exact category structure? What goes wrong when one just considers the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined.

share|improve this question
    
You use both structure to construct the K-theory ring spectrum. Also note that the use of vectors bundles is only appropriate when the scheme is smooth. –  Fernando Muro Jun 1 '12 at 22:19

2 Answers 2

up vote 7 down vote accepted

There is no "universally correct" definition, but $K^0(X)$ usually denotes the K-theory of the exact category of vector bundles over $X$. As you say, there are various approaches for (higher) algebraic K-theory (summarized in Weibel's K-book, for instance), and which one one uses depends on the applications, but also on the realizability. Namely, there are lots of exact sequences of vector bunbles on projective varieties which give you relations in the K-theory, but these exact sequences don't split. Although this makes $K^0(X)$ relatively small compared to $K^0_{\oplus}(X)$ (the K-theory of the monoidal category of vector bundles w.r.t. $\oplus$) and perhaps doesn't contain enough information for what you want, it is at least computable! For example, $K^0(\mathbb{P}^r) \cong \mathbb{Z}[x]/(x^{r+1})$ is well-known (SGA 6), but little is known about $K^0_{\oplus}(\mathbb{P}^r)$ (cf. MO/20444): It is the free abelian group on indecomposable vector bundles on $\mathbb{P}^r$ (Atiyah), but these have not been classified yet.

Grothendieck's celebrated and important resolution theorem won't tell you anything about $K^0_{\oplus}(X)$ because of the appearance of exact seqeunces and alike, so there won't be a nice comparision between K-theory and G-theory. Many other theorems for $K^0$, such as Grothendieck's homotopy-invariance $K^0(X) \cong K^0(X \times \mathbb{A}^1)$ will probably fail for $K^0_{\oplus}$. The upshot is: Although $K^0_{\oplus}$ contains more information, it is not flexible enough for computations.

If $X$ is an affine scheme, the canonical map $K^0_{\oplus}(X) \to K^0(X)$ is an isomorphism. Because of that, many authors define $K_0(R)$ of a ring $R$ to be the free abelian group on iso-classes of f.g. proj. $R$-modules modulo the relation $[P \oplus Q] = [P]+[Q]$, but this should not be seen as the correct definition for the general case of a scheme. In particular I don't agree with your sentence "... the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined."

share|improve this answer

You don't need to go to non-affine schemes to run into trouble with your proposed definition.

In the simplest case, if $R$ is a regular ring and $f$ a non-zero element, you want a long exact localization sequence relating the $K$-theory of $R$ to the $K$-theory of $R[f^{-1}]$. The missing term turns out to be the $K$-theory of the category of finitely generated $R/fR$-modules. In most cases $R/fR$ won't be regular (so you can't just replace this with the category of projective modules), and this category won't have the property that all short exact sequences split. Therefore it matters which $K$-theory you use, and it's the $K$-theory of exact categories that makes the localization theorem true.

share|improve this answer

Your Answer

 
discard

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.