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If I take $\mathcal{A}$ = coherent sheaves on $X$* up to isomorphism, then there are two things I could do which come to mind.

The first is noticing that $(\mathcal{A},\oplus)$ is a monoid and subsequently applying Grothendieck's $K$-functor to it obtaining a group $K(\mathcal{A})$.

The second is to take the free abelian group generated by $\mathcal{A}$ quotiented out by relations

$B = A + C$ for any exact sequence $$0 \to A \to B \to C \to 0$$ obtaining a second group $L(\mathcal{A})$.

Are these two groups $K(\mathcal{A})$ and $L(\mathcal{A})$ isomorphic? If not, which one is the 'correct' one?

*I'm not sure what $X$ should stand for. I'd be happy to assume smooth and projective variety over $\mathbb{C}$, or maybe just locally noetherian scheme, or maybe just locally ringed space or etc.

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This is pretty standard, but amazingly the standard online refs do not seem to have clear answers in obvious places. Answers are "No" and "L(A)" respectively. For fun, one can try computing the groups when $X$ is the projective line. –  t3suji Jul 7 '10 at 14:35
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To see that they're not the same, consider the toy example $R=k[t]/(t2)$. There are only two indecomposable f.g. modules, namely $R$ itself and the residue field $k$. If you're killing relations from exact sequences, you have $[R]=2[k]$ so $L$ is $\mathbb{Z}$. If you're only killing direct−sum relations then you get $\mathbb{Z}^2$. –  Graham Leuschke Jul 7 '10 at 17:50
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1 Answer 1

up vote 6 down vote accepted

The Grothendieck group on $X$ is defined (in your notation) as $K(\mathcal{B})$, where $\mathcal{B}$ is the monoid of f.g. locally free sheaves on $X$, (i.e. vector bundles on $X$.) The group $L(\mathcal{A})$ is usually denoted as $G(X)$, or $K'(X)$. If $X$ is smooth, the embedding $\mathcal{B} \subset \mathcal{A}$ induces an isomorphism from $K(\mathcal{B})$ to $L(\mathcal{A})$ (i.e. from $K_0(X)$ to $G(X)$.)

Chuck Weibel's notes for his K-theory book (which can be found on his website) is a good place find some of this information.

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Thanks for the reference. –  babubba Aug 1 '10 at 17:02
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