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Topological K-theory is usually defined by setting $K(X)$ to be the groupification of the monoid $Vect_\mathbb{C}(X)$ of complex vector bundles over $X$ (with addition given by Whitney sum). However, we can alternatively declare that $[B]\sim [A]+[C]$ whenever $0\rightarrow A \rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence of vector bundles over $X$ (morphisms are required to have locally constant rank): certainly if $B\cong A\oplus C$ then we have such a sequence, and in the other direction we can take a metric on $B$ and identify $C$ with $A^\perp \subseteq B$.

You can take the $K$-theory of any abelian category using this second definition. So, I'm curious to know if people do this for the category of holomorphic vector bundles over a complex manifold. The above splitting construction no longer works since it uses partitions of unity, so assuming we use this more general definition we'd get more equivalence relations. On the other hand, there's all this funny business going on with vector bundles topologically but not holomorphically isomorphic, which means that $K_{hol}(X)$ wouldn't just be a subquotient of $K(X)$. So in the end, I'm not sure whether I should expect this to be a more or less tractable sort of object.

I'm told that the Chow ring might have something to do with this, but the wikipedia page seems to indicate that it's more analogous to singular cohomology than anything else.

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If X is a compact algebraic variety, then Serre's GAGA paper gives you an isomorphism from the algebraic K-theory of X to the holomorphic K-theory of X. – Steven Landsburg Mar 30 '11 at 5:46
People definitely use this (at least for holomorphic v.b. over algebraic varieties). And this gives the setup for Grothendieck Riemann-Roch. On a projective smooth variety K for holomorphic vector bundles is the same as K for the category of coherent sheaves. – Yuhao Huang Mar 30 '11 at 8:02
There is a paper by Ralph Cohen and Paulo Lima Filho: This might also be of intererest: It seems that they study a slightly different object (holomorphic bundles that admit a holomorphic bundle map to the tautological bundle on the Grassmannian). – Johannes Ebert Mar 30 '11 at 8:32
As heretical as this sounds, your non-wikepedia source is correct, it is close to the Chow ring. More precisely, if $X$ is a projective manifold, then what you're calling $K_{hol}(X)$ is rationally the same as Chow. – Donu Arapura Mar 30 '11 at 12:53
@Donu: That's very cool. It sounds like a deep theorem, though; is this a statement about some analogue of the Chern character? – Aaron Mazel-Gee Apr 2 '11 at 7:12
up vote 23 down vote accepted

Grothendieck proved that there is an analytification functor $X \mapsto X^{an}$ from schemes locally of finite type over $\mathbb C$ to the category of (non-reduced!) analytic spaces, which is fully faithful when restricted to proper schemes. This induces isomorphisms from $K-$ groups in the algebraic sense on $X$ to $K-$ groups in the holomorphic sense on $X^{an}$ . This is just a mild generalization of Serre's GAGA principle proved for reduced, projective varieties. So this settles your problem in the compact algebraizable case, by telling algebraic geometers to solve it ( and they actually know quite a lot of the K-theory of schemes !)

In the diametrically opposed case of Stein manifolds, a landmark theorem of Grauert also answers your request. Namely, given a complex manifold there is an obvious forgetful functor $Vecthol(X) \to Vecttop(X)$ from isomorphism classes of holomorphic vector bundles on $X$ to isomorphism classes of topological vector bundles on the underlying topological space $X^{top}$. If $X$ is Stein, Grauert proved that the functor is an isomorphism of monoids : every topological vector bundle has a unique holomorphic structure. ( Results of this nature fit into what is called the "Oka principle". ) There are no extension problems for short exact sequences $0 \to \mathcal E \to \mathcal F \to \mathcal G \to 0$ because they all split: in the Stein case thanks to theorem B and in the topological case because of partitions of unity (theorem B in disguise, actually: fine sheaves are acyclic). So in the Stein case too you can relax and ask topologists to do your work .

Finally, there are complex manifolds between these extreme cases. I am not aware of a general theory there ( of course that proves nothing but my ignorance) . This looks like an interesting topic of investigation, especially in view of Winkelmann's theorem ( link to survey here) that on every compact holomorphic manifold of positive dimension $n $ there exists a non-trivial holomorphic vector bundle of rank $\leq n$.

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Thank you, this is very interesting! I really like this philosophy of complex manifolds bridging the gap between algebraic topology and algebraic geometry, it's been coming up a lot for me recently. – Aaron Mazel-Gee Apr 2 '11 at 6:45
Technically, it's not correct to say that there's an equivalence of Groupoids $Vecthol(C) \cong Vecttop(X)$ since this would imply that every automorphism of topological bundles $E\to E$ of a holomorphic bundle would already respect the holomorphic structure, which is clearly not the case. Hence you can merely say that we've got a bijection of sets. – Dominik Jul 11 '15 at 10:27
@Dominik You are absolutely right: thanks a lot for your comment. I corrected my misstatement – Georges Elencwajg Jul 11 '15 at 10:55

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