It is in some sense folklore that given two arbitrary abelian groups $G,H$ one can find a $C^*$ algebra $A$ such that $K_0(A)=G$ and $K_1(A)=H$. My question is the following: what is known in the case of commutative $C*$-algebra? Which groups can be obtained as $K$ groups of commutative $C^*$ algebra (in other words, $K$ groups for topological spaces)?
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8$\begingroup$ In "Introduction to K-theory for C*-algebras" by Rordam, et.al. in the remark after Exercise 13.2 they mention that all pairs of countable abelian groups can arise as $K_*$ for some abelian C*-algebra. They explicitly point out how to do it in the finitely generated case. My K-theory knowledge is too weak to be of any more use. $\endgroup$– Caleb EckhardtCommented Jun 7, 2014 at 1:49
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$\begingroup$ @truebaran I think that a finite group can Not be isomorphic to $K^{0}(X)$ for a topological space, since $K^{0}(X)$ must contain $\mathbb{Z}$ as a summand(subgroup). So there are some restriction. So a question: Assume that $\mathbb{Z}$ is a summand of an abelian group $G$. Is $G$ isomorphic to $K^{0}(X)$ for some $X$? $\endgroup$– Ali TaghaviCommented Nov 30, 2014 at 18:50
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$\begingroup$ @truebaran Are you considering complex K theory or real K theory? $\endgroup$– Ali TaghaviCommented Nov 30, 2014 at 19:04
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$\begingroup$ Unless you're specifically dealing with real $C^\ast$-algebras, everything in sight will be over the complex numbers. $\endgroup$– Branimir ĆaćićCommented Dec 1, 2014 at 10:00
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2 Answers
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Any pair of abelian groups arises, up to isomorphism, as the $K$-theory groups of a commutative $C^*$-algebra. If the groups were countable, the $C^*$-algebra can be chosen to be separable. This follows from Corollary 23.10.3 in Blackadar's $K$-theory book.
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Using the Chern Character you can solve the problem up to torsion.
For a CW complex X we have $K_0(C(X)) \otimes \mathbb{Q} \cong \bigoplus_n H_{2n}(X; \mathbb{Q})$ and $K_1(C(X)) \otimes \mathbb{Q} \cong \bigoplus_n H_{2n+1}(X; \mathbb{Q})$.
It's pretty easy to find CW complexes with arbitrary rational homology groups by taking wedge sums of spheres of varying dimensions.
The torsion subgroups may be harder, and the order structure on $K_0$ is notoriously tricky for $C(X)$.
