# What properties of a C*-algebra are reflected in their $K_0$ groups?

Title says it all. I've been looking at K-groups of a few C* algebras. My very rough understanding is that these groups reflect (in some appropriate sense) the algebraic structure that we get when we take direct sums of projective modules. This much I have understood by calculating the K0 groups of the complex numbers, complex polynomials in one variable, and cuntz algebras. In the first two cases, the only projective modules are free, so the semi group we get is the natural numbers. I know for cuntz algebras, weirder things happen when you take direct sums of projectives.

I would appreciate if someone could illuminate exactly what the K0 group of a C*-algebra tells us about the structure of that algebra.

• This is a broad question. For $C^{\ast}$-algebras, the ordered $K_{0}$-group is an important invariant. The order unit tells apart $M_{2}(\mathbb{C})$ from $M_{4}(\mathbb{C})$. Scaled ordered $K_{0}$-group is sufficient for classification of $AF$-algebras. For more complicated $C^{\ast}$-algebras, we need to consider $K_{1}$-group. – user78800 May 29 '13 at 22:00
• @user78800 I think when I asked this question some years ago I was under the false illusion that the K-theory of the $C^{*}$ algebra should tell you something specific about the algebra, as opposed to being a coarse algebraic invariant. – jmracek Jan 14 '15 at 1:03
• I'm happy about your question, since I'm having the same "problems" now – Sabrina G. Oct 13 '16 at 20:03

Recall that the commutative Gelfand-Naimark theorem, commutative C*-algebras are precisely the C*-algebras $C(X)$ of continuous complex-valued functions on compact Hausdorff spaces $X$, namely their spectra. So invariants of $C(X)$ are telling us something about the spaces $X$. What does K-theory tell us in particular?
By Swan's theorem, finitely generated projective $C(X)$-modules are the same thing as vector bundles over $X$, so $K_0(C(X))$ turns out to be precisely the topological (complex) K-theory $K^0(X)$ of $X$, and similarly $K_1(C(X))$ turns out to be precisely $K^1(X)$.
Topological K-theory is historically the first example of a generalized cohomology theory, and as such it contains information about spaces $X$ similar to the information given by singular cohomology. In fact the Chern character isomorphism implies that
$$K^0(X) \otimes \mathbb{Q} \cong \prod_n H^{2n}(X, \mathbb{Q})$$ $$K^1(X) \otimes \mathbb{Q} \cong \prod_n H^{2n+1}(X, \mathbb{Q})$$
so, after tensoring with $\mathbb{Q}$, topological K-theory encodes roughly the same information as ordinary cohomology, except that it forgets the $\mathbb{Z}$-grading, retaining only a $\mathbb{Z}_2$-grading. With stronger tools like the Atiyah-Hirzebruch spectral sequence you can attempt to compute $K^0(X)$ itself by relating it to cohomology as well.