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Recall that a real C*-algebra is a Banach $\ast$-algebra $A$ over $\mathbb{R}$ which satisfies the standard C* identity and which also has the property that $1 + a^{\ast}a$ is invertible in the unital …

Here I am referring to Kasparov's KK-theory, a bivariant functor on the category of separable C* algebras. It is well known that $KK(A, \mathbb{C})$ is K-homology and $KK(\mathbb{C}, B)$ is K-theory, …

The main problem with dual algebras for non-separable C*-algebras is that they need not be functorial. Given representations $\rho_A \colon A \to \mathcal{B}(H_A)$ and $\rho_B \colon B \to \mathcal{B …

Before posing my question, let me provide a little background since the Wikipedia page on this stuff is sorely lacking. Let's start with the classical case of the Toeplitz index problem on the circle …

Let $G$ be a discrete group and consider the reduced group C* algebra $C_r^\ast(G)$, viewed as an algebra of bounded operators on $\ell^2(G)$ by the regular representation. The canonical trace on $C_ …

Maybe it will help to see how the algebra $B$ is a special case of the general construction in $2.\alpha$. The compact manifold is the space $Y = \{a, b\}$ consisting of two points. The open cover o …

The answer is basically "yes, because the definitions are rigged to make it so". The point is that you have to be careful both with C*-algebra K-theory in the non-unital case and with topological K-t …

Perhaps the simplest answer is to use the spectral theorem: $L^2(S)$ decomposes as the orthogonal direct sum of $D$-eigenspaces, and $f(D)$ acts on each $\lambda$-eigenspace as multiplication by $f(\l …

KK-theory provides one way to eliminate the asymmetry in the definitions of K-theory and K-homology. A cycle in $KK(A, B)$ is a triple $(H, \rho, F)$ where $H$ is a (adjectives) Hilbert $B$-module, $ …

There are a number of ways to do this calculation, but at risk of shamelessly plugging my own work there is a nice way to see it using a Mayer-Vietoris principle. Decompose $\mathbb{R}$ as the union …

I'm can't claim to have studied the relevant history in a lot of detail, but count me a skeptic of Landsman's claim. Let's take this little paper and the companion that it cites as a test case, which …

One common way to quantify non-commutativity which is especially popular in operator algebra theory and non-commutative geometry is to use the Schatten norms. Given a bounded operator $T$ on a separa …