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I am trying to show that the $K$-theory groups of the following $C^*$-algebra $A$ vanish:

Let $\mathcal{H}$ be a separable Hilbert space. Now consider the subalgebra of $\mathcal{B}(\mathcal{H}\oplus\mathcal{H})$ given by those matrices

\begin{pmatrix} T_{11} & T_{12} \\ T_{21} & T_{22} \end{pmatrix}

where $T_{11}, T_{12}$ and $T_{21}$ are compact and $T_{22}$ is arbitrary but bounded. Schematically this is the algebra

$$A = \begin{pmatrix} \mathcal{K} & \mathcal{K} \\ \mathcal{K} & \mathcal{B} \end{pmatrix} $$

where $\mathcal{K}$ denotes the compact operators of $\mathcal{H}$.

As is said I want to show $K_0(A) = 0 = K_1(A)$.

Unfortunately I do not really know what tools to use here. Any suggestions are appreciated.

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+1 for a question on k-theory. There are not that many. –  Raskolnikov Apr 25 '12 at 17:20
    
to bad there aren't. but thanks anyway :) –  mland Apr 25 '12 at 18:24

1 Answer 1

up vote 6 down vote accepted

Hints: Consider the ideal $$I = \begin{pmatrix} \mathcal{K} & \mathcal{K} \\ \mathcal{K} & \mathcal{K} \end{pmatrix}=\mathcal{K}\otimes\mathbb M_2\mathbb C\cong\mathcal{K} $$ in $A$. What is $A/I$? Write down the six-term exact sequence in K-theory for the extension $$ 0\to I\to A\to A/I\to 0. $$ Plug in what you know about the K-theory of $I$ and $A/I$. The desired result follows.

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of course. thanks. –  mland Apr 25 '12 at 18:23

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