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Hello,

Assume we have $(n+1)$ isometries $S_1,...,S_{n+1}$ in the separable Hilbert space $H$ with the properties that $\sum_{i=1}^{n+1}S_iS_i^*=I, S_i^*S_j=0$ (i.e. $S_i$ are the generators of the Cuntz algebra $O_{n+1}$). In the $C^* $ algebra $C^*(I, S_1,..., S_n)$ consider the closed ideal generated by $P=I-\sum_{i=1}^nS_iS_i^* $. It is easy to prove, using the computation rules in Cuntz algebra, that this ideal is spanned by vectors of the form $S_{\alpha_1}...S_{\alpha_k} P S_{\mu_l}^*...S_{\mu_1}^*$, which we may write in a shortened form as $S_\alpha P S_\mu^*$. It is also easy to see that $S_\alpha P S_\mu^* S_\gamma P S_\beta^*=\delta_{\mu, \gamma} S_\alpha P S_\beta^*$.

Now, in the books and papers I've read regarding the $K_0$ group of $O_n$, the authors say that this means that the elements $S_\alpha P S_\beta^* $ now form a set of matrix units of the ideal generated by $P$ and therefore our ideal is isomorphic to $K(H)$, the ideal of compact generators in $H$ (I guess in the sense of isomorphism of $C^*$ algebras). I have to admit I do not see this isomorphism. I would be really grateful for any help in understanding this part of the proof and/or for a reference where this proof is done in more detail.

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Well, the argument goes roughly as follows. You can define the universal C*-algebra $K$ generated by elements $( e_{i,j} ), i,j\in\mathbb{N}$ with relations $e_{i,j}^*=e_{j,i}$ and $e_{i,j}e_{k,l}=\delta_{j,k}e_{i,l}$. That is, $K$ is a C*-algebra generated by such elements and whenever $A$ is another C*-algebra with a set of elements $(\tilde{e}_{i,j})$

satisfying these relations, then $e_{i,j}\mapsto \tilde{e}_{i,j}$ defines a homomorphism $K\to A$. The existence of such a C*-algebra is a non-trivial question itself, but that's just a fact.

As it turns out, this universal C*-algebra $K$ is simple. The proof is completely analogous to the proof that the compacts are simple. So if $A$ is any C*-algebra generated by elements as above (these are called matrix units), there exists a surjective homomorphism $K\to A$, which is automatically an isomorphism by simplicity of $K$.

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Thank you very much for your answer! –  Walter White Nov 30 '12 at 13:51

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