2
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $K$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.