Ideal spanned by matrix units isomorphic to compact operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:06:50Z http://mathoverflow.net/feeds/question/114539 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114539/ideal-spanned-by-matrix-units-isomorphic-to-compact-operators Ideal spanned by matrix units isomorphic to compact operators Walter White 2012-11-26T15:23:26Z 2012-11-28T15:51:59Z <p>Hello,</p> <p>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 $S_\alpha P S_\mu^* S_\gamma P S_\beta^*=\delta_{\mu, \gamma} S_\alpha P S_\beta^*$.</p> <p>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.</p> http://mathoverflow.net/questions/114539/ideal-spanned-by-matrix-units-isomorphic-to-compact-operators/114780#114780 Answer by Gabor Szabo for Ideal spanned by matrix units isomorphic to compact operators Gabor Szabo 2012-11-28T15:51:59Z 2012-11-28T15:51:59Z <p>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 $(\tilde{e}_{i,j})$ </p> <p>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.</p> <p>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 $K$. </p>