Let $\mathcal{K}(\mathcal{H})$ be the C\*-algebra of compact operators on a Hilbert space $\mathcal{H}$. I am interested in the ($c\_0$-)sum 


$A=\sum \mathcal{K}(\mathcal{H})$ 


of countably many copies of this algebra.


Is it \*-isomorphic to $\mathcal{K}(\mathcal{H})$ itself? Or at least as a Banach space?



$c_0$-direct sum of $\mathcal{K}(\mathcal{H})$
Yes, there is a Banach space isomorphism. The $c\_0$ sum of $\mathcal{K}(H)$ is clearly isometrically isomorphic to its own $c\_0$ sum and contains $\mathcal{K}(H)$ as a norm one complemented subspace, so by the Pelczynski decomposition method it is enough to observe that the $c\_0$ sum of $\mathcal{K}(H)$ embeds into $\mathcal{K}(H)$ as a complemented subspace. Write $H$ as the orthogonal direct sum of orthogonal infinite dimensional subspaces $H\_n$ and $P\_n$ the corresponding orthogonal projections; then $\mathcal{K}(H\_n)\subset \mathcal{K}(H)$ isometrically in an obvious way. Moreover, if $T\_n$ is in $\mathcal{K}(H\_n)$ then $\sum\_{n=1}^N P\_nT\_n P\_n$ in $\mathcal{K}(H)$ has norm the maximum of $\|T\_1\|,\dots \|T\_N\|$. This gives an isometric embedding of the $c\_0$ sum of $\mathcal{K}(H)$ into $\mathcal{K}(H)$. You get a norm one projection onto this subspace of $\mathcal{K}(H)$ by defining $P(T)=\sum P\_nTP\_n$.


