TITLE: $c_0$-direct sum of $\mathcal{K}(\mathcal{H})$
QUESTION [7 upvotes]: 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?

REPLY [11 votes]: 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 $P(T)=\sum P_nTP_n$.