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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?

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    $\begingroup$ For the answer to your first question: note that $K(H)$ has no non-trivial, closed, two-sided ideals, while $A$ has many. The second question is harder (and to me at least more interesting) - perhaps one can apply some version of the Pelczynski decomposition method? $\endgroup$
    – Yemon Choi
    Commented Dec 6, 2011 at 22:49

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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 $P(T)=\sum P_nTP_n$.

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  • $\begingroup$ As ever, Bill Johnson puts my own laziness into sharp relief. It looks like this should also give completely bounded isomorphism. $\endgroup$
    – Yemon Choi
    Commented Dec 6, 2011 at 23:52

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