MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – Yemon Choi Dec 6 '11 at 22:49
up vote 10 down vote accepted

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 $P(T)=\sum P_nTP_n$.

share|cite|improve this answer
Brilliant, thank you. – Habujew Dec 6 '11 at 23:35
As ever, Bill Johnson puts my own laziness into sharp relief. It looks like this should also give completely bounded isomorphism. – Yemon Choi Dec 6 '11 at 23:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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