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Suppose $T: c_{00} \to c_{00}$ is a linear map such that, when regarded as an infinite matrix, there is a uniform bound on the $\ell_1$-norms of its columns, and a uniform bound on the $\ell_1$-norms of its rows. That is, $\newcommand{\Bdd}{{\mathcal B}}\newcommand{\Cst}{{\rm C}^*}$ $\sup_j \sum_i |T_{ij}| <\infty$ and $\sup_i \sum_j |T_{ij}| < \infty \;. $ One can show that if $C$ is the maximum of these two sups, then $\Vert T(\xi)\Vert_p\leq C\Vert\xi\Vert_p$ for all $\xi\in c_{00}$ and all $p\in [1,\infty]$.

Now let $B_{00}$ be the set of all such $T$; this is a subalgebra of ${\rm Lin}(c_{00})$ and by abuse of notation we can regard it as a subalgebra of every $\Bdd(\ell_p)$. In particular it's a $*$-subalgebra of $\Bdd(\ell_2)$, so its norm closure $B_2$ is a $\Cst$-subalgebra of $\Bdd(\ell_2)$.

The algebra $B_2$ is currently somewhat mysterious to me, and I am hoping that people reading MO may recognize it, or know of it under some different name. Note that $B_2$ is not all of $\Bdd(\ell_2)$, because there are operators in $\Bdd(\ell_2)$ that cannot be approximated in norm by $\ell_1$-preserving operators (cf. this old MO question).

If $G$ is any countable discrete group then, regarding $B_2$ as a subalgebra of $\Bdd(\ell_2(G))$, a quick check shows that it contains $\lambda(\ell_1(G))$ and hence contains $\Cst_r(G)$. Since $\Cst_r(G)$ need not be exact, $B_2$ can't be exact (and hence, as was pointed out in the comments, it can't have the cb approximation property).

Q1. Does $B_2$ contain a copy (up to c.b. isomorphism, say) of the $\Cst$-algebraic direct product $\prod_k M_k({\bf C})$? (This would imply that $B_2$ is not an exact $\Cst$-algebra, although we knew this already.)

Q2. Is there some identification of $B_2$ with a "known type" of $\Cst$-algebra? E.g. some kind of crossed product of the form $\ell^\infty(X)\rtimes G$ for some group $G$ acting on a countable set $X$?

Q3. Does $B_2$ have the (metric) approximation property?

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  • $\begingroup$ Probably, you mean $\leq 1$ instead of $< \infty$. $\endgroup$ Jan 1, 2015 at 19:59
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    $\begingroup$ Nice questions. I guess that no one has considered any of them. I asked Szankowski to look at Q3 in case an "easy" modification of his argument for $B(H)$ would show that $B_2$ fails the AP. $\endgroup$ Jan 1, 2015 at 20:09
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    $\begingroup$ Note that if $B_2$ contains a sub-C*-algebra isomorphic to $X=(\bigoplus_{n=1}^\infty M_n)_{\ell_\infty}$ then $B_2$ lacks the AP. Indeed, $X$ is Banach-space isomorphic to $\mathscr{B}(\ell_2)$ and the latter space lacks the AP by a result of Szankowski. On the other hand, $X$ is injective in the category of operator systems, so in particular $B_2$ would contain a complemented subspace without the AP. $\endgroup$ Jan 1, 2015 at 20:26
  • $\begingroup$ I think you already answered parts of Q1 and Q3. You mention that $B_2$ has a non-exact subalgebra. Since exactness passes to subalgebras, $B_2$ can't be exact. Since exactness implies CBAP, $B_2$ cannot have the CBAP. $\endgroup$ Jan 2, 2015 at 5:03
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    $\begingroup$ That's because I said something wrong. CBAP implies exactness. $\endgroup$ Jan 2, 2015 at 5:06

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