Why is $\overline{B(l^2)\odot B(l^2)}^{\| \enspace \|_{op}}\neq B(l^2\otimes l^2)?$

Question

Let $B(l^2)$ be the $C^*$algebra of bounded linear operators on the sequence space $l^2$ and denote with $B(l^2)\odot B(l^2)$ the tensor product of $B(l^2)$ with itself, considered as a $*$algebra (with component-wise multiplication and involution). It is $B(l^2)\odot B(l^2)\subset B(l^2\otimes l^2)$. My question is: how to prove that $$\overline{B(l^2)\odot B(l^2)}^{\| \enspace \|_{op}}\neq B(l^2\otimes l^2)?$$ Here $\| \enspace \|_{op}$ denotes the operator norm on $B(l^2\otimes l^2) $.

First I tried to find an operator $T\in B(l^2\otimes l^2)$ such that there is no sequence $(X_n)\subseteq B(l^2)\odot B(l^2)$ with $\|X_n-T\|_{op}\to 0$ for $n\to \infty$; but I have no idea for an example.

Similary, an other strategy is: If you have an idea of a non-zero continuous function $f:B(l^2\otimes l^2)\to X$ into a suitable space $X$, maybe $X=\mathbb{C}$, such that $f$ vanishes on $B(l^2)\odot B(l^2)$. This would prove that $B(l^2)\odot B(l^2)$ is not dense in $B(l^2\otimes l^2)$ with respect to $\|\enspace \|_{op}$, too. But I have no idea for such an $f$. Could you help me?

Regards

Answer 1

Your two ideas would have been my first attempts; but I have no idea how to make them work (well, for the first one, I would try with $T$ the flip, but I still wouldn't know how to do it).

Let $K\subset B(\ell^2)$ be the compact operators. On $\overline {B(\ell^2)\odot B(\ell^2)}$, consider the ideals $\overline{K\odot B(\ell^2)}$ and $\overline{K\odot K}$. Let $\psi$ be a faithful state on $K$ extended via Hahn-Banach to a state of $B(\ell^2)$, and $\phi$ a state on $B(\ell^2)$ such that $\phi|_K=0$.

Consider the states $\phi\otimes\psi$ and $\psi\otimes\phi$. The state $\phi\otimes \psi$ is nonzero on $\overline {B(\ell^2)\odot B(\ell^2)}$, but it is zero on $\overline{K\odot B(\ell^2)}$. And the state $\psi\otimes\phi$ is nonzero on $\overline{K\odot B(\ell^2)}$ but it is zero on $\overline{K\odot K}$. It follows that the inclusions $$ \overline{K\odot K}\subset \overline{K\odot B(\ell^2)}\subset \overline{B(\ell^2)\odot B(\ell^2)} $$ are both proper. This means that the C$^*$-algebra $\overline {B(\ell^2)\odot B(\ell^2)}$ has two different proper closed ideals, and so it cannot be isomorphic to $B(\ell^2\otimes\ell^2)$.

Answer 2

Martin gave a beautiful answer to your question but let me take this opportunity to advertise a solution to a more general question of which C*-algebras may be written as tensor products of two infinite-dimensional C*-algebras. In particular, it is proved that $B(\ell_2)$ as well as the Calkin algebra cannot be decomposed like that no matter which tensor norms and infinite-dimensional C*-algebras you choose.