A separable $C^*$ algebra which contains all separable $C^*$ algebras.

Question

Is there a unital separable $C^*$ algebra which unitaly contains all unital separable $C^*$ algebras?

The motivation is that the answer is positive in the commutative case since every compact metric space is the image of the Cantor set.

Answer 1

No.

This follows from a paper of Junge and Pisier from the 90s, called "Bilinear forms on exact operator spaces and $B(H)\otimes B(H)$". Here is the arXiv link to this. I will try to sketch the argument:

For operator spaces $E,F$, consider the completely bounded Banach-Mazur distance of these, defined as $d_{cb}(E,F):=\inf\{\|u\|_{cb}\cdot\|u^{-1}\|_{cb}: u:E\to F\text{ is an isomorphism of Banach spaces}\}$. Recall the definition of the completely bounded norm: if $u:E\to F$ is a linear map, then put $u^{(n)}:M_n(E)\to M_n(F)$ for the $n$-th amplification of $u$. We define $\|u\|_{cb}:=\sup_{n\in\mathbb{N}}\|u^{(n)}\|$. Also define $\delta_{cb}(E,F):=\log(d_{cb}(E,F))$.

Fix $n>2$.

Consider $OS_n$ to be the set of equivalence classes of $n$ dimensional operator spaces, where we identify two $n$-dimensional spaces $E$ and $E'$ if they are completely isometric. Then, $(OS_n,\delta_{cb})$ is a non-separable metric space (see theorem 2.3).

Now let $A$ be a separable $C^*$-algebra. Then the set $S_n(A)$ of all $n$-dimensional subspaces of $A$ (where again we identify completely isometric spaces) is a separable subspace of $(OS_n,\delta_{cb})$ (see proposition 2.6).

Suppose that $A$ is a separable $C^*$-algebra so that all unital separable $C^*$-algebras unitally embed in $A$. Then, given any $n$-dimensional operator space $E$, we can embed it in some separable $C^*$-algebra, which in turn embeds unitally inside $A$. But this implies that $S_n(A)=OS_n$; a contradiction to the separability / non-separability statements above.