When can we make $H$ separable in $A\subset B(H)$?

Question

I came across the fundamental result in operator algebras that any $C^*$-algebra can be embedded in some $B(H)$.

Is there some characterization of a $C^*$-algebra (or von Neumann algebra) $A$ such that $H$ can be chosen to be

1. separable?
2. finite dimensional?

How rare/common/special are such situations?

(Apologies if this is a trivial question as I'm not advanced in my studies for this.)

Answer 1

I am only adding details to what has been already stated in the comment section.

If $A$ is finite dimensional, then $A=\bigoplus_{i=1}^nM_{k_i}$, and since $M_k\cong\mathbb{B}(\mathbb{C}^k)$ we have that $A$ can be embedded as the diagonal operators in $\mathbb{B}(\bigoplus_{i=1}^n\mathbb{C}^{k_i})$. On the other hand, if $A\subset\mathbb{B}(\mathcal{H})$ where $\mathcal{H}$ is finite dimensional, then $\mathbb{B}(\mathcal{H})$ is finite dimensional so $A$ is finite dimensional. Thus a $C^*$-algebra admits a faithful representation on a finite dimensional Hilbert space if and only if it is of finite dimension.

The interesting part is question (1). If $A$ is separable for example, then the state space $S(A)$ is separable in the weak-star topology, so, if $\{\sigma_n\}$ is a collection of states that is weak-star dense in $S(A)$, then $\sigma=\sum_{n=1}^\infty\frac{\sigma_n}{2^n}$ defines a faithful state on $A$. It is not hard to verify that the GNS representation associated to $\sigma$, namely $(\mathcal{H}_\sigma,\pi_\sigma)$, is faithful; Also, recall that the GNS space $\mathcal{H}_\sigma$ is defined as the completion of $A$ modulo $\{a\in A:\sigma(a^*a)=0\}\equiv0$ (since $\sigma$ is a faithful state), so $\mathcal{H}_\sigma$ is separable. Thus question (1) is always true for separable $C^*$-algebras.

The converse is not true: not every $C^*$-algebra that is faithfully represented on a separable Hilbert space is separable. This is the easiest thing to see: for example, $\ell^\infty\subset\mathbb{B}(\ell^2)$ as multiplication operators, $$\ell^\infty\ni x:=(x_n)\mapsto M_x:\ell^2\to\ell^2,\;\;\;M_x(y_n)=(x_ny_n)\text{ for all }(y_n)\in\ell^2,$$ but $\ell^\infty$ is well-known to be non-separable. Or, if you prefer, take any $\delta>0$ and an uncountable subset $S$ of $\mathbb{B}(\ell^2)$ so that $\|x-y\|\geq\delta$ for all $x,y\in S$ and simply consider $C^*(S)\subset\mathbb{B}(\ell^2)$; this $C^*$-algebra cannot be separable.