Quasi-Reflexive C*-Algebras

Question

Let $A$ be a C*-algebra. When is it the case that $A$ is quasi-reflexive? i.e. when is $A^{**}/A$ finite dimensional? Of course, this is true when $A$ is finite dimensional itself, but is it true for some infinite dimensional $A?$ If yes, are there conditions guaranteeing when it is definitely not true?

It seems to be generally untrue. Take $A=K(H).$ Then $A^{**}=B(H),$ and so $A^{**}/A$ is the Calkin algebra, which is huge.

Note: I know the James space, which has codimension 1 in its bidual, can be made a Banach algebra. However, it is not $C^*.$

Answer 1

Claim: a $C^*$-algebra $A$ is quasi-reflexive if and only if it is finite-dimensional.

Proof.

For an inclusion of Banach spaces $\iota: X \subset Y$, we have a canonical map $X^{**}/X \to Y^{**}/Y$ given by $\chi +X \mapsto \iota^{**}(\chi)+Y$. This map is well-defined linear and injective as one can easily check.

Let $A$ be a unital C$^*$-algebra. If $A$ is infinite-dimensional, then $A$ contains a self-adjoint element with infinite spectrum. By considering the positive-negative part decomposition, we see that $A$ contains a positive element with infinite spectrum. By adding a scalar of the unit, we conclude that $A$ contains an invertible positive element $a\in A_+$ with infinite spectrum $K:= \sigma(a) \subset (0,\infty)$, and by Gelfand duality we have that C$^*(a)\cong C(K)$. By the preceding observation applied to C$^*(a)$ as $X$ and $A$ as $Y$, we must have that $C(K)^{**}/C(K)$ is finite-dimensional.

Claim. If $C(K)^{**}/C(K)$ is finite-dimensional, then $K$ is a finite set.

Proof of claim. Note that $C(K)^*$ is the space of complex Borel measures on $K$. Denoting by $B_\infty(K)$ the space of bounded Borel measurable functions $K\to\mathbb{C}$ with the supremum norm, we see that we have an embedding $B_\infty(K) \subset C(K)^{**}$ given by $B_\infty(K)\ni u\mapsto \gamma_u\in C(K)^{**}$ where $\gamma_u(\mu)=\int_Xud\mu$. Note that this embedding extends the canonical embedding $C(K)\subset C(K)^{**}$, so if $C(K)^{**}/C(K)$ is finite-dimensional, then $B_\infty(K)/C(K)$ is finite-dimensional. If $K$ is infinite this is impossible: Say $\dim(B_\infty(K)/C(K))=d\in\mathbb{N}$. Let $(x_n)_{n=1}^\infty\subset K$ be a sequence of discrete points converging to some $\bar{x}\in K$. Consider the functions $f_1,\dots,f_{d+1}\in B_\infty(K)$ defined to be $0$ everywhere except on the sequence $(x_j)_{j=1}^\infty$, and defined as $$f_k(x_j)=\begin{cases}1,\quad j \equiv k-1 \mod (d+1) \\0,\quad\text{else}\end{cases},\quad k=1,\dots d+1; j\in\mathbb{N}.$$ The functions $f_k+C(K)$ are linearly independent. Indeed, assume that $\sum_{k=1}^{d+1}\lambda_kf_k+C(K)=0$ for some complex numbers $\lambda_k$, i.e. the function $x\mapsto\sum_{k=1}^{d+1}\lambda_kf_k(x)$ is continuous. In particular, $$\lim_{j\to\infty}\sum_{k=1}^{d+1}\lambda_kf(x_j)=:\lambda$$ exists. We then see that $\lambda=\lambda_1=\dots=\lambda_{d+1}=0$ by looking at the convergence via the subsequences of different residues mod $d+1$.

This shows that any unital, infinite-dimensional C$^*$-algebra $A$ gives an infinite-dimensional quotient $A^{**}/A$.

Now if $A$ is infinite-dimensional and non-unital, we consider the unitization $A^\sim$ which is isomorphic to $A\oplus\mathbb{C}$ as vector spaces and the fact that $(A^\sim)^{**}\cong A^{**}\oplus\mathbb{C}$ so as to see that $\dim((A^\sim)^{**}/A^\sim) = \dim (A^{**}/A)$ and deduce that the same is true for non-unital C$^*$-algebras too.

Answer 2

Based on OP's follow-up question: one can actually even show that $A^{**}/A$ is separable if and only if $A$ is finite-dimensional.

Assume that $A$ is infinite-dimensional and that $A^{**}/A$ is separable. Find $B \subset A$ a separable, infinite-dimensional sub-$C^*$-algebra. We have $B^{**}/B\subset A^{**}/A$ (as in the other answer) so $B^{**}/B$ is separable. Since $B$ and $B^{**}/B$ are separable, we can conclude that $B^{**}$ is separable. Since $B^{**}$ is a von Neumann algebra, (norm) separability implies that $B^{**}$ is finite-dimensional, hence $B$ is finite-dimensional, as $B\subset B^{**}$. This is a contradiction.