Quotients of finite C-star algebras: are they finite?

Question

As I was trying to answer this question, I thought of the following: If $A$ is a unital, finite $C^*$-algebra and $I$ is an ideal in $A$, then is $A/I$ a finite $C^*$-algebra?

Recall that a unital $C^*$-algebra is called finite when the implication $$p\sim_{M-vN}1_A\implies p=1_A$$ holds, i.e. every isometry in $A$ is a unitary. Besides the main question, I realized something: I do not know what $C^*$-algebras are finite or not. For example, I know that abelian $C^*$-algebras are stably finite, as it is (briefly) discussed in this post, and finite dimensional $C^*$-algebras are (stably) finite too, but that's all I know. So a bonus question is: what are some interesting examples of finite $C^*$-algebras besides abelian and finite dimensional ones?

Note that in the answer of the linked question I am using a particular feature of the ideal $\sum_nA_n$ in $\prod_nA_n$, something that we cannot do in the general case.

Answer 1

Consider $$ A = \{f\in C([0, 1], \mathcal T): f(0)\in \mathbb C\cdot 1\}, $$ where $\mathcal T$ is the Toeplitz algebra. Every isometry $f$ in $A$ gives rise to a path of isometries in $\mathcal T$ beginning with a scalar. Observing that all isometries in the Toeplitz algebra have finite Fredholm index, and also that the index is continuous, we can easily show that $A$ is finite. However the map $$ f\in A\mapsto f(1)\in \mathcal T $$ is a surjection from $A$ to a non-finite algebra.

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EDIT. Here is an argument to show the finiteness of $A$ which does not rely on the theory of Fredholm operators:

Lemma. Let $B$ be a unital C*-algebra. Denote by $\mathscr I(B)$ the set of all isometries in $B$, and by $\mathscr U(B)$ the subset formed by the unitary elements. Then $\mathscr U(B)$ is clopen in $\mathscr I(B)$.

Proof. It is well known that the set $\text{GL}(B)$ formed by all invertible elements is open. Since $$ \mathscr U(B) = \mathscr I(B) \cap \text{GL}(B), $$ it follows that $\mathscr U(B)$ is open in $\mathscr I(B)$. To show that $\mathscr U(B)$ is also closed, suppose by contradiction that the sequence $\{u_n\}_n$ of unitaries converges to a proper isometry $s$ (meaning that $ss^*\neq 1$). The element $$ p = 1-ss^* $$ is then a non-trivial projection and, clearly, $ps=0$. We then deduce that $$ 0 = ps = \lim_{n\to \infty }pu_n, $$ but this is a contradiction since $\|pu_n\|=\|p\| = 1$. QED

This said, suppose that $f$ is an isometry in $A$, so we have that $f(t)$ lies in $\mathscr I(\mathcal T)$, for all $t$ in $[0, 1]$.

Since $f(0)$ is necessary a scalar, we have that $f(0)\in \mathscr U(\mathcal T)$, but then $f(t)\in \mathscr U(\mathcal T)$, for every $t$ because $[0,1]$ is connected.

This shows that $f$ is unitary and hence that $A$ is finite.

Answer 2

There is probably an obvious example laying around, but I can't think of one! Here is an overkill answer showing that quotients of finite C*-algebras need not be finite.

There is a characterization of separable exact C*-algebras as subquotients (quotient of a subalgebra) of the CAR algebra, i.e., the UHF algebra $M_{2^{\infty}}$. This algebra is finite and so is any subalgebra since the faithful trace on the CAR algebra restricts to a faithful trace on any smaller algebra. So take $B \subseteq M_{2^{\infty}}$ ($B$ unital) and $I \lhd B$ such that $B/I \simeq \mathcal{O}_2$, the Cuntz algebra generated by 2 isometries. The Cuntz algebra is (purely) infinite.

There is a nice description of finiteness in unital C*-algebras (see Left Invertibility Implies Right Invertibility in Certain $C^*$-algebras). Some examples are UHF algebras (like the CAR algebra), the irrational rotation algebra, the Jiang-Su algebra, etc.

References: This was originally proved in "On subalgebras of the CAR-algebra" by Kirchberg, and later Wassermann gave a "simpler" (take with a grain of salt) proof in "Subquotients of UHF C*-algebras".