Fredholm $C^*$-algebras Let $H$ be a Hilbert space. A vector subspace $W\subset B(H)$ is called a Fredholm subspace if there is an upper bound for the absolute value of  Fredholm index of all Fredholm operators $T$ in $W$.

Is there a classification of all $C^*$-algebras $A$ which admit an irreducible representation $\phi:A \to B(H)$ in some Hilbert space $H$ such that $\phi(A)$ is a Fredholm subspace of $B(H)$?


Is there a classification of all $C^*$-algebras $A$ which admit a faithful  representation $\phi:A \to B(H)$ in some Hilbert space $H$ such that $\phi(A)$ is a Fredholm subspace of $B(H)$?

One can consider the terminology "Fredholm algebra" for any such $C^*$-algebras.
Edit: We add an example according to comment by Yemon Choi.
Put $H=\ell^2$ let $S$ be the shift operator on $\ell^2$ and $n$ be a fixed integer. Then this is a finite dimensional Fredholm subspace of $B(\ell^2)$:
$$\{P(S)\mid \text{P  is  a polynomial of degree at most n}\}.$$
 A: There's a trivial answer to the second question: every C${}^*$-algebra has such a representation. Wlog assume $A \subseteq B(H_0)$ for some Hilbert space $H_0$, then represent $A$ on $H_0 \otimes l^2$ by tensoring everything with the identity on $l^2$. All the Fredholm operators in this representation have index $0$ (in fact they would have to be invertible).
A: Any unital $C^*$-algebra $A$ has an irreducible representation $\phi$ such that every Fredholm operator in $\phi(A)$ has index 0.
To see it, let me first repeat something from Nik Weaver's previous answer: if $\pi$ is a representation of $A$ such that $\pi(A)$ intersects the compact operators trivially, then any Fredholm operator in $\pi(A)$ is actually invertible.
Now to prove my claim, observe that we may assume that $A$ is simple (up to replacing $A$ by $A/I$ where $I$ is a maximal, proper, closed two-sided ideal). Now let $\phi$ be any irreducible representation of $A$. If $\phi$ is finite-dimensional, then the result is clear and was already mentioned. If $\phi$ is infinite-dimensional then $\phi(A)$ intersects the compacts trivially because of simplicity, so Nik's observation above applies.
