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 $W$.



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> 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 $B(H)$?
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> 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 $B(H)$?
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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 $B(\ell^2)$:


$$\{P(S)\mid \text{P is a polynomial of degree at most n}\}.$$



Fredholm $C^*$-algebras
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.


