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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 $B(\ell^2)$:

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

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    $\begingroup$ Do you have an example of a Fredholm subspace? $\endgroup$ – Yemon Choi Sep 6 '20 at 1:09
  • $\begingroup$ @YemonChoi Trivial examples: Finite dimensions. Or in the case of infinite dimension, the scalar 1 dimensional space. or any subspace which does not contain any fredholm operator. As another example the space of $\{P(s)\mid s\text{ is the shift operator, whose index is -1 and P is an arbitrary polynomial of degree at most n\}$ $\endgroup$ – Ali Taghavi Sep 6 '20 at 1:12
  • $\begingroup$ $\{P(s)\mid s\text{ is the shift operator, whose index is -1}\}$ $\endgroup$ – Ali Taghavi Sep 6 '20 at 1:16
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    $\begingroup$ How do you exclude the trivial 1-dim rep for the first question? $\endgroup$ – YCor Dec 12 '20 at 14:34
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    $\begingroup$ No, irreducible is clearly defined (=nonzero and no proper nonzero closed invariant subspace). "Trivial" is slightly more ambiguous since it can be the trivial irrep (1-dimensional), or any trivial representation (which can be in dimension 0, 1, or more). $\endgroup$ – YCor Dec 12 '20 at 18:36
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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).

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    $\begingroup$ Sure, if $T$ has nonzero kernel then $T\otimes I$ has infinite dimensional kernel and hence is not Fredholm. Same for cokernel, so if $T\otimes I$ is Fredholm it must have no kernel or cokernel (and have closed range), which means it must be invertible. $\endgroup$ – Nik Weaver Sep 6 '20 at 2:42
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    $\begingroup$ If $\phi$ is a representation such that $\phi(A)$ is a Fredholm space then every Fredholm operator there has index zero, because $ind(T^n) = n\ ind(T)$. $\endgroup$ – Ruy Sep 6 '20 at 4:31
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    $\begingroup$ I don't understand this question. By definition, the range of any Fredholm operator is closed. If $T$ is not surjective then neither is $T\otimes I$. $\endgroup$ – Nik Weaver Sep 6 '20 at 14:10
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    $\begingroup$ (But of course, this is not really relevant for your answer; I just thought it might be a good idea to point this out anyway.) $\endgroup$ – Jochen Glueck Dec 12 '20 at 13:54
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    $\begingroup$ @JochenGlueck oh, you are right. I'll delete that comment. $\endgroup$ – Nik Weaver Dec 12 '20 at 15:35

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