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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
    Commented Sep 6, 2020 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
    Commented Sep 6, 2020 at 1:12
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    $\begingroup$ How do you exclude the trivial 1-dim rep for the first question? $\endgroup$
    – YCor
    Commented Dec 12, 2020 at 14:34
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    $\begingroup$ Oh, but actually this trivial 1-dimensional rep doesn't always exist... maybe my comment was stupid. I see no reason in particular to single out the trivial 1-dimensional rep among all finite-dimensional irreducible reps. $\endgroup$
    – YCor
    Commented Dec 12, 2020 at 16:14
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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
    Commented Dec 12, 2020 at 18:36

2 Answers 2

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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.

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  • $\begingroup$ Where is the assertion located in Nik Weaver's answer (or thread of comments)? (anyway I think it's true) $\endgroup$
    – YCor
    Commented Oct 21, 2021 at 16:44
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    $\begingroup$ The proof I can think of: (1) if $A\in B(H)$ is Fredholm and has nonzero index, then either $A^*A$ or $A^*A$ is Fredholm, self-adjoint $\ge 0$ and non-invertible. (2) in a $C^*$-algebra, if $M$ is self-adjoint invertible $\ge 0$ then $M^{-1}\in C^*(M)$. Indeed, approximate $x\mapsto x^{-1}$ by polynomials on the spectrum of $M$. (3) If $N$ is self-adjoint $\ge 0$, Fredholm and non-invertible then $N$ is invertible in restriction to the orthogonal of Ker$(N)$, say with inverse $N'$, extended to $0$ on Ker$(N)$, by (2) $N'\in C^*(N)$. So $I-NN'\in C^*(N)$ is nonzero of finite rank. $\endgroup$
    – YCor
    Commented Oct 21, 2021 at 17:09
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    $\begingroup$ @YCor: denote by $q$ the quotient map to the Calkin algebra. If $x\in\pi(A)$ is Fredholm, then $q(x)$ is invertible in the Calkin algebra, hence also in $q(\pi(A))$. As $q$ is injective on $\pi(A)$, the element $x$ is also invertible in $\pi(A)$. $\endgroup$
    – Alain Valette
    Commented Oct 21, 2021 at 17:16
  • $\begingroup$ Ah OK; actually I wasn't aware of this fact that if an element $M$ is invertible in the big $C^*$-algebra then it's invertible in the smaller one, and could only check it ((2) above) under special assumptions. $\endgroup$
    – YCor
    Commented Oct 21, 2021 at 17:24
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    $\begingroup$ @YCor My favorite reference is Proposition 1.3.10 in Dixmier's $C^*$-book. $\endgroup$
    – Alain Valette
    Commented Oct 21, 2021 at 17:50
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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
    Commented Sep 6, 2020 at 2:42
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    $\begingroup$ A fancier reason is that anything nonzero in $A\otimes I$ is noncompact, so factoring out the compacts embeds $A$ in the Calkin algebra $Q(H_0\otimes l^2)$. And Fredholm in $B(H_0\otimes l^2)$ is equivalent to being invertible in $Q(H_0\otimes l^2)$. $\endgroup$
    – Nik Weaver
    Commented Sep 6, 2020 at 3:13
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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
    Commented Sep 6, 2020 at 4:31
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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
    Commented Dec 12, 2020 at 13:54
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    $\begingroup$ @JochenGlueck oh, you are right. I'll delete that comment. $\endgroup$
    – Nik Weaver
    Commented Dec 12, 2020 at 15:35

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