# 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 $$B(\ell^2)$$:

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

• Do you have an example of a Fredholm subspace? Sep 6, 2020 at 1:09
• @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 $\{P(s)\mid s\text{ is the shift operator, whose index is -1 and P is an arbitrary polynomial of degree at most n\}$ Sep 6, 2020 at 1:12
• How do you exclude the trivial 1-dim rep for the first question?
– YCor
Dec 12, 2020 at 14:34
• 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.
– YCor
Dec 12, 2020 at 16:14
• 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).
– YCor
Dec 12, 2020 at 18:36

## 2 Answers

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.

• Where is the assertion located in Nik Weaver's answer (or thread of comments)? (anyway I think it's true)
– YCor
Oct 21, 2021 at 16:44
• 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.
– YCor
Oct 21, 2021 at 17:09
• @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 $\pi(A)$. Oct 21, 2021 at 17:16
• 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.
– YCor
Oct 21, 2021 at 17:24
• @YCor My favorite reference is Proposition 1.3.10 in Dixmier's $C^*$-book. Oct 21, 2021 at 17:50

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).

• 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. Sep 6, 2020 at 2:42
• 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 $Q(H_0\otimes l^2)$. Sep 6, 2020 at 3:13
• If $\phi$ is a representation such that $\phi(A)$ is a Fredholm space then every Fredholm operator there has index zero, because $ind(T^n) = n\ ind(T)$.
– Ruy
Sep 6, 2020 at 4:31
• (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.) Dec 12, 2020 at 13:54
• @JochenGlueck oh, you are right. I'll delete that comment. Dec 12, 2020 at 15:35