# 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? – Yemon Choi Sep 6 '20 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 $\{P(s)\mid s\text{ is the shift operator, whose index is -1 and P is an arbitrary polynomial of degree at most n\}$ – Ali Taghavi Sep 6 '20 at 1:12
• $\{P(s)\mid s\text{ is the shift operator, whose index is -1}\}$ – Ali Taghavi Sep 6 '20 at 1:16
• How do you exclude the trivial 1-dim rep for the first question? – YCor Dec 12 '20 at 14:34
• 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 '20 at 18:36

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. – Nik Weaver Sep 6 '20 at 2:42
• 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 '20 at 4:31
• I don't understand this question. By definition, the range of any Fredholm operator is closed. If $T$ is not surjective then neither is $T\otimes I$. – Nik Weaver Sep 6 '20 at 14:10