1
$\begingroup$

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 $B(\ell^2)$:

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

$\endgroup$
17
  • 2
    $\begingroup$ Do you have an example of a Fredholm subspace? $\endgroup$
    – Yemon Choi
    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 $\{P(s)\mid s\text{ is the shift operator, whose index is -1 and P is an arbitrary polynomial of degree at most n\}$ $\endgroup$ Sep 6, 2020 at 1:12
  • 1
    $\begingroup$ How do you exclude the trivial 1-dim rep for the first question? $\endgroup$
    – YCor
    Dec 12, 2020 at 14:34
  • 1
    $\begingroup$ 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
    Dec 12, 2020 at 16:14
  • 1
    $\begingroup$ 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, 2020 at 18:36

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ Where is the assertion located in Nik Weaver's answer (or thread of comments)? (anyway I think it's true) $\endgroup$
    – YCor
    Oct 21, 2021 at 16:44
  • 1
    $\begingroup$ 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
    Oct 21, 2021 at 17:09
  • 1
    $\begingroup$ @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 $\pi(A)$. $\endgroup$ 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
    Oct 21, 2021 at 17:24
  • 1
    $\begingroup$ @YCor My favorite reference is Proposition 1.3.10 in Dixmier's $C^*$-book. $\endgroup$ Oct 21, 2021 at 17:50
6
$\begingroup$

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).

$\endgroup$
20
  • 1
    $\begingroup$ 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, 2020 at 2:42
  • 1
    $\begingroup$ 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
    Sep 6, 2020 at 3:13
  • 5
    $\begingroup$ 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, 2020 at 4:31
  • 2
    $\begingroup$ (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.) $\endgroup$ Dec 12, 2020 at 13:54
  • 2
    $\begingroup$ @JochenGlueck oh, you are right. I'll delete that comment. $\endgroup$
    – Nik Weaver
    Dec 12, 2020 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.