If $A$ is a $C^*$-algebra, there is a notion of Hilbert $A$-module (which is something like Hilbert space but the inner product takes values in $A$). The standard example is $H_A:=\{(a_n)_{n=1}^{\infty}: \sum_{n=1}^{\infty}a_n^*a_n \ is \ norm \ convergent\}$ with $A$ valued inner product $(a,b):=\sum_{n=1}^{\infty}a_n^*b_n$. The algebra of all adjointable operators on Hilbert module turns out to be a $C^*$-algebra and there is a notion of $A$ compact operator: these are operators which are norm limits of linear combinations of operators of the form $\theta_{x,y}$ where $\theta_{x,y}(z)=x(y,z)$ (our modules are right modules). Having the notion of $A$ compact operator one can speak about $A$ Fredholm operators ($A$-linear). Let us take $A=C(X)$ (continuous functions on some compact space) and let $T:X \to Fred(H)$ be a continuous family of Fredholm operators on $H=\ell^2$.

Why is it true such that $T$ is $A$-Fredholm operator on $H_A$ for $A=C(X)$?

My guess is that if $T$ is a continuos family of Fredholm operators one can view $T$ as an operator $H_A \to H_A$ via $T\xi(x)=T_x\xi(x)$ where $\xi \in H_A$ is a sequence of continuous functions (so $\xi(x)$ makes sense as an element of $\ell^2$). However it is not obvious for me how to show that $T$ is now $A$-Fredholm. I think that the proof boils down to the following fact: if $T:X \to K(H)$ is a family of compact operators on $\ell^2$ then $T$ viewed as a $A$ linear operator in $H_A$ (as explained above) is $A$-compact.


1 Answer 1


The answer involves two arguments (I guess that $Fred (\ell^2)$ does have the norm topology).

  1. If $F: X \to Fred(\ell^2)$ is continuous, then there is another map $G: X \to Fred(\ell^2)$ such that $FG-1$ and $GF-1$ are maps into $\mathcal{K} (\ell^2)$.

  2. If $T:X \to \mathcal{K}(\ell^2)$, then $T$, viewed as an $C(X)$-linear operator on $H_{C(X)}$ is $C(X)$-compact.

Ad 1: Let $\mathcal{Q}(\ell^2) = \mathcal{B}(\ell^2)/\mathcal{K}(\ell^2)$ be the Calkin algebra and let $\pi: \mathcal{B}(\ell^2) \to \mathcal{Q}(\ell^2)$ be the quotient map. It is well-known that $F \in \mathcal{B}(\ell^2)$ is Fredholm iff $\pi(F)$ is invertible, and also that $\mathcal{Q}$ is a Banach algebra. Let $\mathcal{G} \subset \mathcal{Q}(\ell^2) $ be the group of units. Hence the inversion map $\iota:\mathcal{G} \to \mathcal{G}$ is continuous. By the Bartle-Graves theorem, there is a continuous section $\sigma: \mathcal{Q} \to \mathcal{B}(\ell^2)$. With these notations, define $$G:= \sigma \circ \iota \circ \pi \circ F: X \to Fred (\ell^2) \subset \mathcal{B}(\ell^2).$$

Ad 2: Let $T:X \to \mathcal{K}(\ell^2)$ be continuous. Let $\epsilon>0$. For each $x \in X$, pick a finite rank operator $R_x$ so that $\| R_x-T_x\| \leq \epsilon /3$, and pick a neighborhood $U_x$ such that $\| T_x-T_y\| \leq \epsilon/3$ for $y \in U_x$. Take a finite subcover $U_1 , \ldots, U_n$ of $(U_x)_{x \in X}$, and a partition of unity $\lambda_1, \ldots, \lambda_n$. Then $$\| T_x- \sum_{j} \lambda_j (x) R_{x_j} \| \leq \epsilon.$$ The operator $\sum_{j} \lambda_j (x) R_{x_j}$ is of finite rank (in the Hilbert module sense), and it follows that $T$ is compact (in the Hilbert module sense).

  • $\begingroup$ When reading about the Bartle-Graves theorem to fully understand this answer (here looks a good overview: link.springer.com/content/pdf/10.1007/s11228-019-00524-1.pdf), i saw that there was a lot of assumptions and techinicities. I did my best, but could you explain how it is been used here? I think it is by choosing $g: X\times \mathcal B(\ell^2) \longrightarrow \mathcal B(\ell^2), g(x,T) = TF_x$. $\endgroup$ Sep 7, 2022 at 17:46
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    $\begingroup$ There are possibly several theorems called "Bartle-Graves theorem" in the literature. The one I refer to states that a surjective continuous linear map between Banach spaces has a continuous ( but not necessarily linear) right inverse. $\endgroup$ Sep 7, 2022 at 17:57
  • $\begingroup$ Hello Professor @JohannesEbert. Recently, I'm finishing my final project to graduate, and a specific question involving that same construction came to me. I was wondering if you can take a look at my question (link). Even if that isn't possible, I would like to thank you and ask permission to cite the arguments involved in building my project. $\endgroup$
    – Mezzovilla
    Nov 25, 2022 at 19:25

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