0
$\begingroup$

(Not sure if it is appropriate or not, if no I will delete the post)

Recently I am concerned about the number of generator of $C^{*}_{r}(\mathbb{F}_{k})$, the reduced group algebra of the free group, generated by k generators. In this presentation(link), in slide 5, the author mentioned that it is an open problem to determine if $C^{*}_{r}(\mathbb{F}_{k})$ is singly generated for $k\geq2$.

The question I want to ask is, isn't $C^{*}_{r}(\mathbb{F}_{k})$ has at least $k$ generators? If I understand correctly, let $\mathbb{C}G$ denote the group algebra of $\mathbb{F}_{k}$, this is dense subalgebra in $C^{*}_{r}(\mathbb{F}_{k})$. Let $\delta_{g}$ be the indicator function of group element $g$. Then for generators $x_{1},...x_{k}$, the algebra generated by $\delta_{x_{1}},...\delta_{x_{k}}$ is the $\mathbb{C}G$. Now since each $\delta_{x_{i}}$ is algebraically independent, the number of generators is at least $k$. What mistakes did I make in the above?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ There is no proper subset of $\{\delta_{x_1}, \ldots, \delta_{x_k}\}$ that generates, but that does not mean there couldn't be another generating set with fewer elements. $\endgroup$
    – Nik Weaver
    Commented Nov 13, 2021 at 18:44
  • 3
    $\begingroup$ To further illustrate @NikWeaver's point: the non-unital Cstar algebra $c_0$ is singly generated, in the sense that we can find $f\in c_0$ such that the closed subalgebra generated by $f$ is all of $c_0$. (One can take $f(n)= 2^{-n}$.) $\endgroup$
    – Yemon Choi
    Commented Nov 14, 2021 at 0:30
  • $\begingroup$ Take $C^{*}_{r}(\mathbb{F}_{2})$ as an example, if it is singly generated, shouldn't the $\delta_{x}$ and $\delta_{y}$ have some algebraic relation, or $\delta_{y}$ equal to infinite sum of power of $\delta_{x}$? $\endgroup$
    – Ken.Wong
    Commented Nov 15, 2021 at 16:25
  • 1
    $\begingroup$ Belated reply: no, because in this setting the word "generates" means "topologically generates", so it is conceivable that there could be some mixture of delta_x's and delta_y's, call it f, such that combinations of powers of f could approximate delta_x to within arbitrary accuracy. $\endgroup$
    – Yemon Choi
    Commented Jan 10, 2022 at 14:31

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .