2
$\begingroup$

Let $G$ be a countable amenable group. We consider sequences $(z_g)_{g\in G}$ of complex numbers with $|z_g|=1$ for all $g\in G$.

I will say $(z_g)_{g\in G}$ is background noise for a (left-)Følner sequence $(F_N)_N$ if for all $h\in G\setminus\{0\}$ we have

$$\lim_{N\to\infty}\frac{1}{|F_N|}\sum_{g\in F_N}z_{hg}\overline{z_g}\;=0.\qquad\text{(1)}$$

For example, one can check that if $G=\mathbb{Z}$, then the sequence $(z_n)_{n\in\mathbb{Z}}$ given by $z_n=e^{in^2}$ is background noise for any Følner sequence $(F_N)_N$.

Question 1: Given a countable amenable group $G$ with a Følner sequence $(F_N)_N$, can we find a sequence $(z_g)_{g\in G}$ which is background noise for $(F_N)_N$?

Question 2: Given a countable amenable group $G$, can we find a sequence $(z_g)_{g\in G}$ which is background noise for all Følner sequences $(F_N)_N$ in $G$?

We can assume that $G$ is infinite if necessary.

A comment: For a fixed Følner sequence $(F_N)_N$ which does not grow very slowly (e.g. if for all $\alpha\in(0,1)$ we have $\sum_N\alpha^{|F_N|}<1$), then we can create a sequence $(z_g)_g$ by choosing each $z_g$ randomly (and independently) according to the uniform distribution in $\mathbb{S}^1$, and then with probability $1$ the sequence $(z_g)_g$ will be background noise for $(F_N)_N$. So the problems for Question 1 are those Følner sequences which grow very slowly.

$\endgroup$
2
  • $\begingroup$ I think the iid construction you mentioned for Q1 works without any constraints on the Følner sequence — just use the ergodic theorem instead of whatever hands-on concentration proof you were using. $\endgroup$
    – tmh
    Commented May 17 at 23:24
  • 1
    $\begingroup$ Do you mean that for any Følner sequence $(F_N)_N$, the sequence $(z_g)_g$ will with probability $1$ be background noise for $(F_N)_N$ if we choose it as I say above? In that case I don't think that works for a Følner sequence in $\mathbb{Z}$ where the sets $(F_N)_N$ are disjoint intervals and $F_N$ has length $k$ for all $N$ between $2^{2^{k-1}}$ and $2^{2^k}$ (that's not the kind of Følner sequence I am worried about though) $\endgroup$
    – Saúl RM
    Commented May 17 at 23:45

1 Answer 1

0
$\begingroup$

I found an answer to Question 1 little after asking it, but as it was part of the material I was planning to upload to arXiv (and since the proof is long so I prefer to just cite it), I decided to wait until I uploaded it to answer.

Theorem 2.1 of my article says that, in order to prove that there is a sequence $(z_g)_g$ in $\mathbb{S}^1$ such that for all $l\in G$ we have $$\lim_N\frac{1}{|F_N|}\sum_{g\in F_N}z_{lg}\overline{z_g}=0,$$ it is enough to prove that for all $A,L\subseteq G$ finite and for all $\delta>0$ there exists some $K\in\mathbb{N}$ and sequences $(z_{g,k})_{g\in G}$ in $D$, for $k=1,\dots,K$, such that for all $l=1,\dots,L$ we have \begin{equation} \left|\frac{1}{K|A|}\sum_{k=1}^{K}\sum_{g\in A}z_{lg,k}\overline{z_{g,k}}\right|<\delta.\qquad\qquad\qquad (1) \end{equation}

And this is indeed true: for each $g\in G,k\in\mathbb{N}$ choose a value $z_{g,k}$ randomly using the uniform distribution in $\mathbb{S}^1$. Then by the central limit theorem, as $\mathbb{E}(z_{lg,k}\overline{z_{g,k}})=0$ for all $g\in G,l\in G\setminus e$ and $k\in\mathbb{N}$, with probability $1$ we will have that for big enough $K$ Equation (1) above is satisfied.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .