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Group with a translation invariant ultrafilter

Every nontrivial group $G$ satisfies your condition (iii). To see why, note that for any subgroup $H$ of $G$, the left action of $H$ on $G$ is free. So if $X$ is a set of representatives of the orbits …
Saúl RM's user avatar
  • 10.6k
5 votes
0 answers
101 views

Non-monotileable amenable groups

This is crossposted from MSE. We say a subset $A$ of a group $G$ is a monotile for $G$ if $G$ is a disjoint union of right translates of $A$. In his article Monotileable Amenable Groups, B. Weiss give …
Saúl RM's user avatar
  • 10.6k
2 votes
1 answer
146 views

Can we find background noise for every Følner sequence in a countable amenable group?

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 se …
Saúl RM's user avatar
  • 10.6k
1 vote
0 answers
182 views

Does every amenable group $G$ admit a two-sided Folner sequence?

By two-sided Følner sequence I mean a sequence $(F_N)_N$ of subsets of $G$ which is both a left-Følner and a right-Følner sequence. Context: I just came up with this question and surprisingly I haven' …
Saúl RM's user avatar
  • 10.6k
0 votes

Can we find background noise for every Følner sequence in a countable amenable group?

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 wai …
Saúl RM's user avatar
  • 10.6k