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Sep 12, 2019 at 21:40 comment added YCor By the way "probably" of my first comment referred to Q1 (I'm surprised by your negative conclusion). I know that all these examples have $|S^n\cap Z|$ growing exponentially (but I'm not sure its exponential rate $|S^n\cap Z|$ converges, and certainly one should consider $\liminf$, $\limsup$ for it, not $\inf$.
Sep 12, 2019 at 18:31 answer added Sean Eberhard timeline score: 4
Sep 12, 2019 at 14:00 comment added Sean Eberhard I have however managed to convince myself that $|S^n \cap Z|$ grows exponentially, so this probably answers question 3 negatively.
Sep 12, 2019 at 13:55 comment added Sean Eberhard My previous comment was hasty, and the situation is actually rather delicate. It turns out that for most points of $S^n$ the coordinate $z$ is determined by $x$ and $y$, basically because of the geometry of the lamplighter: for most points of $S^n$, the lamplighter made an injective walk, and if you only allow $\pm1$ steps then the walk must be unidirectional. I think you can get around this by allowing bigger steps or by walking in $\mathbf{Z}^2$ instead of $\mathbf{Z}$, but I'm still thinking about it...
Sep 11, 2019 at 11:40 history edited LeechLattice
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Sep 11, 2019 at 9:35 comment added Sean Eberhard With $A = \mathbf{F}_p[t^{\pm1}]$ and the generating set consisting of $\varphi^{\pm1}$ times all matrices with entries of degree $0$, I think you have $|S^n Z/Z| \approx p^{2n}$, $|S^n| \approx p^{4n}$, $|S^n \cap Z| \approx p^{2n}$.
Sep 11, 2019 at 9:32 comment added Sean Eberhard Yes, that's simpler. It's not unlike my suggestion, but with the advantage that by taking a different ring than $\mathbf{Z}$ you can take the automorphism $\varphi$ to be diagonal, which makes it easier on the brain.
Sep 11, 2019 at 9:18 comment added YCor Probably. I'd try Hall's group $G(A)$ of matrices $$\begin{pmatrix}1 & x & z\\0 & t^n & y\\ 0 & 0 & 1\end{pmatrix}$$ with $x,y,z\in A$, $n\in\mathbf{Z}$; here $A$ is the ring generated by $t,t^{-1}$, which leaves several choices, for instance $A=\mathbf{F}_p[t^{\pm}]$ (central extension of the lamplighter) or $A=\mathbf{Z}[1/p]$, $t=p$ (related to Baumslag-Solitar group $\mathrm{BS}(1,p)$).
Sep 11, 2019 at 8:33 history asked Sean Eberhard CC BY-SA 4.0