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Oct 23, 2016 at 23:53 comment added YCor If we consider $\mathbf{Z}^2$ and the matrix $\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$, if $p$ is the projection to the eigenspace for the eigenvalue $>1$ (with kernel the other eigenspace) and we define $d(n,m)=|p(m-n)|+1$ for all $m\neq n$, we get a metric which induces the topology, and in addition is translation invariant, and strictly increases the distance of nonzero elements. Do we want to call such a map "expansive"? I doubt it (it preserves the volume). But as far as the OP doesn't say what he/she requires, the question is unclear / too broad.
Oct 23, 2016 at 19:32 comment added David Handelman Just the existence of some metric yielding the same topology? (This, or anyway a variant of it, is done in actions on Cantor sets, although the definition of expansive is different.) Maybe add to the definition that $\alpha$ is not an isometry, or more extreme, if $x \neq y$, then the inequality is strict. I'm simply asking if this type of definition has been considered.
Oct 23, 2016 at 19:08 comment added YCor @DavidHandelman A metric with what compatibility conditions? Also I understand from the question that the identity is not considered as an expansive automorphism.
Oct 23, 2016 at 19:08 comment added David Handelman You can't just use that there exists a metric $d$ such that $d(\alpha x,\alpha y) \geq d(x,y)$?
Oct 23, 2016 at 19:04 history edited David Handelman CC BY-SA 3.0
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Oct 23, 2016 at 6:29 comment added YCor You have several options but they should depend on what you require. For instance, if $G$ is discrete, what would it mean to be expansive? And if $G$ is compact? (in both cases, all automorphism preserve the volume.) If it's too extreme, maybe specify what you would call an expansive automorphism of $\mathbb{R}\times\mathbb{Z}$ and $\mathbb{R}\times(\mathbb{R}/\mathbb{Z})$.
Oct 23, 2016 at 2:45 history edited Melody CC BY-SA 3.0
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Oct 23, 2016 at 1:21 history edited Melody CC BY-SA 3.0
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Oct 23, 2016 at 1:09 history asked Melody CC BY-SA 3.0