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We say an invertible $n \times n$ matrix with entries in $\Bbb R^n$ is expansive if the absolute values of all of its eigenvalues exceed $1$. An easy calculation also shows that if we consider a ball in $\Bbb R^n$ and act an expansive matrix on it, then the volume becomes bigger. So, expansive makes sense.

Now let $G$ be a locally compact abelian group equipped with a measure $\mu$. (For now, we can assume that the metric is doubling; that is there is a constant $C>0$ such that $\mu(B(x,2r))\leq C \mu(B(x,r))$.) Assume that $\alpha: G\to G$ is an automorphism of $G$. (When $G=\Bbb R^n$, then $\alpha$ is an invertible matrix.) My question is how I should define the notion of expansive for $\alpha$.

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  • $\begingroup$ 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 $\mathbb{R}\times(\mathbb{R}/\mathbb{Z})$. $\endgroup$
    – YCor
    Commented Oct 23, 2016 at 6:29
  • $\begingroup$ You can't just use that there exists a metric $d$ such that $d(\alpha x,\alpha y) \geq d(x,y)$? $\endgroup$
    – David Handelman
    Commented Oct 23, 2016 at 19:08
  • $\begingroup$ @DavidHandelman A metric with what compatibility conditions? Also I understand from the question that the identity is not considered as an expansive automorphism. $\endgroup$
    – YCor
    Commented Oct 23, 2016 at 19:08
  • $\begingroup$ 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. $\endgroup$
    – David Handelman
    Commented Oct 23, 2016 at 19:32
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Oct 23, 2016 at 23:53

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