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Reading the book Almost-Bieberbach Groups: Affine and Polynomial Structures, I have two questions.

  1. The first red underline, I can not understand why they are equivalent.
  2. The second red underline, a uniform discrete subgroup of the semiproduct automatically gives a properly discontinuous action?

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Link
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    Your understanding of proper discontinuity is wrong, it is not the same as a wandering action (although some authors say so by abusing the terminology). – Moishe Kohan Dec 27 '24 at 19:03
  • @MoisheKohan You are right. I confused the other meaning of proper discontinuity. – Link Dec 27 '24 at 19:29
  • @DietrichBurde I is the usual free action. Any nontrivial group element does not fix any point. – Link Dec 27 '24 at 19:31
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    Then it is clear, see this post, because a torsion-free group has no finite subgroup. – Dietrich Burde Dec 28 '24 at 09:27
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    I think the hard part will probably be showing that if the action of $\Pi$ on $G$ is free, then $\Pi$ is torsion-free. If $G = \mathbb{R}^n$, then one can prove this by assuming $\Pi$ has a torsion element $\pi$, and showing that the "average of the orbit of $\pi$" is a fixed by $\pi$, hence the action is not free. I'm not sure if a similar thing can be done for any connected and simply connected Lie group $G$ though... – sTertooy Dec 28 '24 at 21:34
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    The equivalence 1 is false: if $G$ is simply connected but has nontrivial finite subgroups (e.g., $G=\mathrm{SL}_2(\mathbf{C})$ then let $\Pi$ be such a subgroup: then $\Pi$ acts properly ["properly discontinuously" if you like unnecessary additional words to mean the same] but is not torsion-free. However, the other implication is true. And the equivalence is true when $G$ is contractible. – YCor Dec 28 '24 at 21:51
  • I removed my opinion and wiki page link for 'properly discontinuous'. – Link Dec 29 '24 at 06:18
  • @YCor How contractibility works? – Link Dec 30 '24 at 12:51
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    @Link if $G$ is contractible, then $\mathrm{Aut}(G)=\mathrm{Aut}(\mathfrak{g})$ is real algebraic, hence has finitely many components. Also, every maximal compact subgroup of $\mathrm{Aut}(G)$ is maximal compact in $H=G\ltimes\mathrm{Aut}(G)$, and hence every compact subgroup of $H$, being conjugate to a subgroup of $\mathrm{Aut}(G)$, fixes a point. – YCor Dec 30 '24 at 13:32
  • @YCor Thanks, but I can’t understand. I need to study more about this. Please tell me one more. Where are torsion elements used in your comment? – Link Dec 30 '24 at 15:02
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    If the group is not torsion-free, it has a nontrivial finite subgroup, which is a compact subgroup, which by the argument of my comment fixes a point in $G$. – YCor Dec 30 '24 at 15:29
  • Okay. I will try to understand your arguments now. – Link Dec 30 '24 at 15:58