Question: Assume a smooth manifold $M$ does not admit any effective smooth group actions of finite groups $G \neq 1$, does it follow that $M$ also admits no continuous effective group actions of finite groups $G \neq 1$?

Manifolds which do not admit actions of finite groups are interesting because if $M$ is compact and we choose a Riemannian metric $g$ on $M$, then $\text{Isom}(M,g)$ is a compact Lie group by Steenrod-Myers and must be trivial, because otherwise it contains a non-trivial finite group acting on $M$.

For this nice property, it is enough to consider only smooth actions.

However, there are some interesting articles which construct manifolds which do not admit any continuous actions of finite groups. And so I'm wondering if its enough to eliminate smooth actions if the manifold is smooth to get a manifold of this type.

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    $\begingroup$ Restatement of the question using group theory language: If $\mathrm{Diff}^\infty(M)$ is torsion-free, does it follow that $\mathrm{Homeo}(M)$ is torsion-free? $\endgroup$
    – YCor
    Jul 22, 2019 at 14:32
  • $\begingroup$ The linked article of V. Puppe seems to say "manifold" for "(connected) closed manifold". Is the same implicit assumption made in the question? $\endgroup$
    – YCor
    Jul 23, 2019 at 8:12


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