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Edited: Due to work of Raymond and Scott, there exist diffemorphisms (of certain three-dimensional nil-manifolds) whose $n$th power is diffeotopic to the identity, but which are not themselves homotopic to finite order homeomorphisms.

Do they thus provide counterexamples to the claim that that actions by homeomorphisms (on an aspherical three manifold) are conjugate (by a homeomorphism) to isometric actions?

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  • $\begingroup$ Also, the result doesn't seem to make sense (to me), surely a finite order diffeomorphism is a finite order homeomorphism. $\endgroup$
    – Nick L
    Commented May 30, 2022 at 21:32
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    $\begingroup$ Assume the manifold is compact. If $f$ is conjugate to an isometry (say for some Riemannian structure), then $(f^n)_{n\ge 1}$ accumulates to the identity. Hence, just pick a diffeomorphism without this property. [This seems to me quite unrelated to the quoted fact.] $\endgroup$
    – YCor
    Commented May 31, 2022 at 12:35

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Do they thus provide counterexamples to the claim that that actions by homeomorphisms (on an aspherical three manifold) are conjugate (by a homeomorphism) to isometric actions?

Here is a counterexample which I find a bit simpler.

Suppose that $M$ is a closed, connected, oriented hyperbolic three-manifold. Thus $M$ is aspherical, as desired. Also, $\mathrm{Isom}(M)$ is finite. Let $f$ be a homeomorphism of $M$ which is the identity outside of a small ball $B$ and is not the identity on $B$. Then (difficult exercise) $f$ has infinite order. Now suppose that $g$ is any homeomorphism. Then (easy exercise) $gfg^{-1}$ again has infinite order. Thus $f$ is not conjugate to an isometry.

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    $\begingroup$ You do not need the "difficult exercise" or knowing finiteness of the isometry group of a hyperbolic manifold. Just observe that if an isometry of a connected Riemannian manifold has fixed-point set with nonempty interior, then it equals the identity. $\endgroup$
    – Moishe Kohan
    Commented May 31, 2022 at 12:55

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