Skip to main content
7 events
when toggle format what by license comment
May 31, 2022 at 21:56 vote accept Nicolas Boerger
May 31, 2022 at 12:35 comment added YCor 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.]
May 31, 2022 at 10:53 answer added Sam Nead timeline score: 1
May 31, 2022 at 8:02 history edited Sam Nead CC BY-SA 4.0
Fixed ref, spacing, grammar
May 30, 2022 at 21:47 history edited Sam Nead CC BY-SA 4.0
added 35 characters in body
May 30, 2022 at 21:32 comment added Nick L Also, the result doesn't seem to make sense (to me), surely a finite order diffeomorphism is a finite order homeomorphism.
May 30, 2022 at 21:03 history asked Nicolas Boerger CC BY-SA 4.0