I want to know if every smooth (finite)group action on $\mathbb{R}^n$ is conjugate to some linear action.Thank you!
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3 Answers
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Yes in dimensions ≤ 2 (classical). Yes in dimension 3 via the Geometrization Conjecture (with much earlier work in special cases). No in higher dimensions, with the simplest examples perhaps being counterexamples to the "Smith conjecture"-smooth involutions on R^4 (or S^4) with knotted 2-dimensional fixed point set (Giffen, Gordon, Sumners).
answered Apr 29, 2010 at 12:34
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$\begingroup$ Prof.Edmonds,Thank you so much for your answer.It is helpful.May I know where i can find these results and the arguments.Thank you! $\endgroup$– saraCommented Apr 29, 2010 at 13:11
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$\begingroup$ This a vast subject. Are you most interested in the low-dimensional cases? Or the high-dimensional cases? Especially interested in the non-compact case, as compared to actions on S^n? Especially interested in aspects unique to smoothness? As a starter you might try to check out Bredon's book, Introduction to Compact Transformation Groups, and some constructions in Chapter I, Secs. 7-8. $\endgroup$ Commented Apr 29, 2010 at 15:43
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2$\begingroup$ @Professor AE: Speaking of 3-manifolds, I have a book by that name which I believe you lent to me. I haven't finished it yet, but after 13 years I am beginning to think that it may take me a little while to get to it. Would you like it back? $\endgroup$ Commented Apr 29, 2010 at 18:46
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$\begingroup$ @Prof.Edmonds:But aren't there some important papers that deals with this topic and then give the results one after another?I want to trace the history of this problem. $\endgroup$– saraCommented Apr 29, 2010 at 22:49
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2$\begingroup$ @Pete: Ah-hah! My copy of Hempel. Every few years I find myself looking for it for some reason and wondering what became of it. It's not a big deal. If you really aren't using it, then maybe send it along sometime. I've enjoyed reading many of your comments here on MO. $\endgroup$ Commented Apr 30, 2010 at 0:35
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Take a contractible manifold $C$, multiply it by $\mathbb R^n$, and let the finite group act trivially on $C$, and linearly on $\mathbb R^n$ such that $0$ is the unique fixed point. Then $C\times 0$ is the fixed point set. If $C$ is not diffeomorphic to $\mathbb R^n$, the action is not linear, but for sufficiently large $n$ the product $C\times\mathbb R^n$ will be diffeomorphic to a Euclidean space, by Stallings's characterization of Euclidean space as the contactible space that is simply-connected at infinity. In fact, Craig Guilbault proved that $n=1$ suffices (except possibly when $\dim(C)=3$, which I do not quite understand at the moment). See here for Craig's paper.
answered Apr 29, 2010 at 16:08
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For the algebro-geometric setting, that is polynomial automorphisms (and conjugacy in the group of polynomial automorphisms), I recall a talk and several surveys by Hanspeter Kraft where it was stated that it is true for $n=1,2$ and unknown for $n>2$. I am not sure if it gives reasonable intuition for smooth automorphisms though.
answered Apr 29, 2010 at 12:32
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$\begingroup$ I guess the problem with polynomial automorphisms of $\mathbf{R}^n$, with conjugacy in the homeomorphism group, is open as well (in high dimension, maybe in all dimensions $\ge 4$). $\endgroup$– YCorCommented Jan 27, 2022 at 14:09