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Let a finite cyclic group $G = \mathbb Z/n$ act continuously on an open $d$-ball $B^d$. Suppose further that this action extends to the closed ball $\overline{B^d}$. Is there necessarily a fixed point in the interior?

Note that by Brouwer's fixed point theorem, there has to be some fixed point in the closed ball.

It is also true that for $n$ a prime power, every action of $\mathbb Z/n$ on an open ball has a fixed point, this can be seen from Smith theory.

On the other hand, for $n$ not a prime power, there are actions of $\mathbb Z/n$ on open balls without fixed points, see e.g. Theorem 8.3 in Bredon: Introduction to Compact Transformation Groups. But I am not aware of examples of such actions that extend to the closed ball.

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There need not be a fixed point in the interior. As in Bredon's book, p. 61, there exist smooth actions of a cyclic group $C_{r}$, of order $r$, without fixed points on $\mathbb{R}^{n}$ for large enough $n$ (e.g. $n\ge 8$) whenever $r$ is not a prime power. By one-point compactification one obtains a continuous, one-fixed-point action on the $n$-sphere $S^{n}$. Then by coning from the fixed point one obtains a continuous action on the closed ball $D^{n+1}$, smooth on the interior, but with no interior fixed points. One can push $n$ down at least to $7$ (and down to $6$ if $r$ involves at least $3$ primes), but not down to $4$, in this argument. See Richard Haynes, Slawomir Kwasik, Jerrel Mast, and Reinhard Schultz, Periodic maps on $\mathbb{R}^7$ without fixed points, Math. Proc. Cambridge Philos. Soc. 132 (2002), no. 1, 131--136.

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A wealth of information is provided in the answers to this question. In particlar, in the paper of Parris cited in the accepted answer, there are number theoretic conditions on $n,$ which guarantee that an action of $\mathbb{Z}/n\mathbb{Z}$ has fixed points.

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Assume all fixed points are on the boundary. For each fixed point you can find an invariant neighborhood. (You can assume that the $Z/nZ$-action fixes some Riemannian metric. Then any $\epsilon$-ball around the fixed point is invariant.) These sets are homeomorphic to a half-ball.

The complement of the union of these half-balls (over all fixed points) is homeomorphic to a closed ball. The action on this closed ball does not have fixed points, contradicting Brouwer's theorem.

EDIT: The argument works because the cyclic group is finite. (The invariant metric is obtained by averaging any metric over the finite group action.) The argument does not work for infinite cyclic groups because there is no invariant metric and indeed in general no invariant neighborhood for a fixed point.

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  • $\begingroup$ This averaging argument works if your action is smooth, but what if it's only continuous? $\endgroup$ – HJRW Nov 7 '15 at 7:07
  • $\begingroup$ One can still average the path metric (as a metric in the sense of metric spaces, not a Riemannian one). One would have to show that balls in this metric are topological balls, though. $\endgroup$ – ThiKu Nov 7 '15 at 8:54
  • $\begingroup$ To be clear, are you proposing a proof of this fact? $\endgroup$ – HJRW Nov 8 '15 at 6:56
  • $\begingroup$ No, I didn't. An approach might be using Munkres Theorem (Annals '60) that homeomorphisms of the 2- or 3-disk can be approximated by diffeomorphisms. (I don't know what's the status in higher dimensions.) This should imply that the averaged metric can be approximated by Riemannian metrics but it's not clear to me what this means for the $\epsilon$-neighborhoods w.r.t. the limit metric. $\endgroup$ – ThiKu Nov 9 '15 at 8:03

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