Suppose $B$ is a $d$-dimensional ball (for some $d \geq 1$) and $T$ is a homeomorphism from $B$ to itself. Suppose also that $T$ is not of finite order (that is, for no $n \geq 1$ is it the case that $T^n(x)=x$ for all $x$ in $B$). Can we conclude that a nonperiodic point exists (that is, for some $x$ in $B$ it is the case that $T^n(x) \neq x$ for all $n \geq 1$)?

Note that it need not be the case that for each $n \geq 1$ the set of $x$ in $B$ with $T^n(x) \neq x$ is dense. So my question is not just an application of the Baire category theorem (at least under the most simpleminded approach).

Note that the answer becomes "No" if we replace the ball $B$ by a general compact set. For instance, consider the action on $\mathbf{R}/\mathbf{Z} \times \{0,1/2,1/3,1/4,1/5,…\}$ that sends ($x$, $y$) to ($x+y$ mod 1, $y$).

This post is related to Nonperiodic points of piecewise-linear homeomorphisms .


Montgomery showed in 1938 that a "pointwise periodic" homeomorphism of a manifold without boundary is actually periodic (this is in Montgomery and Zippin, page 223, or thereabouts), so if your ball is a closed ball, the result you want follows from this by doubling, and if it's the open ball, you don't even need to double.

  • $\begingroup$ What is the "doubling" trick Rivin is referring to? $\endgroup$ – James Propp Dec 19 '14 at 23:29
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    $\begingroup$ @James Propp Glue two identical pancakes (with identical action, etc.) together and inflate the result to a sphere. :-) $\endgroup$ – fedja Dec 19 '14 at 23:33
  • $\begingroup$ @JamesPropp: Take a homeomorphism $f: B^n\to B^n$ of the closed round ball and extend it to the n-sphere by the formula $J\circ f \circ J$, where $J$ is the inversion in $\partial B^n$. $\endgroup$ – Misha Dec 20 '14 at 0:34
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    $\begingroup$ The original journal reference for this theorem is D. Montgomery. Pointwise periodic homeomorphisms, Amer. J. Math. 59 (1937), 118-120. Thanks! $\endgroup$ – James Propp Dec 20 '14 at 15:17

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