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Suppose $K$ is a compact polytope and $T$ is a piecewise-linear homeomorphism from $K$ 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 $K$). Can we conclude that a nonperiodic point exists (that is, for some $x$ in $K$ 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 $K$ 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).

Ideally I would like to draw stronger conclusions from the stated hypotheses (e.g. that the set of nonperiodic points has positive $d$-dimensional measure where $d$ is the dimension of $K$). Some of the hypotheses on $K$ and $T$ are likely to be red herrings, but I include them since they apply for the class of examples that motivated this question.

I have posted a variant of this question in which the piecewise-linearity constraint is dropped; see Nonperiodic points of homeomorphisms of a ball. Montgomery's theorem solves that problem and therefore solves this one, though I'm still hoping for (and will award a bounty to) the best self- contained, clear, and complete proof for the piecewise-linear case.

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The answer is yes. In other words, if the order of the orbit $x_n=T^n(x)$ of each point $x\in K$ is finite then $T^n$ is identity map for some $n$.

Choose a simplex $\triangle$ of maximal dimension $m$. Set $$E_n=\{\,x\in\triangle\mid T^n(x)=x\,\}.$$

By Baire theorem, $E_n$ has nonempty interior for some $n$; fix such a value $n_1$.

Note that $E_{n_1}$ is a polytope in $\triangle$. If $E_{n_1}\ne \triangle$, we can choose an $m$-simplex $\triangle'\subset \triangle$ such that (1) its base in $E_{n_1}$ (2) the remaining part is outside of $E_{n_1}$ and (3) $T^{n_1}$ is linear on $\triangle'$.

Given $\ell_1$, we can choose a simplex $S_1$ in $\triangle'$ so that (1) $T^{n_1\cdot m}(x)\ne x$ if $x\in S_1$ and $m<\ell_1$ and (2) for some $m_1>\ell_1$, the set $E_{m_1\cdot n_1}\cap S_1$ has nonempty interior but $E_{m_1\cdot n_1}\not\supset S_1$. Indeed (1) follows if entire $S_1$ lies close to the base. Further by Bair theorem $S_1\cap E_{m_1\cdot n_1}$ has nonempty interior for some $m_1>\ell_1$. Note that $x\notin E_{m_1\cdot n_1}$ if $x$ lies terribly close to the base (depending on $m_1$) and we can arrange that there is such point in $S_1$.

Set $n_2=m_1\cdot n_1$ and iterate the construction for an increasing sequence of numbers $\ell_i$. We get a nested sequence of compact set $S_1\supset S_2\supset\dots$ such that $T^n(x)\ne x$ for any $x\in S_n$. If $x\in\bigcap_nS_n$ then $T^n(x)\ne x$ for any $n$, a contradiction.

We proved that the restriction $T^n|_\triangle$ is identity map for some $n$.

Now remove the interior of $\triangle$ from $K$ and pass to the map $T^n$. Repeat this procedure recursively till no simplex is left in $K$.

P.S. The same holds for any continuous map, see the answer in here.

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  • $\begingroup$ I don't understand the sentence "They might have nonempty interiors but the complement contains a controlable open set near the boundary of $E_n$" (partly because I don't know what "controlable" means). $\endgroup$ – James Propp Dec 13 '14 at 4:02
  • $\begingroup$ @JamesPropp: Given a poisitive integer $K$ you can find an open rectangle $\square$ near the base of $\triangle'$ such that $\square\cap E_k=\varnothing$ for any $k<K$. $\endgroup$ – Anton Petrunin Dec 13 '14 at 17:23
  • $\begingroup$ P.S. Nice question, can you tell why did you need it? $\endgroup$ – Anton Petrunin Dec 13 '14 at 17:26
  • $\begingroup$ Anton, your comment about rectangles was very helpful, and I'll explain in a minute why I need this result, but first let me ask why we can conclude that the sets $E_{n \cdot k}$ do not cover $\Delta'$. I don't follow this step. $\endgroup$ – James Propp Dec 15 '14 at 17:31
  • $\begingroup$ In the article-in-progress arxiv.org/abs/1310.5294, David Einstein and I prove that a certain piecewise-linear homeomorphism on the order polytope of the poset $[2] \times [2]$ has infinite order, by constructing arbtrarily long orbit-segments. (Actually, we don't do this in the current draft, but we plan to add details soon.) We'd like to know that nonperiodic points exist, and indeed that "most" point are nonperiodic, in the sense of measure or the sense of category. And of course we'd like to know this for other, similar examples. $\endgroup$ – James Propp Dec 15 '14 at 17:35

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