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Let $(X,f)$ be a topological dynamical system ($f$ continuous, $X$ compact, metric with distance $d$).

Let $C\subseteq X^2$ indicate the chain recurrence relation: $$xCy\iff \forall \epsilon>0\ \exists x_0,\dots,x_n\in X:$$ $$x_0=x,\ x_n=y,\ d(f(x_k),x_{k+1})<\epsilon\ \ (k=0,\dotsm, n-1).$$

The restriction of $C\cap C^{-1}$ to the diagonal of $C$ ($\{x\in X:xCx\}$) is an equivalence relation on $X$, whose classes are usually called chain components. Chain components have a natural structure of directed graph, if one says that $C_1\longrightarrow C_2$ if $x_1Cx_2$ for some $x_1\in C_1, x_2\in C_2$.

Has this structure ever been investigated from the point of view of order theory?

In particular, is it known which posets can be realized as directed graphs of chain components of topological dynamical systems?

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  • $\begingroup$ One observation: I think the only finite posets you can get are those where no two elements compare. $\endgroup$
    – Will Brian
    Commented Oct 24, 2023 at 13:05
  • $\begingroup$ Why? An interval map with two fixed points, one repelling and the other attractive can be a counterexample, no? $\endgroup$ Commented Oct 24, 2023 at 13:18
  • $\begingroup$ Maybe I misunderstood your definition -- do you allow $n=0$ in your definition of $C$? If so (as I assumed) then the map you're describing has $2^{\aleph_0}$ equivalence classes. If not, I think you want to change your description of the poset you're talking about, so that you're not looking at a relation on the whole diagonal of $X$, but only the points $x \in X$ such that $x C x$. $\endgroup$
    – Will Brian
    Commented Oct 24, 2023 at 13:45
  • $\begingroup$ Oh...I see. I meant the diagonal of $C$, indeed. Clarified. $\endgroup$ Commented Oct 24, 2023 at 13:46
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    $\begingroup$ "If you restrict any relation to the diagonal, you get an equivalence relation". No: it has to be transitive and symmetric. However, I wrote that part in a less ambiguous way now, I hope. The equivalence classes are called chain components by P. Kurka in his textbook on topological and symbolic dynamics, for instance. He referred to E.Akin, if I remember well. $\endgroup$ Commented Oct 25, 2023 at 9:20

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Let's stick to zero-dimensional dynamics, it's all I know. In this setting, while I can't think of a reference or a discussion of this idea, one can give a full characterization of the posets up to isomorphism (at least with my interpretation of the question, which I explain below), in terms of closed relations on Cantor space, by standard subshift constructions.

So consider now $X$ a closed subset of Cantor space, $f : X \to X$ continuous. In the definition of $C$, let's assume $n = 0$ is not allowed, because this is the usual definition.

Now the chain-recurrent points $R = \{x \in X \;|\; xCx\}$ form equivalence classes under $[x] \sim [y] \iff xCy \wedge yCx$, and we get a structure of a directed acyclic graph on the equivalence classes. Let's call this the chain-poset of the system $(X, f)$.

I'll characterize the possible graphs that can arise in the zero-dimensional case, in terms of slightly more familiar objects. The following is easy to show:

Lemma. The chain-recurrence relation is closed, and the chain-recurrent points form a closed set.

In particular $R$ is a closed subset of Cantor space, and $(R \times R) \cap C$ is a closed transitive relation on $R$.

Theorem. The following are equivalent for a poset $P$:

  • $P$ arises as the poset of equivalence classes of a closed symmetric transitive relation on a closed subset of Cantor space.
  • $P$ arises as the poset of equivalence classes of a closed symmetric transitive relation on Cantor space.
  • $P$ arises as the chain-poset of a continuous function on a closed subset of Cantor space.
  • $P$ arises as the chain-poset of a two-sided subshift homeomorphic to the Cantor set.

Proof. The only thing to do is to construct a perfect subshift with a prescribed chain-poset. Let $P$ arise from a closed symmetric transitive reflective relation $C'$ on a closed subset $R'$ of Cantor space, say $R' \subset \{0,1\}^{\mathbb{N}}$. We recall the standard Toeplitz coding of $r \in \{0,1\}^{\mathbb{N}}$ as minimal subshift of $\{0,1,2\}^{\mathbb{Z}}$: start from $*^{\mathbb{Z}}$ (we think of $*$ as a position not yet filled), then fill in two out of three symbols periodically $(2 r_0 *)^{\mathbb{Z}}$, and inductively fill the remaining $*$-sequence with the same procedure, using the shift $sigma(r)_i = r_{i+1}$. Finally forbid $*$. Call the resulting minimal subshift $T(r)$. Note that from any point of $T(r)$, we can extract the point $r$.

Now construct a subshift of $\{0,1,2,3\}^{\mathbb{Z}}$ with the following rules: $3$ can only appear once. If it appears, say the point is $x3y$. Then we require that $x$ is the left tail of a point of some $T(r)$, and $y$ the right tail of some $T(r')$, and furthermore $(r, r') \in C'$. Finally, require that any point without $3$ is in some $T(r)$ with $r \in R'$.

Now it is easy to see that no points containing $3$ are chain-recurrent. On the other hand, all points in any $T(r)$ are chain-recurrent since this is a minimal subshift, and $T(r)$ is contained in a single equivalence class of the chain-recurrence relation $C$ for the same reason. Furthermore, if $(r, r') \in C'$ then take any point $x \in T(r)$ and any point $y \in T(r')$, and join their tails with $3$, to get chain-recurrence relation $xCy$ (again recall that $T(r), T(r')$ are minimal). So the chain-poset is exactly the poset of equivalence classes of $(R', C')$, as desired. Square.

Note that this does not directly implement the actual relation, because the equivalence classes are always blown up to a full Cantor space in this construction.

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  • $\begingroup$ You write" the chain-recurrent points $R = \{x \in X \;|\; xCx\}$ form equivalence classes under $[x] \sim [y] \iff xCy$", you probably meant "$\iff xCy$ and $yCx$", right? $\endgroup$ Commented Oct 25, 2023 at 14:55
  • $\begingroup$ Yes, of course. Hard to get right. $\endgroup$
    – Ville Salo
    Commented Oct 25, 2023 at 15:07
  • $\begingroup$ Btw, I suppose that you don't get more posets from general compact metrizable systems, since Cantor space maps onto them. $\endgroup$
    – Ville Salo
    Commented Oct 25, 2023 at 18:45
  • $\begingroup$ I finally found the time to read closely and accepted the answer, thanks. $\endgroup$ Commented Oct 26, 2023 at 12:23

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