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$\DeclareMathOperator\Sym{Sym}$I have a question about what I shall name here "transfinite dynamics" because it involves iterating a topological dynamical system $G \curvearrowright X$ through ordinals. I do not know whether this concept already has a name. If so, I would appreciate being directed to some references.

The intuition behind this construction is that the phase space $X$ will be the "board" of some game and $G$ will be the set of moves that the player is allowed to execute. I wish to understand what happens if the player is allowed to execute transfinitely many moves as long as the board "stabilizes" at limit stages.

To be more precise, let $G$ act on a Hausdorff space $X$. Extend $X$ to $\overline{X}$ by adding a point $\ast \notin X$ as an isolated point and extend the action of $G$ to $\overline{X}$ trivially. Let $\lambda$ be an ordinal and $\mathbf{g}=(g_{\alpha})_{\alpha < \lambda}$ be a sequence of elements of $G$. For each $x \in X$, define $(x_{\alpha})_{\alpha \leq \lambda}$ by transfinite recursion as follows.

  • $x_0=x$,
  • $x_{\alpha+1}=g_{\alpha}\cdot x_{\alpha}$ for all ordinals $\alpha<\lambda$, and
  • $x_{\theta}=\begin{cases} \ast & \text{ if } \lim_{\alpha \rightarrow \theta} x_{\alpha} \text{ does not exist}\\ \lim_{\alpha \rightarrow \theta} x_{\alpha} & \text{ otherwise }\end{cases}$

for every limit ordinal $\theta \leq \lambda$. Let us say that a sequence $(g_{\alpha})_{\alpha < \lambda}$ is valid if $x_{\lambda} \neq \ast$ for all $x \in X$. In other words, valid sequences are those along which every point can be transfinitely iterated.

For each valid sequence $\mathbf{g}=(g_{\alpha})_{\alpha < \lambda}$, consider the map $K_{\mathbf{g}}: X \rightarrow X$ given by $K_{\mathbf{g}}(x)=x_{\lambda}$.

Here are my questions.

Under what (not very restrictive) conditions can we guarantee that the set of $K=\{K_{\mathbf{g}}: \mathbf{g} \text{ is valid of length} <\omega_1\}$ forms a subgroup of $\Sym(X)$?

Can we see this set as some kind of "closure" of the image of $G$ in $\Sym(X)$?

Here are some basic facts that I was able to show

  • The maps $K_{\mathbf{g}}$ are not necessarily bijections. For example, consider the shift action of $\mathbb{Z}$ on the set $X$ of bi-infinite 0-1 sequences that are constant after some index. Then all sequences are mapped to constant sequences under the maps corresponding to the valid sequence $(1,1,1\dots)$ of length $\omega$.
  • If $G$ acts on a compact metric space $X$ by isometries, then each $K_{\mathbf{g}}$ is an isometry. Preservation of distances is trivial and surjectivity can be shown with some effort using transfinite induction and the compactness of the space. However, I cannot show that $K_{\mathbf{g}}^{-1}$ is of the form $K_{\mathbf{h}}$ and so, I cannot guarantee that these maps form a subgroup.

For an example of a non-trivial case where the maps $K_{\mathbf{g}}$ give us more bijections of $X$ than those induced by $G$, consider the action of $\mathbb{Q}$ on $S^1$ by rotations. In this case, the maps $K_{\mathbf{g}}$ induced by valid sequences of length $\omega$ gives all the real rotations because for any real $r$, we can form a sequence $(q_0,q_1,\dots)$ of rationals with $\sum_{i=0}^{\infty} q_i=r$. On the other hand, no other valid sequence gives us a bijection other than these real rotations.

Iterating such a system along ordinals seems very natural to me, however, I was not able to find much on this. As I said, if anything along these lines were investigated before, I would appreciate being directed to the correct references.

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    $\begingroup$ Although distinct, let me point out a slightly related well-studied concept: let $G$ act on a compact space $X$ (usually by homeomorphisms). Let $X^X$ be endowed with the product compact topology. The Ellis semigroup of the action is the closure of (the image of) $G$ in $X^X$. For instance, one can characterize when it's indeed a group of homeomorphisms. In opposite cases such as convergence actions, it consists of $G$ union functions that are constant outside a singleton. $\endgroup$
    – YCor
    Feb 10, 2020 at 14:12
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    $\begingroup$ You're idea reminds me of this paper by Küster where she relates the fixed space of the Koopman operator on a topological dynamical system to transfinite properties of the orbits of the underlying dynamic. $\endgroup$ Feb 10, 2020 at 20:07

1 Answer 1

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Here is a partial answer extracted from a Twitter user's answer that works for actions with uniformly bounded finite orbits, which indeed solves a special case that initially motivated this question.

For every sequence $\mathbf{g}=(g_{\alpha})_{\alpha < \lambda}$ and every $i \in \mathbb{N}$, set $\mathbf{g}^i$ to be the sequence of length $\lambda \cdot i$ given by $g^i_{\lambda j + \alpha}=g_{\alpha}$ for all $\alpha < \lambda$ and $j < i$. In other words, $\mathbf{g}^i$ is the concatenation of $i$-many $\mathbf{g}$'s back to back. Observe that $\mathbf{g}^i$ is valid whenever $\mathbf{g}$ is valid. Moreover, one can show by induction that for every valid sequence $\mathbf{g}$ and every $i \in \mathbb{N}$, we have $K_{\mathbf{g}}^i=K_{\mathbf{g}^i}$.

Lemma: Let $\mathbf{h}$ be a valid sequence such that $K_{\mathbf{h}}: X \rightarrow X$ is a bijection. If there exists $k \in \mathbb{N}$ such that $|\text{Orb}(x)| \leq k$ for all $x \in X$, then the map $K_{\mathbf{h}^{k!-1}}$ is the inverse of $K_{\mathbf{h}}$.

Proof: Assume that $k \in \mathbb{N}$ is a uniform bound for orbit sizes. It follows from the finiteness of orbits that the trajectory of a point under $\mathbf{h}$ must be eventually constant and hence, we must have $K_{\mathbf{h}}(x) \in \text{Orb}(x)$ for every $x \in X$. Consequently, $K_{\mathbf{h}} \upharpoonright \text{Orb}(x) \in \text{Sym}(\text{Orb}(x))$ for every $x \in X$. By Lagrange's theorem, $\left(K_{\mathbf{h}} \upharpoonright \text{Orb}(x) \right)^{k!} = \mathbf{id}_{\text{Orb}(x)}$ for every $x \in X$ and hence $K_{\mathbf{h}}^{k!} = \mathbf{id}_X$. By the previous lemma, $K^{k!-1}_{\mathbf{h}}=K_{\mathbf{h}^{k!-1}}$ is the inverse of $K_{\mathbf{h}}$.

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