A dynamical system $f:X\to X$ is said to be topologically transitive if for any two nonempty open sets $U,V$ there exists $n \in \mathbb{Z}$ such that $f^{\circ n}(U) \cap V \neq \emptyset$. The following two definitions of topological weak mixing can be found in the literature:

  1. $f$ is topologically weak mixing if $f\times f$ is topologically transitive on $X\times X$ (e.g here)
  2. $f$ is topologically weak mixing if it has no non-constant continuous (with respect to the topology) eigenfunctions of the shift operator. (e.g. here)

How are these definitions related? Can they be shown to be equivalent?

  • $\begingroup$ I suppose you want to add that $X$ is a (compact) metric or Hausdorff space and that the dynamical system map $f:X \to X$ is a continuous map. $\endgroup$ Apr 1, 2022 at 7:54
  • $\begingroup$ Please elaborate on what exactly you understand as the eigenfunction of the shift operator? As I understand it, by an eigenfunction of a topological dynamical system an eigenfunction of the corresponding Koopman operator $T_f: \mathcal{F} \to \mathcal{F}$, where $\mathcal{F}$ is a complex linear subspace of $\mathbb{C}^X$. A canonical choice for $\mathcal{F}$ is the Banach space $C(X)$ of complex-valued continuous functions on $X$. But other spaces can be considered as well. However, the Eigenstructure of the Koopman operator may depend on the space $\mathcal{F}$. $\endgroup$ Apr 1, 2022 at 8:03
  • $\begingroup$ I forgot to mention how $T_f$ is defined. It maps $h \in \mathcal{F}$ to $h \circ f$. Certainly, we must assume that $\mathcal{F}$ is closed under the action of $T_f$. $\endgroup$ Apr 1, 2022 at 12:40

1 Answer 1


Which are your regularity assumptions on $f$?

In general there are some subtle differences between the two definitions. If $X$ is a Baire space, then 1. implies (a slightly modified version of) 2. See on this Theorem 2.3 in H.B. Keynes and J.B. Robertson, Eigenvalue Theorems in Topological Transformation Groups.

In the same paper, Theorem 2.5 provides conditions making the two definitions basically equivalent. Theorem 2.8 shows instead a case of non-equivalence.

Notice that:

  1. the mentioned paper concerns the more general setting of topological transformation groups;
  2. the authors mean "almost everywhere" in a topological sense (= everywhere except on a meagre set);
  3. they allow (unless differently specified) a meagre discontinuity set.
  • $\begingroup$ I'm confused, it looks like they allow a meager set of points of discontinuity, am I reading this wrong? $\endgroup$
    – Ville Salo
    Apr 1, 2022 at 5:28
  • $\begingroup$ Yes, they say specifically "continuous transformation group" when continuity is assumed. In Theorem 2.3 meagre discontinuities are allowed. That's why I said "in general" and phrased the following with some caveats. I clarified the issue at the end of the answer. $\endgroup$ Apr 1, 2022 at 7:31
  • $\begingroup$ @AlessandroDellaCorte For the record, the function algebra that Keynes & Robertson use is the algebra of functions with "Baire property" (or "Baire measurable" functions, in other terminology); these are functions that are continuous modulo the $\sigma$-ideal of meager sets (in analogy with ergodicity, we don't get constancy everywhere). To me Tsirelson's account of this in Lecture 7 at tau.ac.il/~tsirel/Courses/MeasCategory/main.html is one of the clearest. $\endgroup$
    – Alp Uzman
    Nov 5, 2022 at 20:30

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