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Let $G$ be a discrete countable infinite group acting on a compact metric space $X$ via homeomorphisms preserving a probability measure $\mu$.

A function $\lambda\colon G\to \mathbb C$ is an eigenvalue of the action of $G$ if there exists a function $f\in L^2(X,\mu)$ such that for every $g\in G$ one has $\lambda(g)\cdot f=f\circ g$.

In this paper: Ergodicity of the Cartesian product by E. Flytzanis, Trans. Amer. Math. Soc. 186 (1973), 171-176 (freely available link at AMS site), there are two results concering the product of two dynamical systems given by $\mathbf Z$-action:

  • a sufficient condition for the ergodicity of the product,
  • a description of the spectrum of their cartesian product.

It is also written there that the above results hold also for $G$-action, but it is not specified what does it exactly mean. Am I right that:

  • A product of $(X,G)$ and $(Y,G)$ is ergodic if the function constantly equal to 1 is the only common eigenvalue for $(X,G)$ and $(Y,G)$?
  • The set of eigenvalues of $X\times Y$ equals the set of all functions of the form $f\cdot g$ (pointwise multiplication), where $f$ is an eigenvalue for $X$ and $g$ is an eigenvalue for $Y$?

I need also two other properties, but I could not find the appropriate references for them (maybe I am wrong that they are true?):

  • The set of eigenvalues of a factor of a dynamical system $(X,G, \mu)$ is contained in the set of eigenvalues of $(X,G,\mu)$?
  • Every eigenvalue of an ergodic dynamical system is simple (beware: I do not assume that $G$ is abelian).
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    $\begingroup$ For your first question the answer is "no". This is already seen when considering the regular representation of a finite group with trivial abelinzation. As a general rule of thumb, it is convenient to replace "eigenvalues" with "finite dimensional subrepresentations" when considering non-commutative groups. Upon such a replacement you'd get a positive answer. Your second question is badly formulated. Are you looking for eigenvalues or eigenvectors? $\endgroup$
    – Uri Bader
    May 19, 2020 at 9:44
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    $\begingroup$ The last two properties that you ask about are correct. For the first, a representation of a factor is a subrepresentation, and for the second, in the particular case of a compact group, every ergodic representation is a factor of the regular. Note that when asking about eigenvalues, the question always reduces to the case of a compact group. $\endgroup$
    – Uri Bader
    May 19, 2020 at 9:45
  • $\begingroup$ Thank you, Uri Bader! Your answer is really helpful. For the first question - could you recommend some reference for this topic? In the second question I meant eigenvalues. If I understand correctly, an eigenvalue is a function from $G$ to $\mathbb C$ so I can consider pointwise multiplication (that is $f\cdot g(x)=f(x)\cdot g(x)$). $\endgroup$
    – Martha Łącka
    May 20, 2020 at 10:46
  • $\begingroup$ I am sorry for the confusion - I used $f$ and $g$ for eigenvalues which is probably not standard. $\endgroup$
    – Martha Łącka
    May 20, 2020 at 10:58
  • $\begingroup$ oh... this is a horrible notation you adopted there. The answer is negative here as well. I'll write it in an answer format. $\endgroup$
    – Uri Bader
    May 21, 2020 at 7:58

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The answer to both question is negative.

Take $G=S_3$, the symmetric group of the set $X=\{1,2,3\}$. Then $L^2(X)$ is decomposed to the trivial representation and a another two dimensional irreducible representation. As a $G$-space, $X\times X\simeq X \cup G$ where $X\subset X\times X$ is the diagonal and $G$ corresponds to the rest. This shows that $X\times X$ is not ergodic, providing a counter example to the first question. It also follows that $L^2(X\times X)$ contains a sub-representation isomorphic to the regular representation $L^2(G)$ and in particular it contains the one dimensional sign representation. This provides a counter example to the second question.

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  • $\begingroup$ Thank you for the answer :) $\endgroup$
    – Martha Łącka
    May 21, 2020 at 11:06

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