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$\newcommand\Cb{C^\text b}$Let $\Cb(\mathbb R)$ be the C*-algebra formed by all bounded, continuous, complex valued functions on $\mathbb R$. Consider the action $\tau $ of $\mathbb R$ on $\Cb(\mathbb R)$ given by $$ \tau _t(f)\mathclose|_s = f(s-t), \quad \forall f\in \Cb(\mathbb R), \quad \forall s,t\in \mathbb R. $$

The spectrum of $\Cb(\mathbb R)$ is well known to be the Stone–Čech compactification $\beta (\mathbb R)$, so we get an action $\hat \tau $ of $\mathbb R$ on $\beta (\mathbb R)$ by duality, which clearly extends the usual action of $\mathbb R$ on itself by translation.

Evidently $\mathbb R$ is a $\hat \tau $-invariant open subset of $\beta (\mathbb R)$, whence the "corona" $$ \partial (\mathbb R)\mathrel{:=} \beta (\mathbb R)\setminus \mathbb R $$ is closed and invariant.

It is easy to see that $\partial (\mathbb R)$ is not minimal among closed invariant subsets because $\partial (\mathbb R)$ splits as the disjoint union of the following two smaller closed invariant subsets: $$ \partial_+ (\mathbb R) = \overline{(0, +\infty )} \setminus \mathbb R, \quad \text{and} \quad \partial_- (\mathbb R) = \overline{(-\infty, 0)} \setminus \mathbb R. $$

Question. Are $\partial_+ (\mathbb R)$ and $\partial_- (\mathbb R)$ minimal? If not, what are examples of minimal closed invariant subsets?

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  • $\begingroup$ I was tempted to say they're not minimal because $\mathbf{R}$ is separable, while these subsets are not. However, I don't think the $\mathbf{R}$-action on $C^\mathrm{b}(\mathbf{R})$ is continuous (consider the action on $f$, where $f(x)=\sin(x^2)$) so the argument doesn't apply. $\endgroup$
    – YCor
    Commented Nov 21, 2020 at 17:37
  • $\begingroup$ That is a very good point. So perhaps I should have emphasized that one should see the acting group as the discrete real line. I suppose one needs to consider the Bohr compactification if one wants a continuous action, right? Do you think the complement of $\mathbb R$ in its Bohr compactification is separable? @YCor $\endgroup$
    – Black
    Commented Nov 21, 2020 at 19:47
  • $\begingroup$ @Black The compactification obtained from the algebra of bounded right uniformly continuous functions should be the biggest over which the left translation action is still continuous. It is $\beta G$ when $G$ is discrete, anche the Samuel compactification of the right uniformity of $G$ in general (the same space with the distinguished point $e$ is the greatest ambit of $G$) $\endgroup$ Commented Nov 23, 2020 at 13:23

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$\newcommand{\Cb}{C^{\text b}}$I think I have a negative answer to my own question. Consider the function $f(x)=\sin(1/x)$, defined for $x$ in $(0,+\infty )$, and let us denote the graph of $f$ by $G$.

Let $\alpha $ be the arc length parametrization of $G$ oriented in such a way that $\alpha (s)$ approaches the vertical axis when $s\to +\infty $.

Next reparametrize $\alpha $ by defining $\gamma (s)=\alpha (\root 3 \of s)$, and observe that $$ \lim_{s\to\infty }\|\gamma '(s)\| = 0. \tag 1 $$

Let us denote the closure of $G$ by $\mathcal R$ and let us consider the action $\sigma $ of $\mathbb R $ on $\mathcal R$, defined to be trivial on the vertical interval $I:= \{0\}\times [-1,1]$ and defined by $\sigma _t(\gamma (s)) = \gamma (t+s)$ on $G$. It is evident that $\sigma $ is continuous on $\mathbb R\times G$, so let us prove that it is also continuous on $\mathbb R\times I$. We therefore suppose that $(t_n,\gamma (s_n))\to (t,x)$, with $x\in I$, and we must prove that $\gamma (t_n+s_n)\to x$. Observing that necessarily $s_n\to+\infty $, this may be easily done employing the mean value Theorem and (1) as follows: \begin{align} \|\gamma (t_n+s_n)-x\| & \leq \|\gamma (t_n+s_n)-\gamma (s_n)\| + \|\gamma (s_n)-x\| \cr & \leq \sup_{t\in [s_n, s_n+t_n]}\|\gamma '(t)\| |t_n| + \|\gamma (s_n)-x\| \cr & \to 0. \end{align}

The map $\phi: \Cb(\mathcal R) \to \Cb(\mathbb R )$, defined by $\phi(f) = f\circ \gamma $, is clearly a covariant *-monomorphism relative to $σ$ and $τ$. By duality it therefore leads to a covariant surjection $$ \hat \phi :\beta(\mathbb R) \to \widehat{\Cb(\mathcal R)}.$$ Picking any $x$ in the vertical interval $I$, hence a fixed point for $\sigma $, the evaluation at $x$ gives an element $\text{ev}_x$ in $\widehat{\Cb(\mathcal R))}$ and $$ \hat\phi^{-1}(\{\text{ev}_x\}) $$ is therefore a closed $\hat\tau$-invariant subset which may be proved to be a proper subset of $\partial _+(\mathbb R)$ (as defined in the original post). Therefore $\partial _+(\mathbb R)$ is not minimal!


The last part of the question, namely asking for examples of minimal closed invariant subsets is still wide open!

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