Let A be a $C^\*$-algebra with unit $I$, and G a locally compact (Hausdorff) group. An action $\alpha$ of G on A is a strongly continuous homomorphism of G into Aut(A), the group of \*-automorphisms of A. We say that $\alpha$ is ergodic if the elements $\lambda I$, with $\lambda$ any complex number are the only elements invariant under $\alpha$.


If A is commutative, i.e. $A\cong C(X)$ where $X$ is a compact Hausdorff space (the spectrum of A), giving an action $\alpha$ on A is equivalent to giving an action $\phi$ of $G$ on the spectrum by homeomorphisms, where for any $f\in C(X)$ we have
$$(\alpha\_g (f))=f(\phi\_g^{-1}(x))\quad \forall x\in X$$


(1) is the strong continuity of $g\mapsto\alpha\_g$ equivalent to continuity of the map $g\mapsto\phi\_g$ if we endow the set of homeomorphisms:$X\rightarrow X$ with the compact-open topology?


It is easy that under the assumption $G$ compact, ergodicity is equivalent to transitivity of the action $\phi$ on the spectrum X (using the averaging operator), but


(2) what can we say about ergodicity in the case where $G$ is just locally compact (and Hausdorff)? Is ergodicity of $\alpha$ equivalent to 'topological' ergodicity, i.e. the only open (equivalently closed) subsets invariant for $\phi$ are X and $\emptyset$


If we have non-constant invariant functions on $X$, then the pre-image of a value taken is a proper invariant closed of $X$, but what about the other implication?


Thank you in advance



algebraic VS topological ergodicity
Given there are no other answers, let me say something about (1). We need only check continuity at the identity of $G$. Let $(g\_i)$ be a net converging to $e\_G$. That $\alpha\_{g\_i}\rightarrow I$ means that for each $f\in C(X)$, we have $\|f\circ\phi\_{g\_i^{-1}} - f\|\_\infty \rightarrow 0$. That $\phi\_{g\_i^{-1}} \rightarrow I$ means that whenever $K\subseteq X, U\subseteq X$ are compact (=closed) and open, respectively, with $K\subseteq U$, we have that $\phi\_{g\_i^{-1}}(K) \subseteq U$ for large $i$. (This is a basic open set about $I$ in the compact-open topology). To show that $\alpha$ continuous implies that $\phi$ is continuous, you can use a simple Urysohn's lemma argument (to find a function $1$ on $K$ and $0$ off $U$).


Conversely, suppose $\phi$ is continuous, and let $f\in C(X)$. For $\epsilon>0$ we can cover $f(X)\subseteq\mathbb C$ by a finite number of closed discs of radius $\epsilon/2$, say $(L\_k)$. Then cover each $L\_k$ by an open disc of slightly larger radius, say $V\_k$. Then $K\_k=F^{-1}(L\_k)$ is closed in $X$, and $U\_k=f^{-1}(V\_k)$ is open, and contains $K\_k$. So for $i$ large, by continuity of $\phi$, we have that $\phi\_{g\_i^{-1}}(K\_k) \subseteq U\_k$. Thus $f(x) \in L\_k \implies x\in K\_k \implies \phi\_{g\_i^{-1}}(x) \in U\_k \implies \alpha\_{g\_i}(f)(x) \in V\_k$. Hence $\|f - \alpha\_{g\_i}(f)\|\_\infty$ is small. So $\alpha$ is continuous.


I think a good reference for the locally compact case is Dana Williams's book "Crossed Products of $C\*$-Algebras". It nicely dots all the is and crossed all the ts.


For (2): let $G=\mathbb Z$, so $\phi$ is generated by a single homeomorphism of $X$. Let $X=\{ z\in\mathbb C : |z|\leq 1\}$ and define $\phi(re^{i\theta}) = r^2 e^{i\theta}$. Then $0$ and all points on the circle are fixed; the orbit of any other point $re^{i\theta}$ has accumulation points $0$ and $e^{i\theta}$. Let $f\in C(X)$ be invariant; translate so $f(0)=0$. Then $f(re^{i\theta}) = \lim\_n f(r^{2n}e^{i\theta})=0$ for all $r<1$, so by continuity $f=0$. Thus the action $\alpha$ is "ergodic", but $\phi$ has non-trivial invariant open and closed sets.


**Edit:** An easier example has $X=[0,1]$ and $\phi(s)=s^2$. Then $\{0\}, \{1\}$ are non-trivial fixed closed sets, and $(0,1)$ is an invariant open set; but again $\alpha$ only leaves the constant functions invariant.


