algebraic VS topological ergodicity 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
 A: 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.
