# 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?

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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.

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I'm not quite sure what "locally closed orbit" means, so I'm not sure how this example interacts with pm's answer... – Matthew Daws Mar 2 '12 at 21:50
Thank you Matthew, but there is still something that I don't understand. Following your argument we prove that every invariant function is the zero function, contradicting the fact that every *-automorphism of $C(X)$ leaves $I$ fixed, i.e. $I$ must be fixed by an automorphic action (which is equivalent to the topological one in the commutative case). Tell me if it's non-sense. Thank you again, Best regards – johnnyblade Mar 3 '12 at 7:22
@johnnyblade: I "translated" $f$ so that $f(0)=0$, and then conclude $f=0$. So actually, the starting $f$ could be anything. It also occurs to me that the same idea works on $X=[0,1]$, just let $\phi(t)=t^2$. – Matthew Daws Mar 3 '12 at 9:17
Of course thank you Matthew, I misread the word 'translate'. Regards – johnnyblade Mar 3 '12 at 10:37