Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.