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Say $\mathfrak{A}$ is a seperable $C^*$ algebra, the space $K$ of states on $\mathfrak{A}$ is compact and convex. Let $\Gamma$ be a countable discrete group acting on $\mathfrak{A}$ via $*$-hom. This gives us an action on $K$, namely $$\gamma.\tau(a)=\tau(\gamma^{-1}a)$$ for every $\tau\in K$. Now given a probability measure $\mu\in Prob(\Gamma)$, we can define the convex compact set of $\mu$-stationary states $$\{\tau\in K:\sum_{\gamma\in\Gamma}\mu(\gamma)\gamma.\tau=\tau\}$$ I am interested in the extreme points of this set. I was not able to find anything online in this regard. One can also define the operator system $$\{a\in\mathfrak{A}:\sum_{\gamma\in\Gamma}\mu(\gamma)\gamma.a=a\}$$ Are there any known relations between the state space of this operator system and restrictions of the $\mu$-stationary states?

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  • $\begingroup$ @Yemonisloggedoutrightnow this is an operator system (closed under star since the action is by * hom). Fixed $\endgroup$ Commented Oct 27 at 16:55
  • $\begingroup$ At least there isn’t a one-to-one correspondence, if that’s what you’re asking for. For example, let $\Gamma=\mathbb{Z}$ act on $\mathfrak{A}=C([0,1])^{\otimes\mathbb{Z}}$ via Bernoulli shift and let $\mu$ be the Dirac delta at one of the generators on $\Gamma$. Then the operator system you wrote down is just $\mathbb{C}$, so it has exactly one state. But there are plenty of $\mu$-stationary measures (for each state on $C([0,1])$ the infinite tensor state on $\mathfrak{A}$ is even $\Gamma$-invariant, for example). $\endgroup$ Commented Oct 27 at 20:12
  • $\begingroup$ Instead of considering the operator system of $\mu$-stationary elements of $\mathfrak{A}$, you should consider the operator system of $\mu$-stationary elements of the double dual of $\mathfrak{A}$. The space of normal states on that operator system is actually naturally in one-to-one correspondence with $\mu$-stationary elements of $K$ ($\mu$-stationary measures in the terminology you used). $\endgroup$ Commented Oct 27 at 20:22
  • $\begingroup$ @DavidGao Have you got a reference? $\endgroup$ Commented Oct 28 at 6:57
  • $\begingroup$ I don't have a reference. I just thought of an argument in my head. But now that I thought about this again, I realized my argument is wrong. I mistakenly thought that there would be a normal conditional expectation from $\mathfrak{A}^{\ast\ast}$ onto the sub-operator system consisting of $\mu$-stationary elements, but that isn't correct. There is a conditional expectation, but it doesn't have to be normal. So, the correct statement should have been the space of $\mu$-stationary states on $\mathfrak{A}$ is only a retract (via affine maps) of the space of normal states on the operator system... $\endgroup$ Commented Oct 31 at 7:08

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