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I am wondering whether a state $\omega$ on a $C^*$-algebra which is KMS (http://en.wikipedia.org/wiki/KMS_state) with respect to the group of automorphisms $\tau^t$, $t\in\mathbb{R}$, and at a given inverse temperature $\beta$ can be non-faithful. ($\omega$ is faithful iff $\omega(A^*A)=0$ for some element $A$ of the $C^*$-algebra implies $A=0$). Can anyone give me an example?

As a particular case: is there an example of a tracial state $\omega$ on a $C^*$-algebra which is not faithful? Tracial means that $\omega(AB)=\omega(BA)$ for all $A,B$, and since a tracial state is trivially KMS with respect to any group of automorphisms $\{\tau^t, t\in\mathbb{R}\}$, this would be an example of non-faithful KMS.

user162526
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Every non-simple finite-dimensional C$^*$-algebra has non-faithful traces.

For the simplest case, let $A=\mathbb C\oplus\mathbb C$, $\varphi(a,b)=a$. Then $\varphi$ is tracial, non-faithful.

You can generalize the same idea to any direct sum of tracial algebras. If $A_1,A_2,\ldots$ are C$^*$-algebras with tracial states $\varphi_1,\varphi_2,\ldots$ respectively, then $$ \varphi(a_1,a_2,\ldots)=\varphi_1(a_1) $$ is a non-faithful tracial state on $\bigoplus_nA_n$. Another way to describe this situation is the case of $A$ with trace $\varphi$ and $p\in A$ a non-trivial central projection. Then $a\mapsto \varphi(pa)$ is a non-faithful trace.

Martin Argerami
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  • Pro Argerami,I was troubled by your 2nd construction.If we let $\tau(a)=\phi(pa)$ ,how to check that there exists a nonzero elment $a_0$ such that $\tau(a_0^*a_0)=0?$ – math112358 Mar 21 '19 at 04:25
  • $a_0=1-p$ $\ \ \ \ $ – Martin Argerami Mar 21 '19 at 13:14
  • Thanks,but I don't understand why the non-trivial central projection must exist. – math112358 Mar 21 '19 at 16:01
  • I'm not sure what you are asking. Must exist where? I wrote that if $A$ has a trace and a non-trivial central projection, it has a non-faithful trace. – Martin Argerami Mar 21 '19 at 18:19
  • Pro Argerami,does there exist a connecttion between trace and projetion of a $C^$ algebra?If a $C^$ algebra $A$ have no projetions,can we conclude that $A$ has no traces? – math112358 Apr 10 '19 at 12:51
  • No, of course not. $C_0(\mathbb R)$ is projectionless, but it is abelian so every state is a trace. And you have projectionless group algebras like $C_r^*(\mathbb F_2)$; group algebras always have at the very least a canonical trace. – Martin Argerami Apr 10 '19 at 13:51