Suppose $A$ has a tracial state $\psi$, I want to prove $A/ker(\pi_{\psi})$ has a faithful tracial state,where $\pi_{\psi}$ is the $GNS$ respresentation with respect to $\psi$. My thought: define $\tau(a+ker(\pi_{\psi}))=\psi(a)$,it is well defined, in order to prove it is faithful, let $a\in A/ker(\pi_{\psi})$.

Then $\psi(a*a)=\langle\pi_{\psi}(a^*a)e_{\psi},e_{\psi}\rangle=0$, where $e_{\psi}$ is the cyclic vector of $\psi$. How to show that $\pi_{\psi}(a)=0$?


1 Answer 1


As you said, assuming you take $b=a+ker(\pi_\psi)\in A/ker(\pi_\psi)$ such that $\tau(b)=0$, then $0 = \tau(bb^*) = \tau(b^*b)=\psi(a^*a) =\langle\pi_{\psi}(a^*a)e_{\psi},e_{\psi}\rangle=\langle\pi_{\psi}(a)\pi_\psi(a)^*e_{\psi},e_{\psi}\rangle=\langle\pi_\psi(a)^*e_\psi,\pi_\psi(a)^*e_\psi\rangle$, so $\pi_\psi(a)^*e_\psi = 0$, using the symmetry of the trace and the fact that $\pi_\psi$ is a *-morphism. I claim that $a\in ker_(\pi)$ automatically follows.

Indeed, you want to show that $\pi_\psi(a)\pi(A)e_\psi = 0$, and it will imply (because $e_\psi$ is cyclic for $\pi$) that $\pi_\psi(a)H_\pi = 0$, hence $\pi_\psi(a)=0$, so $a\in ker(\pi_\psi)$, hence $b=0$.

But this is immediate, since $\vert\vert\pi_\psi(a)\pi(c)e_\psi\vert\vert^2=\langle\pi_\psi(a)\pi_\psi(c)e_\psi,\pi_\psi(a)\pi_\psi(c)e_\psi\rangle=\langle\pi_\psi(c^*a^*ac)e_\psi,e_\psi\rangle = \psi((c^*a^*)(ac))=\psi((ac)(c^*a^*))=\langle\pi_\psi(a)\pi_\psi(c)\pi_\psi(c^*)\pi_\psi(a^*)e_\psi,e_\psi\rangle = 0$.


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