extension of a trace on the positive cone of a $*$-algebra

Question

Let $A$ be a $*$-algebra. Denote $A^+$ by the positive self-adjoint elements in $A$. If $\tau$ is a tracial state of $A^+$, can we construct a tracial state $\tilde{\tau}$ on $A$ such that $\tilde{\tau}|_{A}=\tau$?

Answer 1

$\newcommand{\Asa}{A_{\text{sa}}}\newcommand{\tt}{\tilde \tau }$There are few strong results on positivity that apply to general $^*$-algebras, so let me restrict to C*-algebras here. Among the crucial tools at our disposal is the fact that every self-adjoint element decomposes as a difference os positive elements.

Definition. Given a C$^*$-algebra $A$, a state on $A_+$ is a function $\tau :A_+\to {\mathbb R}_+$ that is additive ($\tau (a+b)=\tau (a)+\tau (b)$) and positively homogeneous $(\tau (\lambda a)=\lambda \tau (a)$ for $\lambda \geq 0$). If moreover $\tau (a^*a)=\tau (aa^*)$ for all $a$ in $A$, then $\tau $ is called a tracial state.

Theorem. Any state $\tau $ on $A_+$ admits a unique linear extension $\tt$ to $A$. If $\tau $ is moreover a tracial state, then $\tt$ is a trace.

Proof. Denote by $\Asa$ the set of all self-adjoint elements in $A$. For each $a$ in $\Asa$, write $a=b-c$, with $b,c\in A_+$, and set $$ \varphi (a) = \tau (b)-\tau (c). $$ If $b'$ and $c'$ satisfy the same properties as above then $$ b-c=b'-c' \quad\Rightarrow\quad b+c'=b'+c \quad\Rightarrow\quad \tau (b)-\tau (c) = \tau (b')-\tau (c'), $$ so we see tht $\varphi $ is well defined and it is easy to see that it becomes a real-linear functional $$ \varphi :\Asa\to \mathbb R. $$

Now, given any $a$ in $A$, observe that both $$ \text{Re}(a):= {a+a^*\over 2}, \quad \text {and} \quad \text{Im}(a):= {a-a^*\over 2i} $$ lie in $\Asa$, so we may define $$ \tt (a) = \varphi (\text{Re}(a)) + i \varphi (\text{Im}(a)), $$ and one may then prove that $\tt $ is a complex linear functional on $A$ extending $\tau $. The uniqueness of $\tt $ is clear.

Assuming that $\tau $ is a tracial state, we need to show that $\tt (ab)= \tt (ba)$, for every $a$ and $b$ in $A$.

For this, observe that if $u$ is a unitary element in the multiplier of $A$, and $a\in A_+$, then $$ \tt(u^*au) = \tt(u^*a^{1/2}a^{1/2}u) = \tt(a^{1/2}uu^*a^{1/2})= \tt(a). $$ By linearity we therefore also have that $$ \tt(u^*au) = \tt(a), $$ for every element $a$ in $A$. It $u$ is again assumed to be a unitary multiplier, then $$ \tt(ua) = \tt(u^*(ua)u) = \tt(au). $$ Any element $b$ in $A$ may be written as the linear combination of four unitary multipliers, so it follows that $$ \tt(ba) = \tt(ab). $$ QED.