Could you give an example of a unital simple $C^*$-algebra that $\tau (p) = \tau (q)$ for all normalized traces does not imply $p \sim q$?
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$\begingroup$ Is that true for $n \times n$ complex matrices? $\endgroup$– Gerald EdgarCommented Feb 7, 2021 at 19:50
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3$\begingroup$ @GeraldEdgar Yes. $\endgroup$– David HandelmanCommented Feb 7, 2021 at 21:33
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1 Answer
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If a simple $C^*$-algebra admits an infinite projection $p$ (ie a projection that is equivalent to a proper subprojection $q$), then it does not carry any tracial trace and in particular it provides an example of the kind you are looking for. Indeed a trace $\tau$ would vanish on the nonzero projection $p-q$, and therefore vanish everywhere by simplicity, as $\{x | \tau(x^*x)=0\}$ is an ideal.
For an explicit example, take the Cuntz algebra $\mathcal{O}_2$.
answered Feb 7, 2021 at 21:19
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8$\begingroup$ I suspect that the OP wanted a simple C*-algebra which has a (necessarily faithful) trace. This is actually quite easy. Take any simple dimension group $G$ with infinitesimals; as a specific example, $Q^2$ with positive cone given by $0 \cup \{(a,b)| a + b > 0\}$; this has unique trace up to scalar multiple $(a,b) \mapsto a+b$; take as distinguished order unit (1,1)). There exists an AF C*algebra whose ordered K$_0$ group is $G$, and (1/2,-1/2) can be represented as [p-q] in K_0 where $p$ and $q$ are projections, e.g., represented by (1/2,0) - (0,1/2). The trace vanishes on this element. $\endgroup$ Commented Feb 7, 2021 at 21:33
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$\begingroup$ @DavidHandelman I do not know what the OP wanted, but your interpretation of the question is indeed imore nteresting. Thanks for the argument. $\endgroup$ Commented Feb 8, 2021 at 10:05