Could you give an example of a unital simple $C^\*$-algebra that $\tau (p) = \tau (q)$ for all normalized traces does not imply $p \sim q$?



$\tau (p) = \tau (q)$ for all normalized traces does not imply $p \sim q$
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 $\mathcal{O}\_2$.


