In their paper: "Addition of $C^\*$-algebra extensions", G. A. Elliott and D. E. Handelman have discussed some relation between traces and equivalence of projections in $M(A)$, where $M(A)$ is the multiplier algebra of $A$. One result is the following:


2.4. COROLLARY. Let $A$ be a separable AF algebra, and let $e$ and
$f$ be projections in $M(A)$. Suppose that neither $eAe$ nor $fAf$ has a
**nonzero unital quotient**. The following two conditions are equivalent:


(i) $e$ is equivalent to $f$ in $M(A)$.


(iii) $\tau(e) = \tau(f)$ whenever $\tau$ is a semifinite lower semicontinllous
trace on $A\_{+}$.


In their paper, they did not give the definition of "non zero unital quotient". I can't find the definition in the literature. Do it mean that the quotient algebra is not unital? specially, when it is simple(has no ideal)? I don't know.


But I also learn the following result in the literature:


If $A$ is a separable nonunital matroid algebra(this is also a special AF algebras), then there always exist projections $p \in M(A)\backslash A$ and $q \in A$ such that $\tau(p) = \tau(q)$, p and q can not be equivalent. where $\tau$ is the (unique) trace of $A$.


Do we have some contradiction between the two results above? Since the matroid algebra is simple. Hope some help!



when does a $C^*$-algebra have no nonzero unital quotient?
It seems that "$A$ has a nonzero unital quotient" should mean: there exists a proper ideal $J$ of $A$ (possibly $J=0$) such that $A/J$ is unital. You would be correct to say that, if $A$ is nonunital and simple, then it has no nonzero unital quotient.


Elliott-Handelman's result does not contradict the second result that you cited, because in that case, $qAq$ is unital. 


