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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!

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up vote 4 down vote accepted

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.

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Thank you very much! If that is the definition, then there is no problem, also no contradiction. But can you suggest me some reference about this concept? – Aviv Mar 10 '12 at 15:02
Although I can't offer a reference, I would justify the definition I gave based simply on the wording of "$A$ has a nonzero unital quotient". A quotient of $A$ is simply $A/J$ for some ideal $J$, and the quotient is nonzero so long as $J$ is a proper ideal. – Aaron Tikuisis Mar 10 '12 at 19:01
Some basic examples of classes of $C^∗$-algebras which have no nonzero unital quotients are: nonunital simple $C^∗$-algebras and stable $C^∗$-algebras (i.e. $C^∗$-algebras of the form $A \otimes \mathcal{K}$, where $\mathcal{K}$ denotes the algebra of compact operators of a separable Hilbert space). – Aaron Tikuisis Mar 10 '12 at 19:12
Oh, I am very sorry for my delay! Thank you very much, Aaron! I think I have understanded this concept by your interpretation. I see in another paper written by Prof. K. R. Goodearl.(I am sure you know which paper.) In his paper, Goodearl called this cencept for "no nonzero unital $C^*$-homomorphic images", and under some additional assumption, he generalized many results in Elliott-Handelman's paper. Thanks again! – Aviv Mar 18 '12 at 4:44

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