# Can anyone give an example of two stably equivalent projections that are not Murray Von Neumann equivalent?

Two projections $P$, $Q$ are MvN equivalent in $C^*$-algebra $A$ when there is an element $u\in A$ such that $uu^*=P$ and $u^*u=Q$, and two projections $P$, $Q$ are stably equivalent if $P\oplus 1_n\overset{MvN}{\sim}Q\oplus1_n$, is there an example of two projections stably equivalent but not MvN equivalent?

• I think I found an answer here math.stackexchange.com/questions/8795/… Mar 30, 2014 at 4:05
• But I am not able to understand the last comment there, can anybody explain it a little bit? Mar 30, 2014 at 6:42
• I got it, but I am not able to construct a concrete example here Mar 30, 2014 at 7:20

Note that Murray-von Neumann equivalence depends on the algebra where you consider it. Usually stably equivalence of projections is considered on $(A\otimes K(H))^{**}$, and you require $P\otimes1\sim Q\otimes1$.
So let $A=M_3 (\mathbb C)$, $P=E_{22}+E_{33}$, $Q=E_{33}$ (or any other rank-one projection for that matter), which are non equivalent as they have different rank. We have $$(A\otimes K(H))^{**}=M_3 (K(H))^{**}=M_3(B(H))=B(H\oplus H\oplus H).$$ Here, $$P\otimes1=0\oplus1\oplus1\sim0\oplus0\oplus1=Q\otimes 1.$$ The necessary partial isometry is induced by the isomorphism $H\oplus H\simeq H$.
If you want to require $P\oplus 1\sim Q\oplus 1$, it is not clear to me over what algebra you would consider the equivalence. If you want $A\oplus B(H)$, then stable equivalence is exactly the original equivalence. And if you don't, you need to specify how you will embed $A$ in some direct summand, in order for $P\oplus 1$ to make sense.