# $V(A)$ semi group of equivalent projections in $M_∞(A)$ cancelative?

I found in the book of Murphy, C*- Algebras and Operator Theory, the Theorem 7.1.2 :

Let A be an unital C* algebra, the semi group $V(A)$ of equivalent projections (under Murray Von Neumann equivalence) in $M_∞(A)$ is cancellative.

For the proof he proceeds as follow :

• take some projection $p, q, r$ such that $[p] \oplus [r] = [q] \oplus [r]$
• add to it $[I_n - r]$ if $r\in M_n(A)$ to obtain $[p] \oplus [I_n] = [q] \oplus [I_n]$

And he concludes with that, and I don't see why that should be true that one can simplify $[I_n]$.

Furthermore, one knows that there exists C*algebras where the $K^0$ group (the Grothendieck group of $V(A)$ ) is of Torsion, for exemple, the Cuntz algebra with $K^0$ group $\mathbb{Z} / n\mathbb{Z}$. For this algebra the morphism of semi group $V(A) \rightarrow K^0(A)$ cannot be injective and $V(A)$ cannot be cancellative, right ?

So my question is who is right ? Me or Murphy ^^ ?

• Cancellative does not mean torsion-free, so I'm not sure that your example of ${\mathcal O}_n$ has much bearing on your actual question Feb 11, 2016 at 12:13
• but $V(A)$ is always torsion free right ? The only projection equivalent to $0$ is $0$ itself. Feb 11, 2016 at 13:08

Theorem 7.1.2 in Murphy actually says that for $A$ a unital C$^*$-algebra the semigroup $K_0(A)^+$ is cancellative. But $K_0(A)^+$ is the semigroup of stably equivalent projections in $M_\infty(A)$ not just Murray von Neumann equivalent.
As a reminder $P,Q$ projections in $M_\infty(A)$ are stably equivalent if and only if there exists $n$ such that $P\oplus I_n$ is Murray-von Neumann equivalent to $Q\oplus I_n$.
• Oh ! Ok, you're right, thanks a lot :) I had a look first in the book of Bruce Blackadar. He took the Murray Von Naumann equivalence to define $V(A)$. However those two semi groups lead to the same group and to the same K theory. Feb 11, 2016 at 15:59
Let $R$ be a projection in $B(l^2)$ onto a subspace whose dimension and codimension are both infinite. Then $[P] \oplus [R] = [Q] \oplus [R]$ for any projections $P$ and $Q$ in $B(l^2)$, but not all such $P$ and $Q$ are equivalent.