# When is the corner algebra $PM_n(A)P$ isomorphic to $A$?

In the algebra of matrices $M_n(A)$ over a $C^*$ algebra $A$, consider the corner algebra $PM_n(A)P$ for a Hermitian projection $P\in M_n(A)$. Is there any condition known for $P$ to make $PM_n(A)P$ isomorphic to $A$? If not, where should I look for what is known on the structure of $PM_n(A)P$? (This question is aimed at linking the peoperties of $P$ to $PM_n(A)P$ in general, rather than for the case of specific $C^*$ algebras.)

The background: A Morita context from $A$ to itself can be viewed as a left finitely generated projective $A$-module $L$ whose left module maps are themselves isomorphic to $A$ - thus giving a bimodule structure. If $L$ has associated projection matrix $P$ in $K$-theory, then the left module maps are given by the corner algebra $PM_n(A)P$, thus the question of when this is isomorphic to $A$. (This is the noncommutative analogue of a line bundle in geometry.)

• Larry Brown's paper, Stable isomorphism of hereditary subalgebras of C*-algebras. Pacific J. Math. 71 (1977), no. 2, 335–348, is probably a good place to start. – Caleb Eckhardt Jun 9 '14 at 22:33
• Just got a copy of Brown's paper - I will read it! – Edwin Beggs Jun 10 '14 at 13:23

If $A$ is unital, then an obvious sufficient condition would be that $P$ is Murray-von-Neumann equivalent to the unit $1_A$ of $A$. (Where we view $A$ as sitting canonically in $M_n(A)$ as the upper left corner, or more algebraically, $A$ is identified with $e_{11}\otimes A\subset M_n(A)$).
Because if $v\in M_n(A)$ is a partial isometry with $v^*v=1_A$ and $vv^*=P$, then the map $A\to PM_n(A)P$ defined by $x\mapsto vxv^*$ yields an isomorphism.
• This seems wrong to me. The map you define gives an isomorphism $M_n(A) \to PM_n(A)P$, rather than $A \to PM_n(A)P$. More generally, if two projections $P$ and $Q$ are Murray-von Neumann equivalent, then a partial isometry inducing the equivalence should induce an isomorphism between the corners $PM_n(A)P$ and $QM_n(A)Q$ (and there is nothing terribly special about the algebra $M_n(A)$ here). – Manny Reyes Jun 9 '14 at 20:41
• Note the identification of the unit $1_A$ with $e_{11}\otimes 1_A$, as I mentioned right at the beginning of my answer. You have $(e_{11}\otimes 1_A)M_n(A)(e_{11}\otimes 1_A)\cong A$, so what you say is exactly the point. – Gabor Szabo Jun 10 '14 at 7:10
• Oops! My apologies for misreading $1_A$ for $1_{M_n(A)}$! – Manny Reyes Jun 10 '14 at 11:26