Murray V.N equivalence of projections in finite C* algebras.

Question

In general, Murray von Neumann equivalence of projections is not the same as unitary equivalence. In fact in a unital infinite C* algebra 1 is Murray von Neumann equivalent to a proper subprojection p and so these are going to be murray von Neumann equivalent but certainly not unitarily equivalent. I was wondering if in a finite C* algebra these equivalence relations are the same? I certainly can't find a counterexample but I struggle to show in general that if p Murray von Neumann equivalent to q the 1-p is Murray von Neumann equivalent to 1-q when the C* algebra is finite.

Answer 1

Let $\mathbb K$ be either the field of real or complex numbers. A $\mathbb K$-vector bundle $\eta $ over a compact Hausdorff space $X$ is said to be stably trivial if there are integers $k, n\geq 0$, such that $$ \xi _k\oplus \eta = \xi _n, \tag 1 $$ where $\xi _k$ and $\xi _n$ are the trivial bundles of rank $k$ and $n$, respectively.

In case $\mathbb K = \mathbb R$, an example is the tangent bundle of the sphere $S^2$ which is not trivial by the hairy sphere Theorem, but is stably trivial since its direct sum with the (trivial) normal bundle is the trivial bundle of rank 3.

Assuming we are in the situation of (1), consider the algebra $A=M_n\big (C_{\mathbb K}(X)\big )$. Then we may find pairwise orthogonal projections $p,q\in A$, corresponding to $\xi _k$ and $\eta $, respectively, such that $p+q=1_n$, where $1_n$ is the identity of $A$.

One then has that $p$ is Murray-von Neumann equivalent to $1_k$, but $1_n-p$ is not Murray-von Neumann equivalent to $1_n-1_k=1_{n-k}$ because $1_n-p=q$, and $\eta $ is not the trivial bundle.