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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.

sirjoe
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  • Can you prove it for matrix algebras? Don't finite dimensional C*-algebras have a representation theorem? – Berci Jan 06 '21 at 23:30
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    By finite I mean that the identity is a finite projection so that it is not Murray von neumann equivalent to a proper subprojection. I do not mean finite dimensional. For finite dimensional C* algebras its clear because projections are Murray von neumann equivalent iff they have the same rank iff they are unitarily equivalent. – sirjoe Jan 06 '21 at 23:40
  • Ah ok. Maybe you can clarify this in the post as well. Also, please add the definition of 'unitary equivalence' and 'Murray von Neumann equivalence' to make it more self coherent. – Berci Jan 06 '21 at 23:44
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    Are you really asking about comparison of projections on arbitrary C$^*$-algebras, and not on von Neumann algebras? – Martin Argerami Jan 07 '21 at 00:12
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    Well yes, I understand the factor setting. But basically I want to understand when Murray von Neumann equivalence implies unitary equivalence in C* algebras but the only counterexamples I know are in the infinite case. – sirjoe Jan 07 '21 at 09:50

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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.

Ruy
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  • Thank you! So in this case $1_n-p$ and $1_n-q$ are not comparable as projections, right? I.e, neither is Murray von Neumann subequivalent to the other? – sirjoe Jan 07 '21 at 16:21
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    Correct! As vector bundles $1_n-p$ and $1_n-q$ have the same rank so they can only be comparable if equivalent, which they are not. – Ruy Jan 07 '21 at 17:18