Does countable decomposable von Neumann algebra will necessary imply that hilbert space has to be separable where the von Neumann algebra acts, if not give a counterexample.
2026-03-25 14:32:41.1774449161
Countable decomposable von Neumann algebra
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Counterexample: $\mathbb C\,I\subset B(\ell^2(\mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".
This example is non-degenerate.