# Decomposition of a von Neumann Algebra into Factors

I know that a von Neumann algebra on a separable Hilbert space can be (uniquely) decomposed into factors. But is there any non-trivial example showing how we can explicitly compute it?

Non-trivial: I mean it is not a factor itself, not abelian or of finite dimension.

• You should also consider the finite-dimensional case as trivial, otherwise the following gives an answer: en.wikipedia.org/wiki/Finite-dimensional_von_Neumann_algebra – Sebastien Palcoux Aug 23 '17 at 21:19
• I'm puzzled as to what you might mean by "explicitly computing" a factor decomposition. – Nik Weaver Aug 24 '17 at 0:32
• @SebastienPalcoux thanks for your comment. I have updated my question. – Fan Aug 24 '17 at 0:49
• @NikWeaver Similar to the theorem given in the finite-dimensional case, to identify the type of each factor. – Fan Aug 24 '17 at 1:03
• @Fan: the word type has a specific meaning in the factor theory: there are type ${\rm I}_n$, ${\rm I}_{\infty}$, ${\rm II}_1$, ${\rm II}_{\infty}$, ${\rm III}_{0}$, ${\rm III}_{\lambda}$, ${\rm III}_{\infty}$. The type does not characterize completely the factor, except in the amenable case (${\rm III}_{0}$ excepted). – Sebastien Palcoux Aug 24 '17 at 6:24