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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.

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  • $\begingroup$ 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 $\endgroup$
    – Sebastien Palcoux
    Commented Aug 23, 2017 at 21:19
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    $\begingroup$ I'm puzzled as to what you might mean by "explicitly computing" a factor decomposition. $\endgroup$
    – Nik Weaver
    Commented Aug 24, 2017 at 0:32
  • $\begingroup$ @SebastienPalcoux thanks for your comment. I have updated my question. $\endgroup$
    – Fan
    Commented Aug 24, 2017 at 0:49
  • $\begingroup$ @NikWeaver Similar to the theorem given in the finite-dimensional case, to identify the type of each factor. $\endgroup$
    – Fan
    Commented Aug 24, 2017 at 1:03
  • $\begingroup$ @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). $\endgroup$
    – Sebastien Palcoux
    Commented Aug 24, 2017 at 6:24

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