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Let $\mathcal M$ be a type $\mathrm{III}$ von Neumann algebra. Is it true that for all $n\geq 1,$ $\mathcal M$ contains a copy of $M_n$ as a von Neumann subalgebra? By Theorem 9.24 of these lecture notes, I just need to show that for all $n\in\ N$ there are mutually orthogonal projections $p_1,\dots,p_n$ such that $p_1+\dots+p_n=1.$

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    $\begingroup$ you mean... $M_n$ is a square matrix ring over the field for the algebra? $\endgroup$
    – rschwieb
    Commented Sep 29, 2022 at 17:23
  • $\begingroup$ Yes! Matrix ring over complex field. $\endgroup$
    – A beginner mathmatician
    Commented Sep 29, 2022 at 18:04
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    $\begingroup$ Any two nontrivial nonidentity projections are equivalent, and so $M $ is isomorphic to $M_n M$, so in particular contains $M_n C $ as a unital subalgebra. $\endgroup$
    – David Handelman
    Commented Sep 30, 2022 at 2:28

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