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Let $M$ be a von Neumann algebra which acts faithfully on a Hilbert space of density character $\kappa$ but does not on a Hilbert space of density character $\lambda<\kappa$ (that is, the density character of the predual $M_0$ is $\kappa$. Does $M$ contain a subalgebra *-isomorphic to $\ell^\infty(\kappa)$? Does $M_0$ contain a complemented subspace isomorphic as a Banach space to $\ell_1(\kappa)$?

A density character is the minimal cardinality of a dense set.

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    $\begingroup$ What is the density character of a Hilbert space? $\endgroup$
    – Alain Valette
    Commented Apr 2, 2012 at 21:37
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    $\begingroup$ The smallest cardinality that a dense set can have, which for an infinite dimensional Hilbert space is the same as its orthonormal dimension. $\endgroup$
    – Bill Johnson
    Commented Apr 2, 2012 at 22:48

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This is not true. Popa showed in "Orthogonal pairs of ∗-subalgebras in finite von Neumann algebras" (1983), that if $F$ is a free group with arbitrary cardinality than any abelian von Neumann subalgebra of the group von Neumann algebra $LF$ must have separable predual.

Edit: This doesn't even hold when $M$ is abelian since $\ell^\infty(\mathbb R)$ has no faithful state and hence does not embed into any $\sigma$-finite von Neumann algebra.

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  • $\begingroup$ This brings to my mind the following question: Is always the cardinality of pure states on a given von Neumann algebra greater than the cardinality of the algebra itself? $\endgroup$
    – Jan Veselý
    Commented Apr 3, 2012 at 11:37

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