Abelian sub-W*-algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:22:00Z http://mathoverflow.net/feeds/question/92922 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92922/abelian-sub-w-algebras Abelian sub-W*-algebras J. Grass 2012-04-02T20:06:30Z 2012-04-03T05:55:28Z <p>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 $\lambda&lt;\kappa$ (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 $\ell_1(\kappa)$?</p> <hr> <p>A density character is the minimal cardinality of a dense set.</p> http://mathoverflow.net/questions/92922/abelian-sub-w-algebras/92947#92947 Answer by Jesse Peterson for Abelian sub-W*-algebras Jesse Peterson 2012-04-03T00:38:01Z 2012-04-03T05:55:28Z <p>This is not true. Popa showed in "<a href="http://www.ams.org/mathscinet-getitem?mr=703810" rel="nofollow">Orthogonal pairs of ∗-subalgebras in finite von Neumann algebras</a>" (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.</p> <p>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.</p>