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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$, with the [[adjoint|6HSJBJ8TTQ]] $T \\mapsto T^*$.\nA *von Neumann algebra* on $H$ is a subset $M \\subseteq B(H)$ such that:\n\n1. $M$ is a subalgebra of $B(H)$ that contains the identity operator $I$;\n2. $T^* \\in M$ for every $T \\in M$;\n3. $M$ is closed in the [[weak operator topology|4XM1VN1C1N]].\n\n## Remarks\n\nExample: let $(X, \\mu)$ be a $\\sigma$-finite measure space and $H = L^2(X, \\mu)$.\nFor $f \\in L^\\infty(X, \\mu)$ let $M_f$ be multiplication by $f$.\nThe set $\\set{M_f : f \\in L^\\infty}$ contains $I = M_1$, is closed under products and under $M_f^* = M_{\\bar f}$, and one checks that it equals its own commutant, so it is weakly closed.\nOther examples are $B(H)$ itself and the matrix algebras $M_n(\\C)$ acting on $\\C^n$.\n\nNon-example: on $H = L^2[0,1]$, the multiplication operators $M_f$ with $f$ continuous satisfy clauses 1 and 2 and form a C*-algebra.\nBut clause 3 fails.\nThe indicator function of a subinterval is a pointwise limit of uniformly bounded continuous functions.\nBy dominated convergence, its multiplication operator is a weak limit of elements of the algebra, but it is not continuous.\nThe weak closure of this algebra is the whole multiplication algebra of $L^\\infty[0,1]$.\n\nBy the von Neumann bicommutant theorem, clause 3 can be replaced by closedness in the strong operator topology, or by the condition $M = M''$, where $M'$ is the commutant.\nEvery von Neumann algebra is a C*-algebra.\nBy Sakai's theorem, an abstract C*-algebra is isomorphic to a von Neumann algebra exactly when it is a dual Banach space.\nSuch algebras are called W*-algebras.\n",
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