Von Neumann bicommutant theorem
In mathematics, specifically functional analysis, the von Neumann bicommutant theorem relates the closure of a set of bounded operators on a Hilbert space in certain topologies to the bicommutant of that set. In essence, it is a connection between the algebraic and topological sides of operator theory.
The formal statement of the theorem is as follows:
This algebra is called the von Neumann algebra generated by.
There are several other topologies on the space of bounded operators, and one can ask what are the *-algebras closed in these topologies. If is closed in the norm topology then it is a C*-algebra, but not necessarily a von Neumann algebra. One such example is the C*-algebra of compact operators. For most other common topologies the closed *-algebras containing 1 are von Neumann algebras; this applies in particular to the weak operator, strong operator, *-strong operator, ultraweak, ultrastrong, and *-ultrastrong topologies.
It is related to the Jacobson density theorem.
Proof
Let be a Hilbert space and the bounded operators on. Consider a self-adjoint unital subalgebra of .The theorem is equivalent to the combination of the following three statements:
where the and subscripts stand for closures in the weak and strong operator topologies, respectively.
Proof of (i)
For any and in, the map T → <Tx, y> is continuous in the weak operator topology, by its definition. Therefore, for any fixed operator, so is the mapLet S be any subset of, and S′ its commutant. For any operator in S′, this function is zero for all O in S. For any not in S′, it must be nonzero for some O in S and some x and y in. By its continuity there is an open neighborhood of for the weak operator topology on which it is nonzero, and which therefore is also not in S′. Hence any commutant S′ is closed in the weak operator topology. In particular, so is ; since it contains, it also contains its weak operator closure.
Proof of (ii)
This follows directly from the weak operator topology being coarser than the strong operator topology: for every point in, every open neighborhood of in the weak operator topology is also open in the strong operator topology and therefore contains a member of ; therefore is also a member of.Proof of (iii)
Fix. We must show that, i.e. for each h ∈ H and any, there exists T in with.Fix h in. The cyclic subspace is invariant under the action of any T in. Its closure in the norm of H is a closed linear subspace, with corresponding orthogonal projection : H → in L. In fact, this P is in, as we now show.
By definition of the bicommutant, we must have XP = PX. Since is unital,, and so. Hence. So for each, there exists T in with, i.e. is in the strong operator closure of.