Rigid body

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. Let M be an algebra of bounded operators on a Hilbert space H, containing the identity operator and closed under taking adjoints. Then the closures of M in the weak operator topology and the strong operator topology are equal, and are in turn equal to the bicommutant M′′ of M.Template:Clarify This algebra is the von Neumann algebra generated by M.

There are several other topologies on the space of bounded operators, and one can ask what are the *-algebras closed in these topologies. If M 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 (on an infinite dimensional Hilbert space). For most other common topologies the closed *-algebras containing 1 are still 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 H be a Hilbert space and L(H) the bounded operators on H. Consider a self-adjoint subalgebra M of L(H). Suppose also, M contains the identity operator on H.

As stated above, the theorem claims the following are equivalent:

i) M = M′′.

ii) M is closed in the weak operator topology.

iii) M is closed in the strong operator topology.

The adjoint map T → T* is continuous in the weak operator topology. So the commutant S’ of any subset S of L(H) is weakly closed. This gives i) ⇒ ii). Since the weak operator topology is weaker than the strong operator topology, it is also immediate that ii) ⇒ iii). What remains to be shown is iii) ⇒ i). It is true in general that S ⊂ S′′ for any set S, and that any commutant S′ is strongly closed. So the problem reduces to showing M′′ lies in the strong closure of M.

For h in H, consider the smallest closed subspace Mh that contains {Mh| M ∈ M}, and the corresponding orthogonal projection P.

Since M is an algebra, one has PTP = TP for all T in M. Self-adjointness of M further implies that P lies in M′. Therefore for any operator X in M′′, one has XP = PX. Since M is unital, h ∈ Mh, hence Xh∈ Mh and for all ε > 0, there exists T in M with ||Xh - Th|| < ε.

Given a finite collection of vectors h₁,...h_n, consider the direct sum

$\tilde{H}=\underset{1}{\overset{n}{\oplus }}H.$

The algebra N defined by

$\mathbf{N}=\{\underset{1}{\overset{n}{\oplus }}M|M\in \mathbf{M}\}\subset L(\tilde{H})$

is self-adjoint, closed in the strong operator topology, and contains the identity operator. Given a X in M′′, the operator

$\underset{1}{\overset{n}{\oplus }}X\in L(\tilde{H})$

lies in N′′, and the argument above shows that, all ε > 0, there exists T in M with ||Xh₁ - Th₁||,...,||Xh_n - Th_n|| < ε. By definition of the strong operator topology, the theorem holds.

Non-unital case

The algebra M is said to be non-degenerate if for all h in H, Mh = {0} implies h = 0. If M is non-degenerate and a sub C*-algebra of L(H), it can be shown using an approximate identity in M that the identity operator I lies in the strong closure of M. Therefore the bicommutant theorem still holds.

References

• W.B. Arveson, An Invitation to C*-algebras, Springer, New York, 1976.