Von Neumann bicommutant theorem

From Wikipedia, the free encyclopedia

In mathematics, the von Neumann bicommutant theorem in functional analysis 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. 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.

[edit] Proof

As is usually the case when relating algebraic and topological properties, one makes use of continuity and denseness arguments.

Let H be a Hilbert space and L(H) the bounded operators on H. Consider a self-adjoint subalgebra M of L(H). For h in H, the smallest closed subspace that contains {Mh| M ∈ M} will be denoted by Mh. The algebra M is said to be non-degenerate if for all h in H, Mh = {0} implies h = 0.

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).

Assume M is strongly closed and that, without loss of generality, it is non-degenerate. It is true in general that S ⊂ S’ ’ for any family S of bounded operators. So it is sufficient to show M’’ lies in M, i.e. M’’ lies in the strong closure of M. Given M’ ’ ∈ M’’, a typical ε-neighborhood of M’ ’ in the strong operator topology takes the form

$\{ T \in L(H) \;|\; \| Th_i - M''h_i \| < \epsilon \} \quad \mbox{for some} \quad h_1 \cdots h_n \in H \;.$

Suppose now n = 1. We need to find M ∈ M that lies in the above ε-neighborhood. In other words, find g in the subspace {Mh₁| M ∈ M} such that ||g - M’ ’ h₁ || < ε. One can show that both M h₁ and its orthogonal complement are invariant under any M ∈ M. Coupled with the non-degeneracy assumption, this means h₁ must lie in M h₁. It can also be shown that M h₁ is invariant under M’ ’. Therefore M’ ’ h₁ lies in M h₁. The denseness of {Mh₁| M ∈ M} in M h₁ proves the theorem for n = 1.

For n > 1, consider the direct sum of n copies

$H' = \oplus_1 ^n H.$

The subalgebra N of L(H' ) defined by

${\mathbf N} = \{\oplus_1 ^n M \; | \; M \in {\mathbf M} \}$

is still self-adjoint, nondegenerate, and closed in the strong operator topology. In the argument for n = 1, replacing M’ ’ by

$N'' = \oplus_1 ^n M'' = \begin{bmatrix} M'' & & \\ & \ddots & \\ & & M'' \end{bmatrix}$

and h₁ by

$h_1 \oplus \cdots \oplus h_n$

proves the general case.

Note To see why both M h₁ and its orthogonal complement are invariant under any M ∈ M, notice first that because M is an algebra, M h₁ is invariant under any M ∈ M. Self-adjointness of M means the same holds for the orthogonal complement of M h₁. So the orthogonal projection P whose range is M h₁ lies in the commutant of M. By assumption, PM’ ’ = M’ ’ P, i.e. M h₁ (and its orthogonal complement) is invariant under M’ ’.

[edit] Degenerate case

The above argument shows that M’’ is the weak (therefore strong) closure of M when M is a nondegenerate self-adjoint algebra. In general, if M is only assumed to be self-adjoint, M’’ is the weak closure of the algebra generated by M ∪ {I}. This is why, when defining von Neumann algebras, M is required to contain the identity operator.

The proof given can be readily modified for the nondegenerate case. The difference between degeneracy and nondegeneracy can be spelled out more explicitly. Let G ⊂ H be the subspace G = {h ∈ H | Mh = 0, for all M in M}. G is invariant under every M in M. By self-adjointness, same holds for its orthogonal complement G`. Consider the self-adjoint algebra N = {M restricted to G` | M in M} ⊂ L(G`). By construction, N is nondegenerate. M can be viewed as the direct sum