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:

:Von Neumann bicommutant theorem. Let be an algebra consisting of bounded operators on a Hilbert space , containing the identity operator, and closed under taking adjoints. Then the closures of in the weak operator topology and the strong operator topology are equal, and are in turn equal to the bicommutant of .

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 (on an infinite dimensional Hilbert space). 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 (this means that contains the adjoints of its members, and the identity operator on ).

The theorem is equivalent to the combination of the following three statements:

:(i)
:(ii)
:(iii)

where the and subscripts stand for closures in the weak and strong operator topologies, respectively.

Proof of (i)

By definition of the weak operator topology, for any and in , the map ''T'' → <''Tx'', ''y''> is continuous in this topology. Therefore, for any operator (and by substituting once and once ), so is the map

:$T \to \langle Tx, O^*y\rangle - \langle TOx, y\rangle = \langle OTx, y\rangle - \langle TOx, y\rangle.$

Let ''S'' be any subset of , and ''S''′ its commutant. For any operator not in ''S''′, <''OTx'', ''y''> - <''TOx'', ''y''> is nonzero for some ''O'' in ''S'' and some ''x'' and ''y'' in . By the continuity of the abovementioned mapping, there is an open neighborhood of in the weak operator topology for which this is nonzero, therefore this open neighborhood is also not in ''S''′. Thus ''S''′ is closed in the weak operator, i.e. ''S''′ is ''weakly closed''. Thus every commutant is weakly closed, and so is ; since it contains , it also contains its weak 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 will show .

Fix an open neighborhood of in the strong operator topology. By definition of the strong operator topology, ''U'' contains a finite intersection ''U''(''h''₁,ε₁) ∩...∩''U''(''h''_n,ε_n) of subbasic open sets of the form ''U''(''h'',ε) = , where ''h'' is in ''H'' and ε > 0.

Fix ''h'' in . Consider the closure of with respect to the norm of ''H'' and equipped with the inner product of ''H''. It is a Hilbert space (being a closed subspace of a Hilbert space ), and so has a corresponding orthogonal projection which we denote . is bounded, so it is in . Next we prove:

:Lemma. .

:Proof. Fix . Then , so it is the limit of a sequence with in for all . Then for all , is also in and thus its limit is in . By continuity of (since it is in and thus Lipschitz continuous), this limit is . Since , ''PTPx'' = ''TPx''. From this it follows that ''PTP'' = ''TP'' for all in .

:By using the closure of under the adjoint we further have, for every in and all :

::$\langle x,TPy\rangle = \langle x,PTPy\rangle = \langle Px,TPy\rangle = \langle T^*Px,Py\rangle = \langle PT^*Px,y\rangle = \langle T^*Px,y\rangle = \langle Px,Ty\rangle = \langle x,PTy\rangle$

:thus ''TP'' = ''PT'' and ''P'' lies in .

By definition of the bicommutant ''XP'' = ''PX''. Since is unital, , hence . Thus for every , there exists ''T'' in with . Then ''T'' lies in ''U''(''h'',ε).

Thus in every open neighborhood of in the strong operator topology there is a member of , and so is in the strong operator topology closure of .

Non-unital case

A C*-algebra acting on H is said to act ''non-degenerately'' if for ''h'' in , implies . In this case, it can be shown using an approximate identity in that the identity operator ''I'' lies in the strong closure of . Therefore, the conclusion of the bicommutant theorem holds for .

References

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

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