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Von Neumann bicommutant theorem

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Inmathematics, specificallyfunctional analysis, thevon Neumann bicommutant theorem relates theclosure of a set ofbounded operators on aHilbert space in certaintopologies to thebicommutant of that set. In essence, it is a connection between thealgebraic and topological sides ofoperator theory.

The formal statement of the theorem is as follows:

Von Neumann bicommutant theorem. LetM be analgebra consisting of bounded operators on a Hilbert spaceH, containing the identity operator, and closed under takingadjoints. Then theclosures ofM in theweak operator topology and thestrong operator topology are equal, and are in turn equal to thebicommutantM′′ ofM.

This algebra is called thevon Neumann algebra generated byM.

There are several other topologies on the space of bounded operators, and one can ask what are the *-algebras closed in these topologies. IfM is closed in thenorm topology then it is aC*-algebra, but not necessarily a von Neumann algebra. One such example is the C*-algebra ofcompact 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 theJacobson density theorem.

Proof

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LetH be a Hilbert space andL(H) the bounded operators onH. Consider a self-adjoint unitalsubalgebraM ofL(H) (this means thatM contains the adjoints of its members, and the identity operator onH).

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

(i)clW(M) ⊆M′′
(ii)clS(M) ⊆ clW(M)
(iii)M′′ ⊆ clS(M)

where theW andS subscripts stand forclosures in theweak andstrong operator topologies, respectively.

Proof of (i)

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For anyx andy inH, the mapT → <Tx,y> is continuous in the weak operator topology, by its definition. Therefore, for any fixed operatorO, so is the map

T(OTTO)x,y=Tx,OyTOx,y{\displaystyle T\to \langle (OT-TO)x,y\rangle =\langle Tx,O^{*}y\rangle -\langle TOx,y\rangle }

LetS be any subset ofL(H), andS′ itscommutant. For any operatorT inS′, this function is zero for allO inS. For anyT not inS′, it must be nonzero for someO inS and somex andy inH. By its continuity there is an open neighborhood ofT for the weak operator topology on which it is nonzero, and which therefore is also not inS′. Hence any commutantS′ isclosed in the weak operator topology. In particular, so isM′′; since it containsM, it also contains its weak operator closure.

Proof of (ii)

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This follows directly from the weak operator topology being coarser than the strong operator topology: for every pointx inclS(M), every open neighborhood ofx in the weak operator topology is also open in the strong operator topology and therefore contains a member ofM; thereforex is also a member ofclW(M).

Proof of (iii)

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FixXM′′. We must show thatX ∈ clS(M), i.e. for eachhH and anyε > 0, there existsT inM with||XhTh|| <ε.

Fixh inH. The cyclic subspaceMh = {Mh :MM} is invariant under the action of anyT inM. Itsclosurecl(Mh) in the norm ofH is a closed linear subspace, with correspondingorthogonal projectionP :Hcl(Mh) inL(H). In fact, thisP is inM, as we now show.

Lemma.PM.
Proof. FixxH. AsPx ∈ cl(Mh), it is the limit of a sequenceOnh withOn inM. For anyTM,TOnh is also inMh, and by the continuity ofT, this sequence converges toTPx. SoTPx ∈ cl(Mh), and hencePTPx =TPx. Sincex was arbitrary, we havePTP =TP for allT inM.
SinceM is closed under the adjoint operation andP isself-adjoint, for anyx,yH we have
x,TPy=x,PTPy=(PTP)x,y=PTPx,y=TPx,y=Px,Ty=x,PTy{\displaystyle \langle x,TPy\rangle =\langle x,PTPy\rangle =\langle (PTP)^{*}x,y\rangle =\langle PT^{*}Px,y\rangle =\langle T^{*}Px,y\rangle =\langle Px,Ty\rangle =\langle x,PTy\rangle }
SoTP =PT for allTM, meaningP lies inM.

By definition of thebicommutant, we must haveXP =PX. SinceM is unital,hMh, and soh =Ph. HenceXh =XPh =PXh ∈ cl(Mh). So for eachε > 0, there existsT inM with||XhTh|| <ε, i.e.X is in the strong operator closure ofM.

Non-unital case

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A C*-algebraM acting onH is said to actnon-degenerately if forh inH,Mh = {0} impliesh = 0. In this case, it can be shown using anapproximate identity inM that the identity operatorI lies in the strong closure ofM. Therefore, the conclusion of the bicommutant theorem holds forM.

References

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  • W.B. Arveson,An Invitation to C*-algebras, Springer, New York, 1976.
  • M. Takesaki,Theory of Operator Algebras I, Springer, 2001, 2nd printing of the first edition 1979.

Further reading

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