Von Neumann bicommutant theorem

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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 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.[clarification needed] 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.


Let H be a Hilbert space and L(H) the bounded operators on H. Consider a self-adjoint unital subalgebra M of L(H). (this means that M contains the adjoints of its members, and the identity operator on H)

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

Proof of (i)[edit]

By definition of the weak operator topology, for any x and y in H, the map T → <Tx, y> is continuous in this topology. Therefore, for any operator O (and by substituting once yOy and once xOx), 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 L(H), and S’ its commutant. For any operator T not in S’, <OTx, y> - <TOx, y> is nonzero for some O in S and some x and y in H. By the continuity of the abovementioned mapping, there is an open neighborhood of T 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 M′′; since it contains M, it also contains its weak closure.

Proof of (ii)[edit]

This follows directly from the weak operator topology being coarser than the strong operator topology: for every point x in clS(M), every open neighborhood of x in the weak operator topology is also open in the strong operator topology and therefore contains a member of M; therefore x is also a member of clW(M).

Proof of (iii)[edit]

Fix XM′′. We will show X ∈ clS(M).

Fix an open neighborhood U of X in the strong operator topology. By definition of the strong operator topology, U contains a finite intersection U(h11) ∩...∩U(hnn) of subbasic open sets of the form U(h,ε) = {OL(H): ||Oh - Xh|| < ε}, where h is in H and ε > 0.

Fix h in H. Consider the closure cl(Mh) of Mh = {Mh : MM} with respect to the norm of H. It is a vector space that is complete (being a closed subset of a complete space H), and so has a corresponding orthogonal projection which we denote P. P is bounded, so it is in L(H). Next we prove:

Lemma. PM.
Proof. Fix xH. Then Px ∈ cl(Mh), so it is the limit of a sequence Onh with On in M for all n. Then for all TM, TOnh is also in Mh and thus its limit is in cl(Mh). By continuity of T (since it is in L(H) and thus Lipschitz continuous), this limit is TPx. Since TPx ∈ cl(Mh), PTPx = TPx. From this it follows that PTP = TP for all T in M.
By using the closure of M under the adjoint we further have, for every T in M and all x, yH:
\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 M.

By definition of the bicommutant XP = PX. Since M is unital, hMh, hence Xh = XPh = PXh ∈ cl(Mh). Thus for every ε > 0, there exists T in M with ||XhTh|| < ε. Then T lies in U(h,ε).[clarification needed]

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

Non-unital case[edit]

An algebra M[clarification needed] acting on H is said to act non-degenerately if for h in H, Mh = {0} implies h = 0. If M acts non-degenerately, and is a sub C*-algebra of L(H),[clarification needed] it can be shown using an approximate identity in M that the identity operator I lies in the strong closure of M. Therefore the conclusion of the bicommutant theorem still holds.


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