# 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.

## Proof

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)

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, and so it has a nonzero open neighborhood. 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 this topology (i.e. it 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)

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)

Let XM′′, we will show X ∈ clS(M). Every open neighborhood U of X in the strong operator topology, is the preimage of V, an open neighborhood of $\|Xh\|$ for some h in H, so that for every O in L(H), O is in U if and only if $\|Oh\|$ is in V. Since V is open, it contains an open ball of radius d > 0 centered at $\|Xh\|$.

Consider the closure cl(Mh) of Mh = {Mh : MM}. 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. For every xH, Px ∈ cl(Mh) so it is the limit of a series Onh with On in M for all n, hence 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) and for every ε > 0, there exists T in M with ||XhTh|| < ε. In particular, for ε = d, there is T0 in M such that ||XhT0h|| < d. By the triangle inequality $|\|Xh\| - \|T_0h\|| < d$ and T0 is in U.

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

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.