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

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)

cl W (M) \subseteq M''

(ii)

cl S (M) \subseteq cl W (M)

(iii)

M'' \subseteq cl S (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 $y \to O * y$ and once $x \to Ox$ ), 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)

This follows directly from the weak operator topology being coarser than the strong operator topology: for every point $x$ in $cl S (M)$ , 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 $cl W (M)$ .

Proof of (iii)

Let $X \in M''$ , we will show $X \in cl S (M)$ . 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 $cl(M h)$ of $M h = {Mh : M \in M}.$ 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.

P \in M'

.

Proof. For every

x \in H

,

Px \in cl(M h)

so it is the limit of a series

O n h

with

O n

in

M

for all

n

, hence for all

T \in M

,

TO n h

is also in

M h

and thus its limit is in

cl(M h)

. By continuity of

T

(since it is in

L (H)

and thus Lipschitz continuous), this limit is

TPx

. Since

TPx \in cl(M h)

, 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, y \in H

:

\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, $h \in M h$ , hence $Xh = XPh = PXh \in cl(M h)$ and for every $ε > 0$ , there exists T in $M$ with $|| Xh - Th || < ε$ . In particular, for $ε = d$ , there is $T 0$ in $M$ such that $|| Xh - T 0 h || < d$ . By the triangle inequality $|\|Xh\|-\|T_{0}h\||<d$ and $T 0$ 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

An algebra $M$ ^{[clarification needed]} acting on H is said to act non-degenerately if for h in $H$ , $M h = {0}$ 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 bicommutant theorem still holds.

References

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