Commutants and the Bicommutant Theorem

The bicommutant theorem identifies an algebraic closure operation with two operator-topological closures. For a unital star subalgebra of bounded operators, taking the commutant twice gives exactly its strong and weak operator closure. This is why von Neumann algebras can be defined either through limits or through commutation.

Required background. Von Neumann Algebras introduces the closure problem. Tracking Topologies gives the finite-vector tests defining strong neighborhoods. Direct Sum Decompositions gives a symmetry application of commutants; this page proves the general closure theorem.

Commutants are closed under the relevant limits

For $\mathcal S\subset\mathcal B(\mathcal H)$, define

$$\mathcal S'=\{T:TS=ST\text{ for all }S\in\mathcal S\}.$$

It always contains $I$ and is closed under sums and products. If $\mathcal S$ is star closed, so is its commutant: commutation with $S^*$ can be adjointed to give commutation of $T^*$ with $S$.

For each fixed bounded $S$, multiplication on either side by $S$ is weak-operator continuous. Thus the equations $TS-ST=0$ define a weakly closed set. Consequently the commutant of a star-closed set is a von Neumann algebra.

We also have $\mathcal S\subseteq\mathcal S''$, order reversal under taking commutants, and $\mathcal S'''=\mathcal S'$. These elementary identities do not require a topology. The substantive theorem identifies what $\mathcal A''$ contains beyond $\mathcal A$.

The theorem equates three closures

Let $\mathcal A\subset\mathcal B(\mathcal H)$ be a unital star subalgebra, not necessarily norm closed. Then

$$\overline{\mathcal A}^{\,\mathrm{SOT}} =\overline{\mathcal A}^{\,\mathrm{WOT}} =\mathcal A''.$$

Since $\mathcal A''$ is weakly closed and contains $\mathcal A$,

$$\overline{\mathcal A}^{\,\mathrm{SOT}} \subseteq\overline{\mathcal A}^{\,\mathrm{WOT}} \subseteq\mathcal A''.$$

The remaining direction is to approximate $T\in\mathcal A''$ strongly by elements of $\mathcal A$. A strong neighborhood specifies finitely many vectors simultaneously. Approximating one vector at a time without this simultaneous control would leave a gap.

First approximate on a single vector

Fix $\xi\in\mathcal H$ and let $\mathcal K=\overline{\mathcal A\xi}$. The subspace is invariant under $\mathcal A$ and its adjoints, so its orthogonal projection $P$ commutes with $\mathcal A$. Thus $P\in\mathcal A'$.

Because $T\in\mathcal A''$, it commutes with $P$. Unitality gives $\xi\in\mathcal K$, hence

$$T\xi=TP\xi=PT\xi\in\mathcal K.$$

By the definition of $\mathcal K$, for every $\epsilon>0$ there exists $a\in\mathcal A$ with $\|(T-a)\xi\|<\epsilon$. The argument is geometric: the cyclic subspace is a reducing subspace that every bicommutant element must preserve.

Amplification supplies simultaneous approximation

For vectors $\xi_1,\ldots,\xi_n$, work on $\mathcal H^{\oplus n}$ with the diagonal algebra

$$\mathcal A^{(n)} =\{\operatorname{diag}(a,\ldots,a):a\in\mathcal A\}.$$

An operator on this direct sum is an $n\times n$ matrix of bounded operators. It commutes with $\mathcal A^{(n)}$ precisely when each matrix entry lies in $\mathcal A'$. Therefore $T^{(n)}=\operatorname{diag}(T,\ldots,T)$ commutes with every such matrix and belongs to $(\mathcal A^{(n)})''$.

Apply the single-vector argument to $\Xi=(\xi_1,\ldots,\xi_n)$. It gives one $a\in\mathcal A$ such that

$$\sum_{j=1}^n\|(T-a)\xi_j\|^2<\epsilon^2.$$

This one element approximates $T$ on all specified vectors. Every strong neighborhood of $T$ therefore meets $\mathcal A$, completing the proof.

The proof constructs approximants directed by finite sets of vectors and tolerances. It does not automatically give a sequence on an arbitrary Hilbert space, nor does it bound the norms of the approximating operators. Controlled approximation within the unit ball is the stronger Kaplansky density theorem.

Why the star and identity hypotheses matter

The upper-triangular matrices in $M_2(\mathbb C)$ form a unital algebra closed in every finite-dimensional topology. Their commutant is scalar, so their bicommutant is all of $M_2$. They violate the conclusion because they are not star closed.

Likewise, the zero algebra on a nonzero Hilbert space has bicommutant $\mathbb CI$, not zero. Omitting the identity requires an appropriate nondegenerate formulation, not silent use of the unital theorem.

Finally, the result concerns bounded operators. Formal commutation of unbounded differential expressions does not define the same commutant. Spectral projections or bounded unitary groups are the appropriate inputs when studying self-adjoint unbounded observables.

Exercise: finite rank generates every bounded operator strongly

For separable infinite-dimensional $\mathcal H$, let $P_n\uparrow I$ be finite-rank projections. Show directly that $P_nTP_n\to T$ strongly for every bounded $T$. Does this imply norm density of finite-rank operators?

Solution

For each $\xi$, $\|(P_nTP_n-T)\xi\|\leq\|T\|\|(P_n-I)\xi\|+\|(P_n-I)T\xi\|\to0$. Each approximant has finite rank and norm at most $\|T\|$. This proves strong density. Norm density fails: the identity has distance one from every finite-rank operator on an infinite-dimensional Hilbert space. The norm closure is the compact operators.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987, doi:10.1007/978-3-662-02520-8 — commutants and density theorems.
• Jones, Vaughan F. R., Von Neumann Algebras, lecture notes, 2010, author’s notes, Chapter 3 — the bicommutant theorem and operator topologies.