mathematics

Von Neumann bicommutant theorem

Von Neumann bicommutant theorem is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Von Neumann bicommutant theorem rather than just read about it. In short: 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.

Key takeaways

Von Neumann bicommutant theorem belongs to mathematics; place it in that map before memorising details.
Learn the definition first, then one example that makes the definition concrete.
Connect Von Neumann bicommutant theorem to a quantity you can measure, compute or draw — that is where exam questions come from.
Reproduce the core statement of Von Neumann bicommutant theorem from memory before moving on to harder problems.

Reference excerpt

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 consisting 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. This algebra is called 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 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 closure in the weak and strong operator topologies, respectively.

Proof of (i) For any x and y in H, the map T → <Tx, y> is continuous in the weak operator topology, by its definition. Therefore, for any fixed operator O, so is the map

T → ⟨ ( O T − T O ) x , y ⟩ = ⟨ T x , O ∗ y ⟩ − ⟨ T O x , y ⟩ {\displaystyle T\to \langle (OT-TO)x,y\rangle =\langle Tx,O^{*}y\rangle -\langle TOx,y\rangle }

Let S be any subset of L(H), and S′ its commutant. For any operator T in S′, this function is zero for all O in S. For any T not in S′, it must be nonzero for some O in S and some x and y in H. By its continuity there is an open neighborhood of T for the weak operator topology on which it is nonzero, and which therefore is also not in S′. Hence any commutant S′ is closed in the weak operator topology. In particular, so is M′′; since it contains M, it also contains its weak operator 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) Fix X ∈ M′′. We must show that X ∈ clS(M), i.e. for each h ∈ H and any ε > 0, there exists T in M with ||Xh − Th|| < ε. Fix h in H. The cyclic subspace Mh = {Mh : M ∈ M} is invariant under the action of any T in M. Its closure cl(Mh) in the norm of H is a closed linear subspace, with corresponding orthogonal projection P : H → cl(Mh) in L(H). In fact, this P is in M′, as we now show.

Lemma. P ∈ M′. Proof. Fix x ∈ H. As Px ∈ cl(Mh), it is the limit of a sequence Onh with On in M. For any T ∈ M, TOnh is also in Mh, and by the continuity of T, this sequence converges to TPx. So TPx ∈ cl(Mh), and hence PTPx = TPx. Since x was arbitrary, we have PTP = TP for all T in M. Since M is closed under the adjoint operation and P is self-adjoint, for any x, y ∈ H we have

⟨ x , T P y ⟩ = ⟨ x , P T P y ⟩ = ⟨ ( P T P ) ∗ x , y ⟩ = ⟨ P T ∗ P x , y ⟩ = ⟨ T ∗ P x , y ⟩ = ⟨ P x , T y ⟩ = ⟨ x , P T y ⟩ {\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 }

So TP = PT for all T ∈ M, meaning P lies in M′. By definition of the bicommutant, we must have XP = PX. Since M is unital, h ∈ Mh, and so h = Ph. Hence Xh = XPh = PXh ∈ cl(Mh). So for each ε > 0, there exists T in M with ||Xh − Th|| < ε, i.e. X is in the strong operator closure of M.

Non-unital case A C*-algebra M acting on H is said to act non-degenerately if for h in H, Mh = {0} implies h = 0. In this case, 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 holds for M.

References 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 Jacob Lurie's lecture notes on a von Neumann algebra at https://www.math.ias.edu/~lurie/261y.html

Worked examples

Example 1 — a first encounter with Von Neumann bicommutant theorem

Start with the simplest possible case. Write down what Von Neumann bicommutant theorem claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Von Neumann bicommutant theorem before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Von Neumann bicommutant theorem ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Von Neumann bicommutant theorem

In research: Von Neumann bicommutant theorem appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry: Engineering practice reuses Von Neumann bicommutant theorem in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom: Von Neumann bicommutant theorem is common in secondary-school and first-year university syllabi. It links to neighbouring topics Operator theory, Theorems in functional analysis, Von Neumann algebras, so understanding it makes those chapters shorter.
In everyday life: Look for Von Neumann bicommutant theorem outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

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How to study Von Neumann bicommutant theorem in 20 minutes

Read the reference excerpt below once, without taking notes.
Close the page and write down what Von Neumann bicommutant theorem means in your own words.
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Work through the three examples above with pen and paper.
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Frequently asked questions

What is Von Neumann bicommutant theorem in simple terms?

Why does Von Neumann bicommutant theorem matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Von Neumann bicommutant theorem?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Von Neumann bicommutant theorem.