- Von Neumann bicommutant theorem
In
mathematics , the von Neumann bicommutant theorem infunctional analysis relates the closure of a set ofbounded operator s on aHilbert space in certain topologies to thebicommutant of that set. In essence, it is a connection between thealgebra ic and topological sides ofoperator theory .The formal statement of the theorem is as follows. 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 thestrong operator topology are equal, and are in turn equal to thebicommutant M’’ of M. This algebra is thevon 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 aC*-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 theJacobson density theorem .Proof
Let "H" be a Hilbert space and "L"("H") the bounded operators on "H". Consider a self-adjoint subalgebra M of "L"("H"). Suppose also, M contains the identity operator on "H".
As stated above, the theorem claims the following are equivalent:
:i) M = M′′.:ii) M is closed in the
weak operator topology .:iii) M is closed in thestrong operator topology .The adjoint map "T" → "T*" is continuous in the weak operator topology. So the commutant "S’" of any subset "S" of "L"("H") is weakly closed. This gives i) ⇒ ii). Since the weak operator topology is weaker than the strong operator topology, it is also immediate that ii) ⇒ iii). What remains to be shown is iii) ⇒ i). It is true in general that "S" ⊂ "S′′" for any set "S", and that any commutant "S′" is strongly closed. So the problem reduces to showing M′′ lies in the strong closure of M.
For "h" in "H", consider the smallest closed subspace M"h" that contains {"Mh"| "M" ∈ M}, and the corresponding orthogonal projection "P".
Since "M" is an algebra, one has "PTP" = "TP" for all "T" in M. Self-adjointness of M further implies that "P" lies in M′. Therefore for any operator "X" in M′′, one has "XP" = "PX". Since "M" is unital, "h" ∈ M"h", hence "Xh"∈ M"h" and for all ε > 0, there exists "T" in M with ||"Xh - Th"|| < ε.
Given a finite collection of vectors "h1",..."hn", consider the direct sum
:
The algebra N defined by
:
is self-adjoint, closed in the strong operator topology, and contains the identity operator. Given a "X" in M′′, the operator
:
lies in N′′, and the argument above shows that, all ε > 0, there exists "T" in M with ||"Xh"1 - "Th"1||,...,||"Xhn - Thn"|| < ε. By definition of the strong operator topology, the theorem holds.
Non-unital case
The algebra M is said to be "non-degenerate" if for all "h" in "H", M"h" = {0} implies "h" = 0. If M is non-degenerate and a sub
C*-algebra of "L"("H"), it can be shown using anapproximate 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.
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