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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 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)

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)

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\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 $cl S (M)$ , 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)

Fix $X \in M''$ . We must show that $X \in cl S (M)$ , 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 $M h = {Mh : M \in M$ } is invariant under the action of any T in $M$ . Its closure $cl(M h)$ in the norm of H is a closed linear subspace, with corresponding orthogonal projection $P$ : H → $cl(M h)$ in L(H). In fact, this P is in $M'$ , as we now show.

Lemma.

P \in M'

.

Proof. Fix

x \in H

. As

Px \in cl(M h)

, it is the limit of a sequence

O n h

with

O n

in

M

. For any

T \in M

,

TO n h

is also in

M h

, and by the continuity of

T

, this sequence converges to

TPx

. So

TPx \in cl(M h)

, 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 \in H

we have

\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 \in M

, meaning P lies in

M'

.

By definition of the bicommutant, we must have XP = PX. Since $M$ is unital, $h \in M h$ , and so $h = Ph$ . Hence $Xh = XPh = PXh \in cl(M h)$ . 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 $H$ , $M h = {0}$ 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.

@@ Line 1: / Line 1: @@
+{{cleanup|reason=As mentioned on the talk page, the proof of item (iii) is incomplete.|date=September 2018}}
 In [[mathematics]], specifically [[functional analysis]], the '''von Neumann bicommutant theorem''' relates the [[closure (mathematics)|closure]] of a set of [[bounded operator]]s on a [[Hilbert space]] in certain [[operator topology|topologies]] to the [[bicommutant]] of that set. In essence, it is a connection between the [[algebra]]ic and topological sides of [[operator theory]].
+The formal statement of the theorem is as follows:
-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 the [[strong operator topology]] are equal, and are in turn equal to the [[bicommutant]] '''M&prime;&prime;''' of '''M'''.{{clarify|Algebra of all bounded operators is undefined|date=February 2013}} This algebra is the [[von Neumann algebra]] generated by '''M'''.
+:'''Von Neumann bicommutant theorem.''' Let {{math|'''M'''}} be an [[Operator algebra|algebra]] consisting of bounded operators on a Hilbert space {{mvar|H}}, containing the identity operator, and closed under taking [[Hermitian adjoint|adjoint]]s. Then the [[closure (topology)|closure]]s of {{math|'''M'''}} in the [[weak operator topology]] and the [[strong operator topology]] are equal, and are in turn equal to the [[bicommutant]] {{math|'''M'''′′}} of {{math|'''M'''}}.
+This algebra is called the [[von Neumann algebra]] generated by {{math|'''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 operator on Hilbert space|compact operator]]s (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 topology|ultraweak]], [[ultrastrong topology|ultrastrong]], and *-ultrastrong topologies.
+There are several other topologies on the space of bounded operators, and one can ask what are the *-algebras closed in these topologies. If {{math|'''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 operator on Hilbert space|compact operator]]s (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 topology|ultraweak]], [[ultrastrong topology|ultrastrong]], and *-ultrastrong topologies.
 It is related to the [[Jacobson density theorem]].
 == Proof ==
+Let {{mvar|H}} be a Hilbert space and {{math|''L''(''H'')}} the bounded operators on {{mvar|H}}. Consider a self-adjoint unital [[subalgebra]] {{math|'''M'''}} of {{math|''L''(''H'')}} (this means that {{math|'''M'''}} contains the adjoints of its members, and the identity operator on {{mvar|H}}).
+The theorem is equivalent to the combination of the following three statements:
-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''.
+:(i) {{math|cl<sub>''W''</sub>('''M''') ⊆ '''M'''′′}}
-As stated above, the theorem claims the following are equivalent:
+:(ii) {{math|cl<sub>''S''</sub>('''M''') ⊆ cl<sub>''W''</sub>('''M''')}}
+:(iii) {{math|'''M'''′′ ⊆ cl<sub>''S''</sub>('''M''')}}
+where the {{mvar|W}} and {{mvar|S}} subscripts stand for [[Closure (topology)|closure]]s in the [[weak operator topology|weak]] and [[strong operator topology|strong]] operator topologies, respectively.
-:i) '''M''' =  '''M&prime;&prime;'''.
-:ii) '''M''' is closed in the [[weak operator topology]].
-:iii) '''M''' is closed in the [[strong operator topology]].
+===Proof of (i)===
-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&prime;&prime;'' for any set ''S'', and that any commutant ''S&prime;'' is strongly closed. So the problem reduces to showing '''M&prime;&prime;''' lies in the strong closure of '''M'''.
+For any {{mvar|x}} and {{mvar|y}} in {{mvar|H}}, the map ''T'' → <''Tx'', ''y''> is continuous in the weak operator topology, by its definition. Therefore, for any fixed operator {{mvar|O}}, so is the map
+:<math>T \to \langle (OT - TO)x, y\rangle = \langle Tx, O^*y\rangle - \langle TOx, y\rangle </math>
-For ''h'' in ''H'', consider the smallest closed subspace '''M'''''h'' that contains {''Mh''| ''M'' &isin; '''M'''}, and the corresponding orthogonal projection ''P''.
+Let ''S'' be any subset of {{math|''L''(''H'')}}, and ''S''′ its [[commutant]]. For any operator {{mvar|T}} in ''S''′, this function is zero for all ''O'' in ''S''. For any {{mvar|T}} not in ''S''′, it must be nonzero for some ''O'' in ''S'' and some ''x'' and ''y'' in {{mvar|H}}. By its continuity there is an open neighborhood of {{mvar|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 set|closed]] in the weak operator topology.  In particular, so is {{math|'''M'''′′}}; since it contains {{math|'''M'''}}, it also contains its weak operator closure.
-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&prime;'''. Therefore for any operator ''X'' in '''M&prime;&prime;''', one has ''XP'' = ''PX''. Since ''M'' is unital, ''h'' &isin; '''M'''''h'', hence ''Xh''∈ '''M'''''h'' and for all ε > 0, there exists ''T'' in '''M''' with ||''Xh - Th''|| < ε.
+===Proof of (ii)===
-Given a finite collection of vectors ''h<sub>1</sub>'',...''h<sub>n</sub>'', consider the direct sum
+This follows directly from the weak operator topology being coarser than the strong operator topology: for every point {{mvar|x}} in {{math|cl<sub>''S''</sub>('''M''')}}, every open neighborhood of {{mvar|x}} in the weak operator topology is also open in the strong operator topology and therefore contains a member of {{math|'''M'''}}; therefore {{mvar|x}} is also a member of {{math|cl<sub>''W''</sub>('''M''')}}.
+===Proof of (iii)===
-:<math> {\tilde H} = \oplus_1 ^n H.</math>
+Fix {{math|''X'' ∈ '''M'''′′}}. We must show that {{math|''X'' ∈ cl<sub>''S''</sub>('''M''')}}, i.e. for each ''h'' ∈ ''H'' and any {{math|''ε'' > 0}}, there exists ''T'' in {{math|'''M'''}} with {{math|{{!!}}''Xh'' − ''Th''{{!!}} < ''ε''}}.
+Fix ''h'' in {{mvar|H}}. The [[Cyclic subspace | cyclic subspace]] {{math|'''M'''''h'' {{=}} {''Mh'' : ''M'' ∈ '''M'''}}} is invariant under the action of any ''T'' in {{math|'''M'''}}. Its [[Closure (topology)|closure]] {{math|cl('''M'''''h'')}} in the norm of ''H'' is a closed linear subspace, with corresponding [[orthogonal projection]] {{mvar|P}} : ''H'' → {{math|cl('''M'''''h'')}} in ''L''(''H''). In fact, this ''P'' is in {{math|'''M'''′}}, as we now show.
-The algebra '''N''' defined by
+:'''Lemma.''' {{math|''P'' ∈ '''M'''′}}.
-:<math>{\mathbf N} = \{\oplus_1 ^n M \; | \; M \in {\mathbf M} \} \sub L({\tilde H})</math>
+:'''Proof.''' Fix {{math|''x'' ∈ ''H''}}. As {{math|''Px'' ∈ cl('''M'''''h'')}}, it is the limit of a sequence {{mvar|O<sub>n</sub>h}} with {{mvar|O<sub>n</sub>}} in {{math|'''M'''}}. For any {{math|''T'' ∈ '''M'''}}, {{mvar|TO<sub>n</sub>h}} is also in {{math|'''M'''''h''}}, and by the continuity of {{mvar|T}}, this sequence converges to {{mvar|TPx}}. So {{math|''TPx'' ∈ cl('''M'''''h'')}}, and hence ''PTPx'' = ''TPx''. Since ''x'' was arbitrary, we have ''PTP'' = ''TP'' for all {{mvar|T}} in {{math|'''M'''}}.
-is self-adjoint, closed in the strong operator topology, and contains the identity operator. Given a ''X'' in '''M&prime;&prime;''', the operator
+:Since {{math|'''M'''}} is closed under the adjoint operation and ''P'' is [[self-adjoint operator|self-adjoint]], for any {{math|''x'', ''y'' ∈ ''H''}} we have
-:<math> \oplus_1 ^n X \in L( {\tilde H} ) </math>
+::<math>\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</math>
-lies in '''N&prime;&prime;''', and the argument above shows that,  all ε > 0, there exists ''T'' in '''M''' with ||''Xh''<sub>1</sub> - ''Th''<sub>1</sub>||,...,||''Xh<sub>n</sub> - Th<sub>n</sub>''|| < ε. By definition of the strong operator topology, the theorem holds.
+:So ''TP'' = ''PT'' for all {{math|''T'' ∈ '''M'''}}, meaning ''P'' lies in {{math|'''M'''′}}.
-=== 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 an [[approximate identity]] in '''M''' that the identity operator ''I'' lies in the strong closure of '''M'''. Therefore the bicommutant theorem still holds.
+By definition of the [[bicommutant]], we must have ''XP'' = ''PX''. Since {{math|'''M'''}} is unital, {{math|''h'' ∈ '''M'''''h''}}, and so {{math| ''h'' {{=}} ''Ph''}}. Hence {{math|''Xh'' {{=}} ''XPh'' {{=}} ''PXh'' ∈ cl('''M'''''h'')}}. So for each {{math|''ε'' > 0}}, there exists ''T'' in {{math|'''M'''}} with {{math|{{!!}}''Xh'' − ''Th''{{!!}} < ''ε''}}, i.e. {{mvar|X}} is in the strong operator closure of {{math|'''M'''}}.
-== References ==
+=== Non-unital case ===
+A C*-algebra {{math|'''M'''}} acting on '''H''' is said to act ''non-degenerately'' if for ''h'' in {{mvar|H}}, {{math|'''M'''''h'' {{=}} {0} }} implies {{math|''h'' {{=}} 0}}. In this case, it can be shown using an [[approximate identity]] in {{math|'''M'''}} that the identity operator ''I'' lies in the strong closure of {{math|'''M'''}}. Therefore, the conclusion of the bicommutant theorem holds for {{math|'''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
+{{Functional analysis}}
 [[Category:Operator theory]]