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In algebra , the bicommutant of a subset S of a semigroup (such as an algebra or a group ) is the commutant of the commutant of that subset. It is also known as the double commutant or second commutant and is written
S
′
′
{\displaystyle S^{\prime \prime }}
.
The bicommutant is particularly useful in operator theory , due to the von Neumann double commutant theorem , which relates the algebraic and analytic structures of operator algebras . Specifically, it shows that if M is a unital, self-adjoint operator algebra in the C*-algebra B(H) , for some Hilbert space H , then the weak closure , strong closure and bicommutant of M are equal. This tells us that a unital C*-subalgebra M of B(H) is a von Neumann algebra if, and only if,
M
=
M
′
′
{\displaystyle M=M^{\prime \prime }}
, and that if not, the von Neumann algebra it generates is
M
′
′
{\displaystyle M^{\prime \prime }}
.
The bicommutant of S always contains S . So
S
′
′
′
=
(
S
′
′
)
′
⊆
S
′
{\displaystyle S^{\prime \prime \prime }=\left(S^{\prime \prime }\right)^{\prime }\subseteq S^{\prime }}
. On the other hand,
S
′
⊆
(
S
′
)
′
′
=
S
′
′
′
{\displaystyle S^{\prime }\subseteq \left(S^{\prime }\right)^{\prime \prime }=S^{\prime \prime \prime }}
. So
S
′
=
S
′
′
′
{\displaystyle S^{\prime }=S^{\prime \prime \prime }}
, i.e. the commutant of the bicommutant of S is equal to the commutant of S . By induction, we have:
S
′
=
S
′
′
′
=
S
′
′
′
′
′
=
…
=
S
2
n
−
1
=
…
{\displaystyle S^{\prime }=S^{\prime \prime \prime }=S^{\prime \prime \prime \prime \prime }=\ldots =S^{2n-1}=\ldots }
and
S
⊆
S
′
′
=
S
′
′
′
′
=
S
′
′
′
′
′
′
=
…
=
S
2
n
=
…
{\displaystyle S\subseteq S^{\prime \prime }=S^{\prime \prime \prime \prime }=S^{\prime \prime \prime \prime \prime \prime }=\ldots =S^{2n}=\ldots }
for n > 1.
It is clear that, if S 1 and S 2 are subsets of a semigroup,
(
S
1
∪
S
2
)
′
=
S
1
′
∩
S
2
′
.
{\displaystyle \left(S_{1}\cup S_{2}\right)'=S_{1}'\cap S_{2}'.}
If it is assumed that
S
1
=
S
1
″
{\displaystyle S_{1}=S_{1}''\,}
and
S
2
=
S
2
″
{\displaystyle S_{2}=S_{2}''\,}
(this is the case, for instance, for von Neumann algebras ), then the above equality gives
(
S
1
′
∪
S
2
′
)
″
=
(
S
1
″
∩
S
2
″
)
′
=
(
S
1
∩
S
2
)
′
.
{\displaystyle \left(S_{1}'\cup S_{2}'\right)''=\left(S_{1}''\cap S_{2}''\right)'=\left(S_{1}\cap S_{2}\right)'.}
See also