- Bicommutant
In
algebra , the bicommutant of asubset "S" of asemigroup (such as an algebra or a group) is thecommutant of the commutant of that subset. It is also known as the double commutant or second commutant and is written S^{prime prime}.The bicommutant is particularly useful in
operator theory , due to thevon Neumann double commutant theorem , which relates the algebraic and analytic structures ofoperator algebra s. Specifically, it shows that if "M" is a unital, self-adjoint operator algebra in theC*-algebra "B(H)", for someHilbert 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 avon Neumann algebra if, and only if, M = M^{prime prime}, and that if not, the von Neumann algebra it generates is M^{prime prime}.The bicommutant of "S" always contains "S". So S^{prime prime prime} = (S^{prime prime})^{prime} subseteq S^{prime}. On the other hand, S^{prime} subseteq (S^{prime})^{prime prime} = S^{prime prime prime}. So 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^{prime} = S^{prime prime prime} = S^{prime prime prime prime prime} = ldots = S^{2n-1} = ldots
and
: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 cup S_2 )' = S_1 ' cap S_2 ' .
If it is assumed that S_1 = S_1" , and S_2 = S_2", (this is the case, for instance, for
von Neumann algebra s), then the above equality gives:S_1' cup S_2')" = (S_1 " cap S_2 ")' = (S_1 cap S_2)' .
See also
*
von Neumann double commutant theorem
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