Inalgebra, thebicommutant of asubsetS of asemigroup (such as analgebra or agroup) is thecommutant of the commutant of that subset. It is also known as the double commutant or second commutant and is written${\displaystyle S^{\mathrm{\prime }\mathrm{\prime }}}$.

The bicommutant is particularly useful inoperator theory, due to thevon Neumann double commutant theorem, which relates the algebraic and analytic structures ofoperator algebras. Specifically, it shows that ifM is a unital, self-adjoint operator algebra in theC*-algebraB(H), for someHilbert spaceH, then theweak closure,strong closure and bicommutant ofM are equal. This tells us that a unitalC*-subalgebraM ofB(H) is avon Neumann algebra if, and only if,${\displaystyle M=M^{\mathrm{\prime }\mathrm{\prime }}}$, and that if not, the von Neumann algebra it generates is${\displaystyle M^{\mathrm{\prime }\mathrm{\prime }}}$.

The bicommutant ofS always containsS. So${\displaystyle S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}={\left(S^{\mathrm{\prime }\mathrm{\prime }}\right)}^{\mathrm{\prime }}\subseteq S^{\mathrm{\prime }}}$. On the other hand,${\displaystyle S^{\mathrm{\prime }}\subseteq {\left(S^{\mathrm{\prime }}\right)}^{\mathrm{\prime }\mathrm{\prime }}=S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}}$. So${\displaystyle S^{\mathrm{\prime }}=S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}}$, i.e. the commutant of the bicommutant ofS is equal to the commutant ofS. By induction, we have:

${\displaystyle S^{\mathrm{\prime }}=S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=\cdots =S^{2n-1}=\cdots }$

and

${\displaystyle S\subseteq S^{\mathrm{\prime }\mathrm{\prime }}=S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=S^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=\cdots =S^{2n}=\cdots }$

forn > 1.

It is clear that, ifS₁ andS₂ are subsets of a semigroup,

${\displaystyle {\left(S_1\cup S_2\right)}^{\prime }=S_1^{\prime }\cap S_2^{\prime }.}$

If it is assumed that${\displaystyle S_1=S_1^{″}}$ and${\displaystyle S_2=S_2^{″}}$ (this is the case, for instance, forvon Neumann algebras), then the above equality gives

${\displaystyle {\left(S_1^{\prime }\cup S_2^{\prime }\right)}^{″}={\left(S_1^{″}\cap S_2^{″}\right)}^{\prime }={\left(S_1\cap S_2\right)}^{\prime }.}$

• von Neumann double commutant theorem

• J. Dixmier,Von Neumann Algebras, North-Holland, Amsterdam, 1981.

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