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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク bicommutant ： ウィキペディア英語版 bicommutant 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 $S^$. 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, $M\; =\; M^$, and that if not, the von Neumann algebra it generates is $M^$. The bicommutant of ''S'' always contains ''S''. So $S^\; =\; (S^)^\; \backslash subseteq\; S^$. On the other hand, $S^\; \backslash subseteq\; (S^)^\; =\; S^$. So $S^\; =\; S^$, i.e. the commutant of the bicommutant of ''S'' is equal to the commutant of ''S''. By induction, we have: :$S^\; =\; S^\; =\; S^\; =\; \backslash ldots\; =\; S^\; =\; \backslash ldots$ and :$S\; \backslash subseteq\; S^\; =\; S^\; =\; S^\; =\; \backslash ldots\; =\; S^\; =\; \backslash ldots$ for ''n'' > 1. It is clear that, if ''S''1 and ''S''2 are subsets of a semigroup, :$(\; S\_1\; \backslash cup\; S\_2\; )\text{'}\; =\; S\_1\; \text{'}\; \backslash cap\; S\_2\; \text{'}\; .$ If it is assumed that $S\_1\; =\; S\_1\text{'}\text{'}\; \backslash ,$ and $S\_2\; =\; S\_2\text{'}\text{'}\backslash ,$ (this is the case, for instance, for von Neumann algebras), then the above equality gives :$(S\_1\text{'}\; \backslash cup\; S\_2\text{'})\text{'}\text{'}\; =\; (S\_1\; \text{'}\text{'}\; \backslash cap\; S\_2\; \text{'}\text{'})\text{'}\; =\; (S\_1\; \backslash cap\; S\_2)\text{'}\; .$ ==See also== * von Neumann double commutant theorem 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「bicommutant」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.