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en>David Eppstein: /* In specific graphs */ authorlink

2014-12-25T06:43:03Z

In specific graphs: authorlink

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	~~{{Unreferenced\|date=December 2009}}~~		Last week I woke up and realised - At the moment I have also been solitary for a little while and following much intimidation from friends I today find myself signed-up for web dating. They assured me that there are plenty [http://lukebryantickets.omarfoundation.org lukebrya] of ordinary, pleasant and entertaining individuals to meet, so the pitch is gone by here!<br>My household and friends are wonderful and hanging out together at bar gigabytes or dishes is obviously critical. As I find that one may never own a good dialog using the [http://browse.Deviantart.com/?qh=&section=&global=1&q=sound+I sound I] have never been into clubs. In addition, I got two very cunning and undoubtedly [http://www.cinemaudiosociety.org lukebryan.com] cheeky puppies that are always ready to meet fresh people.<br>I strive to keep as physically fit as potential coming to the gym several times per week. I appreciate my sports and strive to perform or see because many a possible. Being winter I'll often at Hawthorn matches. Notice: I have observed the carnage of fumbling suits at stocktake sales, Supposing that you really considered buying an athletics [http://lukebryantickets.sgs-suparco.org luke bryan show] I really don't brain.<br><br>
	In [~~[algebra]~~], the ~~'''bicommutant''' of~~ a [~~[subset~~]~~] ''S'' of a~~ [~~[semigroup~~]~~] (such~~ as ~~an [[algebra over a field\|algebra]]~~ or a ~~[[group (mathematics)\|group]]) is~~ the [~~[commutant~~]~~] of the commutant of that subset~~. ~~It is also known as the double commutant or second commutant and is written~~ <~~math~~>~~S^{\prime \prime}~~<~~/math~~>.

	~~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 algebra]]s. 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 operator topology\|weak closure]], [[Strong operator topology\|strong closure]] and bicommutant of ''M'' are equal. This tells us that a unital [[C-algebra\|C*-subalgebra]] ''M'' of ''B(H)'' is a [[von Neumann algebra]] if, and only if, <math>M = M^{\prime \prime}</math>, and that if not, the von Neumann algebra it generates is <math>M^{\prime \prime}</math>.		my homepage [http://www.netpaw.org luke bryan to]

	The bicommutant of ''S'' always contains ''S''. So <math>S^{\prime \prime \prime} = (S^{\prime \prime})^{\prime} \subseteq S^{\prime}</math>. On the other hand, <math>S^{\prime} \subseteq (S^{\prime})^{\prime \prime} = S^{\prime \prime \prime}</~~math>. So <math>S^{\prime} = S^{\prime \prime \prime}<~~/~~math>, i~~.e. ~~the commutant of the bicommutant of ''S'' is equal~~ to ~~the commutant of ''S''. By induction, we have:~~

	~~:<math>S^{\prime} = S^{\prime \prime \prime} = S^{\prime \prime \prime \prime \prime} = \ldots = S^{2n-1} = \ldots</math>~~

	~~and~~

	~~:<math>S \subseteq S^{\prime \prime} = S^{\prime \prime \prime \prime} = S^{\prime \prime \prime \prime \prime \prime} = \ldots = S^{2n} = \ldots</math>~~

	~~for ''n'' > 1.~~

	~~It is clear that, if ''S''<sub>1</sub> and ''S''<sub>2</sub> are subsets of a semigroup,~~

	~~:<math>( S_1 \cup S_2 )' = S_1 ' \cap S_2 ' .</math>~~

	~~If it is assumed that <math>S_1 = S_1'' \,</math> and <math>S_2 = S_2''\,</math> (this is the case, for instance, for [[von Neumann algebra]]s), then the above equality gives~~

	~~:<math>(S_1' \cup S_2')'' = (S_1 '' \cap S_2 '')' = (S_1 \cap S_2)' .</math>~~

	~~==See also==~~
	* [[von Neumann double commutant theorem]]

	~~[[Category:Group theory]~~]

en>Tobias Bergemann: Reverted 1 edit by 103.21.52.74 (talk): Undo unexplained removal of content. (TW)

2013-12-02T13:29:00Z

Reverted 1 edit by 103.21.52.74 (talk): Undo unexplained removal of content. (TW)

← Older revision		Revision as of 15:29, 2 December 2013
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	~~Some person who wrote all~~ of the ~~article~~ is ~~called Leland but it's not this most masucline name on~~ the ~~internet~~. ~~Managing people~~ is ~~very much where his primary wage comes from. His wife and him live~~ in ~~Massachusetts~~ and ~~he owns everything that he conditions there~~. ~~Base jumping is something~~ that ~~he or she~~ is ~~been doing~~ for ~~many~~. ~~He has been~~ [~~http:~~//~~Www~~.~~Wikipedia~~.~~org~~/~~wiki~~/~~running running] and maintaining one specific blog here~~: ~~http~~://~~circuspartypanama~~.~~com~~<br><br>~~my blog post [http~~://~~circuspartypanama~~.~~com clash of clans hack no survey no password~~]		{{Unreferenced\|date=December 2009}}
			In [[algebra]], the '''bicommutant''' of a [[subset]] ''S'' of a [[semigroup]] (such as an [[algebra over a field\|algebra]] or a [[group (mathematics)\|group]]) is the [[commutant]] of the commutant of that subset. It is also known as the double commutant or second commutant and is written <math>S^{\prime \prime}</math>.

			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 algebra]]s. 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 operator topology\|weak closure]], [[Strong operator topology\|strong closure]] and bicommutant of ''M'' are equal. This tells us that a unital [[C-algebra\|C*-subalgebra]] ''M'' of ''B(H)'' is a [[von Neumann algebra]] if, and only if, <math>M = M^{\prime \prime}</math>, and that if not, the von Neumann algebra it generates is <math>M^{\prime \prime}</math>.

			The bicommutant of ''S'' always contains ''S''. So <math>S^{\prime \prime \prime} = (S^{\prime \prime})^{\prime} \subseteq S^{\prime}</math>. On the other hand, <math>S^{\prime} \subseteq (S^{\prime})^{\prime \prime} = S^{\prime \prime \prime}</math>. So <math>S^{\prime} = S^{\prime \prime \prime}</math>, i.e. the commutant of the bicommutant of ''S'' is equal to the commutant of ''S''. By induction, we have:

			:<math>S^{\prime} = S^{\prime \prime \prime} = S^{\prime \prime \prime \prime \prime} = \ldots = S^{2n-1} = \ldots</math>

			and

			:<math>S \subseteq S^{\prime \prime} = S^{\prime \prime \prime \prime} = S^{\prime \prime \prime \prime \prime \prime} = \ldots = S^{2n} = \ldots</math>

			for ''n'' > 1.

			It is clear that, if ''S''<sub>1</sub> and ''S''<sub>2</sub> are subsets of a semigroup,

			:<math>( S_1 \cup S_2 )' = S_1 ' \cap S_2 ' .</math>

			If it is assumed that <math>S_1 = S_1'' \,</math> and <math>S_2 = S_2''\,</math> (this is the case, for instance, for [[von Neumann algebra]]s), then the above equality gives

			:<math>(S_1' \cup S_2')'' = (S_1 '' \cap S_2 '')' = (S_1 \cap S_2)' .</math>

			==See also==
			* [[von Neumann double commutant theorem]]

			[[Category:Group theory]]

en>David Eppstein: Reverted edits by 112.109.18.67 (talk) to last version by David Eppstein

2012-08-29T15:24:04Z

Reverted edits by 112.109.18.67 (talk) to last version by David Eppstein

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