Spanning tree: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
← Older edit
en>Tobias Bergemann
Reverted 1 edit by 103.21.52.74 (talk): Undo unexplained removal of content. (TW)
en>David Eppstein
 
Line 1: Line 1:
{{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]]

Latest revision as of 08:43, 25 December 2014

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 lukebrya of ordinary, pleasant and entertaining individuals to meet, so the pitch is gone by here!
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 sound I have never been into clubs. In addition, I got two very cunning and undoubtedly lukebryan.com cheeky puppies that are always ready to meet fresh people.
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 luke bryan show I really don't brain.

my homepage luke bryan to

Retrieved from "https://en.formulasearchengine.com/index.php?title=Spanning_tree&oldid=230692"