Let A = C &#x2217;<!-- ∗ --> </msup> ( 1 , T 1 </msub> ,

Quintacj

Quintacj

Answered question

2022-05-25

Let A = C ( 1 , T 1 , T 2 , . . . | T i = T i , | | T i | | 1 ) - universal enveloped C -algebra of countable family of selfadjoint operators. I want to know as more as possible about that algebra, so, any links are welcome.
In that C ∗-algebra we have a family of Banach subspaces B n which defines like B n = {{ All words generated by T i which contain at least one T n and does not contain T n + 1 }.
Let e n B n some system of vectors. It is obviously linear independent, but moreover, it is Schauder base system. It is exist simple argument: let π n - is representation of A which seems like π n ( T k ) = T k if k n and π n ( T k ) = 0 otherwise. So, we have inequality
( a n ) C n N m n k = 1 m a k e k π m ( k = 1 n a k e k ) k = 1 n a k e k
and it is sufficient condition to be e n Schauder basic system.
So, my question: does this base system have some good properties?

Answer & Explanation

Kaiden Porter

Kaiden Porter

Beginner2022-05-26Added 10 answers

Your question is unclear to me but I will comment on it anyway. What generating relations do you impose on your generators? Note that it is an open problem whether the reduced group algebra of the free group on two generators has a Schauder basis. Before Enflo showed that there exist Banach spaces without the AP, people had looked at this object as a natural candidate for a space without a basis. Eventually, Haagerup proved that it has (some stronger form of) the AP.
Also, if you have a commutative C -algebra that has an unconditional basis, then it is isomorphic to C 0 [ 0 , α ) for some ordinal α < ω ω . Similar restrictions apply to nicely behaved non-commutative C -algebras and I think it is known that the existence of an unconditional basis forces the algebra to be of type I, hence uninteresting.

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