On Schauder basic systems in universal enveloped algebra of system of countable family of bounded selfadjoint operators

Question

Let $A = C^*(1,T_1,T_2, ... | T_i^* = T_i, ||T_i|| \leqslant 1)$ - 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 $e_n \in B_n$ some system of vectors. It is obviously linear independent, but moreover, it is Schauder base system. It is exist simple argument: let $\pi_n$ - is representation of $A$ which seems like $\pi_n(T_k) = T_k$ if $k \leqslant n$ and $\pi_n(T_k) = 0$ otherwise. So, we have inequality $$ \forall (a_n)\subset \mathbb{C}\quad \forall n\in\mathbb{N}\quad\forall m\leq n \quad\left\Vert \sum\limits_{k=1}^m a_k e_k\right\Vert\leq\left\Vert\pi_m(\sum\limits_{k=1}^n a_k e_k)\right\Vert\leq \left\Vert \sum\limits_{k=1}^n a_k e_k\right\Vert $$ and it is sufficient condition to be $e_n$ Schauder basic system.

So, my question: does this base system have some good properties? For example: boundedly completeness, unconditionally...

I was create another question: Universal $C^*$-algebra of countable family of self-adjoint operators have boundedly complete standard Schauder basis but question on the link it is not exactly what I want to ask. So I slightly edit my question and now it correct.

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

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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 $C_0[0,\alpha)$ for some ordinal $\alpha<\omega^\omega$. 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.

A space with a boundedly complete basis is isomorphic to a dual space. This is impossible for an infinite-dimensional, separable C*-algebra.

edited Apr 13 '17 at 12:21

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answered Jun 02 '16 at 21:18

Tomasz Kania

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I dont have a Schauder basis, but a Shchauder basic system (Schauder basis in linear span of that system) - which just Schauder basis on it own linear span. But I have a banach subspace which is not actually C*-algebra, maybe it will help? – Mykola Pochekai Jun 02 '16 at 21:24
@kp9r4d, every separable Banach space embeds into $C[0,1]$; I suppose you need to restrict yourself a bit. – Tomasz Kania Jun 02 '16 at 21:39
I dont really understand. What says that fact about $e_n$? – Mykola Pochekai Jun 02 '16 at 21:47
@kp9r4d it is not clear what are the relations defining $T_k$, hence $B_k$ and so $e_k$ can be quite arbitrary. – Tomasz Kania Jun 02 '16 at 22:19
It seems like that, thanks for your answer. – Mykola Pochekai Jun 02 '16 at 23:01
Can you give me your email or some another contacts? I have several clarifying questions. – Mykola Pochekai Jun 07 '16 at 16:34
@kp9r4d, you will find a link to my webpage in my profile here. – Tomasz Kania Jun 07 '16 at 16:49

On Schauder basic systems in universal enveloped algebra of system of countable family of bounded selfadjoint operators

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