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Let $A = C^*(T_1,T_2,T_3,... | T_i=T_i^*, ||T_i|| \leqslant 1)$ - universal enveloped $C^*$-algebra of countable family of self-adjoint operators. $A$ have standard Schauder basis, which contains all words generated by $T_i$ (i.e. $T_1 T_3 T_2 T_5$ - is basis vector). Is it true, that it is boundedly complete Schauder basis or monotonically boundedly complete Schauder basis?

Boundedly complete Schauder basis $e_i \in X$ it is such basis, that for every sequence $\alpha_i \in \mathbb{C}$ such the partial sums $V_N = \sum_{i=1}^N \alpha_i e_i$ are bounded in $X$, the sequence $V_N$ are converge in $X$.

Monotonically boundedly complete Schauder basis is the same, but we consider only $\alpha_i$ which converge to zero.

UPD: I found some interesting theorem here (page 2), maybe it will help to answering on that question.

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  • $\begingroup$ I don't see why the set of non-commutative monomials is a Schauder basis. What would be the argument to justify that? $\endgroup$ – Martin Argerami May 29 '16 at 5:25
  • $\begingroup$ @MartinArgerami Let $D$ is C*-subalgebra generated by noncomutative monomials. Then null map $0 : A \to A/D$ and factormap $F: A \to A/D$ is 2 maps which makes triangle $D \to A \to A/D$ commutative. But $A$ is universal, so there exist only one such map, so, factormap equals 0 and $D=A$. $\endgroup$ – kp9r4d May 29 '16 at 5:38
  • $\begingroup$ As far as I can tell, that doesn't show that the monomials form a Schauder basis for $D$. $\endgroup$ – Martin Argerami May 29 '16 at 5:41
  • $\begingroup$ It show that monomials is total system (it is some hole in my previous reasoning: we dont know that $D$ is ideal, but it is fixable). And it is linear independent, because they linear independent in left regular representation. Every monomial act on formal space $L_2 (Monomials)$. $\endgroup$ – kp9r4d May 29 '16 at 5:55
  • $\begingroup$ Are you saying "linearly independent plus dense span = Schauder basis"? Because I don't think that's true. $\endgroup$ – Martin Argerami May 29 '16 at 5:58
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I claim that if $A$ is an infinite-dimensional C*-algebra then it does not have a boundedly complete basis.

Proof. Assume that $A$ has such a basis. A space with a basis is separable, so $A$ is separable. Moreover, $A$ being infinite-dimensional, contains an infinite-dimensional abelian sub-C*-algebra. That abelian subalgebra will contain a sequence of norm-one orthogonal elements and such elements span a Banach-space copy of $c_0$. This copy of $c_0$ will be complemented in $A$ by Sobczyk's theorem. However, a space with a boundedly complete basis is isomorphic to a dual space. $c_0$ is not complemented in any isomorphic copy of a dual space, because it would be complemented in its bidual (see the diagram on p. 22 here) but by the Phillips–Sobczyk theorem (or by the Grothendieck property of $\ell_\infty$), no copy of $c_0$ is complemented in $\ell_\infty$.

The above argument shows that there is no infinite-dimensional, separable C*-algebra isomorphic to a dual space as a Banach space. In particular, there is no separable, infinite dimensional von Neumann algebra.

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  • $\begingroup$ Tomek, it is not true that any infinite-dimensional C$^*$-contains a copy of $c_0$. Many infinite-dimensional separable C$^*$-algebras are projectionless (examples: $C[0,1]$, $C_r(\mathbb F_n)$). $\endgroup$ – Martin Argerami Jun 3 '16 at 1:59
  • $\begingroup$ If you are only talking linear, I don't know what you mean by "norm-one orthogonal elements". $\endgroup$ – Martin Argerami Jun 3 '16 at 5:40
  • $\begingroup$ @MartinArgerami, in $C_0(X)$ take a family of disjointly supported functinos. This is what I mean. They span (an isometric) copy of $c_0$. $\endgroup$ – Tomek Kania Jun 3 '16 at 6:03
  • $\begingroup$ Yes, I get it now. $\endgroup$ – Martin Argerami Jun 3 '16 at 6:10
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Since the properties involved are about boundedness and convergence, they survive $*$-homomorphisms, which means that they pass to quotients. And any separable C$^*$-algebra is a quotient of $C^*(\mathbb F_\infty)$. In particular, $C[0,1]$ is, so $C^*( \mathbb F_\infty)$ does not have a boundedly complete Schauder basis by the article you linked to.

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  • $\begingroup$ Thanks, it seems like that. But what about monotonically boundedly completeness? If we take in terms of theorem in this article $N_k = k, m_0 = 1, P = 1/2, F_k = TrivialRepresentation$ it will be true? $\endgroup$ – kp9r4d May 29 '16 at 5:46
  • $\begingroup$ Martin, I am not sure if I understand your claim. Certainly $\ell_1$ has a boundedly complete basis but every separable Banach space is a quotient of $\ell_1$, so this property does not pass to (linear) quotients. Since having a b.c. basis is a purely linear property, I see no reason why *-homomorphisms would have to preserve it. $\endgroup$ – Tomek Kania Jun 3 '16 at 0:25
  • $\begingroup$ I think you are right. Let me think about it. $\endgroup$ – Martin Argerami Jun 3 '16 at 1:57

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