Let $A$ be an involutive $\mathbb{C}$-algebra, and let $C$ be a class of $C^*$-algebras.

I'm looking for the name and information about the property

$\forall a \in A$, $\exists B \in C, \exists \pi : A \rightarrow B$ $\quad \vert \vert \pi(a) \vert \vert = \displaystyle \sup\limits_{\substack{B \in C\\ \pi : A \rightarrow B}} \vert \vert \pi(a) \vert \vert$

in the special case where $C$ is the class of finite-dimensional $C^*$-algebras, or in general.

I would also like to know if the algebra of a product of free groups has this property - I am aware that a strenghthened version of Kirchberg's conjecture would imply this, but I am not interested, for the moment, in knowing if the above $sup$ is the maximal norm or not, which, I think, is what Kirchberg's conjecture is really about.

EDIT : If $A$ is the $\mathbb{C}$-algebra of a discrete group, then, for each $*$-morphism $\pi : A \rightarrow B$ with $B$, $C^*$-algebra, for each $a \in A$, $\vert \vert \vert \pi(a) \vert \vert \vert \leq \Vert a \Vert_1$, so the $\sup$ is finite in the cases I'm interested in.


1 Answer 1


Let $C$ be a class of $C^*$-algebras. Consider a sequence $(B_n)_{n \in \mathbb{N}}$ of $C^*$-agebras in $C$. For an involutive $\mathbb{C}-$ algebra $A$, let $\pi:A\to B_n.$ Set $a_n:=\pi(a)\in B_n.$ Then there is no guarantee that the sequence $(\|a_n\|)_{n\in \mathbb{N}}$ be bounded.

Therefore $\|\pi(a)\|=\sup_{B_n\in C}\|a_n\|$ may not exist at all. Hence your definition of $\|\pi(a)\|$ is invalid.

  • $\begingroup$ I did not define $\Vert \pi(a)\Vert$... And if $\sup_{B \in C} \Vert \pi(a) \Vert$ is infinite, then the algebra does not possess the property I'm interested in. $\endgroup$
    – Plop
    Nov 22, 2017 at 18:10

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