1
$\begingroup$

Let $A$ be a simple unital $C^{*}$ algebra with invertible elements $G(A)$. Assume that $A^{*}$ is its dual space, which is equipped with the weak star topology.

Is there a continuous map $\alpha:A^{*}-\{0\}\to G(A)$ with $\alpha(\phi)\in \ker \phi$, in the other words: $\phi(\alpha_{\phi})=0$ where $\alpha_{\phi}=\alpha(\phi)$?

If the answer is yes, is there an extension of $\phi$ to whole $A^{*}$?

Note that there is a (possible) non continuous $\phi$ as above.(As a consequence of the Gleason Kahane Zelazko theorem)

What about if we consider the norm topology for the dual space?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Hmm... I guess you’re imposing the simplicity condition on $ A $ so as to rule out the existence of non-trivial characters on $ A $ that would make the assertion false? $\endgroup$
    – Transcendental
    Commented Feb 23, 2015 at 20:26
  • $\begingroup$ @Transcendental exactly for this reason that you said. $\endgroup$
    – Ali Taghavi
    Commented Feb 23, 2015 at 20:29
  • 1
    $\begingroup$ You would also have to exclude the case $ A = \mathbb{C} $, which is unital and simple. The non-zero continuous linear functionals on $ \mathbb{C} $ are precisely those defined by $ \phi_{\lambda}: z \mapsto \lambda z $ for $ \lambda \in \mathbb{C}^{\times} $. As $$ G(\mathbb{C}) \cap \ker(\phi_{\lambda}) = \mathbb{C}^{\times} \cap \{ 0_{\mathbb{C}} \} = \varnothing $$ for each $ \lambda \in \mathbb{C}^{\times} $, the answer to your question is negative in this special case. $\endgroup$
    – Transcendental
    Commented Feb 26, 2015 at 0:14

0

Reset to default

You must log in to answer this question.