A continuous choice of invertible elements

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?

• 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? – Transcendental Feb 23 '15 at 20:26
• @Transcendental exactly for this reason that you said. – Ali Taghavi Feb 23 '15 at 20:29
• 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. – Transcendental Feb 26 '15 at 0:14