Let $G$ be a torsion-free amenable group. Consider, $\mathbf M$, the collection of all multiplicative functionals on $\mathbb CG$, the complex group algebra of $G$. So, $\ker\phi$ is an ideal of $\mathbb CG$, for all $\phi\in\mathbf M$. Has the ideal $\mathcal I_G=\bigcap_{\phi\in\mathbf M}\ker\phi$ been studied before? Is there a class of torsion-free amenable groups $G$ for which $\mathcal I_G=0$?
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2$\begingroup$ Yes, there is such a class, but I don't know if it is what you had in mind. When $G$ is discrete and abelian ${\bf M}$ is naturally in bijection with the dual group $\widehat{G}$ and so your ideal ${\mathcal I}_G$ is zero, since characters on (locally compact) abelian groups separate points $\endgroup$– Yemon ChoiCommented Sep 3, 2019 at 13:41
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2$\begingroup$ Conversely, it's clear that if $I_G=0$ then $G$ is abelian. Hence for any group $G$ (regardless of the restriction to torsion-free amenable groups) $I_G=0$ iff $G$ is abelian. $\endgroup$– YCorCommented Sep 3, 2019 at 14:06
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$\begingroup$ I've removed the cstar and von neumann tags and functional analysis tags because as it stands I don't see how they are relevant to your question $\endgroup$– Yemon ChoiCommented Sep 3, 2019 at 15:00
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2$\begingroup$ $I_G$ should just be the kernel of the natural map $\mathbb CG$ to $\mathbb CG^{ab}$ where $G^{ab}$ is the abelianization. $\endgroup$– Benjamin SteinbergCommented Sep 3, 2019 at 15:27
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$\begingroup$ @YemonChoi Thanks for your comment. By your comment, what I have now in mind is this: If $A$ is a closed subalgebra of a Banach algebra $B$, and $\pi$ is a representation of $A$ on a Hilbert space $H$, under what condition(s) we can extend $\pi$ to all of $B$? $\endgroup$– MSMalekanCommented Sep 4, 2019 at 7:59
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