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Sep 4, 2019 at 21:04 comment added Yemon Choi Meisam, it is better to ask separate questions rather than keep shifting the goalposts after you have posted a question
Sep 4, 2019 at 8:09 comment added MSMalekan @YemonChoi In particular I want to know the answer in the case where $B=\ell^1(G)$ and $A$ is a closed subalgebra of $B$ generated by an element in $\mathbb CG$?
Sep 4, 2019 at 7:59 comment added MSMalekan @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 $B$?
Sep 3, 2019 at 15:27 comment added Benjamin Steinberg $I_G$ should just be the kernel of the natural map $\mathbb CG$ to $\mathbb CG^{ab}$ where $G^{ab}$ is the abelianization.
Sep 3, 2019 at 15:00 history edited Yemon Choi
edited tags
Sep 3, 2019 at 15:00 comment added Yemon Choi 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
Sep 3, 2019 at 14:06 comment added YCor 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.
Sep 3, 2019 at 14:06 history edited YCor CC BY-SA 4.0
fixed language
Sep 3, 2019 at 13:41 comment added Yemon Choi 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
Sep 3, 2019 at 11:01 history asked MSMalekan CC BY-SA 4.0