Timeline for What do we know about this ideal of the group algebra?
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
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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 |
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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 |
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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 |