What is known about idempotents in Lie group algebras (such as on the classical Lie groups)? Specifically the self-adjoint ones. Is there anything interesting to say? I haven't been able to find much written about non-discrete ones. It could be about either one of the C*-algebras or just the L^1 convolution algebra.
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3$\begingroup$ Which algebra are you looking at? The group ring, the L^1-algebra or C*-algebra? $\endgroup$– Benjamin SteinbergSep 8, 2018 at 1:57
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1$\begingroup$ The latter is not simpler! Trying to understand self-adjoint idempotents in L^1-algebras is going to be harder than for the Cstar algebras because you have fewer tools (restricted functional calculus, etc) $\endgroup$– Yemon ChoiSep 8, 2018 at 3:08
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1$\begingroup$ For non-discrete groups the following vague paradigm is worth keeping in mind. Suppose $G$ is LCA. Then non-zero idempotents in ${\rm C}^*(G) \cong C_0(\widehat{G})$ will correspond to compact open subsets of $\widehat{G}$, and in particular if $G={\bf R}^d$ there are no such idempotents. For well-behaved locally compact groups $G$ where we have a good notion of non-abelian Fourier transform, you can then try to do something similar, but new technical problems and phenomena arise $\endgroup$– Yemon ChoiSep 8, 2018 at 3:18
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1$\begingroup$ I guess at a higher level of abstraction, if you are mostly interested in the phenomena for discrete groups that have something to do with $K_0({\rm C}^*_r(G))$, then some words to look up for the Lie group setting might be "Connes--Kasparov conjecture" $\endgroup$– Yemon ChoiSep 8, 2018 at 3:25
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1$\begingroup$ @YemenChoi Well I seem to have calculated them for the two dimensional affine group, or at least a large class of them, and I can find some on the group U(n), but I'm curious to know if much about them is known, for example do they even often exist. I guess more specifically, some of the idempotents are equivalent to states (ie. linear functionals on the algebra) and I'm wondering if anything about an application of the GNS construction to these states is known. $\endgroup$– Josh LackmanSep 8, 2018 at 5:54
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