on Williams Crossed product book,on page 198, it is mentioned that there is only one regular representation for C_c(G), and that is the left regular representation. I know that this representation is one of the regular representations, but why it is the only one. Indeed I know in the process of constructing the reduced group C*-algebra, realizing as the reduced crossed product of (C, G, id), we start with a representation of C (the complex number) on H, and that will be the identity representation ,and finally we have the induced rep. of C*(G) on B(L^2(G, H)). I mean each rep of G on different H, gives us a different rep of C*(G), so why there is only one such a representation.
1 Answer
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It looks like he's using the term "regular representation" to mean: a representation of $A \rtimes G$ on $L^2(G; H)$ constructed by combining a given representation of $A$ on $H$ with the left action of $G$ on $L^2(G;H)$. If you start with $A = {\bf C}$ then there is "only one" representation of $A$ to start with, so there is only one regular representation of ${\bf C}\rtimes G$. (Not technically true, and maybe this is your point, since we can actually represent ${\bf C}$ on any Hilbert space $H$, but all that changes about the construction is to tensor everything with $H$.)
answered Jan 8, 2013 at 17:18
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$\begingroup$ Incidentally, can anyone tell me how to TeX the crossed product symbol? $\endgroup$ Commented Jan 8, 2013 at 17:19
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1$\begingroup$ I believe it's just "\rtimes": $A \rtimes G$. $\endgroup$ Commented Jan 8, 2013 at 17:28