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There is something that is nagging me, I thought that the operator norm of the group ring inside the $C^*_r(G)$ was the same as it's $l^2$ norm but this seems to not be true, clearly the $l^2$ norm is less than the operator norm by evaluating by the function which is one at the identity and 0 elsewhere. I just realised that this could not be the case for the norms to be the same as the definition of RD group given in Josillaint's paper below would not be interesting.

https://www.jstor.org/stable/2001458?seq=1#metadata_info_tab_contents

could anyone give me an example of why this is not true? I feel like I am missing something.

sirjoe
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  • This is true, I was just confused with this, is it in general know what the operator norm of an element in the group algebra is? Also is it easy to see that $||a||=sup|a_i|$ in the finite case? ( I can see why it's at least the supremum but not why it's bounded by this) – sirjoe Apr 17 '20 at 20:16

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Suppose $g\in G$ is a non-trivial torsion element, and let $n>1$ be the smallest number such that $g^n=1$. Then the element $p=\frac1n\sum_{k=0}^{n-1}g^k$ of $\mathbb CG$ is a projection (i.e. $p^2=p=p^*$), so $\|p\|_{op}=1$. But we can compute directly that $\|p\|_2=\frac{1}{\sqrt n}$, so the two norms are not equal (in general).

If instead $G$ is torsion-free, and $g\neq1$, then we have $$\|1+g\|_2^2=2<\sqrt 6=\|(1+g^{-1})(1+g)\|_2\leq\|(1+g^{-1})(1+g)\|_{op}=\|1+g\|_{op}^2.$$

Aweygan
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