Timeline for Sums of unitaries with small norm in full group $C^*$-algebras

Current License: CC BY-SA 3.0

7 events

when toggle format	what		by	license	comment
May 12, 2012 at 5:01	vote	accept	Mike Jury
May 12, 2012 at 18:14
May 12, 2012 at 4:54	history	edited	Nik Weaver	CC BY-SA 3.0	added 88 characters in body
May 12, 2012 at 4:53	comment	added	Nik Weaver		Or if you want an example with infinite order generators, take $G = \langle a,b: ab^{-1} = ba^{-1}\rangle$ and again let $S = \{a,b\}$ and $x = 1a + ib$. Again the $l^1$ norm of $x$ is 2 but its C* norm is at most $\sqrt{2}$.
May 12, 2012 at 4:36	comment	added	Nik Weaver		For example, let $G = ({\bf Z}/2)^2$ with standard generators $a$ and $b$ and let $S = \{a,b\}$ . Then we can take $x = 1a + ib$, so the $l^1$ norm of $x$ is 2 but the $l^1$ norm of $x^x = (1a - ib)(1a + ib) = 2e$ is 2, so that the C norm of $x$ can be at most $\sqrt{2}$.
May 12, 2012 at 4:16	comment	added	Nik Weaver		Right. But actually checking that something is an example might still be hard if you have to compute norms. Here's an idea: the C* norm of $x = \sum a_jg_j$ squared equals the C* norm of $x^x = \sum a_j\bar{a}_k g_jg_k^{-1}$. If the $l^1$ norm of the latter is less than the $l^1$ norm of $x$ squared, then we know the C norm of $x$ is less than its $l^1$ norm. This can give us an easy way to generate examples where the C* norm doesn't equal the $l^1$ norm.
May 12, 2012 at 3:27	comment	added	Mike Jury		Ah, duh. I hadn't thought very hard about the conditions on $S$. It is probably natural to impose the condition that $S$ be "minimal" in some sense. Say, no proper subset of $S$ is generating? (I think I've been tacitly assuming this in all the examples I've looked at.) I'll edit the question to include this condition. Even so, now that you say it seems obvious that there should be (lots of?) examples, just on the grounds that the C*-norm won't equal the $\ell^1$ norm. Thanks.
May 12, 2012 at 3:07	history	answered	Nik Weaver	CC BY-SA 3.0