Timeline for Comparison between the operator norm and the $L^1$ norm on group algebras
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2023 at 12:50 | comment | added | Onur Oktay | @YemonChoi thanks for your kind words. Yes indeed, although it is a pretty standard argument: if $A$ is a $C^*$-algebra, $x\in A$ is normal, then $C(\sigma(x))$ is a closed subalgebra of $A$, which can be wsc iff the spectrum $\sigma(x)$ is a finite set. Thus, if $A$ is wsc, then the spectrum of every self-adjoint element in $A$ is finite, so $A$ must be finite-dimensional. | |
| Nov 10, 2023 at 1:12 | comment | added | Yemon Choi | @OnurOktay's very nice argument is actually showing the stronger result that there is no Banach space isomorphism between $\ell^1(G)$ and ${\rm C}_r^\ast(G)$. (It may be worth adding that WSC passes to closed subspaces and $c_0$ is not WSC, which I assume is how one shows that infinite-dimensional ${\rm C}^\ast$-algebras are not WSC) | |
| Nov 8, 2023 at 14:25 | comment | added | Onur Oktay | Q1 has a relatively easy answer. $\ell^1(G)$ is weakly sequentially complete (wsc). $\|.\|_{op}$ is a $C^*$-algebra norm, and the completion of $\mathbb{C}[G]$ with this norm is the reduced group $C^*$-algebra $C^*_r(G)$. Were the two norms equivalent, then $C^*_r(G)$ would be a wsc $C^*$-algebra. Only wsc $C^*$-algebras are finite dimensional ones. $C^*_r(G)$ is finite dimensional only when $G$ is a finite group. | |
| Nov 6, 2023 at 16:29 | vote | accept | David Gao | ||
| Nov 6, 2023 at 9:22 | answer | added | Stefaan Vaes | timeline score: 4 | |
| Nov 5, 2023 at 22:53 | answer | added | Stefaan Vaes | timeline score: 4 | |
| Nov 5, 2023 at 14:04 | answer | added | Yemon Choi | timeline score: 6 | |
| Nov 5, 2023 at 13:08 | comment | added | Yemon Choi | On further reflection, I think one can use classical examples from Fourier analysis to show that the "weakest form" of Q2 has a negative answer for $G={\mathbb Z}$, and then this should bootstrap to give a negative answer for any $G$ that contains an element of infinite order. I suspect that the answer is negative for any infinite $G$. | |
| Nov 5, 2023 at 12:48 | comment | added | Yemon Choi | Have you tried to see if the inequality in Q2 holds for $G={\mathbb Z}$ when we can study things as functions on ${\mathbb T}$? | |
| Nov 5, 2023 at 12:47 | comment | added | Yemon Choi | Your guess is correct for Q1 - the norms are inequivalent for every infinite G - but currently the proof I have invokes machinery (namely, Arens regularity) that is surely unnecessary and over-the-top, and which doesn't seem to help with Q2. | |
| Nov 5, 2023 at 4:58 | history | asked | David Gao | CC BY-SA 4.0 |