Timeline for Characterizing Herz-Schur multipliers using coefficient functions of uniformly bounded representations
Current License: CC BY-SA 3.0
9 events
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| Feb 6, 2018 at 16:55 | comment | added | Adrián González Pérez | Sadly, I think I was confused in my previous comment. You are right, unitarizability implies that $B(G) = \cup_{c} B_c(G)$ but does not immediately give that $B(G) = M_{cb}A(G)$. | |
| Feb 6, 2018 at 16:36 | comment | added | Mahmood Al | @AdriánGonzález-Pérez So based on your comment, if $G$ is unitarizable then $B(G)=M_{cb}A(G)$. But this implies amenability! Ruan (in an unpublished manuscript generalizing the argument of Losert in [Proc. Amer. Math. Soc. 92 (1984), no. 3, 347–354]) proved that $G$ is amenable if and only if $B(G)=M_{cb}A(G)$. All of this can be found in Pisier's book on similarity on page 54 (2nd edition). So your remark would solve the conjecture. Would you tell us where you have seen the proof? | |
| Feb 5, 2018 at 16:33 | comment | added | Mahmood Al | @YemonChoi: The proof is using radial cb-multipliers. I have a copy of the paper which I will send you via email. (I personally do not see any indication of Leinert sets there, but I might be naive.) | |
| Feb 5, 2018 at 16:28 | comment | added | Yemon Choi | Mahmood, do you know what these "exotic" cb-multipliers look like for the case of free groups? Is this another trick with Leinert sets or similar? | |
| Feb 5, 2018 at 16:25 | history | edited | Mahmood Al | CC BY-SA 3.0 |
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| Feb 5, 2018 at 16:23 | comment | added | Mahmood Al | @YemonChoi: Thanks I definitely meant that! | |
| Feb 5, 2018 at 16:18 | comment | added | Yemon Choi | Small correction to 1st paragraph: I assume you wish to say that $\Vert \pi(g)\Vert\leq c$ for all $g\in G$? | |
| Feb 5, 2018 at 16:06 | history | edited | Mahmood Al | CC BY-SA 3.0 |
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| Feb 5, 2018 at 15:50 | history | asked | Mahmood Al | CC BY-SA 3.0 |