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Feb 20, 2021 at 17:43 vote accept Edwin Beggs
Feb 19, 2021 at 1:35 comment added Narutaka OZAWA @Edwin Beggs: I learned it from Kirchberg's paper (Invent Math 1993) and recorded in my QWEP survey (IJM 2004). A correction on the previous comment: $p$ is in the center of the C*-algebra generated by the range of $f$. In any case, by the Jordan multiplicativity, $f(xy)$ equals to either $f(x)f(y)$ or $f(y)f(x)$ for each $x,y \in G$. The rest is an algebra/combinatorics problem...
Feb 18, 2021 at 22:20 answer added Matthew Daws timeline score: 2
Feb 18, 2021 at 21:05 comment added Edwin Beggs Sorry that is the paper: On Positive Linear Maps Preserving Invertibility - just found that.
Feb 18, 2021 at 19:57 comment added Edwin Beggs How does the hypothesis imply that we have a Jordan homomorphism? This is very interesting, and this sounds like the critical point!
Feb 18, 2021 at 0:57 comment added Yemon Choi @NarutakaOZAWA Indeed something about the original question reminded me of M. Walter's theorem concerning isometric isomorphism of Fourier algebras, which would lead to the kind of conclusion that you suggest.
Feb 18, 2021 at 0:51 comment added Narutaka OZAWA The hypothesis implies that $f\colon {\mathbb C}G\to{\mathbb C}H$ is a Jordan homomorphism. Hence by Stormer's theorem, there is a central projection $p$ in ${\mathbb C}H$ such that $f(\,\cdot\,)p$ is a homomorphism and $f(\,\cdot\,)(1-p)$ is an anti-homomorphism (as suggested in Yemon's answer). Probably, this would lead to that $f$ itself is either a homomorphism or an anti-homomorphism?
Feb 18, 2021 at 0:35 answer added Yemon Choi timeline score: 6
Feb 17, 2021 at 23:11 history edited Francesco Polizzi CC BY-SA 4.0
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Feb 17, 2021 at 22:09 history asked Edwin Beggs CC BY-SA 4.0