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Jun 23, 2020 at 18:46 comment added Yemon Choi @MatthewDaws to show contractive forces M to be commutative, maybe show things go wrong for 2 by 2 matrices, and then argue that a noncommutative M contains a copy of ${\bf M}_2$. But as you say it's not clear if M commutative is even sufficient to get $\Delta(A) \subseteq A \hat\otimes A$
Jun 23, 2020 at 18:43 comment added Yemon Choi @MatthewDaws 8 years is fine, I'm still one day going to solve the problems from my PhD...
Jun 23, 2020 at 9:17 comment added Matthew Daws I am coming to this 8 years too late... But Quigg's paper (mathscinet.ams.org/mathscinet-getitem?mr=806641) shows exactly this, in Section 5. However, this only shows that the coproduct $\Delta$ on $A$ is bounded iff $M$ is as stated. The original question wanted this map to be "short", that is, contractive. Surely then you only get commutative vn algs, but I don't quite see the proof. Similarly, for a full answer, you then need to character which commutative vn algs give $\Delta$ mapping into $A\widehat\otimes A$, and not something larger.
Aug 26, 2012 at 8:43 history answered Yemon Choi CC BY-SA 3.0