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May 21, 2021 at 11:49 vote accept Ali Taghavi
May 21, 2021 at 10:16 comment added Ali Taghavi Yes I see The irreducible representation works otherwise the range of idempotent $1_A$ would be $A$-invariant.
May 21, 2021 at 9:39 comment added Mateusz Wasilewski You can choose the representation to satisfy this. The usual procedure involving the GNS construction works, for example.
May 21, 2021 at 9:05 comment added Ali Taghavi Thanks for your answer. In this representation is is it obvious that $1_A=1_{B(H)}$?
May 21, 2021 at 8:26 history answered Mateusz Wasilewski CC BY-SA 4.0