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Feb 11, 2016 at 15:59 vote accept InfiniteLooper
Feb 11, 2016 at 15:59 comment added InfiniteLooper Oh ! Ok, you're right, thanks a lot :) I had a look first in the book of Bruce Blackadar. He took the Murray Von Naumann equivalence to define $V(A)$. However those two semi groups lead to the same group and to the same K theory.
Feb 11, 2016 at 15:43 history answered Chris Ramsey CC BY-SA 3.0