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Jul 31, 2020 at 16:38 comment added LSpice No, yours is right and mine (which I have now deleted) is wrong; I was incorrectly treating it as a homomorphism $A \times A \times A \to \mathbb C^\times$.
Jul 31, 2020 at 14:52 comment added Bipolar Minds @LSpice I got $t(a,b,a)=\frac{t(a,a+b,a+b)}{t(a,a,a+b)t(a,b,b)}$ but probably its the same as yours
Jul 31, 2020 at 14:40 comment added Bipolar Minds @LSpice I think it follows from the other two conditions and $t$ being a homomorphism
Jul 31, 2020 at 14:37 history edited LSpice CC BY-SA 4.0
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Jul 31, 2020 at 14:37 comment added LSpice Do you also want $t(a, b, a) = 1$ for all $t$?
Jul 31, 2020 at 12:09 history asked Bipolar Minds CC BY-SA 4.0