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Jul 26, 2020 at 12:57 comment added Tobias Diez The action on $C^2$ is by conjugation: $(\phi \cdot c) (x, y) = \phi \, c(\phi^{-1} x, \phi^{-1}y)$.
Jul 26, 2020 at 7:17 comment added Tobias Diez Maybe I'm missing something, but I think they do compose to zero: $J(\phi \cdot [\,,]) = \phi \cdot J([\,,]) = 0$.
Jul 25, 2020 at 20:09 history asked Tobias Diez CC BY-SA 4.0