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Sep 10, 2015 at 10:24 comment added eric @Hans Schoutens: Isn't that just the definition of the Hasse invariant?
Sep 10, 2015 at 2:45 comment added Hans Schoutens I guessed that something like that must hold, but I couldn't find a reference. Do you have any?
Sep 10, 2015 at 2:45 vote accept Hans Schoutens
Sep 10, 2015 at 0:52 comment added grghxy Strictly speaking, $A$ is the Hasse invariant attached to the pair $(E, \omega)$ where $\omega$ is the global 1-form dual to $\delta$.
Sep 9, 2015 at 20:13 history answered Felipe Voloch CC BY-SA 3.0