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Dec 25, 2015 at 18:34 comment added Lisa S. @NoamD.Elkies: I just mean that the action is "defined over $k$" as you say, i.e., $X$ is a $k$-scheme, $G$ is also a $k$-scheme (a "constant" one), there is a $k$-scheme morphism $G \times_k X \rightarrow X$ such that blah blah blah, etc.
Dec 25, 2015 at 4:24 comment added Noam D. Elkies What does it mean for the action of $G$ on a curve to be "$k$-linear"? Or do you just mean that the action is defined over $k$?
Dec 24, 2015 at 19:30 vote accept Lisa S.
Dec 24, 2015 at 16:17 answer added Jason Starr timeline score: 8
Dec 24, 2015 at 10:48 answer added Niels timeline score: 7
Dec 24, 2015 at 8:09 comment added Philip Engel Yes, the pullback of a form from the quotient is invariant, so you need only check that an invariant form is the pullback of a form. Given an invariant form on $X$, one can define a form locally on $Y$ away from the ramification locus by defining it as the value on one of the covering sheets. It is easy to check that this form extends holomorphically to all of $Y$. For instance, using Riemann-Hurwitz (or really just the local description of holomorphic maps $z\mapsto z^n$) to verify that the extension is holomorphic.
Dec 24, 2015 at 7:44 answer added R. van Dobben de Bruyn timeline score: 4
Dec 23, 2015 at 21:40 history asked Lisa S. CC BY-SA 3.0