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Say I have an algebraic curve $C$ over a field $k$ and a group $G$ acting on $C$. Under what conditions on $C$ and/or the action of $G$ on $C$ can one conclude that $H^0(C,\Omega^1_C)^G = H^0(C/G,\Omega^1_{C/G})$ holds?

For example, if the morphism $C \to C/G$ has some ramification, what can one say? Could it be that as long as both $C$ and $C/G$ are smooth and projective, this equality holds, for example?

Evariste
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  • I'm pretty sure this is true as long as the map $C\rightarrow C/G$ is affine. In particular, if $G$ is a finite group, you should be okay. – oxeimon May 20 '14 at 17:29
  • Could you say a few words about why this should be true? – Evariste May 20 '14 at 21:18
  • Is $G$ finite? Or is it an algebraic group? In general, it is not clear how $C/G$ has the structure of a variety. – Jesko Hüttenhain May 21 '14 at 06:02
  • It is a theorem that $C/G$ always has a unique structure of variety such that the projection is a morphism (if $C$ is projective and smooth). – rfauffar May 21 '14 at 19:26
  • generally, whenever you have an affine map, and something about the map is true whenever the source and target are affine, then it's always true. I'm pretty sure what you say is true when C is affine, since then C/G is also affine. If G is finite, then C -> C/G is a finite map, which is affine. – oxeimon May 21 '14 at 22:21
  • checking if what you say is true when C is affine is just a matter of commutative algebra which I don't feel like doing right now. Just note that if C = Spec A, then C/G is just Spec A^G (spec of the G-invariants of A) – oxeimon May 21 '14 at 23:05
  • Let's say $G$ is finite... thanks for your comments. – Evariste May 21 '14 at 23:26

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