# Is it true that $\forall G\leq Aut(\mathbb{P}^1)=PGL_2(\mathbb{C})$ the map $\mathbb{P}^1\rightarrow \mathbb{P}^1/G$ is defined over $\mathbb{Q}$?

Is it true that for every finite $G\leq Aut(\mathbb{P}^1_{\mathbb{C}})=PGL_2(\mathbb{C})$ the morphism $\mathbb{P}^1_{\mathbb{C}}\rightarrow \mathbb{P}^1_{\mathbb{C}}/G$ descends as a morphism (not nec. with the group actions) to $\mathbb{Q}$?

My intuition is going haywire here. For a while I think it's true, and then I think it's not. Do you have a decisive answer?

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Doesn't this follow from the explicit form of the invariant subalgebras of those groups (as given, among many other places, in Klein's book on the icosahedron or Dolgushev's notes on the McKay correspondence)? –  Mariano Suárez-Alvarez Sep 3 '11 at 3:41
I'm not aware of this literature, and I only vaguely remember hearing about the McKay correspondence. Are you saying it is true? –  Nicole Sep 3 '11 at 3:43
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