3
$\begingroup$

I starded studying Git theory, and I am stuck with the follwoing problem.

Let $\textbf{Sch}$ be the category os Schemes of a field (it can be algebraically closed if needed), and $\textbf{Sets}$ be the category of sets. Let $X \in \textbf{Sch}$ and Let G be a algebraic group (G is a group object in $\textbf{Sch}$)acting on $X$.

For each $T \in \textbf{Sch}$ Consider the action of $Hom(T,G)$ on Hom(T,X) as follwoing, for each morphism $g : T \to G$ and each morphism $x:T \to X$, we have $Hom(T,G) \times Hom(T,X) \to Hom(T,X)$, $(g,x) \to (g(t)x(t))$. We say that two morphisms $x,y : T \to X$ are in the same class if there exist one $g \in Hom(T,G)$ such that $x(t) = g(t)y(t)$ for all t.

Define the functor and $\mathcal{F} : \textbf{Sch} \to \textbf{Sch} $ that sends each scheme T to the set of classes of equivalences of $Hom(T,X)$ defined as before.

I saw in some notes that I am not able to find that is it possible to say that the space of orbits X/G represents the functor $\mathcal{F}$, if G is a reductive group, but I am not able to find such notes, neither prove this fact, any references are welcome.

Thank you

As suggested, I posted this question on Mathoverflow.

https://mathoverflow.net/questions/262268/naive-question-on-quotients-of-algebraic-groups-and-moduli-spaces

  • 3
    By an algebraic group you mean a group object in the category of schemes? So you are not saying that $G$ is an algebraic group over a field or anything? – 2017-02-13
  • 0
    You right, I should edit the question, by category of schemes I should say category of schemes over a field, and G a group object in this category. Thank you. – 2017-02-13
  • 0
    I have no idea how to do this problem. You might have better luck asking on mathoverflow. – 2017-02-13
  • 0
    Thank you for your suggestion, I will transfer this question to there, and put a link here. – 2017-02-15

0 Answers 0