# Naive question on quotients of algebraic groups and moduli spaces

I asked this question on MathStackExchange and was suggested to ask here.

I started studying GIT theory, and I am stuck with the following problem.

Let $\textbf{Sch}$ be the category of schemes over 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 $\text{Hom}(T,G)$ on $\text{Hom}(T,X)$: for each morphism $g : T \to G$ and each morphism $x:T \to X$, we have $\text{Hom}(T,G) \times \text{Hom}(T,X) \to \text{Hom}(T,X)$, where $(g,x) \to (g(t), x(t))$. We say that two morphisms $x$, $y : T \to X$ are in the same class if there exist one $g \in \text{Hom}(T,G)$ such that $x(t) = g(t)y(t)$ for all $t$.

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

• If $G$ acts freely on $X$ this is not true unless you take the fppf sheafification of $\mathcal F$. Your $\mathcal F$ is the presheaf quotient of the sheaf represented by $X$ by the sheaf represented by $G$, and that is not usually a sheaf. – Marc Hoyois Feb 15 '17 at 13:12
• Dear @MarcHoyois, sorry I made I mistake up there, I want the functor $\mathcal{F}$ to be a moduli functor and not a sheaf. The fact that $\mathcal{F}$ is the quotient of the sheaf represented by $X$ and the sheaf represented by G is what I am failing to prove. – User43029 Feb 15 '17 at 13:22
• A moduli functor has to be a sheaf if you want it to be representable... – Marc Hoyois Feb 15 '17 at 13:52
• I see it, thank you, I was forgoting the fact the $\mathcal{F}$ must be a sheaf in the Zariski topology, that admits an open covering. The fact that $F$ is a pre-sheaf quotient of the sheaf represented by X and the sheaf represented by G came from the very definition of $\mathcal{F}$ right? So there should be examples where $F$ is a sheaf and where is not ? – User43029 Feb 16 '17 at 14:59
• @Marc: Sometimes people only want their moduli functors $\mathcal{F}$ to be "corepresentable", in the sense that there is a scheme $Y$ and a functor $\mathcal{F}\to\underline{Y}$ that for any scheme $T$, the induced map $\operatorname{Hom}(\underline{Y},\underline{T})\to\operatorname{Hom}(\mathcal{F},\underline{T})$ is bijective. (Here $\underline{T}$ means the image of $T$ under the Yoneda embedding.) In this case, it's okay for $\mathcal{F}$ not to be a sheaf; in fact, $\mathcal{F}$ is corepresentable if and only if its sheafification is. – Ingo Blechschmidt Jun 5 '17 at 21:43