Let $X$ be an affine variety and let $G$ be a reductive algebraic group acting on $X$. Let $U \subset X$ be a $G$-invariant open set.

*Under what hypothesis there exists a categorical quotient of $U$ by $G$?*

Some comments to the question:

- Since the GIT quotient of $X$ exists, $X//G = Spec(\mathcal{O}(X))^G$, it is a categorical quotient for $X$ with the induced morphism $\pi: X \to X//G$. If $U$ is saturated for $\pi$ (i.e. if there exists an open set $V \subset X//G$ such that $U = \pi^{-1}(V))$ then the restriction $\pi|_U: U \to V$ is a categorical quotient for $U$.
- Consider the action of $G = k^* = \mathbb{G}_m$ on $X =k^2$ by $\alpha \cdot(x,y) = (\alpha x, \alpha^{-1}y)$ for $\alpha \in G$ and $(x,y) \in X$ and let $U=k^2 - \left\{(0,0)\right\}$. Since $\mathcal{O}(U) = \mathcal{O}(X)$, every regular function $f: U \to Y$ extends to a regular function $\tilde{f}: X \to Y$. In particular, if $f$ is $G$-invariant, then it factorices uniquely through $X//G$. Thus, in this case, $X//G$ is the categorical quotient of $U$ by $G$. In fact, the same argument applies if $X$ is a normal variety and $Z = X - U$ has codimension at least two, since, in that case every regular morphism $f: U \to Y$ extends to a morphism $\tilde{f}: X \to Y$ (a kind of Hartog's theorem).

From these comments, it seems like the categorical quotient should be constructed by attaching to $U$ the closure of its orbits and taking the image under $\pi$ of that set. Does it make sense? Is there some example where such procedure does not work?

Thank you in advance!

Geometric Invariant Theoryby Mumford, Fogarty, Kirwan. $\endgroup$ – Jason Starr Mar 9 '17 at 13:06