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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!

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  • $\begingroup$ In your second example, the categorical quotient of $U$ by $G$ is not $X//G = \text{Spec}\ k[xy]$. Rather it is two copies of this scheme glued together along the complement of the origin, i.e., the categorical quotient is a line with doubled origin. $\endgroup$ – Jason Starr Mar 8 '17 at 21:06
  • $\begingroup$ For a $k$-scheme $U$ that has an open covering by $G$-invariant open affines, you can construct the categorical quotient of $U$ by glueing together the categorical quotients of those open affines. However, there are examples when a $G$-invariant quasi-affine is not covered by $G$-invariant opens, e.g., for the conjugation action of $G=\textbf{GL}_2$ on the irreducible $2$-dimensional affine scheme $X$ of $2\times 2$ matrices with determinant and trace $0$, for $U$ the complement of the origin, $U$ is a single $G$-orbit. $\endgroup$ – Jason Starr Mar 8 '17 at 21:24
  • $\begingroup$ @JasonStarr: Thank you so much for your response! I understand what you say. So, it is not expected that the categorical quotient (if exists) will be a (quasi)-algebraic variety, but an scheme obtained by pasting the GIT quotients of an affine open cover of $U$... $\endgroup$ – a_g Mar 9 '17 at 0:16
  • $\begingroup$ @JasonStarr: However, I can't see what is wrong in my argument for the second example... In that case, $X//G$ factorices every $G$-invariant function from $U$, isn't it? Maybe the point is that this condition only guarantees that every $G$-invariant function $f:U \to Y$, with $Y$ affine, factorices through $X//G$, but the same is not true for more general $Y$, like projective varieties. In that case, $X//G$ is a kind of 'affine categorical quotient', isn't it? Anycase, this 'affine categorical quotient' will be unique, right? $\endgroup$ – a_g Mar 9 '17 at 0:21
  • $\begingroup$ I recommend that you read Proposition 1.9, Section 4, Chapter 1, p. 37, of Geometric Invariant Theory by Mumford, Fogarty, Kirwan. $\endgroup$ – Jason Starr Mar 9 '17 at 13:06

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