# GIT quotients of open subsets

Let X be a projective variety on with a action of reductive group G. Let L be a G Linearised ample line bundle on X. Let U be a G stable open subset of X. Let $U^{ss}:=X^{ss}\cap U$. Is it true that $U^{ss}//G\subset X^{ss}//G$?

What I feel is that there is only a morphism $U^{ss}//G\rightarrow X^{ss}//G$? It may not be a subset in general.

I would appreciate an argument if i am wrong.

I would like to reformulate my question:

Let $U\hookrightarrow X$ be a G invariant open immersion. Suppose X has a Good quotient X//G. Suppose that U also has a good quotient U//G. Is it true in general that U//G is a sub-variety of X//G?

• If you define $U^{ss}//G$ as a geometric quotient of $U^{ss}$, then the morphism need not be injective. Consider the action of $\mathbb{G}_m$ on $\text{Proj}\ k[t_-,t_0,t_+]$ by $\lambda\ast[t_-,t_0,t_+] = [\lambda^{-1} t_-,\lambda^0t_0, \lambda^{+1} t_+].$ The properly unstable locus consists of the two points $[1,0,0]$ and $[0,0,1]$. The GIT quotient of the semistable locus is the morphism $\text{Proj}\ k[t_-,t_0,t_+] \to \text{Proj}\ k[r,s]$ by $[t_-,t_0,t_+]\mapsto [t_0^2,t_-t_+].$ Let $U$ be the open complement of $[0,1,0]$. The geometric quotient is the line with doubled origin. Sep 23, 2017 at 10:36

The answer to the reformulated question is no. Let $G=\mathbf G_m$ act on $X=\mathbf A^n$ by scalar multiplication with $n\ge2$. Then a good quotient $X//G$ exists and is a point. On the other hand, $U=X\setminus\{0\}$ has a good quotient $U//G$ which is $\mathbf P^{n-1}$.
The problem arises because orbits which are closed in $U$ maybe non-closed in $X$. For your first question it might be better to ask: which open subsets $U\subseteq X^{ss}$ are pull-backs from open subsets of $x^{ss}//G$? For those obviously $U//G\to X^{ss}//G$ is an open embedding. The answer is that $U$ has to be saturated in the following sense: the orbit closure of any $x\in U$ in $X^{ss}$ lies in $U$: $$\overline{Gx}\cap X^{ss}\subseteq U.$$