Geometric invariant theory

Remarks

The ring $R^G$ is finitely generated because $G$ is reductive, so the quotient is a projective variety. The map $\pi$ is a good categorical quotient: two semistable points have the same image exactly when the closures of their orbits meet in $X^{ss}$. On the stable locus $X^s$, which is open, $\pi$ restricts to a geometric quotient $X^s \to X^s / G$ whose fibres are single orbits.

Example: let $G = \mathbb{G}_m$ act on $X = \mathbb{P}^1$ by $t \cdot [x : y] = [t x : t^{-1} y]$, linearized on $L = \mathcal{O}(1)$ by the weights $1$ and $-1$. A monomial $x^a y^b$ of degree $m = a + b$ is invariant exactly when $a = b$. So the invariant sections are the multiples of powers of $xy$, and the semistable points are those with $xy \ne 0$. They form a single orbit, which is closed in $X^{ss}$, and the stabilizer of $[1 : 1]$ is $\set{\pm 1}$, so these points are stable and the quotient is a point.

Non-example: the point $[1 : 0]$ is unstable, since every invariant section of positive degree is a multiple of $xy$ and vanishes there.

For an affine variety $X = \operatorname{Spec} A$ with the trivial linearization, the quotient is $X /\!\!/ G = \operatorname{Spec} A^G$.