Proper morphism

Question

Maybe this could be a silly question, but I am considering the following problem.

Let $G$ be a connected reductive group and $X$ be an affine $G$-variety, i.e., $G$ acts on $X$. (You can assume the smoothness of $X$ if necessary.) Furthermore, let us assume that the centre $Z(G)$ of $G$ acts trivial on $X$, i.e., $z.x=x$ for all $z\in Z(G)$.

Then we can consider the following stable locus subset $Y:= \{ x\in X| x\ \text{is stable} \}$. Here, $x$ is stable means that the stabiliser $G_x/Z(G)$ is finite and $G.x$ is closed.

Then is the canonical map $f: Y/\!/G \rightarrow X/\!/G$ proper?

I appreciate any comments in advance.

Answer 1

This is false. Let $G = \mathrm{GL}_2$ and $X$ be the vector space of binary quartic forms $q \in \mathbb{C}[x,y]$, with action given by linearly substituting and dividing by the square of the determinant $g\cdot q = q((x,y)\cdot g)/\det(g)^2$. The determinant factor ensures that the center acts trivially.

The ring of $G$-invariant polynomials is generated by two invariants $I$ and $J$, so $X//G$ is isomorphic to $\mathbb{A}^2$. On the other hand, the stable locus $Y\subset X$ equals the subset of quartics $q$ that are products of four distinct linear factors. The discriminant $\Delta$ of $q$ is a certain linear combination of $I^3$ and $J^2$, and has the property that $\Delta(q) \neq 0$ if and only if $q$ is stable. Therefore $Y//G$ is the subset of $X//G = \mathbb{A}^2$ where $\Delta$ (seen as a polynomial in $I,J$) does not vanish. This is an open and dense subset of an affine space, so the inclusion is not proper.

The following paper summarizes the invariant theory of binary quartics:

Bhargava, Manjul; Shankar, Arul, Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves, Ann. Math. (2) 181, No. 1, 191-242 (2015). ZBL1307.11071.