For simplicity, work over an algebraically closed field of characteristic $0$. Let $$\begin{aligned} X &= \text{a smooth projective variety,} \\ G &= \text{a reductive group acting linearly on $X$,} \\ H &= \text{a finite subgroup of $G$,}\\ N(H) &= \text{the normalizer of $H$ in $G$.}\\ \end{aligned}$$ For $x \in X$, let $$\begin{aligned} \text{Stab}(x) &= \{ g \in G : g \cdot x = x \},\\ X_H &= \{ x \in X : \text{Stab$(x)$ contains $H$ }\} \end{aligned}$$

Then $G$ doesn't necessarily act on $X_H$, but $N(H)$ does, so we can look at the $N(H)$-stable and semi-stable points of $X_H$ in the sense of GIT.

Questions (1) Are the $N(H)$-semi-stable points of $X_H$ automatically $G$-semi-stable points of $X$? Ditto for stable points. (This seems unlikely, but I don't know a counterexample.)

(2) If not, is there some simple criterion for determining the points $x\in X_H$ such that $x$ is $G$-stable when viewed as a point of $X$?

If it helps, the specific situation I have is: $$\begin{aligned} X &= \mathbb P^N \\ G &= \text{PGL}_k~\text{or SL}_k \\ N(H) &= \text{group of diagonal matrices in $G$.}\\ \end{aligned}$$ In particular, $N(H)$ is a maximal torus.


The answer to (1) is affirmative for semistable points. This is basically a result of Luna in his paper "Adhérences d'orbite et invariants. Invent. Math. 29 (1975), 231–238". More precisely:

We only assume that $H$ is reductive (not necessarily finite). Let $\tilde X$ be the affine cone over $X$. For any $x\in X^H$ ($=X_H$ in the Op's notation) let $\tilde x\in\tilde X$ be a lift. Then, by definition, $x$ is semistable for $G$ (or $N$) if the orbit $G\tilde x$ (or $N\tilde x$) does not contain the vertex $0$ in its closure. This already shows that $\tilde x$ is $N$-semistable if it is $G$-semistable.

Conversely assume that $x$ is $N$-semistable. By replacing $\tilde x$ by a point in the (unique) closed $N$-orbit of $\overline{N\tilde x}$ we may assume that $N\tilde x$ closed and $\ne\{0\}$.

The group $H$ acts on the line $k\tilde x$ by a character $\chi$. If $\chi$ is of infinite order then $x$ is not $N$-semistable. Thus, $S:=\chi(H)\subseteq\mathbf G_m$ is a finite group. Put $\tilde G:=G\times S$ and $\tilde H:=\Delta(H)$ where $\Delta(h):=(h,\chi(h)^{-1})$. The point is that $\tilde G$ acts on $\tilde X$ and $\tilde x$ is an $\tilde H$-fixed point. Observe that the normalizer $\tilde N$ of $\tilde H$ in $\tilde G$ is of finite index in $N\times S$. In particular, $\tilde N\tilde x$ is closed and $\ne\{0\}$.

Now we invoke Luna's result, Cor. 1 in op.cit., which says that "$\tilde N\tilde x$ closed" implies "$\tilde G\tilde x$ closed". In particular $0\not\in \overline{G\tilde x}$, i.e. $\tilde x$ is $G$-semistable. qed.

Preservation of stability cannot be expected, in general. For example assume assume $H$ is contained in an infinite reductive subgroup $F$ of $G$ such that $N_F(H)$ is finite. Then $x=eF\in X=G/F$ has a finite $N$-stabilizer but the $G$-stablizer is $F$.

On the positive side, Luna proves something different which might be useful: the morphism $X^H/\!/N\to X/\!/G$ is always finite.

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  • $\begingroup$ Thanks. That's exactly what I was looking for. When you say that "$\tilde G$ acts on $X$ and $\tilde x$ is an $\tilde H$-fixed point", did you mean to say that $\tilde G$ acts on $\tilde X$? Also, probably a minor point, but I don't see why $\tilde N$ isn't exactly equal to $N\times S$, rather than simply being of finite index. $\endgroup$ – Joe Silverman Jul 5 '16 at 17:37
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    $\begingroup$ The $X$ is a typo. I fixed it. Moreover, $\tilde N$ is basically the stabilizer of the character $\chi$ in $N$. One more remark: the whole thing works only in characteristic zero. $\endgroup$ – Friedrich Knop Jul 5 '16 at 18:08

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