# Orbit-stabilizer for algebraic groups

I'm currently trying to studying quotients of algebraic groups, but I find this topic very confusing. I think all my doubts boil down to the following question:

Let $$G$$ be an algebraic group, and suppose it acts non-trivially on a projective variety $$X$$, that is I have a map $$G\times X\to X$$. Given a point $$p\in X$$, I can consider the isotropy group $$Iso(p)=\{g\in G\mid gp=p\}$$ and the orbit of $$p$$ under $$G$$, i.e. $$Gp=\{gp\mid g\in G\}.$$

Then when, and if yes under which hypothesis, do the following isomorphisms hold? $$\frac{G}{Iso(p)}\overset{(1)}{\simeq} Gp \overset{(2)}{\simeq} X?$$

What are the condition on $$G$$, $$X$$ in order that $$(1)$$ and $$(2)$$ holds? I apologize for the vagueness of the question and I hope this naive question.

• (1) always holds, and (2) is always a locally closed embedding, so if $Gp=X$ on the level of points (e.g. let's say we we're working over a field $F$, then if $Gp=X$ on the level of $\overline{F}$) points and $X$ is reduced then the second equality holds. Commented Aug 12, 2020 at 7:37
• Dear @AlexYoucis, thanks a lot for your answer... if you want to expand a bit more your comment, providing a proof (or a reference for $(1)$, and your claim: reductive $\implies (2)$), I'll surely accept it! Moreover, for $(2)$, I was implicitly working on $\mathbb{C}$, but nevertheless I don't understand what you mean by "at the level of points"... thanks again and have a nice day!
– user618650
Commented Aug 12, 2020 at 10:29

I'll write the answer in the case of a general ground field $$F$$ since this may be of interest to people in more generality, and then explain at the end what simplifications happen if $$F=\mathbb{C}$$.

So, let $$F$$ be an arbitrary field and let $$G$$ be a finite type group scheme over $$F$$ and let $$X$$ be a finite type scheme over $$F$$. Suppose that

$$\mu:G\times X\to X$$

is an algebraic action. Let us now fix $$x$$ in $$X(F)$$.

We need to define what we mean by the '$$G$$-orbit' of $$x$$ in $$X$$. One possible answer is the following. We have a natural $$G$$-equivariant map of varieties with $$G$$-action

$$\mu_x:G\to X,\qquad G(R)\ni g\mapsto gx\in X(R)$$

(here $$R$$ is any $$F$$-algebra) where $$G$$ is given the left multiplication action and $$X$$ is given its $$G$$-action. It is then true that the image $$\mu_x(|G|)$$ (where $$|\cdot |$$ denotes the underlying space of a scheme) is a locally closed subset of $$X$$ (e.g. see [1, Proposition 1.65(b)] with $$X=G$$, $$Y=X$$, and $$f=\mu_x$$). Since $$\mu_x(|X|)$$ is locally closed it has a natural reduced scheme structure (e.g. see [2, Tag0F2L]), and we denote the resulting reduced locally closed subscheme of $$X$$ by $$O(x)$$ and call it the $$G$$-orbit of $$x$$. Notice, for example, that for any field $$L$$ containing $$F$$ one has that

$$O(x)(L)=\{gx:g\in G(L)\}\subseteq X(L)$$

as one would expect.

Thus, we can understand your questions really as follows:

(Q1) When does the orbit map $$\mu_x:G\to O(x)$$ define an isomorphism $$G/G_x\to O(x)$$?

(Q2) When is $$O(x)$$ equal to $$X$$?

In 1. by $$G_x$$ I mean the isotropy subgroup associated to $$x$$ whose $$R$$-points, for an $$F$$-algebra $$R$$, are given

$$G_x(R):=\{g\in G(R):gx=x\}$$

as one would expect.

(A1) This essentially always holds true. Namely, suppose that $$X$$ and $$G$$ are both separated and geometrically reduced, then $$\mu_x:G\to O(x)$$ induces an isomorphism $$G/G_x\to O(x)$$ whenever $$G/G_x$$ exists. For a proof see [1, Corollary 7.13]. Note that $$G/G_x$$ essentially always exists (e.g. see [1, Theorem 5.28] and [1, Theorem B.37])

(A2) If we're already in the situation of (A1) then there is quite a simple answer: when $$G(\overline{F})$$ acts transitively on $$X(\overline{F})$$. Indeed, in this case we see that

$$O(x)(\overline{F})=\{gx:g\in G(\overline{F})\}=X(\overline{F})$$

But, this then implies that $$|O(x)|=|X|$$. Indeed, write $$|O(x)|=U\cap Z$$ where $$U$$ is open and $$Z$$ is closed. Since $$Z(\overline{F})\supseteq (U\cap Z)(\overline{F})$$ and $$U(\overline{F})\supseteq (U\cap Z)(\overline{F})$$ one has that $$U=X$$ and $$Z=X$$ since the $$\overline{F}$$-points of $$X$$ are very dense in $$X$$ (e.g. see [3, Definition 3.34, Proposition 3.35, and Corollary 3.36]) and so $$|X|=|O(x)|$$ as desired. But, since $$X$$ is reduced, this implies that $$X=O(x)$$ as schemes (e.g. use the unicity in [2, Tag01J3]).

So, what does this all mean when $$F=\mathbb{C}$$? Presumably (if not let me know) you're working with the classical perspective of algebraic geometry for example as in the notion of prevarieties in [3, Chapter 1].

In this light, we can summarize the above discussion as follows. Let $$G$$ be a group prevariety over $$\mathbb{C}$$ acting on the prevariety $$X$$ over $$\mathbb{C}$$.

Then, we have the following first important fact:

Fact 1: For all $$x$$ in $$X$$ the subset

$$O(x):=\{gx:g\in G\}$$

is a locally closed subset of $$X$$ and thus naturally a subprevariety of $$X$$.

Of course, you are just wrting $$O(x)$$ as $$Gx$$.

Now, we have a natural map

$$\mu_x:G\to X:g\mapsto gx$$

and the second fact we need is the following:

Fact 2: The map $$\mu_x$$ induces an isomorphism $$G/G_x\to O(x)$$.

Here $$G_x$$ is the isotropy subgroup

$$G_x:=\{g\in G:gx=x\}$$

which is just $$Iso(x)$$ in your language.

And, the last final fact you need is the following:

Fact 3: One has that $$O(x)=X$$ if and only if they are equal as sets which is true if and only if $$G$$ acts transitively on $$X$$.

This is obvious.

References:

[1] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[2] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/

[3] Görtz, U. and Wedhorn, T., 2010. Algebraic geometry. Wiesbaden: Vieweg+ Teubner.

• Your answer was extremely exahustive and interesting, again thanks a lot for the patience in writing it! Yes, I am studying it from an algebraic geometry point of view, so I'm glad you specify the discussion in the complex case. I hope to recive other answer from you, thangs a lot and have a nice day sir!
– user618650
Commented Aug 12, 2020 at 19:47