5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ (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. $\endgroup$ Commented Aug 12, 2020 at 7:37
  • $\begingroup$ 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! $\endgroup$
    – user618650
    Commented Aug 12, 2020 at 10:29

1 Answer 1

4
$\begingroup$

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.

The answers to your questions then are:

(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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – user618650
    Commented Aug 12, 2020 at 19:47

You must log in to answer this question.