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Suppose I have a linear algebraic group $G$ over some field $k$ that is algebraically closed, and $G$ is acting on a variety $X$. If I take some $x \in X$, then can we view the orbit $G \cdot x$ as an (affine algebraic) variety? I came across a talk that took this as a fact, and this doesn't seem obvious to me at all. As a result, I'm hoping for some help to see how this is "obvious".

hm2020
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user1438127
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2 Answers2

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Q: "If I take some x∈X, then can we view the orbit G⋅x as an (affine algebraic) variety? I came across a talk that took this as a fact, and this doesn't seem obvious to me at all. As a result, I'm hoping for some help to see how this is "obvious"."

A: This is not true in general - If $V:=k\{e_0,e_1\}$ and $P \subseteq SL(V)$ is the subgroup fixing the line $v$ spanned by $e_0$ it follows $SL(V)v \cong SL(V)/P$ where $P$ is a parabolic subgroup and this quotient is not affine. There is an explicit isomorphism

$$ f: SL(V)/P \cong C:=\mathbb{P}^1_k$$

and a canonical "projection map"

$$\pi: SL(V) \rightarrow C$$

where $\pi(g)$ is the first column of the matrix $g$. This example is interesting for students for the following reason: If you let $U_i:=D(x_i) \subseteq C$ be the basic open subsets, it follows there is an isomorphism $f_i: \pi^{-1}(U_i) \cong U_i \times_k P$. Hence the map $\pi$ is a principal $P$-bundle (in the Zariski topology) commuting with the projection map: There is an "equality" of maps

$$pr_{U_i} \circ f_i = \pi.$$

C: "Thank you for the reply! This answer clears up things. I guess I slightly lied by omission in the question because I had assumed that it would be true, but my confusion stems from the idea of assigning a "dimension" to an orbit, as seen in the talk I saw. My naive idea was to say that dim(G⋅x)=dim(G)−dim(Stabilizer) but I don't think that's true now for linear algebraic groups. Maybe connectedness is required. – user1438127 Commented 6 mins ago"

A: @user1438127 - I believe the following holds: The quotient $SL(V)/H$ is a projective variety iff $H$ is a parabolic subgroup. In general the quotient $SL(V)/H$ will be a smooth quasi projective variety of finite type over $k$. You find this proved in a book on linear algebraic groups (Springer, Humphreys or Borel). There are "obvious" instances where the quotient is affine: If $H$ is a closed normal subgroup, it follows $SL(V)/H$ is an affine algebraic group of finite type over $k$ and hence affine.

There is a book by Akhiezer that may contain some of the above results. My claim about normal subgroups may be found (in the language of schemes) in the Waterhouse GTM book.

hm2020
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  • Thank you for the reply! This answer clears up things. I guess I slightly lied by omission in the question because I had assumed that it would be true, but my confusion stems from the idea of assigning a "dimension" to an orbit, as seen in the talk I saw. My naive idea was to say that $\text{dim}(G\cdot x) = \text{dim}(G) - \text{dim}(\text{Stabilizer})$ but I don't think that's true now for linear algebraic groups. Maybe connectedness is required. – user1438127 Oct 14 '25 at 10:29
  • @user1438127 - Note: Over a non algebraically closed field you may not use the classical books on linear algebraic groups. These books assume the base field is algebraically closed. – hm2020 Oct 14 '25 at 13:27
  • I appreciate that you rewrote your answer to include additional details from the comment. Comments are ephemeral and may disappear, but by editing it into your post you've made it more permanent. – davidlowryduda Oct 14 '25 at 13:48
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    @davidlowryduda - I have been critiziced for this but the reason I include the comment and my response is that this makes the post easier to understand for others. – hm2020 Oct 14 '25 at 14:00
  • @hm2020 Yes, that's also why I make it a point to comment on posts like these. Sometimes the diamond next to a name helps act as a signal. – davidlowryduda Oct 14 '25 at 14:03
  • @davidlowryduda - since you are a moderator I have a question: On this site (and the MO site) I frequently find that someone (a moderator?) delets many comments that are informative and interesting for other users - why do you do this? – hm2020 Oct 14 '25 at 16:10
  • @hm2020 I'm not sure what the MO story is, but it's rare for moderators here to delete comments (unless there are, say, 5+ comments on a post). But there is a relatively organized group of users here who team up and group flag/downvote/delete content they think lowers the quality of the site. Users cannot delete comments, but if a certain threshold of users all flag it then it is automatically deleted. Whether this is a good thing or a useful thing is complicated. Some of these users are particularly annoyed when users provide answers in comments, or at interactions with perceived bad posts. – davidlowryduda Oct 14 '25 at 19:17
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The orbit is the quotient variety $G/H$ which can be given the structure of an algebraic variety -- this can be found, for example, in Humphrey's textbook on Linear Algebraic Groups. However the orbit need not be an affine variety. In fact when the closed subgroup $H$ (the isotropy subgroup of $x$, in your notation) contains a maximal connected solvable subgroup (Borel subgroup) of $G$, the orbit would be a projective variety. There are cases when $G/H$ is rich in global regular functions ($G/H$ being a quasi-affine variety. I think it is called an observable subgroup in some work of Grosshans). Borel has studied the other extreme cases where the only global regular functions on $G/H$ are constants -- in which case he calls $H$ an epimorphic subgroup of $G$. They cover all parabolic subgroups.

P Vanchinathan
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