# Orbit-stabilizer theorem for Lie groups?

Let $G$ be a finite-dimensional lie group, with a transitive action on the points of a smooth finite-dimensional manifold $S$. Let $p$ be some point of $S$ and let $T$ be the stabilizer of $p$ in $G$. Suppose that the action is compatible with the smooth structure of $S$, in the sense that for fixed $g\in G$, $g:S\rightarrow S$ defined by $x\mapsto gx$ is a smooth map, and for fixed $x\in S$, $x:G\rightarrow S$ defined by $g\mapsto gx$ is a smooth map. (I assume this is a standard concept but I do not know its name.)

I have very little background in Lie groups, but it seems clear to me that the above forces $T$ to be a submanifold of $G$. (Initial question: is this correct?) Assuming this,

Primary question: Is it true that $\dim S + \dim T = \dim G$?

Secondary questions, if the answer is "yes":

• What is the argument?

• Can the assumptions be relaxed?

• Is there an analogous theorem in other settings? For example if $G$ and $S$ have the structure of algebraic varieties over some algebraically closed field $k$, can we replace all references to "smooth map" with "morphism" and get the same result?

Motivation: Feel free to ignore this part but comments on it are welcome. The question occurred to me when I was working on a problem (in Miles Reid's introductory algebraic geometry text) about putting an arbitrary cubic curve in $\mathbb{P}^2$ possessed of an inflection point into normal form $y^2z=x^3+ax^2z+bxz^2+cz^3$. It was necessary to make use of the fact that the group of projective transformations of $\mathbb{P}^2$, which I believe is an 8-dimensional quasi-projective variety, acts transitively on the set of pairs (line, point on that line), which set forms a 3-dimensional variety. I grew curious about what the stabilizer of a particular pair (line, point on that line) looked like. Based on intuition from basic group theory and basic linear algebra, I expected it to be a 5-dimensional subvariety of the group of projective transformations (because $3+5=8$), and this was true: for example if the point is $(1:0:0)$ and the line is spanned by this and $(0:1:0)$ then the stabilizer in question is the quotient group of the (6-dimensional) subgroup of upper triangular $3\times 3$ matrices by its (1-dimensional) center. I got curious if this relationship was true in some general settings. It seemed to me it should be at least in the lie group context, because of something like this: perhaps the derivative of the map $g\mapsto gx$ mentioned in the first paragraph has as kernel the tangent space of $T$? And then the desired result becomes the rank-nullity theorem? But I do not know enough to convince myself that this works.

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Suppose that $G$ acts on a (topological) space $S$ (by homeomorphisms). Then restricting $G$'s action to a single orbit, say $\mathrm{orb}(x_0)=X$, we get a (continuous) transitive action of $G$ on $X$. This makes $X$ a homogeneous space. One can prove that the stabilizer $\mathrm{stab}(x_0)=H$ is a closed subgroup group $G$. So that the coset space $G/H$ is homeomorphic to $X$. The dimension of $G/H$ is $\mathrm{dim}(G)-\mathrm{dim}(H)$ and the result in question follows.
Trying to fill in the details: $\mathrm{stab}(x_0)$ is closed because it is the inverse image of $x_0$ under the map $G\rightarrow X$ defined by $g\mapsto gx$, which is continuous by def. of a continuous action, right? Then doesn't this depend on the topology on $X$ being such that singletons like $\{x_0\}$ are closed? Or is there another reason $\mathrm{stab}(x_0)$ should be closed? In both Euclidean and Zariski topologies this is true, so there is no issue for manifolds or varieties or probably any other setting in which dimension makes sense, but I am just trying to understand fully. –  Ben Blum-Smith Jan 13 '12 at 15:12
Second question: I am assuming that we are assuming $G$ is a topological group. What is the argument that the quotient topology on $G/H$ makes it homeomorphic with $X$? How does it rely on $H$ being closed? –  Ben Blum-Smith Jan 13 '12 at 15:19