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.

Solution 1:

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.